[{"Name":"1. Introduction And Deriving The Equation","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Intro","Duration":"16m 8s","ChapterTopicVideoID":8106,"CourseChapterTopicPlaylistID":5426,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/8106.jpeg","UploadDate":"2016-12-08T11:59:33.1070000","DurationForVideoObject":"PT16M8S","Description":null,"MetaTitle":"Intro: Video + Workbook | Proprep","MetaDescription":"Variable mass - 1. Introduction and Deriving the Equation. Watch the video made by an expert in the field. Download the workbook and maximize your learning.","Canonical":"https://www.proprep.uk/general-modules/all/physics-1-mechanics-waves-and-thermodynamics/variable-mass/1.-introduction-and-deriving-the-equation/vid8222","VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.920","Text":"Hello. In this section,"},{"Start":"00:01.920 ","End":"00:05.369","Text":"we\u0027re going to be speaking about variable mass systems."},{"Start":"00:05.369 ","End":"00:08.100","Text":"First, we\u0027re going to do a little introduction."},{"Start":"00:08.100 ","End":"00:11.370","Text":"First of all, when speaking about variable mass system,"},{"Start":"00:11.370 ","End":"00:13.665","Text":"we\u0027re using Newton\u0027s second law,"},{"Start":"00:13.665 ","End":"00:15.869","Text":"but when speaking about momentum,"},{"Start":"00:15.869 ","End":"00:19.170","Text":"which as we know, is dp by dt."},{"Start":"00:19.170 ","End":"00:20.475","Text":"Just as a reminder,"},{"Start":"00:20.475 ","End":"00:27.510","Text":"we get our equation of force equals mass times acceleration by adding in over here."},{"Start":"00:27.510 ","End":"00:29.355","Text":"We say that our p,"},{"Start":"00:29.355 ","End":"00:33.445","Text":"our momentum is mass times velocity,"},{"Start":"00:33.445 ","End":"00:40.925","Text":"then we\u0027ll have d by dt of mass times velocity, and then,"},{"Start":"00:40.925 ","End":"00:44.330","Text":"because we know that our mass isn\u0027t dependent on"},{"Start":"00:44.330 ","End":"00:48.650","Text":"time necessarily for what we\u0027ve learned up until now,"},{"Start":"00:48.650 ","End":"00:54.655","Text":"we\u0027ll have that m is dv by dt,"},{"Start":"00:54.655 ","End":"00:57.815","Text":"and then dv by dt, as we know,"},{"Start":"00:57.815 ","End":"01:01.745","Text":"is acceleration, and then we get mass times acceleration."},{"Start":"01:01.745 ","End":"01:05.585","Text":"Now, because we are speaking about a varying mass system,"},{"Start":"01:05.585 ","End":"01:09.005","Text":"that means that the mass varies over time."},{"Start":"01:09.005 ","End":"01:15.800","Text":"Which means that going from this step over to this step is no more."},{"Start":"01:15.800 ","End":"01:19.415","Text":"Now, we can put a big X on this,"},{"Start":"01:19.415 ","End":"01:22.265","Text":"because in varying mass systems,"},{"Start":"01:22.265 ","End":"01:25.360","Text":"the mass is dependent on time as well."},{"Start":"01:25.360 ","End":"01:28.085","Text":"From now on, never use"},{"Start":"01:28.085 ","End":"01:32.450","Text":"this equation unless you know in the question that the mass is constant."},{"Start":"01:32.450 ","End":"01:34.744","Text":"Unless stated that the mass is constant,"},{"Start":"01:34.744 ","End":"01:37.285","Text":"this equation is incorrect."},{"Start":"01:37.285 ","End":"01:39.870","Text":"How can we solve these questions?"},{"Start":"01:39.870 ","End":"01:43.625","Text":"The first way is by using differential equations,"},{"Start":"01:43.625 ","End":"01:48.605","Text":"and the second way is by rearranging Newton\u0027s second law and adding a term"},{"Start":"01:48.605 ","End":"01:53.920","Text":"to account for the momentum carried by mass entering or leaving the system."},{"Start":"01:53.920 ","End":"02:00.070","Text":"The equation of a variable mass motion is written like so."},{"Start":"02:00.070 ","End":"02:02.720","Text":"What this equation is saying is that the sum of"},{"Start":"02:02.720 ","End":"02:06.620","Text":"all the external forces acting on the body is equal to"},{"Start":"02:06.620 ","End":"02:13.705","Text":"the mass as an equation of time multiplied by the change in velocity during the time,"},{"Start":"02:13.705 ","End":"02:20.630","Text":"dv by dt, plus the relative velocity multiplied by the change of mass."},{"Start":"02:20.630 ","End":"02:22.070","Text":"Now in future lessons,"},{"Start":"02:22.070 ","End":"02:24.800","Text":"we\u0027re going to go over this a lot more in detail."},{"Start":"02:24.800 ","End":"02:26.465","Text":"Now from my experience,"},{"Start":"02:26.465 ","End":"02:28.475","Text":"we said that there\u0027s 2 ways of solving this"},{"Start":"02:28.475 ","End":"02:31.470","Text":"through this equation or differential equations,"},{"Start":"02:31.470 ","End":"02:32.810","Text":"and from my experience,"},{"Start":"02:32.810 ","End":"02:36.875","Text":"the differential equations are slightly harder and slightly more confusing."},{"Start":"02:36.875 ","End":"02:41.225","Text":"I\u0027m going to spend a little bit more time going over this method."},{"Start":"02:41.225 ","End":"02:44.810","Text":"Even if you haven\u0027t learned this method and this equation in class,"},{"Start":"02:44.810 ","End":"02:47.330","Text":"this is still a correct way to solve it,"},{"Start":"02:47.330 ","End":"02:49.685","Text":"and so you can use this."},{"Start":"02:49.685 ","End":"02:55.210","Text":"Having said that, we will solve some questions through the differential equation."},{"Start":"02:55.210 ","End":"02:58.880","Text":"Now what we\u0027re going to do is we\u0027re going to go over how to"},{"Start":"02:58.880 ","End":"03:02.525","Text":"derive this equation because then you\u0027ll understand the equation a lot better."},{"Start":"03:02.525 ","End":"03:08.515","Text":"It will be easier to understand how to use it and where it comes from and to remember it."},{"Start":"03:08.515 ","End":"03:16.340","Text":"Now, let\u0027s speak about this equation that dm by dt section,"},{"Start":"03:16.340 ","End":"03:21.300","Text":"it\u0027s the rate of emission."},{"Start":"03:22.820 ","End":"03:26.310","Text":"So rate of emission of the substance,"},{"Start":"03:26.310 ","End":"03:31.490","Text":"and then the v_rel, the relative velocity,"},{"Start":"03:31.490 ","End":"03:37.070","Text":"is the relative velocity of the mass being emitted,"},{"Start":"03:37.070 ","End":"03:42.280","Text":"the substance being emitted relative to the body moving."},{"Start":"03:42.280 ","End":"03:47.975","Text":"For instance, if you have a rocket and there\u0027s an emission of fuel outside."},{"Start":"03:47.975 ","End":"03:51.420","Text":"Let\u0027s say it looks like this."},{"Start":"03:52.820 ","End":"03:55.950","Text":"There\u0027s fuel coming out."},{"Start":"03:55.950 ","End":"03:59.810","Text":"The rate of emission is how much fuel is leaving the rocket"},{"Start":"03:59.810 ","End":"04:03.095","Text":"per second and the relative velocity is,"},{"Start":"04:03.095 ","End":"04:05.435","Text":"if here\u0027s some fuel going out,"},{"Start":"04:05.435 ","End":"04:15.000","Text":"what is the velocity of this fuel relative to the velocity that the rocket is moving at."},{"Start":"04:15.320 ","End":"04:18.995","Text":"The v_rel, the relative velocity,"},{"Start":"04:18.995 ","End":"04:24.050","Text":"is the velocity of the emitted substance relative to the body,"},{"Start":"04:24.050 ","End":"04:26.165","Text":"as in the rocket,"},{"Start":"04:26.165 ","End":"04:28.895","Text":"the rocket would be the body."},{"Start":"04:28.895 ","End":"04:32.660","Text":"Now notice when speaking about this equation,"},{"Start":"04:32.660 ","End":"04:36.560","Text":"you can also say that it\u0027s for some body which is losing mass,"},{"Start":"04:36.560 ","End":"04:38.465","Text":"such as a rocket losing fuel,"},{"Start":"04:38.465 ","End":"04:41.780","Text":"but you can also use it when working out."},{"Start":"04:41.780 ","End":"04:46.190","Text":"For instance, if you have some bucket and it\u0027s"},{"Start":"04:46.190 ","End":"04:47.840","Text":"raining onto the bucket or you have"},{"Start":"04:47.840 ","End":"04:51.620","Text":"some cart with sand falling on it at a constant rate,"},{"Start":"04:51.620 ","End":"04:54.920","Text":"so you can also use this equation."},{"Start":"04:54.920 ","End":"04:59.795","Text":"Also if the body is losing mass and if it\u0027s gaining mass."},{"Start":"04:59.795 ","End":"05:05.705","Text":"Now the easiest way to derive this equation is by looking at the rocket example,"},{"Start":"05:05.705 ","End":"05:07.670","Text":"a body losing mass."},{"Start":"05:07.670 ","End":"05:09.815","Text":"Now, in our rocket example,"},{"Start":"05:09.815 ","End":"05:16.250","Text":"there\u0027s fuel inside the rocket and the rocket burns the fuel and emits all the gases out."},{"Start":"05:16.250 ","End":"05:17.860","Text":"That\u0027s how the rocket works."},{"Start":"05:17.860 ","End":"05:24.060","Text":"The first thing that we need to know is our change in mass over time,"},{"Start":"05:24.060 ","End":"05:26.225","Text":"our dm by dt."},{"Start":"05:26.225 ","End":"05:28.565","Text":"Now let\u0027s start our derivation."},{"Start":"05:28.565 ","End":"05:32.885","Text":"Now the first thing that we have to do is we have to turn to Newton\u0027s second law,"},{"Start":"05:32.885 ","End":"05:37.610","Text":"which is the sum of all of the forces is equal to"},{"Start":"05:37.610 ","End":"05:44.055","Text":"the change in momentum and the change in time."},{"Start":"05:44.055 ","End":"05:47.810","Text":"Now, I\u0027ve changed the direction of the rocket to make"},{"Start":"05:47.810 ","End":"05:50.690","Text":"this calculation in this derivation a lot easier"},{"Start":"05:50.690 ","End":"05:56.635","Text":"and we\u0027re going to say that the rocket is moving in the y-direction."},{"Start":"05:56.635 ","End":"06:03.270","Text":"I can say that my f external in the y-direction is equal to the change"},{"Start":"06:03.270 ","End":"06:10.860","Text":"in momentum in the y-direction and by the change in time."},{"Start":"06:10.860 ","End":"06:12.765","Text":"As you can see,"},{"Start":"06:12.765 ","End":"06:16.785","Text":"there\u0027s nothing happening in the x-direction so we don\u0027t have to talk about it."},{"Start":"06:16.785 ","End":"06:22.335","Text":"Now, when we say the change in the momentum in the y-direction,"},{"Start":"06:22.335 ","End":"06:24.960","Text":"let\u0027s not look at the dt for a second,"},{"Start":"06:24.960 ","End":"06:28.185","Text":"just the dp in the y-direction,"},{"Start":"06:28.185 ","End":"06:38.520","Text":"is the same as saying that P in the y-direction as a function of t plus dt."},{"Start":"06:38.520 ","End":"06:40.620","Text":"What is dt? A small,"},{"Start":"06:40.620 ","End":"06:43.230","Text":"infinitely small change in time"},{"Start":"06:43.230 ","End":"06:49.680","Text":"minus the momentum in the y-direction at time t. What does this mean?"},{"Start":"06:49.680 ","End":"06:52.890","Text":"Actually, this means that at time t,"},{"Start":"06:52.890 ","End":"06:54.480","Text":"I check what the momentum is,"},{"Start":"06:54.480 ","End":"06:57.780","Text":"and then at time t plus dt,"},{"Start":"06:57.780 ","End":"07:03.405","Text":"which means that just after time t, one moment after,"},{"Start":"07:03.405 ","End":"07:05.295","Text":"I check the momentum again,"},{"Start":"07:05.295 ","End":"07:11.650","Text":"and then I work out the difference between these two values."},{"Start":"07:11.650 ","End":"07:15.280","Text":"Now what I want to do is I want to see what"},{"Start":"07:15.280 ","End":"07:19.435","Text":"my momentum is of the entire rocket at a certain time,"},{"Start":"07:19.435 ","End":"07:22.255","Text":"at time t. I want to start at a general time,"},{"Start":"07:22.255 ","End":"07:30.490","Text":"naught t=0 or something just t. I say that my momentum at time t is equal to"},{"Start":"07:30.490 ","End":"07:34.795","Text":"my mass at time t multiplied by"},{"Start":"07:34.795 ","End":"07:41.095","Text":"my velocity at time t. My mass and my velocity are unknown."},{"Start":"07:41.095 ","End":"07:45.715","Text":"So all that we just spoke about happened at time t. Now,"},{"Start":"07:45.715 ","End":"07:48.250","Text":"at time t plus dt."},{"Start":"07:48.250 ","End":"07:50.868","Text":"A moment later,"},{"Start":"07:50.868 ","End":"07:58.510","Text":"we have our rocket and it [inaudible] out some gas of the fuel,"},{"Start":"07:58.510 ","End":"08:02.680","Text":"and this way is dm."},{"Start":"08:02.680 ","End":"08:08.275","Text":"Some mass, a small section of the mass, of the total mass."},{"Start":"08:08.275 ","End":"08:10.030","Text":"For example, you could say,"},{"Start":"08:10.030 ","End":"08:11.935","Text":"in the time that it takes dt,"},{"Start":"08:11.935 ","End":"08:13.720","Text":"say half a second,"},{"Start":"08:13.720 ","End":"08:18.910","Text":"1 kilogram of mass was emitted, for example."},{"Start":"08:18.910 ","End":"08:20.125","Text":"Now what\u0027s happened?"},{"Start":"08:20.125 ","End":"08:23.524","Text":"Because now the rocket is slightly lighter,"},{"Start":"08:23.524 ","End":"08:25.525","Text":"its velocity, here,"},{"Start":"08:25.525 ","End":"08:28.900","Text":"its velocity was v as a function of t. Now,"},{"Start":"08:28.900 ","End":"08:31.390","Text":"it\u0027s velocity will be v as a function of t,"},{"Start":"08:31.390 ","End":"08:35.785","Text":"the same velocity, plus some dv."},{"Start":"08:35.785 ","End":"08:38.440","Text":"We\u0027ll be going slightly faster."},{"Start":"08:38.440 ","End":"08:43.855","Text":"Now, my momentum at time t plus dt,"},{"Start":"08:43.855 ","End":"08:47.305","Text":"so a moment after this, is equal to,"},{"Start":"08:47.305 ","End":"08:52.300","Text":"now my mass is no longer M as a function of t. It\u0027s"},{"Start":"08:52.300 ","End":"08:58.525","Text":"M as a function of t minus dm y minus,"},{"Start":"08:58.525 ","End":"09:01.675","Text":"because this dm,"},{"Start":"09:01.675 ","End":"09:05.005","Text":"was emitted out, it\u0027s no longer part of the rockets."},{"Start":"09:05.005 ","End":"09:08.740","Text":"This is the mass multiplied velocity,"},{"Start":"09:08.740 ","End":"09:10.810","Text":"but it\u0027s no longer v(t),"},{"Start":"09:10.810 ","End":"09:17.274","Text":"it\u0027s some initial velocity plus our dv."},{"Start":"09:17.274 ","End":"09:23.500","Text":"I added velocity from the fact that the rocket is now lighter and so got a boost."},{"Start":"09:23.500 ","End":"09:28.030","Text":"Now, we know that there\u0027s conservation of momentum."},{"Start":"09:28.030 ","End":"09:33.220","Text":"Here, the whole system was with fuels still in the rocket,"},{"Start":"09:33.220 ","End":"09:35.140","Text":"and now the fuel is out of the rocket,"},{"Start":"09:35.140 ","End":"09:36.865","Text":"but because of conservation of momentum,"},{"Start":"09:36.865 ","End":"09:42.100","Text":"we have to find out the momentum of this fuel that has been emitted."},{"Start":"09:42.100 ","End":"09:44.935","Text":"We do plus,"},{"Start":"09:44.935 ","End":"09:50.830","Text":"and then our momentum for this fuel is our dm, which is our mass,"},{"Start":"09:50.830 ","End":"09:54.295","Text":"our change in mass, or a small piece of mass that came out,"},{"Start":"09:54.295 ","End":"09:57.175","Text":"multiplied by the velocity of it."},{"Start":"09:57.175 ","End":"10:00.385","Text":"We don\u0027t know what that is. I\u0027ll just write v_gas."},{"Start":"10:00.385 ","End":"10:03.640","Text":"Now, an important point here is when I\u0027m working"},{"Start":"10:03.640 ","End":"10:06.910","Text":"out whatever the velocity of my emitted gas is,"},{"Start":"10:06.910 ","End":"10:09.250","Text":"I do it relative to the origin."},{"Start":"10:09.250 ","End":"10:11.425","Text":"Here, my origin was the ground,"},{"Start":"10:11.425 ","End":"10:16.120","Text":"because I\u0027m assuming that my rocket was launched from the ground from earth,"},{"Start":"10:16.120 ","End":"10:19.705","Text":"which means that the velocity of the gas here"},{"Start":"10:19.705 ","End":"10:24.295","Text":"has to be also relative to my same starting point."},{"Start":"10:24.295 ","End":"10:28.210","Text":"That\u0027s very important. Now,"},{"Start":"10:28.210 ","End":"10:31.795","Text":"my next step is to find the change in momentum."},{"Start":"10:31.795 ","End":"10:33.655","Text":"How do I do that?"},{"Start":"10:33.655 ","End":"10:36.400","Text":"We\u0027ve seen here, my change in momentum,"},{"Start":"10:36.400 ","End":"10:45.295","Text":"my dp here is just my momentum at a time t plus dt minus my original momentum,"},{"Start":"10:45.295 ","End":"10:50.065","Text":"my momentum at time t. Now,"},{"Start":"10:50.065 ","End":"10:56.050","Text":"I\u0027m going to open all of these brackets and then I\u0027m just going to do this arithmetic."},{"Start":"10:56.050 ","End":"11:01.900","Text":"Let\u0027s do that. We have my change in momentum in y-direction,"},{"Start":"11:01.900 ","End":"11:05.260","Text":"because right now there\u0027s no x-direction, so it doesn\u0027t matter,"},{"Start":"11:05.260 ","End":"11:12.877","Text":"so it equals to M as a function of t,"},{"Start":"11:12.877 ","End":"11:18.415","Text":"v as a function of t plus M as a function of t,"},{"Start":"11:18.415 ","End":"11:27.895","Text":"dv minus dm multiplied by v as a function of t,"},{"Start":"11:27.895 ","End":"11:29.980","Text":"and minus"},{"Start":"11:29.980 ","End":"11:40.840","Text":"dmdv plus this dmv_gas,"},{"Start":"11:40.840 ","End":"11:45.655","Text":"and then minus my p(t),"},{"Start":"11:45.655 ","End":"11:52.450","Text":"which will just be minus M(t)v(t)."},{"Start":"11:52.450 ","End":"11:54.970","Text":"Now I can see that I have M(t)v(t) in here."},{"Start":"11:54.970 ","End":"11:56.620","Text":"I have negative M(t)v(t),"},{"Start":"11:56.620 ","End":"11:58.810","Text":"so this I can cross out,"},{"Start":"11:58.810 ","End":"12:00.535","Text":"and this I can cross out."},{"Start":"12:00.535 ","End":"12:01.945","Text":"Then over here,"},{"Start":"12:01.945 ","End":"12:05.080","Text":"I have dm multiplied by dv."},{"Start":"12:05.080 ","End":"12:06.670","Text":"Now, as we\u0027ve said before,"},{"Start":"12:06.670 ","End":"12:10.525","Text":"dm and dv are infinitely small numbers."},{"Start":"12:10.525 ","End":"12:15.490","Text":"Then if I have an infinitely small number multiplied by an infinitely small number,"},{"Start":"12:15.490 ","End":"12:19.030","Text":"it\u0027s really a very small number."},{"Start":"12:19.030 ","End":"12:26.170","Text":"I can also cross out this because relative to all the other factors in the equation,"},{"Start":"12:26.170 ","End":"12:28.645","Text":"it\u0027s so small that it\u0027s insignificant."},{"Start":"12:28.645 ","End":"12:32.320","Text":"Here, I can\u0027t cross it out because it\u0027s"},{"Start":"12:32.320 ","End":"12:36.555","Text":"an infinitely small number multiplied by a normal number,"},{"Start":"12:36.555 ","End":"12:40.020","Text":"and here as well, which means that I still have to take it into account."},{"Start":"12:40.020 ","End":"12:44.470","Text":"But if I have an infinitely small numbers squared, okay,"},{"Start":"12:44.470 ","End":"12:47.215","Text":"like we had a similar situation here,"},{"Start":"12:47.215 ","End":"12:53.260","Text":"then we can just cross it out because it\u0027s almost equal to 0."},{"Start":"12:53.260 ","End":"12:58.585","Text":"Now, the next thing that we\u0027re going to focus on is the v_gas,"},{"Start":"12:58.585 ","End":"13:00.010","Text":"the velocity of the gas."},{"Start":"13:00.010 ","End":"13:01.510","Text":"Now, as we just said,"},{"Start":"13:01.510 ","End":"13:04.390","Text":"the v_gas is relative to the origin."},{"Start":"13:04.390 ","End":"13:05.755","Text":"In this example here,"},{"Start":"13:05.755 ","End":"13:08.290","Text":"it\u0027s relative to the ground."},{"Start":"13:08.290 ","End":"13:12.925","Text":"However, because it\u0027s the rocket which is emitting this gas,"},{"Start":"13:12.925 ","End":"13:16.900","Text":"it will be easier to find out the velocity relative to"},{"Start":"13:16.900 ","End":"13:21.625","Text":"the rocket because it will make the whole equation a lot simpler."},{"Start":"13:21.625 ","End":"13:24.340","Text":"What we can do is we can say that"},{"Start":"13:24.340 ","End":"13:30.280","Text":"the relative velocity of the gas relative to the rocket is equal"},{"Start":"13:30.280 ","End":"13:39.085","Text":"to the velocity of the gas minus the velocity of the rocket,"},{"Start":"13:39.085 ","End":"13:41.290","Text":"which we just called v(t)."},{"Start":"13:41.290 ","End":"13:47.200","Text":"Now, because I want to substitute in a different value into my v_gas,"},{"Start":"13:47.200 ","End":"13:50.695","Text":"I\u0027m going to isolate out my v_gas,"},{"Start":"13:50.695 ","End":"13:59.095","Text":"so I\u0027ll get that by v_gas is equal to my v_rel plus my v(t)."},{"Start":"13:59.095 ","End":"14:02.200","Text":"Just isolated this out, rearranged them."},{"Start":"14:02.200 ","End":"14:09.880","Text":"Then I can just substitute that back in into my equation for my change in momentum."},{"Start":"14:09.880 ","End":"14:14.830","Text":"My dp in the y-direction is equal to"},{"Start":"14:14.830 ","End":"14:24.385","Text":"M(t)dv minus dmv(t) plus"},{"Start":"14:24.385 ","End":"14:27.100","Text":"dm multiplied by this,"},{"Start":"14:27.100 ","End":"14:36.525","Text":"so dm multiplied by v_rel plus dm multiplied by v(t)."},{"Start":"14:36.525 ","End":"14:41.060","Text":"Then we can see that I have negative dmv(t) plus"},{"Start":"14:41.060 ","End":"14:46.655","Text":"dmv(t) over here so we can cross out this and cross out this."},{"Start":"14:46.655 ","End":"14:51.190","Text":"Then finally, all I\u0027m left with is"},{"Start":"14:51.190 ","End":"15:00.560","Text":"M(t)dv plus dmv relative."},{"Start":"15:01.020 ","End":"15:07.460","Text":"Now, I can just play around with the mathematics a bit and divide it by dt."},{"Start":"15:07.460 ","End":"15:10.580","Text":"Because remember, we want divided by dt,"},{"Start":"15:10.580 ","End":"15:13.550","Text":"so then I can just pretend like this as numbers."},{"Start":"15:13.550 ","End":"15:16.505","Text":"If I divided this side of the equal sign by dt,"},{"Start":"15:16.505 ","End":"15:21.350","Text":"I have to divide all of my variables over here also by dt,"},{"Start":"15:21.350 ","End":"15:27.480","Text":"so dv by dt and dm by dt."},{"Start":"15:28.020 ","End":"15:32.675","Text":"Then we\u0027ll notice that if we look back up,"},{"Start":"15:32.675 ","End":"15:37.635","Text":"we have this exact equation over here."},{"Start":"15:37.635 ","End":"15:42.890","Text":"Now we can see that we got our exact equation that we wanted to find."},{"Start":"15:42.890 ","End":"15:50.120","Text":"We have our mass at time t multiplied by the change in"},{"Start":"15:50.120 ","End":"15:54.440","Text":"velocity plus our changing mass of"},{"Start":"15:54.440 ","End":"15:59.015","Text":"the emitted mass multiplied by the relative velocity of it,"},{"Start":"15:59.015 ","End":"16:02.580","Text":"which is exactly what we had written above."},{"Start":"16:02.630 ","End":"16:08.730","Text":"Let\u0027s move on to some more explanations and some more examples."}],"ID":8222}],"Thumbnail":null,"ID":5426},{"Name":"2. Using The Equation","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Using The Equation In Emission","Duration":"16m 42s","ChapterTopicVideoID":9220,"CourseChapterTopicPlaylistID":5427,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.710","Text":"Hello. In this lesson,"},{"Start":"00:01.710 ","End":"00:04.004","Text":"we\u0027re going to see how to use this equation,"},{"Start":"00:04.004 ","End":"00:07.890","Text":"which we spoke about a lot in the introduction."},{"Start":"00:07.890 ","End":"00:11.400","Text":"We\u0027re going to focus on emission."},{"Start":"00:11.400 ","End":"00:17.105","Text":"One of the most popular examples and questions that like to be asked when"},{"Start":"00:17.105 ","End":"00:22.700","Text":"dealing with variable mass systems is the rocket,"},{"Start":"00:22.700 ","End":"00:31.940","Text":"which is traveling straight in an upwards direction and it\u0027s emitting gases."},{"Start":"00:31.940 ","End":"00:40.055","Text":"They usually want us to find some equation of velocity as a function of time."},{"Start":"00:40.055 ","End":"00:43.925","Text":"The velocity as a function of time of this rocket."},{"Start":"00:43.925 ","End":"00:49.670","Text":"Usually, they\u0027ll give us dm by dt."},{"Start":"00:49.670 ","End":"00:51.940","Text":"It will usually be given. We\u0027ll call it Alpha."},{"Start":"00:51.940 ","End":"00:53.585","Text":"It will also usually be constant."},{"Start":"00:53.585 ","End":"00:56.150","Text":"Not always, but in the meantime, let\u0027s say it\u0027s constant."},{"Start":"00:56.150 ","End":"01:00.470","Text":"Say that the starting mass of the rocket is M_0,"},{"Start":"01:00.470 ","End":"01:06.330","Text":"that the starting velocity V_0 is equal to 0,"},{"Start":"01:06.330 ","End":"01:07.980","Text":"so it starts from rest,"},{"Start":"01:07.980 ","End":"01:13.805","Text":"and that the velocity with which the gases are being emitted,"},{"Start":"01:13.805 ","End":"01:16.895","Text":"which fuel is being emitted from the rocket,"},{"Start":"01:16.895 ","End":"01:20.460","Text":"we\u0027ll say that it\u0027s U_0."},{"Start":"01:20.800 ","End":"01:27.835","Text":"Now let\u0027s see how we use this equation in order to find this equation."},{"Start":"01:27.835 ","End":"01:33.965","Text":"The first thing that I\u0027m going to do is find out the sum of all of my external forces."},{"Start":"01:33.965 ","End":"01:37.430","Text":"I have to say, which is my positive direction,"},{"Start":"01:37.430 ","End":"01:41.785","Text":"which I\u0027ll say is this direction in the y-axis."},{"Start":"01:41.785 ","End":"01:46.043","Text":"Then I have to find out what all of my external forces are,"},{"Start":"01:46.043 ","End":"01:48.710","Text":"so that means not taking into consideration"},{"Start":"01:48.710 ","End":"01:53.600","Text":"the forces acting between the rocket and the fuel that is emitted."},{"Start":"01:53.600 ","End":"01:59.690","Text":"Now the only force that I can say which is working also on the gas is being emitted and"},{"Start":"01:59.690 ","End":"02:06.160","Text":"also on the rocket as a whole is my mass times gravity."},{"Start":"02:06.740 ","End":"02:13.675","Text":"I\u0027m going to write M as a function of t times gravity."},{"Start":"02:13.675 ","End":"02:15.740","Text":"Now, just as a note,"},{"Start":"02:15.740 ","End":"02:19.355","Text":"I\u0027m going to write large M when referring to the mass of the rocket,"},{"Start":"02:19.355 ","End":"02:23.045","Text":"and m when referring to the mass of the gases."},{"Start":"02:23.045 ","End":"02:29.005","Text":"Now, what I could have written here is M as a function of t plus dm"},{"Start":"02:29.005 ","End":"02:35.315","Text":"of this multiplied by g to also take into account the mass of the gas is being emitted."},{"Start":"02:35.315 ","End":"02:38.990","Text":"But, I\u0027m not going to write that because this dm is"},{"Start":"02:38.990 ","End":"02:40.805","Text":"such an infinitely small number"},{"Start":"02:40.805 ","End":"02:44.860","Text":"that it\u0027s really insignificant in comparison to this number,"},{"Start":"02:44.860 ","End":"02:47.869","Text":"and also especially later on in the calculation,"},{"Start":"02:47.869 ","End":"02:50.329","Text":"you\u0027ll see that it\u0027s really insignificant"},{"Start":"02:50.329 ","End":"02:53.305","Text":"and there\u0027s no point even writing it at the beginning."},{"Start":"02:53.305 ","End":"02:57.300","Text":"Now I can start rewriting my equation."},{"Start":"02:57.300 ","End":"03:02.135","Text":"My sum of all the forces which are external is going to be"},{"Start":"03:02.135 ","End":"03:07.385","Text":"negative Mt multiplied by g. Why negative?"},{"Start":"03:07.385 ","End":"03:10.100","Text":"Because it\u0027s acting in the downwards direction and we said that"},{"Start":"03:10.100 ","End":"03:14.220","Text":"this direction is the positive direction."},{"Start":"03:14.220 ","End":"03:17.755","Text":"This equals to Mt."},{"Start":"03:17.755 ","End":"03:20.480","Text":"We still don\u0027t know this, so we\u0027ll write it,"},{"Start":"03:20.480 ","End":"03:22.955","Text":"multiplied by dv,"},{"Start":"03:22.955 ","End":"03:26.475","Text":"in the y-direction, by dt."},{"Start":"03:26.475 ","End":"03:32.915","Text":"Now another way that I could have written this section over here is just like"},{"Start":"03:32.915 ","End":"03:39.905","Text":"my mass times acceleration in the y-direction of my rocket."},{"Start":"03:39.905 ","End":"03:42.020","Text":"I could have just written this as well."},{"Start":"03:42.020 ","End":"03:45.140","Text":"This is like my normal ma if I wasn\u0027t dealing"},{"Start":"03:45.140 ","End":"03:49.310","Text":"with the changing mass due to the emission of gases."},{"Start":"03:49.310 ","End":"03:53.555","Text":"Now, as discussed in a previous lesson in the introduction,"},{"Start":"03:53.555 ","End":"03:59.085","Text":"we said that our v_rel is the velocity of the emission relative to the body."},{"Start":"03:59.085 ","End":"04:04.460","Text":"That means the velocity that the gas is traveling at relative to the rocket,"},{"Start":"04:04.460 ","End":"04:08.670","Text":"and our dm by dt is our rate of emission."},{"Start":"04:08.670 ","End":"04:13.730","Text":"So how much fuel, how much gas is emitted per second,"},{"Start":"04:13.730 ","End":"04:15.760","Text":"per unit of time."},{"Start":"04:15.760 ","End":"04:23.750","Text":"Now we can also say that v_rel sometimes it will be unknown,"},{"Start":"04:23.750 ","End":"04:27.110","Text":"but here we were told that it\u0027s U_0."},{"Start":"04:27.110 ","End":"04:28.594","Text":"Now, just be careful."},{"Start":"04:28.594 ","End":"04:32.000","Text":"We\u0027re told that our U_0 was traveling in the downwards direction,"},{"Start":"04:32.000 ","End":"04:33.770","Text":"and because we\u0027ve said that this is"},{"Start":"04:33.770 ","End":"04:36.710","Text":"our positive direction going in the upwards direction,"},{"Start":"04:36.710 ","End":"04:43.545","Text":"so our v_rel is going to be negative U_0, not U_0."},{"Start":"04:43.545 ","End":"04:47.910","Text":"We can add in a negative here just to make that obvious."},{"Start":"04:48.110 ","End":"04:52.265","Text":"Now I can complete my equation by saying plus v_rel,"},{"Start":"04:52.265 ","End":"04:53.840","Text":"which is negative U_0."},{"Start":"04:53.840 ","End":"04:57.875","Text":"Negative U_0 multiplied by dm by dt,"},{"Start":"04:57.875 ","End":"05:01.815","Text":"which here we\u0027ve said is Alpha."},{"Start":"05:01.815 ","End":"05:07.710","Text":"Now what I need to do is I have to find out what my Mt is."},{"Start":"05:08.050 ","End":"05:17.735","Text":"What I can say is that my M as a function of time is equal to my starting mass,"},{"Start":"05:17.735 ","End":"05:22.130","Text":"which is M_0, minus this Alpha."},{"Start":"05:22.130 ","End":"05:27.199","Text":"What is my Alpha? The amount of mass taken out per second,"},{"Start":"05:27.199 ","End":"05:35.015","Text":"so minus Alpha times t. At 5 seconds if I want to know what the mass of my rocket,"},{"Start":"05:35.015 ","End":"05:37.640","Text":"it\u0027s going to be my starting mass,"},{"Start":"05:37.640 ","End":"05:44.260","Text":"minus 5 times my Alpha, 5 times this."},{"Start":"05:44.260 ","End":"05:46.820","Text":"That\u0027s the simple way to look at it."},{"Start":"05:46.820 ","End":"05:49.497","Text":"Now there\u0027s a slightly more complicated way to look at it,"},{"Start":"05:49.497 ","End":"05:50.570","Text":"so we\u0027re going to look at it now,"},{"Start":"05:50.570 ","End":"05:53.420","Text":"and I hope it won\u0027t confuse anyone too much."},{"Start":"05:53.420 ","End":"05:56.855","Text":"Now, this method that we\u0027re going to speak about now is if"},{"Start":"05:56.855 ","End":"06:00.860","Text":"the rate that the mass is being emitted,"},{"Start":"06:00.860 ","End":"06:05.210","Text":"that the gases being emitted, isn\u0027t a constant."},{"Start":"06:05.210 ","End":"06:09.350","Text":"How we did it over here is great if Alpha is a constant."},{"Start":"06:09.350 ","End":"06:10.939","Text":"If it\u0027s not a constant,"},{"Start":"06:10.939 ","End":"06:14.848","Text":"then we do it this way through a differential equation,"},{"Start":"06:14.848 ","End":"06:16.880","Text":"but an easy one. Don\u0027t panic."},{"Start":"06:16.880 ","End":"06:21.485","Text":"We can write that our rate of change"},{"Start":"06:21.485 ","End":"06:26.660","Text":"of the mass of the rocket as a function of time. That\u0027s what this means."},{"Start":"06:26.660 ","End":"06:28.010","Text":"The rate of change,"},{"Start":"06:28.010 ","End":"06:30.245","Text":"the mass of the rocket as a function of time,"},{"Start":"06:30.245 ","End":"06:33.050","Text":"is equal to negative Alpha."},{"Start":"06:33.050 ","End":"06:34.625","Text":"Why negative?"},{"Start":"06:34.625 ","End":"06:37.984","Text":"Because we know that it\u0027s being emitted."},{"Start":"06:37.984 ","End":"06:39.950","Text":"We\u0027re not gaining mass,"},{"Start":"06:39.950 ","End":"06:41.465","Text":"we\u0027re taking mass away,"},{"Start":"06:41.465 ","End":"06:43.710","Text":"so it\u0027s negative Alpha."},{"Start":"06:44.000 ","End":"06:46.740","Text":"This is the rate of change."},{"Start":"06:46.740 ","End":"06:50.385","Text":"Then all we have to do is integrate."},{"Start":"06:50.385 ","End":"06:57.080","Text":"Then we\u0027ll just have that our M as a function of t is equal to"},{"Start":"06:57.080 ","End":"07:04.870","Text":"negative Alpha and then we integrate with relation to t. We can say dt,"},{"Start":"07:04.870 ","End":"07:12.660","Text":"which equals negative Alpha t plus C, our constant."},{"Start":"07:12.660 ","End":"07:16.685","Text":"We can find out what our constant is by using our initial conditions."},{"Start":"07:16.685 ","End":"07:20.355","Text":"We know that Mt=0,"},{"Start":"07:20.355 ","End":"07:24.430","Text":"our starting time is equal to M_0."},{"Start":"07:24.430 ","End":"07:32.650","Text":"Then we know that therefore at Mt=0,"},{"Start":"07:32.650 ","End":"07:35.390","Text":"we\u0027ll have negative Alpha times 0,"},{"Start":"07:35.390 ","End":"07:38.585","Text":"which will equal 0 plus C,"},{"Start":"07:38.585 ","End":"07:43.000","Text":"meaning that C has to equal M_0."},{"Start":"07:45.690 ","End":"07:50.845","Text":"That means that therefore C is equal to M_0,"},{"Start":"07:50.845 ","End":"07:54.820","Text":"and then we get the exact same equation,"},{"Start":"07:54.820 ","End":"07:56.935","Text":"so if I sub in here M_0,"},{"Start":"07:56.935 ","End":"08:00.410","Text":"so we get this exact equation."},{"Start":"08:00.780 ","End":"08:04.390","Text":"A quick little tip when doing it like this,"},{"Start":"08:04.390 ","End":"08:08.875","Text":"remember we said that V_rel is going to be, here we go."},{"Start":"08:08.875 ","End":"08:10.765","Text":"V_rel is negative U_0,"},{"Start":"08:10.765 ","End":"08:13.720","Text":"the opposite sign to what\u0027s written here."},{"Start":"08:13.720 ","End":"08:15.235","Text":"Then the same over here,"},{"Start":"08:15.235 ","End":"08:21.040","Text":"we have dm by dt is equal to Alpha because that\u0027s our rate of emission,"},{"Start":"08:21.040 ","End":"08:26.365","Text":"which means that we have to have negative Alpha over here because we\u0027re losing this mass,"},{"Start":"08:26.365 ","End":"08:30.925","Text":"or rate of change of mass is losing,"},{"Start":"08:30.925 ","End":"08:33.745","Text":"so that\u0027s our second tip."},{"Start":"08:33.745 ","End":"08:40.510","Text":"That was the most confusing section of how to solve this question so now let\u0027s carry on."},{"Start":"08:40.510 ","End":"08:46.000","Text":"Now we just have to do a little bit of arithmetic,"},{"Start":"08:46.000 ","End":"08:47.680","Text":"a little bit of algebra."},{"Start":"08:47.680 ","End":"08:52.885","Text":"I can divide everything by M(t) and then"},{"Start":"08:52.885 ","End":"08:57.880","Text":"I\u0027ll move this expression to the other side of the equal sign,"},{"Start":"08:57.880 ","End":"09:04.085","Text":"so what I\u0027ll have is U_0 Alpha divided by"},{"Start":"09:04.085 ","End":"09:13.150","Text":"M(t) minus g because I\u0027ve divided by my M(t) which will equal to,"},{"Start":"09:13.150 ","End":"09:20.335","Text":"again divided by M(t) here it\u0027ll equal to dV in my y-direction divided by dt,"},{"Start":"09:20.335 ","End":"09:26.269","Text":"which is just equal to my acceleration in the y direction."},{"Start":"09:26.520 ","End":"09:34.000","Text":"Now, when I substitute in what my M(t) is into here,"},{"Start":"09:34.000 ","End":"09:42.640","Text":"then I get that my a_y is equal to U_0 multiplied by Alpha divided by my M(t) which is"},{"Start":"09:42.640 ","End":"09:46.915","Text":"M_0 minus Alpha t minus"},{"Start":"09:46.915 ","End":"09:54.860","Text":"g. Now I have my acceleration in the y-direction as a function of time."},{"Start":"09:54.900 ","End":"09:59.920","Text":"As we know, if we want to get our velocity,"},{"Start":"09:59.920 ","End":"10:09.445","Text":"we have to integrate so then we can say that our V in the y direction as a function of t,"},{"Start":"10:09.445 ","End":"10:11.500","Text":"this is meant to be in the y-direction,"},{"Start":"10:11.500 ","End":"10:15.700","Text":"is equal to the integral of my acceleration in the y-direction,"},{"Start":"10:15.700 ","End":"10:21.955","Text":"which is equal to the integral of U_0 Alpha divided"},{"Start":"10:21.955 ","End":"10:29.960","Text":"by M_0 minus Alpha t minus gdt."},{"Start":"10:30.360 ","End":"10:38.005","Text":"I\u0027m going to help you solve this integral because it\u0027s a little bit hard,"},{"Start":"10:38.005 ","End":"10:40.580","Text":"and a little bit unexpected."},{"Start":"10:41.460 ","End":"10:48.370","Text":"The integral on of this is equal to,"},{"Start":"10:48.370 ","End":"10:52.660","Text":"so I\u0027ll have U_0 multiplied by Alpha,"},{"Start":"10:52.660 ","End":"10:58.870","Text":"multiplied by ln of M_0 minus"},{"Start":"10:58.870 ","End":"11:03.655","Text":"Alpha t. This can be"},{"Start":"11:03.655 ","End":"11:06.445","Text":"curved because we know that this will always be a positive"},{"Start":"11:06.445 ","End":"11:09.925","Text":"because our mass won\u0027t ever be negative in here."},{"Start":"11:09.925 ","End":"11:16.435","Text":"Then divide this whole thing by the inner derivative of this,"},{"Start":"11:16.435 ","End":"11:21.520","Text":"which has just divided by negative Alpha."},{"Start":"11:21.520 ","End":"11:24.760","Text":"Then this Alpha and this Alpha can cross off,"},{"Start":"11:24.760 ","End":"11:26.440","Text":"so that\u0027s of this section."},{"Start":"11:26.440 ","End":"11:29.950","Text":"Then we have g so that\u0027s a constant,"},{"Start":"11:29.950 ","End":"11:32.170","Text":"so it\u0027s negative gt,"},{"Start":"11:32.170 ","End":"11:38.620","Text":"and then plus C. Don\u0027t forget our plus C because we have an indefinite integral here."},{"Start":"11:38.620 ","End":"11:43.405","Text":"Now I have to find out what my C is. How do I find that?"},{"Start":"11:43.405 ","End":"11:45.760","Text":"Find that through my initial conditions."},{"Start":"11:45.760 ","End":"11:48.805","Text":"Let\u0027s go back to the diagram."},{"Start":"11:48.805 ","End":"11:55.720","Text":"We can see that our initial condition is that V_0,"},{"Start":"11:55.720 ","End":"11:59.590","Text":"initial velocity is equal to 0."},{"Start":"11:59.590 ","End":"12:02.560","Text":"At time t = 0,"},{"Start":"12:02.560 ","End":"12:05.215","Text":"our velocity is equal to 0,"},{"Start":"12:05.215 ","End":"12:08.860","Text":"so then if I say that V(t"},{"Start":"12:08.860 ","End":"12:17.620","Text":"=0) = V_0 which here an example because I said at the top that V_0 = 0."},{"Start":"12:17.620 ","End":"12:21.730","Text":"It might not necessarily be so remember it\u0027s not always zero,"},{"Start":"12:21.730 ","End":"12:30.805","Text":"but here it is equals 0 and that equals this equation when we substitute in t equals 0."},{"Start":"12:30.805 ","End":"12:38.440","Text":"So we\u0027ll have negative U_0 multiplied by ln of, because t is zero,"},{"Start":"12:38.440 ","End":"12:44.350","Text":"we just have ln of M_0 multiplied by negative g multiplied by 0,"},{"Start":"12:44.350 ","End":"12:54.470","Text":"which is 0 plus C. So then we can just rearrange it and say that our C is equal to"},{"Start":"13:03.240 ","End":"13:07.850","Text":"positive U_0 ln M_0."},{"Start":"13:07.880 ","End":"13:14.635","Text":"Then we can just substitute this back in to our equation,"},{"Start":"13:14.635 ","End":"13:21.985","Text":"and then we get that our velocity in the y-direction as a function of time is equal to"},{"Start":"13:21.985 ","End":"13:27.945","Text":"negative U_0ln(M_0 ) minus"},{"Start":"13:27.945 ","End":"13:34.360","Text":"Alpha t minus gt plus C,"},{"Start":"13:34.360 ","End":"13:40.000","Text":"So plus U_0, ln(M_0)."},{"Start":"13:40.000 ","End":"13:45.220","Text":"Then we can say that U_0 is a common factor,"},{"Start":"13:45.220 ","End":"13:48.190","Text":"so we can say U_0 multiplied by ln,"},{"Start":"13:48.190 ","End":"13:53.575","Text":"and then we have ln of M_0 minus ln of M_0 minus Alpha t,"},{"Start":"13:53.575 ","End":"14:00.175","Text":"so we can just write that as M_0 divided by M_0 minus"},{"Start":"14:00.175 ","End":"14:07.360","Text":"Alpha t, then negative gt."},{"Start":"14:07.360 ","End":"14:11.545","Text":"Now we have it, this is the answer to our question."},{"Start":"14:11.545 ","End":"14:15.730","Text":"We were originally asked to find what is the equation for"},{"Start":"14:15.730 ","End":"14:21.415","Text":"velocity as a function of time for a rocket when it\u0027s emitting fuel."},{"Start":"14:21.415 ","End":"14:23.770","Text":"Now we\u0027ve gotten to our answer."},{"Start":"14:23.770 ","End":"14:27.955","Text":"Now remember here the only reason why I could take out"},{"Start":"14:27.955 ","End":"14:32.695","Text":"my V_0 was because my V_0 I said was equal to 0."},{"Start":"14:32.695 ","End":"14:34.735","Text":"This is our final answer."},{"Start":"14:34.735 ","End":"14:36.595","Text":"Now let\u0027s quickly do a recap,"},{"Start":"14:36.595 ","End":"14:38.875","Text":"and then that is the end of our lesson."},{"Start":"14:38.875 ","End":"14:43.600","Text":"What we were asked to do is find an equation for velocity with"},{"Start":"14:43.600 ","End":"14:48.760","Text":"a function of time when we\u0027re given a rocket which is emitting gases."},{"Start":"14:48.760 ","End":"14:55.180","Text":"Now, we decided in which direction was our positive direction,"},{"Start":"14:55.180 ","End":"15:00.610","Text":"so we set up the y-axis and we were given that initial mass,"},{"Start":"15:00.610 ","End":"15:02.530","Text":"the starting mass of the whole rocket with"},{"Start":"15:02.530 ","End":"15:05.995","Text":"all the fuel and still inside the rocket was M_0,"},{"Start":"15:05.995 ","End":"15:08.590","Text":"and that the rocket starts from rest."},{"Start":"15:08.590 ","End":"15:10.510","Text":"So our V_0 is equal to 0."},{"Start":"15:10.510 ","End":"15:13.975","Text":"We\u0027re given our rate of emission and we were given"},{"Start":"15:13.975 ","End":"15:20.545","Text":"the relative velocity of the emitted gases relative to the rocket."},{"Start":"15:20.545 ","End":"15:23.268","Text":"Then we used this equation,"},{"Start":"15:23.268 ","End":"15:27.430","Text":"and we said that external force is the force"},{"Start":"15:27.430 ","End":"15:32.050","Text":"or the forces that are acting on both the emitted gases and the rocket,"},{"Start":"15:32.050 ","End":"15:37.530","Text":"and the only force that is acting on both is mass times gravity."},{"Start":"15:37.530 ","End":"15:40.970","Text":"Then we set up our equation putting"},{"Start":"15:40.970 ","End":"15:44.450","Text":"in all our variables according to our equation up here."},{"Start":"15:44.450 ","End":"15:47.990","Text":"Then we found out what our M as a function"},{"Start":"15:47.990 ","End":"15:51.500","Text":"of time is because here we knew that it was constant,"},{"Start":"15:51.500 ","End":"15:53.840","Text":"we can just write it straight out."},{"Start":"15:53.840 ","End":"15:56.180","Text":"However, we said that if it isn\u0027t"},{"Start":"15:56.180 ","End":"15:59.330","Text":"constant or even if it is constant in other way of doing it,"},{"Start":"15:59.330 ","End":"16:04.100","Text":"is saying that our rate of change of mass with regards to"},{"Start":"16:04.100 ","End":"16:09.830","Text":"time is equal to negative the rate of emission,"},{"Start":"16:09.830 ","End":"16:15.170","Text":"and then we solve this differential equation and got an answer."},{"Start":"16:15.170 ","End":"16:17.165","Text":"Then we just substituted this n,"},{"Start":"16:17.165 ","End":"16:20.390","Text":"got another differential equation,"},{"Start":"16:20.390 ","End":"16:27.785","Text":"integrated then we use our initial conditions in order to find out what our C is,"},{"Start":"16:27.785 ","End":"16:31.665","Text":"what are constant is because this was an indefinite integral,"},{"Start":"16:31.665 ","End":"16:34.820","Text":"and then we substituted everything in and"},{"Start":"16:34.820 ","End":"16:38.660","Text":"rearranged it in order to find our final equation."},{"Start":"16:38.660 ","End":"16:42.120","Text":"That\u0027s the end of this lesson."}],"ID":10610},{"Watched":false,"Name":"Using The Formula In Accretion","Duration":"22m 50s","ChapterTopicVideoID":9221,"CourseChapterTopicPlaylistID":5427,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.320 ","End":"00:03.194","Text":"Hello. In this example,"},{"Start":"00:03.194 ","End":"00:07.304","Text":"we\u0027re going to be seeing how to use the formula in the case of accretion."},{"Start":"00:07.304 ","End":"00:08.895","Text":"Now what does accretion mean?"},{"Start":"00:08.895 ","End":"00:11.400","Text":"It just means gaining mass."},{"Start":"00:11.400 ","End":"00:15.470","Text":"In the rocket, we saw emission when we\u0027re losing mass,"},{"Start":"00:15.470 ","End":"00:18.844","Text":"and here, there\u0027s accretion, gaining mass."},{"Start":"00:18.844 ","End":"00:20.240","Text":"What\u0027s happening here,"},{"Start":"00:20.240 ","End":"00:24.815","Text":"there\u0027s a cart moving on the floor, frictionless."},{"Start":"00:24.815 ","End":"00:29.630","Text":"There\u0027s no friction when it\u0027s moving and rain is falling into the cart at a rate"},{"Start":"00:29.630 ","End":"00:35.855","Text":"of dm by dt is equal to Alpha."},{"Start":"00:35.855 ","End":"00:39.005","Text":"That\u0027s the rate that the rain is falling into the cart."},{"Start":"00:39.005 ","End":"00:43.250","Text":"The cart is moving at a velocity v(t),"},{"Start":"00:43.250 ","End":"00:49.595","Text":"and it has an initial mass of m_0 of the cart,"},{"Start":"00:49.595 ","End":"00:57.600","Text":"and it\u0027s starting velocity at time t equals 0 is equal to v_0."},{"Start":"00:58.770 ","End":"01:03.550","Text":"Now, let\u0027s take a look at our equation and see how it"},{"Start":"01:03.550 ","End":"01:08.870","Text":"changes between the accretion and emission examples."},{"Start":"01:08.870 ","End":"01:14.200","Text":"We have the sum of all of our forces on the x-axis."},{"Start":"01:14.200 ","End":"01:19.450","Text":"Let\u0027s look first which equals to the sum of our external forces."},{"Start":"01:19.450 ","End":"01:21.310","Text":"Because we said that this is frictionless,"},{"Start":"01:21.310 ","End":"01:23.020","Text":"so that equals 0."},{"Start":"01:23.020 ","End":"01:26.140","Text":"There\u0027s no forces working on the x-axis."},{"Start":"01:26.140 ","End":"01:30.500","Text":"Then we say that this equals to M,"},{"Start":"01:30.500 ","End":"01:32.920","Text":"my mass as a function of time,"},{"Start":"01:32.920 ","End":"01:41.180","Text":"multiplied by my dv on the x-axis divided by dt."},{"Start":"01:41.580 ","End":"01:45.920","Text":"In a second, we\u0027re going to find out what this function is,"},{"Start":"01:45.920 ","End":"01:47.180","Text":"M as a function of time,"},{"Start":"01:47.180 ","End":"01:52.340","Text":"and then we have plus our v relative."},{"Start":"01:52.340 ","End":"01:54.215","Text":"Now what is our v relative?"},{"Start":"01:54.215 ","End":"02:02.710","Text":"It\u0027s the relative velocity of the rain going into the carts relative to the cart."},{"Start":"02:02.710 ","End":"02:04.625","Text":"Now, when we were dealing with a rocket,"},{"Start":"02:04.625 ","End":"02:10.850","Text":"it was a lot easier because the rocket was emitting the fuel or the gases,"},{"Start":"02:10.850 ","End":"02:14.450","Text":"so the velocity of the fuel of the gases exiting"},{"Start":"02:14.450 ","End":"02:19.775","Text":"the rocket was always going to be constant and constant relative to the rocket."},{"Start":"02:19.775 ","End":"02:25.655","Text":"However, here the rain is falling into the cart and it isn\u0027t relative to the cart,"},{"Start":"02:25.655 ","End":"02:29.670","Text":"it\u0027s relative to the ground."},{"Start":"02:30.550 ","End":"02:32.825","Text":"Now, in our example,"},{"Start":"02:32.825 ","End":"02:37.790","Text":"we\u0027re currently assuming that the rain is falling downwards,"},{"Start":"02:37.790 ","End":"02:39.140","Text":"along the y-axis,"},{"Start":"02:39.140 ","End":"02:42.470","Text":"and so it doesn\u0027t have an x-direction."},{"Start":"02:42.470 ","End":"02:47.375","Text":"Later, we\u0027ll change the angle and everything and they will be in the x and y-direction,"},{"Start":"02:47.375 ","End":"02:49.310","Text":"and we\u0027ll see how to solve it."},{"Start":"02:49.310 ","End":"02:51.550","Text":"But in the meantime right now,"},{"Start":"02:51.550 ","End":"03:00.310","Text":"our velocity of the rain in the x-direction is equal to 0."},{"Start":"03:00.800 ","End":"03:07.340","Text":"Now, this is our rain relative to the ground in the x-direction."},{"Start":"03:07.340 ","End":"03:15.300","Text":"Now, we want to find what the velocity of the rain is relative to our cart."},{"Start":"03:15.300 ","End":"03:18.240","Text":"Let\u0027s remember how we do this."},{"Start":"03:18.240 ","End":"03:22.440","Text":"The equation for this is v-tag."},{"Start":"03:22.440 ","End":"03:26.510","Text":"This means the velocity relative to something."},{"Start":"03:26.510 ","End":"03:35.495","Text":"The velocity of the rain relative to the cart is equal to the velocity relative"},{"Start":"03:35.495 ","End":"03:44.585","Text":"to the ground minus the velocity of the cart relative to the ground."},{"Start":"03:44.585 ","End":"03:49.385","Text":"We know that the velocity of the cart relative to the ground is equal to v(t) and that"},{"Start":"03:49.385 ","End":"03:56.280","Text":"the velocity of the rain relative to the ground is equal to 0."},{"Start":"03:56.280 ","End":"04:04.760","Text":"What I\u0027ve done is I\u0027ve added in this equation over here and it\u0027s important to use"},{"Start":"04:04.760 ","End":"04:07.880","Text":"this whenever we use this equation as well"},{"Start":"04:07.880 ","End":"04:12.830","Text":"because this v_rel will always be equal to this,"},{"Start":"04:12.830 ","End":"04:18.710","Text":"where this is the emitted material,"},{"Start":"04:18.710 ","End":"04:22.389","Text":"the material relative to the object."},{"Start":"04:22.389 ","End":"04:28.745","Text":"Here, it would be the rain relative to the cart or the fuel relative to the rocket,"},{"Start":"04:28.745 ","End":"04:30.215","Text":"whatever it would be,"},{"Start":"04:30.215 ","End":"04:37.220","Text":"equals this v, which will be the material relative to the ground."},{"Start":"04:37.220 ","End":"04:39.860","Text":"Here, it\u0027s the rain relative to the ground,"},{"Start":"04:39.860 ","End":"04:44.810","Text":"minus the velocity of the body relative to the ground."},{"Start":"04:44.810 ","End":"04:50.010","Text":"Here is the velocity of the cart or the velocity of the rocket or whatever it will be."},{"Start":"04:51.380 ","End":"04:56.784","Text":"Now, these two equations I\u0027ve put in red,"},{"Start":"04:56.784 ","End":"05:01.040","Text":"because if you\u0027re allowed to bring in an equation sheet,"},{"Start":"05:01.040 ","End":"05:02.810","Text":"these equations should definitely be written on"},{"Start":"05:02.810 ","End":"05:07.320","Text":"that sheet and they\u0027re important to be very familiar with."},{"Start":"05:07.550 ","End":"05:10.455","Text":"Back to here."},{"Start":"05:10.455 ","End":"05:13.640","Text":"Now I know that my v relative,"},{"Start":"05:13.640 ","End":"05:15.935","Text":"my rain relative to my cart,"},{"Start":"05:15.935 ","End":"05:18.433","Text":"is equal to 0 minus v,"},{"Start":"05:18.433 ","End":"05:22.460","Text":"t. I can add here 0 minus v,"},{"Start":"05:22.460 ","End":"05:24.440","Text":"t. Now as you know that will just be v(t),"},{"Start":"05:24.440 ","End":"05:29.470","Text":"but I\u0027ve written it in long so that you understand where it comes from."},{"Start":"05:29.470 ","End":"05:35.495","Text":"Now, the next thing I have to do is to multiply by this dm by dt."},{"Start":"05:35.495 ","End":"05:40.085","Text":"Now here comes the great change between"},{"Start":"05:40.085 ","End":"05:45.544","Text":"the example of accretion versus the example of emission."},{"Start":"05:45.544 ","End":"05:47.585","Text":"What it is, is that here,"},{"Start":"05:47.585 ","End":"05:50.450","Text":"when we derived this equation,"},{"Start":"05:50.450 ","End":"05:53.210","Text":"we did it with the example of the rocket,"},{"Start":"05:53.210 ","End":"05:56.525","Text":"with an example of emission when the mass"},{"Start":"05:56.525 ","End":"06:00.530","Text":"of the initial objects is reducing, is going down."},{"Start":"06:00.530 ","End":"06:02.870","Text":"However here, the mass is increasing."},{"Start":"06:02.870 ","End":"06:10.470","Text":"We have to multiply over here by negative dm by dt,"},{"Start":"06:10.470 ","End":"06:14.430","Text":"which here we said was Alpha."},{"Start":"06:14.430 ","End":"06:16.373","Text":"I\u0027ll just rub this out now."},{"Start":"06:16.373 ","End":"06:19.990","Text":"So negative Alpha."},{"Start":"06:20.060 ","End":"06:23.330","Text":"That is what\u0027s really important."},{"Start":"06:23.330 ","End":"06:26.593","Text":"When you\u0027re gaining mass in your object,"},{"Start":"06:26.593 ","End":"06:30.770","Text":"you have to make sure that the dm by dt here is"},{"Start":"06:30.770 ","End":"06:36.997","Text":"represented as a negative expression because we\u0027re gaining mass."},{"Start":"06:36.997 ","End":"06:40.550","Text":"In the rocket, it was shooting out the fuel so we were losing mass,"},{"Start":"06:40.550 ","End":"06:43.105","Text":"it would just be dm by dt,"},{"Start":"06:43.105 ","End":"06:44.420","Text":"and here we\u0027re gaining mass,"},{"Start":"06:44.420 ","End":"06:47.550","Text":"it\u0027s negative dm by dt."},{"Start":"06:47.560 ","End":"06:54.589","Text":"The main differences between emission and accretion is our v relative."},{"Start":"06:54.589 ","End":"06:59.060","Text":"In emission, the fuel or"},{"Start":"06:59.060 ","End":"07:00.860","Text":"whatever it is which is being emitted is"},{"Start":"07:00.860 ","End":"07:04.115","Text":"relative to the rocket so you don\u0027t have to change anything."},{"Start":"07:04.115 ","End":"07:06.889","Text":"However, when you\u0027re gaining mass,"},{"Start":"07:06.889 ","End":"07:08.554","Text":"whatever is being gained,"},{"Start":"07:08.554 ","End":"07:10.610","Text":"the velocity at which it is happening isn\u0027t"},{"Start":"07:10.610 ","End":"07:13.450","Text":"relative to the object but it\u0027s relative to the ground."},{"Start":"07:13.450 ","End":"07:18.670","Text":"You have to use this equation in order to translate the values,"},{"Start":"07:18.670 ","End":"07:23.810","Text":"and the second change with accretion is that instead of dm by dt,"},{"Start":"07:23.810 ","End":"07:26.770","Text":"you have negative dm by dt."},{"Start":"07:26.770 ","End":"07:30.635","Text":"Just in case that was a bit confusing what I said,"},{"Start":"07:30.635 ","End":"07:32.315","Text":"I wrote it down."},{"Start":"07:32.315 ","End":"07:34.340","Text":"You can also write this as a little note in"},{"Start":"07:34.340 ","End":"07:39.125","Text":"your formula sheets if it helps you when you\u0027re doing the questions."},{"Start":"07:39.125 ","End":"07:44.353","Text":"Now, let\u0027s get back to our equation."},{"Start":"07:44.353 ","End":"07:46.190","Text":"Now we\u0027ve sorted this all out,"},{"Start":"07:46.190 ","End":"07:50.545","Text":"and now we just have to find out what this M(t) is equal to."},{"Start":"07:50.545 ","End":"07:53.115","Text":"Let\u0027s take a look."},{"Start":"07:53.115 ","End":"07:59.630","Text":"We have that our mass as a function of time is equal to our starting mass,"},{"Start":"07:59.630 ","End":"08:01.505","Text":"which we said was M_0,"},{"Start":"08:01.505 ","End":"08:03.425","Text":"the mass that we start off with,"},{"Start":"08:03.425 ","End":"08:08.450","Text":"plus our change in mass times the time."},{"Start":"08:08.450 ","End":"08:12.305","Text":"We saw that our change in mass is dm by dt,"},{"Start":"08:12.305 ","End":"08:16.200","Text":"which is equal to Alpha."},{"Start":"08:16.400 ","End":"08:19.410","Text":"Why Alpha not negative Alpha?"},{"Start":"08:19.410 ","End":"08:21.870","Text":"Because we\u0027re adding mass in,"},{"Start":"08:21.870 ","End":"08:25.875","Text":"Alpha multiplied by the time."},{"Start":"08:25.875 ","End":"08:31.370","Text":"We have our beginning mass plus our amount that"},{"Start":"08:31.370 ","End":"08:37.430","Text":"our rate that the mass is increasing multiplied by the time."},{"Start":"08:37.430 ","End":"08:43.410","Text":"Then we just substitute this into M(t) over here,"},{"Start":"08:43.410 ","End":"08:49.240","Text":"and then what we\u0027re going to get is we\u0027re going to get that 0 is equal to M(t)."},{"Start":"08:49.240 ","End":"08:56.440","Text":"It\u0027s equal to M_0 plus Alpha t multiplied by d"},{"Start":"08:56.440 ","End":"09:04.495","Text":"of our velocity in the x-direction by dt now as you know that v-dot is equal to a."},{"Start":"09:04.495 ","End":"09:11.630","Text":"You remember that? That\u0027s multiplied by a in the x-direction plus,"},{"Start":"09:11.630 ","End":"09:17.080","Text":"and then we have negative v(t) multiplied by negative Alpha so it becomes a positive."},{"Start":"09:17.080 ","End":"09:19.769","Text":"That\u0027s the positive here,"},{"Start":"09:19.769 ","End":"09:24.330","Text":"so v(t) multiplied by Alpha."},{"Start":"09:24.840 ","End":"09:28.885","Text":"Now, we can see that this equation is a lot more"},{"Start":"09:28.885 ","End":"09:35.785","Text":"complicated when we think about the equation that we had for the emission."},{"Start":"09:35.785 ","End":"09:40.000","Text":"The only way that we can solve this is via"},{"Start":"09:40.000 ","End":"09:45.130","Text":"differential equations and using the separation variables method."},{"Start":"09:45.130 ","End":"09:49.285","Text":"Now, if you saw the lesson about frictional force,"},{"Start":"09:49.285 ","End":"09:55.105","Text":"I show how to work out a differential equation via this method, separation of variables."},{"Start":"09:55.105 ","End":"09:56.575","Text":"However, I\u0027m going to do it now,"},{"Start":"09:56.575 ","End":"10:00.290","Text":"if you don\u0027t understand go back to that lesson."},{"Start":"10:00.900 ","End":"10:07.010","Text":"Now the first thing that you have to do in this method is when"},{"Start":"10:07.010 ","End":"10:12.515","Text":"you see variables that are derivatives of other variables,"},{"Start":"10:12.515 ","End":"10:15.650","Text":"you have to put them altogether in 1 side,"},{"Start":"10:15.650 ","End":"10:19.175","Text":"and then when you have some expression"},{"Start":"10:19.175 ","End":"10:23.915","Text":"that doesn\u0027t have to do anything with derivatives of other variables,"},{"Start":"10:23.915 ","End":"10:25.730","Text":"then you take it to the other side."},{"Start":"10:25.730 ","End":"10:27.425","Text":"Here, we have this a_x,"},{"Start":"10:27.425 ","End":"10:30.125","Text":"which we know is the derivative of velocity."},{"Start":"10:30.125 ","End":"10:33.815","Text":"We\u0027re going to keep this expression over here and we\u0027re going to move this expression"},{"Start":"10:33.815 ","End":"10:37.250","Text":"over to the other side of the equal sign,"},{"Start":"10:37.250 ","End":"10:42.180","Text":"so we\u0027re going to have negative v(t) multiplied by Alpha,"},{"Start":"10:42.180 ","End":"10:51.251","Text":"which will equal to M_0 plus Alpha t, a_x."},{"Start":"10:51.251 ","End":"10:54.550","Text":"Then instead of a_x,"},{"Start":"10:54.550 ","End":"10:56.980","Text":"I\u0027m going to write what we had over here,"},{"Start":"10:56.980 ","End":"11:02.240","Text":"which was dv_x by dt."},{"Start":"11:02.820 ","End":"11:06.115","Text":"Now, I\u0027m just going to do normal algebra."},{"Start":"11:06.115 ","End":"11:08.800","Text":"I have this dt here in the denominator,"},{"Start":"11:08.800 ","End":"11:12.325","Text":"so I\u0027m going to multiply both sides by this dt."},{"Start":"11:12.325 ","End":"11:16.180","Text":"Then I\u0027m going to get, and I\u0027m going to just rearrange this a little bit,"},{"Start":"11:16.180 ","End":"11:23.868","Text":"negative Alpha v_x(t),"},{"Start":"11:23.868 ","End":"11:33.440","Text":"dt is equal to M_0 plus Alpha t dv_x."},{"Start":"11:35.940 ","End":"11:38.440","Text":"Now, as we can see,"},{"Start":"11:38.440 ","End":"11:45.970","Text":"I have a variable of my velocity and variable with time over here. This is velocity."},{"Start":"11:45.970 ","End":"11:48.100","Text":"Now what I want to do is I want to get my variable with"},{"Start":"11:48.100 ","End":"11:51.160","Text":"velocity to the one side where my dv"},{"Start":"11:51.160 ","End":"11:57.130","Text":"is and I want to get my variable of time to the side where my dt is."},{"Start":"11:57.130 ","End":"12:01.720","Text":"What I\u0027m going to do is here\u0027s my dt and here\u0027s my variable of t,"},{"Start":"12:01.720 ","End":"12:07.375","Text":"so I\u0027m going to divide both sides by M_0 plus Alpha t. Then over here,"},{"Start":"12:07.375 ","End":"12:10.210","Text":"I have my v as a function of t,"},{"Start":"12:10.210 ","End":"12:13.540","Text":"but it\u0027s a v and I want to get it to my dv over here,"},{"Start":"12:13.540 ","End":"12:21.880","Text":"so I\u0027m going to divide both sides by v_x as a function of t. Let\u0027s do this."},{"Start":"12:21.880 ","End":"12:26.045","Text":"I\u0027m dividing both sides by this and by this,"},{"Start":"12:26.045 ","End":"12:30.850","Text":"by v_x t and by M_0 plus"},{"Start":"12:30.850 ","End":"12:38.800","Text":"Alpha t. Then I\u0027m going to get negative Alpha divided by this over here,"},{"Start":"12:38.800 ","End":"12:44.035","Text":"which is M_0 plus Alpha t dt,"},{"Start":"12:44.035 ","End":"12:55.760","Text":"which is going to equal 1 over vt dv."},{"Start":"12:56.640 ","End":"13:03.625","Text":"Now, I have these expressions and I\u0027m just going to integrate on both sides."},{"Start":"13:03.625 ","End":"13:07.224","Text":"Now we can do definite integrals and indefinite integrals."},{"Start":"13:07.224 ","End":"13:10.750","Text":"Here, I\u0027m going to do a definite integral."},{"Start":"13:10.750 ","End":"13:14.739","Text":"Now, why am I doing it definite instead of indefinite?"},{"Start":"13:14.739 ","End":"13:19.865","Text":"Because I want to get my answer without my extra added constant"},{"Start":"13:19.865 ","End":"13:25.750","Text":"when you do plus C. What are my bounds going to be?"},{"Start":"13:25.750 ","End":"13:27.535","Text":"With regards to time,"},{"Start":"13:27.535 ","End":"13:32.410","Text":"I\u0027m going to be starting from t equals 0 until some time t,"},{"Start":"13:32.410 ","End":"13:35.995","Text":"so from 0 to t. Then with my velocity,"},{"Start":"13:35.995 ","End":"13:40.644","Text":"I\u0027m going to be starting from v(t) equals 0,"},{"Start":"13:40.644 ","End":"13:43.120","Text":"which is equal to v_0."},{"Start":"13:43.120 ","End":"13:45.430","Text":"Let\u0027s just scroll back up here."},{"Start":"13:45.430 ","End":"13:49.945","Text":"Remember, v(t=0), our starting velocity is going to be v_0."},{"Start":"13:49.945 ","End":"13:55.525","Text":"I put that in up until v(t),"},{"Start":"13:55.525 ","End":"13:57.445","Text":"up until my velocity of some time,"},{"Start":"13:57.445 ","End":"14:01.885","Text":"which will be the same velocity at time t. Now,"},{"Start":"14:01.885 ","End":"14:05.545","Text":"all I have to do is I have to do my integral."},{"Start":"14:05.545 ","End":"14:11.755","Text":"I\u0027m going to have just negative Alpha divided by Alpha,"},{"Start":"14:11.755 ","End":"14:13.690","Text":"which will obviously just be 1,"},{"Start":"14:13.690 ","End":"14:16.930","Text":"but I\u0027m just writing it so that you can see where I\u0027m getting this from."},{"Start":"14:16.930 ","End":"14:21.310","Text":"ln of M_0 plus"},{"Start":"14:21.310 ","End":"14:30.205","Text":"Alpha t. Then my bounds are from 0 to t. We\u0027ll substitute these in one moment,"},{"Start":"14:30.205 ","End":"14:31.915","Text":"which equals to,"},{"Start":"14:31.915 ","End":"14:39.775","Text":"this will be ln of v_x."},{"Start":"14:39.775 ","End":"14:43.889","Text":"The bounds will also be v_0 and v(t)."},{"Start":"14:43.889 ","End":"14:47.330","Text":"We\u0027re going to substitute them now."},{"Start":"14:47.400 ","End":"14:50.440","Text":"Now, substituting in our bounds,"},{"Start":"14:50.440 ","End":"14:53.635","Text":"so this goes to 1."},{"Start":"14:53.635 ","End":"14:58.225","Text":"Remember that when you\u0027re dealing with lns or logs,"},{"Start":"14:58.225 ","End":"15:02.170","Text":"when you have log of something minus log of another,"},{"Start":"15:02.170 ","End":"15:06.565","Text":"then you can just do log of the something divided by another."},{"Start":"15:06.565 ","End":"15:16.260","Text":"Here, it\u0027s just going to be negative ln of M_0 plus Alpha t divided by,"},{"Start":"15:16.260 ","End":"15:19.688","Text":"then substitute 0 into this t over here,"},{"Start":"15:19.688 ","End":"15:22.450","Text":"so divided by M_0,"},{"Start":"15:22.680 ","End":"15:24.970","Text":"which is going to equal,"},{"Start":"15:24.970 ","End":"15:26.500","Text":"again, the same thing,"},{"Start":"15:26.500 ","End":"15:32.770","Text":"ln of v(t) divided by v_0."},{"Start":"15:32.770 ","End":"15:36.700","Text":"Because it\u0027s ln of v(t) minus ln of v_0,"},{"Start":"15:36.700 ","End":"15:40.610","Text":"so it becomes ln of v(t) divided by v_0."},{"Start":"15:41.910 ","End":"15:45.580","Text":"Now, to make things simpler,"},{"Start":"15:45.580 ","End":"15:49.960","Text":"I\u0027m going to get rid of this negative sign because I\u0027m going to turn it into a positive,"},{"Start":"15:49.960 ","End":"15:53.695","Text":"and then I\u0027m going to put here negative 1."},{"Start":"15:53.695 ","End":"15:56.119","Text":"Because remember, with logs,"},{"Start":"15:56.119 ","End":"15:59.455","Text":"if you have any number,"},{"Start":"15:59.455 ","End":"16:01.940","Text":"in fact, 3, 100,"},{"Start":"16:01.940 ","End":"16:03.640","Text":"17, whatever it is,"},{"Start":"16:03.640 ","End":"16:05.785","Text":"multiplying a log,"},{"Start":"16:05.785 ","End":"16:11.545","Text":"then you can take it away from being the coefficient and put it as a power."},{"Start":"16:11.545 ","End":"16:14.500","Text":"Here, my coefficient was negative 1,"},{"Start":"16:14.500 ","End":"16:19.525","Text":"so I put these brackets to the power of negative 1."},{"Start":"16:19.525 ","End":"16:24.070","Text":"Then I can just exponent this,"},{"Start":"16:24.070 ","End":"16:27.700","Text":"so I can do e to the power of this and e to the power of that,"},{"Start":"16:27.700 ","End":"16:29.350","Text":"and that gets rid of the ln,"},{"Start":"16:29.350 ","End":"16:33.685","Text":"and then I\u0027ll flip this over because it\u0027s to the negative 1 so I can turn it around."},{"Start":"16:33.685 ","End":"16:38.785","Text":"Then I\u0027ll just have M_0 divided by M_0 plus"},{"Start":"16:38.785 ","End":"16:46.840","Text":"Alpha t is going to be equal to v(t) divided by v_0."},{"Start":"16:46.840 ","End":"16:48.865","Text":"Now, if you remember,"},{"Start":"16:48.865 ","End":"16:51.460","Text":"I want my velocity as a function of time."},{"Start":"16:51.460 ","End":"16:54.670","Text":"All I have to do is I have to isolate out my v(t)."},{"Start":"16:54.670 ","End":"16:57.280","Text":"If I multiply both sides by v_0,"},{"Start":"16:57.280 ","End":"16:59.350","Text":"I\u0027ll get my v(t)."},{"Start":"16:59.350 ","End":"17:06.655","Text":"Then I can just erase my v_0 here and multiply it here."},{"Start":"17:06.655 ","End":"17:11.570","Text":"Then this is the final answer."},{"Start":"17:12.330 ","End":"17:18.250","Text":"This is the velocity as a function of time."},{"Start":"17:18.250 ","End":"17:21.460","Text":"Let\u0027s do a short little recap."},{"Start":"17:21.460 ","End":"17:25.465","Text":"What we did was we said that we had a cart that was moving at"},{"Start":"17:25.465 ","End":"17:31.114","Text":"an initial velocity of v_0 and was then moving at some kind"},{"Start":"17:31.114 ","End":"17:35.815","Text":"of v as a function of t. We were asked to find what this v as a function of t is"},{"Start":"17:35.815 ","End":"17:42.205","Text":"when we knew that rain was falling into the cart at a rate of dm by dt,"},{"Start":"17:42.205 ","End":"17:44.995","Text":"which we said was just going to be equal to Alpha."},{"Start":"17:44.995 ","End":"17:49.900","Text":"Then what we did is we just used this equation that we know and love."},{"Start":"17:49.900 ","End":"17:52.885","Text":"We took into account that here,"},{"Start":"17:52.885 ","End":"17:55.900","Text":"because mass is being added to the cart,"},{"Start":"17:55.900 ","End":"18:02.490","Text":"that our v relative is going to be this according to this formula,"},{"Start":"18:02.490 ","End":"18:07.060","Text":"where this v was the rain relative to the ground,"},{"Start":"18:07.060 ","End":"18:11.340","Text":"this v(t) was the cart relative to the ground,"},{"Start":"18:11.340 ","End":"18:12.745","Text":"and our v_rel was"},{"Start":"18:12.745 ","End":"18:18.670","Text":"the velocity of the rain relative to the cart when we said that the rain"},{"Start":"18:18.670 ","End":"18:22.360","Text":"was falling just straight downwards just"},{"Start":"18:22.360 ","End":"18:26.695","Text":"in the y-direction with an angle of 90 degrees to the ground."},{"Start":"18:26.695 ","End":"18:30.010","Text":"Then we said that the important thing to remember"},{"Start":"18:30.010 ","End":"18:33.700","Text":"was when adding in this v_rel to use this equation,"},{"Start":"18:33.700 ","End":"18:37.285","Text":"and that when mass is being added to the cart,"},{"Start":"18:37.285 ","End":"18:40.990","Text":"that we have to have this value,"},{"Start":"18:40.990 ","End":"18:42.610","Text":"this dm by dt,"},{"Start":"18:42.610 ","End":"18:45.040","Text":"as a negative value."},{"Start":"18:45.040 ","End":"18:48.025","Text":"Then we wrote this down to remind us."},{"Start":"18:48.025 ","End":"18:50.590","Text":"Now, what we\u0027re going to do is we\u0027re going to look at"},{"Start":"18:50.590 ","End":"18:53.559","Text":"the rain as if it\u0027s not falling perpendicular"},{"Start":"18:53.559 ","End":"18:58.751","Text":"to the ground but in a horizontal manner."},{"Start":"18:58.751 ","End":"19:00.055","Text":"So like this."},{"Start":"19:00.055 ","End":"19:06.500","Text":"Imagine each one of these droplets falling at an angle Theta."},{"Start":"19:08.610 ","End":"19:15.520","Text":"Now, the angle that the rain is falling at with regards to the x-axis,"},{"Start":"19:15.520 ","End":"19:18.835","Text":"so imagine that this is the x-axis,"},{"Start":"19:18.835 ","End":"19:21.385","Text":"is going to be Theta,"},{"Start":"19:21.385 ","End":"19:26.125","Text":"and the rain is falling with the size of its velocity,"},{"Start":"19:26.125 ","End":"19:27.940","Text":"so its speed, if you will,"},{"Start":"19:27.940 ","End":"19:30.940","Text":"is going to be u_0."},{"Start":"19:30.940 ","End":"19:39.580","Text":"Now let\u0027s see how we can change over this equation in order to match this new system."},{"Start":"19:39.580 ","End":"19:44.380","Text":"Now, seeing that all of the other information remains the same,"},{"Start":"19:44.380 ","End":"19:46.460","Text":"our v(t) is the same,"},{"Start":"19:46.460 ","End":"19:48.636","Text":"our time 0 is the same,"},{"Start":"19:48.636 ","End":"19:50.830","Text":"our initial mass is the same,"},{"Start":"19:50.830 ","End":"19:55.890","Text":"our rate of increase in mass or dm by dt is also the same,"},{"Start":"19:55.890 ","End":"20:00.780","Text":"so the only things we would have to change in this equation are as so."},{"Start":"20:00.780 ","End":"20:05.820","Text":"Our v_x, so this over here."},{"Start":"20:05.820 ","End":"20:11.810","Text":"Our v_x will not be 0 anymore."},{"Start":"20:11.810 ","End":"20:16.080","Text":"It will be equal to u_0."},{"Start":"20:17.240 ","End":"20:24.040","Text":"Then because we want the adjacent to the angle because we want on the x-axis,"},{"Start":"20:24.040 ","End":"20:28.165","Text":"we want the adjacent so it will be cosine of Theta."},{"Start":"20:28.165 ","End":"20:33.955","Text":"Then that means that in this equation over here,"},{"Start":"20:33.955 ","End":"20:40.720","Text":"instead of 0, we would have u_0 cosine of Theta."},{"Start":"20:40.720 ","End":"20:44.529","Text":"Then you just do the exact same maths,"},{"Start":"20:44.529 ","End":"20:52.600","Text":"but just by substituting in u_0 cosine Theta minus v(t) instead of just minus v(t)."},{"Start":"20:52.600 ","End":"20:57.059","Text":"Our negative Alpha remains exactly the same and our negative vt remains exactly the same."},{"Start":"20:57.059 ","End":"20:58.311","Text":"We don\u0027t do anything,"},{"Start":"20:58.311 ","End":"21:03.430","Text":"we just substitute in this over here,"},{"Start":"21:03.990 ","End":"21:10.255","Text":"this and this, and then you\u0027ll see that after doing all of the maths,"},{"Start":"21:10.255 ","End":"21:13.060","Text":"you\u0027ll get this exact same answer."},{"Start":"21:13.060 ","End":"21:23.240","Text":"The only difference is that you\u0027ll get here plus u_0 cosine of Theta."},{"Start":"21:23.240 ","End":"21:28.470","Text":"Your final answer when the rain is falling at an angle Theta and in"},{"Start":"21:28.470 ","End":"21:33.595","Text":"an initial speed of u_0 to the ground,"},{"Start":"21:33.595 ","End":"21:35.620","Text":"then you\u0027ll get the exact same answer,"},{"Start":"21:35.620 ","End":"21:40.755","Text":"but with an added u_0 cosine of Theta."},{"Start":"21:40.755 ","End":"21:45.260","Text":"Now it\u0027s important to note that specifically in this answer, well,"},{"Start":"21:45.260 ","End":"21:48.890","Text":"we can just add in this u_0 cosine of Theta,"},{"Start":"21:48.890 ","End":"21:51.370","Text":"because when we integrated,"},{"Start":"21:51.370 ","End":"21:54.590","Text":"we integrated by dt and dv,"},{"Start":"21:54.590 ","End":"21:56.370","Text":"so not by d Theta."},{"Start":"21:56.370 ","End":"22:00.815","Text":"However, sometimes your change over here,"},{"Start":"22:00.815 ","End":"22:04.385","Text":"your velocity of the rain relative to the ground,"},{"Start":"22:04.385 ","End":"22:10.235","Text":"won\u0027t be in relation to Theta but might be in relation to v or to t, in which case,"},{"Start":"22:10.235 ","End":"22:13.670","Text":"it would be changed in the integration,"},{"Start":"22:13.670 ","End":"22:16.415","Text":"and then you won\u0027t get this final answer."},{"Start":"22:16.415 ","End":"22:19.260","Text":"In this case specifically, it\u0027s like this."},{"Start":"22:19.260 ","End":"22:22.835","Text":"Now we\u0027ve seen what happens in the basic case of"},{"Start":"22:22.835 ","End":"22:27.035","Text":"rain falling downwards or drops of water or a rope,"},{"Start":"22:27.035 ","End":"22:33.380","Text":"whatever it might be falling straight down like on the y-axis,"},{"Start":"22:33.380 ","End":"22:35.570","Text":"just straight along the y-axis into a cart."},{"Start":"22:35.570 ","End":"22:41.105","Text":"We\u0027ve seen what happens if something is falling at an angle Theta."},{"Start":"22:41.105 ","End":"22:43.490","Text":"We\u0027ve seen how to deal with this and how to use"},{"Start":"22:43.490 ","End":"22:47.775","Text":"differential equations and our main equation over here."},{"Start":"22:47.775 ","End":"22:50.690","Text":"That\u0027s it for this lesson."}],"ID":10611},{"Watched":false,"Name":"Ejection and Accretion together","Duration":"17m 24s","ChapterTopicVideoID":9222,"CourseChapterTopicPlaylistID":5427,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.290 ","End":"00:03.525","Text":"Hello. Now, in this question,"},{"Start":"00:03.525 ","End":"00:06.445","Text":"we\u0027re going to be dealing with double trouble."},{"Start":"00:06.445 ","End":"00:10.380","Text":"Here, we\u0027re going to be dealing with ejection and accretion."},{"Start":"00:10.380 ","End":"00:12.510","Text":"Now ejection is just like a mission in"},{"Start":"00:12.510 ","End":"00:15.795","Text":"the rocket and accretion is just like in the previous example,"},{"Start":"00:15.795 ","End":"00:20.265","Text":"where we saw that rain was collecting in a cut."},{"Start":"00:20.265 ","End":"00:22.875","Text":"We have our usual equation,"},{"Start":"00:22.875 ","End":"00:29.130","Text":"which is sum of all the external forces is equal to mass as a function of time,"},{"Start":"00:29.130 ","End":"00:31.830","Text":"multiplied by acceleration plus"},{"Start":"00:31.830 ","End":"00:37.245","Text":"the relative velocity multiplied by the rate of change of mass."},{"Start":"00:37.245 ","End":"00:45.520","Text":"Now here, the only difference is that we\u0027re going to be adding a few other expressions."},{"Start":"00:45.520 ","End":"00:47.485","Text":"Now what are these other expressions?"},{"Start":"00:47.485 ","End":"00:50.605","Text":"This remains exactly the same, this section."},{"Start":"00:50.605 ","End":"00:53.630","Text":"But here, we have our v_rel,"},{"Start":"00:53.630 ","End":"00:55.740","Text":"which will be of the rain,"},{"Start":"00:55.740 ","End":"00:58.485","Text":"so I\u0027ll label it 1,"},{"Start":"00:58.485 ","End":"01:01.230","Text":"so this is 1."},{"Start":"01:01.230 ","End":"01:10.605","Text":"Then we also have to do v relative dm by dt of the water being ejected out."},{"Start":"01:10.605 ","End":"01:20.865","Text":"We\u0027re going to also add a v_rel of 2 and the water is 2,"},{"Start":"01:20.865 ","End":"01:29.230","Text":"multiplied by dm_2, divided by dt."},{"Start":"01:30.110 ","End":"01:34.415","Text":"All we\u0027ve done over here is add on"},{"Start":"01:34.415 ","End":"01:38.855","Text":"this extra expression which represents the water being ejected,"},{"Start":"01:38.855 ","End":"01:47.200","Text":"and we\u0027ve stated that this expression represents the rain being accreted."},{"Start":"01:47.450 ","End":"01:54.365","Text":"Now what we\u0027re going to do is we\u0027re going to label that dm_1 by dt,"},{"Start":"01:54.365 ","End":"01:56.030","Text":"which is referring to the rain,"},{"Start":"01:56.030 ","End":"01:58.470","Text":"is equal to Alpha_1."},{"Start":"01:58.470 ","End":"02:03.315","Text":"We\u0027ll have dm_1 by dt is equal to Alpha_1,"},{"Start":"02:03.315 ","End":"02:05.215","Text":"and then dm_2 by dt,"},{"Start":"02:05.215 ","End":"02:08.735","Text":"which is referring to the water being ejected,"},{"Start":"02:08.735 ","End":"02:12.855","Text":"will be equal to Alpha_2."},{"Start":"02:12.855 ","End":"02:17.360","Text":"Now let\u0027s use the equation, and of course,"},{"Start":"02:17.360 ","End":"02:24.790","Text":"before I forget, that our velocity of the water exiting is u_0."},{"Start":"02:25.310 ","End":"02:32.270","Text":"Now we\u0027re going to start looking at the x-axis because everything\u0027s on the x-axis."},{"Start":"02:32.270 ","End":"02:36.180","Text":"Now, I forgot to mention there\u0027s no friction when the cart is moving."},{"Start":"02:36.680 ","End":"02:43.775","Text":"We have the sum of all the forces on the x-axis is equal to the 0,"},{"Start":"02:43.775 ","End":"02:47.030","Text":"sum of all external forces. Why is it equal to 0?"},{"Start":"02:47.030 ","End":"02:49.765","Text":"Because I said there\u0027s no friction."},{"Start":"02:49.765 ","End":"02:53.645","Text":"This equals to mt,"},{"Start":"02:53.645 ","End":"02:55.460","Text":"which we don\u0027t know what this is."},{"Start":"02:55.460 ","End":"02:59.417","Text":"We\u0027ll deal with that in just a minute, dv,"},{"Start":"02:59.417 ","End":"03:07.829","Text":"on the x-axis of course, by dt."},{"Start":"03:07.829 ","End":"03:10.450","Text":"My v_1 relative."},{"Start":"03:10.450 ","End":"03:12.335","Text":"Now, as you remember,"},{"Start":"03:12.335 ","End":"03:15.280","Text":"we have our equation."},{"Start":"03:15.280 ","End":"03:19.335","Text":"We have this equation for our v relative."},{"Start":"03:19.335 ","End":"03:25.275","Text":"Our v relative for number 1,"},{"Start":"03:25.275 ","End":"03:32.930","Text":"so v_1 rel is going to equal to our v,"},{"Start":"03:32.930 ","End":"03:40.905","Text":"which we said was the velocity of the rain with regards to the ground on the x-axis."},{"Start":"03:40.905 ","End":"03:49.380","Text":"We know that that is 0 because it\u0027s perpendicular to the x-axis, minus v(t),"},{"Start":"03:49.380 ","End":"03:53.925","Text":"which is the velocity that our cart is moving at,"},{"Start":"03:53.925 ","End":"03:56.160","Text":"so we have v(t) over here,"},{"Start":"03:56.160 ","End":"03:59.950","Text":"so we have minus v(t)."},{"Start":"04:01.400 ","End":"04:05.100","Text":"This information we can sub in to here,"},{"Start":"04:05.100 ","End":"04:08.695","Text":"I won\u0027t write the 0 because it doesn\u0027t matter."},{"Start":"04:08.695 ","End":"04:11.095","Text":"This negative does matter."},{"Start":"04:11.095 ","End":"04:15.730","Text":"Then I\u0027ll have negative v(t)"},{"Start":"04:15.730 ","End":"04:23.640","Text":"multiplied by negative Alpha_1."},{"Start":"04:23.640 ","End":"04:25.160","Text":"Why negative Alpha?"},{"Start":"04:25.160 ","End":"04:26.635","Text":"Because remember our law,"},{"Start":"04:26.635 ","End":"04:28.795","Text":"when mass is being gained,"},{"Start":"04:28.795 ","End":"04:32.935","Text":"its negative Alpha when mass is being lost is plus Alpha,"},{"Start":"04:32.935 ","End":"04:35.470","Text":"so we have negative Alpha_1."},{"Start":"04:35.470 ","End":"04:38.950","Text":"Then we have our v_2 rel."},{"Start":"04:38.950 ","End":"04:43.795","Text":"That is going to be plus."},{"Start":"04:43.795 ","End":"04:47.870","Text":"Our v_2 rel is u_0."},{"Start":"04:47.870 ","End":"04:49.905","Text":"That\u0027s the relative velocity of"},{"Start":"04:49.905 ","End":"04:57.030","Text":"the water with regards to the cart when it\u0027s being ejected or emitted."},{"Start":"04:57.030 ","End":"04:59.700","Text":"It\u0027s u_0, however,"},{"Start":"04:59.700 ","End":"05:04.665","Text":"it\u0027s going to be negative u_0 because it\u0027s going in this direction,"},{"Start":"05:04.665 ","End":"05:08.480","Text":"and it\u0027s quite obvious that the cart is moving in this direction,"},{"Start":"05:08.480 ","End":"05:14.460","Text":"so we\u0027ll say that this is the x-axis in the positive direction."},{"Start":"05:16.990 ","End":"05:25.670","Text":"It\u0027s negative u_0 and then we\u0027re going to have multiplied by Alpha_2."},{"Start":"05:25.670 ","End":"05:29.585","Text":"It\u0027s positive Alpha_2 because mass is being lost,"},{"Start":"05:29.585 ","End":"05:32.725","Text":"which means that it\u0027s positive Alpha_ 2."},{"Start":"05:32.725 ","End":"05:36.720","Text":"Now what we\u0027re going to deal with is our M(t),"},{"Start":"05:36.720 ","End":"05:39.315","Text":"our mass as a function of time."},{"Start":"05:39.315 ","End":"05:42.920","Text":"What we can do is we can say that our m as"},{"Start":"05:42.920 ","End":"05:47.465","Text":"a function of t is equal to a starting mass M_0."},{"Start":"05:47.465 ","End":"05:53.590","Text":"We\u0027re going to say that cart has a starting mass of M_0."},{"Start":"05:54.320 ","End":"06:02.050","Text":"Then it\u0027s going to be plus Alpha_1"},{"Start":"06:02.300 ","End":"06:11.720","Text":"multiplied by t minus Alpha_ 2 multiplied by t. Why is this?"},{"Start":"06:11.720 ","End":"06:15.350","Text":"Because it has its initial mass that we know."},{"Start":"06:15.350 ","End":"06:18.870","Text":"Then we have a rate of change of mass,"},{"Start":"06:18.870 ","End":"06:23.085","Text":"an increase in mass of Alpha_1."},{"Start":"06:23.085 ","End":"06:25.035","Text":"Then we multiply it by time."},{"Start":"06:25.035 ","End":"06:26.765","Text":"As time goes on,"},{"Start":"06:26.765 ","End":"06:28.670","Text":"this expression will get bigger."},{"Start":"06:28.670 ","End":"06:30.565","Text":"More mass will be inside."},{"Start":"06:30.565 ","End":"06:33.995","Text":"Then we also have Alpha_2,"},{"Start":"06:33.995 ","End":"06:40.195","Text":"the rate of change dm_2 by dt going out, being thrown out."},{"Start":"06:40.195 ","End":"06:45.290","Text":"Then we have minus that rate of change times the time."},{"Start":"06:45.290 ","End":"06:53.800","Text":"Now, another way of looking at it is that we can say that dM(t) by dt."},{"Start":"06:53.800 ","End":"07:02.595","Text":"The rate of change of the mass will be equal to Alpha_1 minus Alpha _2,"},{"Start":"07:02.595 ","End":"07:08.858","Text":"because we have Alpha_1 going in and Alpha_2 going out,"},{"Start":"07:08.858 ","End":"07:14.765","Text":"and then what we would do is if we just started off with this expression,"},{"Start":"07:14.765 ","End":"07:19.812","Text":"we could just multiply both sides by dt and then integrate,"},{"Start":"07:19.812 ","End":"07:28.720","Text":"so we would have M(t) equals Alpha_1 minus Alpha_2 dt."},{"Start":"07:28.720 ","End":"07:31.510","Text":"Then if we integrate according to t,"},{"Start":"07:31.510 ","End":"07:39.045","Text":"we\u0027ll see that we\u0027ll then get that Alpha_1 multiplied by t minus"},{"Start":"07:39.045 ","End":"07:49.050","Text":"Alpha_2 multiplied by t plus a constant and this constant is equal to our starting mass,"},{"Start":"07:49.050 ","End":"07:50.880","Text":"our initial mass, which is M_0,"},{"Start":"07:50.880 ","End":"07:54.280","Text":"so we\u0027ll get to the same answer."},{"Start":"07:55.130 ","End":"08:00.230","Text":"Now what we\u0027re going to do is we\u0027re going to sub in this M as"},{"Start":"08:00.230 ","End":"08:04.850","Text":"a function of t into our equation and see what happens."},{"Start":"08:04.850 ","End":"08:09.920","Text":"We\u0027re going to have 0=M(t),"},{"Start":"08:09.920 ","End":"08:19.290","Text":"which is M_0 plus Alpha_1t minus Alpha_2t,"},{"Start":"08:19.570 ","End":"08:29.750","Text":"multiplied by dv_x by dt plus a negative times a negative is a positive,"},{"Start":"08:29.750 ","End":"08:35.608","Text":"so plus v as a function of t multiplied by Alpha_1,"},{"Start":"08:35.608 ","End":"08:43.140","Text":"and then negative u_0, Alpha_2."},{"Start":"08:43.650 ","End":"08:48.640","Text":"A little side note before I carry on to solve this,"},{"Start":"08:48.640 ","End":"08:53.510","Text":"so here, side note."},{"Start":"08:53.670 ","End":"09:01.010","Text":"A thing that they could ask is they could ask what our v final will be."},{"Start":"09:01.950 ","End":"09:05.485","Text":"A way to do it, if we have v final,"},{"Start":"09:05.485 ","End":"09:12.865","Text":"it means that our acceleration equals to 0 because our velocity will be constant."},{"Start":"09:12.865 ","End":"09:17.560","Text":"Then we can say that 0 is equal"},{"Start":"09:17.560 ","End":"09:25.375","Text":"to Alpha_1v minus Alpha_2 u_0."},{"Start":"09:25.375 ","End":"09:28.135","Text":"Just to clarify how I got this,"},{"Start":"09:28.135 ","End":"09:30.070","Text":"if my acceleration is 0,"},{"Start":"09:30.070 ","End":"09:34.341","Text":"another way of writing acceleration is dv/dt,"},{"Start":"09:34.341 ","End":"09:39.685","Text":"taking the derivative of the velocity."},{"Start":"09:39.685 ","End":"09:42.670","Text":"That means that we\u0027ll substitute in here."},{"Start":"09:42.670 ","End":"09:46.180","Text":"This will be equal to 0 in v final."},{"Start":"09:46.180 ","End":"09:51.490","Text":"Then this crosses out and then we have Alpha_v as a function of"},{"Start":"09:51.490 ","End":"09:58.250","Text":"t minus Alpha_2 u_0 is equal to 0."},{"Start":"09:58.470 ","End":"10:09.475","Text":"Then all we have to do is we can say that Alpha_1v is equal to Alpha_2 u_0."},{"Start":"10:09.475 ","End":"10:14.260","Text":"Then we can get that v as a function of t or our v final"},{"Start":"10:14.260 ","End":"10:20.150","Text":"is equal to Alpha_2 divided by Alpha_1 u_0."},{"Start":"10:21.720 ","End":"10:26.380","Text":"That\u0027s another example of a question that they could ask."},{"Start":"10:26.380 ","End":"10:30.460","Text":"Carry on. Back to our question at hand."},{"Start":"10:30.460 ","End":"10:34.825","Text":"This can lead to a lot of complicated mathematics."},{"Start":"10:34.825 ","End":"10:37.270","Text":"In order to shorten the time over here,"},{"Start":"10:37.270 ","End":"10:40.047","Text":"but really you can do it on your own,"},{"Start":"10:40.047 ","End":"10:42.940","Text":"it just follows the same recipe,"},{"Start":"10:42.940 ","End":"10:45.090","Text":"the same step-by-step guide,"},{"Start":"10:45.090 ","End":"10:50.395","Text":"what I\u0027ll say is that here that Alpha 1 will equal Alpha 2,"},{"Start":"10:50.395 ","End":"10:52.540","Text":"which is equal to Alpha."},{"Start":"10:52.540 ","End":"11:00.115","Text":"Now, this is something that you\u0027ll see a lot in homework and also in texts."},{"Start":"11:00.115 ","End":"11:08.140","Text":"It means that the rate that the rain is coming in equals the rate that it\u0027s going out."},{"Start":"11:08.140 ","End":"11:10.120","Text":"This is a very common question,"},{"Start":"11:10.120 ","End":"11:14.440","Text":"so this could appear exactly this case in your exam,"},{"Start":"11:14.440 ","End":"11:17.920","Text":"but this is private case and should not be"},{"Start":"11:17.920 ","End":"11:23.170","Text":"taken as a constant for every single question like this that you\u0027ll be asked."},{"Start":"11:23.170 ","End":"11:26.455","Text":"What I\u0027m going to do is first I\u0027m going to write down what"},{"Start":"11:26.455 ","End":"11:29.650","Text":"the final answer is before I solve it."},{"Start":"11:29.650 ","End":"11:35.656","Text":"We\u0027ll get that v as a function of t is equal to u_0,"},{"Start":"11:35.656 ","End":"11:41.065","Text":"1 minus e to the negative Alpha divided by"},{"Start":"11:41.065 ","End":"11:50.720","Text":"m_0 multiplied by t. This is the final answer."},{"Start":"11:50.910 ","End":"11:54.055","Text":"Then if we say that t,"},{"Start":"11:54.055 ","End":"11:57.070","Text":"the limits of t goes to infinity,"},{"Start":"11:57.070 ","End":"12:00.625","Text":"then we\u0027ll see that this expression will just equal"},{"Start":"12:00.625 ","End":"12:05.740","Text":"u_0 and we know from here that our v final,"},{"Start":"12:05.740 ","End":"12:11.020","Text":"our v_f is equal"},{"Start":"12:11.020 ","End":"12:16.090","Text":"to Alpha_2 over Alpha_1 u_ 0 and we know that Alpha_2 is equal to Alpha_1,"},{"Start":"12:16.090 ","End":"12:19.015","Text":"so this will just equal 1."},{"Start":"12:19.015 ","End":"12:24.700","Text":"We\u0027ll just get that it equals to u_0 so we can check that it makes sense."},{"Start":"12:24.700 ","End":"12:30.640","Text":"Another way is that if we have Alpha_1 is equal to Alpha_2,"},{"Start":"12:30.640 ","End":"12:36.919","Text":"then we can say in this equation,"},{"Start":"12:36.919 ","End":"12:40.435","Text":"we\u0027ll just have Alpha_t minus Alpha_t."},{"Start":"12:40.435 ","End":"12:46.670","Text":"We\u0027ll just get that mass as a function of time is equal to a constant is equal to m_0."},{"Start":"12:48.660 ","End":"12:53.680","Text":"Then it really reduces down this whole expression."},{"Start":"12:53.680 ","End":"12:58.210","Text":"Now let\u0027s see how we actually solve it though."},{"Start":"12:58.210 ","End":"13:04.690","Text":"If I substitute in Alpha_1 equals to Alpha_2 into the equation,"},{"Start":"13:04.690 ","End":"13:09.730","Text":"we will get that m_0 multiplied"},{"Start":"13:09.730 ","End":"13:15.640","Text":"by dv/dt is equal to."},{"Start":"13:15.640 ","End":"13:18.175","Text":"Now because it equals 0, I can move"},{"Start":"13:18.175 ","End":"13:22.500","Text":"this expression under the other side of my equal sign,"},{"Start":"13:22.500 ","End":"13:25.930","Text":"so I\u0027ll get equals negative Alpha because I said that"},{"Start":"13:25.930 ","End":"13:29.650","Text":"Alpha_1 and Alpha_2 equals Alpha and I\u0027ll take it out"},{"Start":"13:29.650 ","End":"13:38.120","Text":"as a common factor multiplied by v as a function of t minus u_0."},{"Start":"13:40.170 ","End":"13:47.080","Text":"Now what I\u0027m going to do is I\u0027m going to multiply both sides by dt."},{"Start":"13:47.080 ","End":"13:55.745","Text":"Then I\u0027ll have m_0 multiplied by dv equals negative Alpha"},{"Start":"13:55.745 ","End":"14:05.560","Text":"v(t) minus u_0 dt."},{"Start":"14:05.560 ","End":"14:10.750","Text":"Now what we\u0027ll do is we\u0027ll divide both sides by m_0 and divide both sides by v(t)"},{"Start":"14:10.750 ","End":"14:16.465","Text":"minus u_0 in order to"},{"Start":"14:16.465 ","End":"14:22.420","Text":"put like variables with our dv\u0027s and our dt\u0027s and then we\u0027ll integrate both sides,"},{"Start":"14:22.420 ","End":"14:32.240","Text":"so we\u0027ll integrate dv as a function of x divided by v(t) minus u_0."},{"Start":"14:32.250 ","End":"14:42.890","Text":"Integrate that, which will equal negative the integral of Alpha over m_0 dt."},{"Start":"14:43.200 ","End":"14:48.115","Text":"Then again, we\u0027ll do a definite integral."},{"Start":"14:48.115 ","End":"14:55.240","Text":"We\u0027ll say that our time will start from 0 till t and that our velocity,"},{"Start":"14:55.240 ","End":"14:57.490","Text":"we\u0027ll say that our starting velocity,"},{"Start":"14:57.490 ","End":"15:01.960","Text":"so here we\u0027ll say that our v(t) equals 0."},{"Start":"15:01.960 ","End":"15:03.400","Text":"Let\u0027s just say it equals to 0."},{"Start":"15:03.400 ","End":"15:06.190","Text":"You could also sub n equals to v_0 and then"},{"Start":"15:06.190 ","End":"15:09.430","Text":"you\u0027ll get a much more general but more complex answer."},{"Start":"15:09.430 ","End":"15:15.130","Text":"For here, we\u0027ll just substitute in 0 until some v(t)."},{"Start":"15:15.130 ","End":"15:25.615","Text":"Then the answer that we\u0027ll get in the end is Lan of v(t) minus u_0"},{"Start":"15:25.615 ","End":"15:31.105","Text":"divided by u_0 which will equal to"},{"Start":"15:31.105 ","End":"15:39.880","Text":"negative Alpha divided by m_0 t. Then I can get rid of the Lan."},{"Start":"15:39.880 ","End":"15:49.990","Text":"We\u0027ll go to v(t) minus u_0 divided by u_0 is equal"},{"Start":"15:49.990 ","End":"15:57.280","Text":"to e to the negative Alpha over m_0 multiplied by t. Then I\u0027ll multiply"},{"Start":"15:57.280 ","End":"16:05.590","Text":"both sides by u_0 and add u_0 to both sides in order to isolate out this v(t)."},{"Start":"16:05.590 ","End":"16:10.300","Text":"I\u0027m just going to isolate out the v(t) and then it will just"},{"Start":"16:10.300 ","End":"16:15.800","Text":"go and you\u0027ll see that you will get this answer."},{"Start":"16:16.740 ","End":"16:25.090","Text":"The important point to take away from this lesson is that if you"},{"Start":"16:25.090 ","End":"16:29.260","Text":"have your equation and you know that mass is being"},{"Start":"16:29.260 ","End":"16:33.635","Text":"added and mass is being extracted simultaneously,"},{"Start":"16:33.635 ","End":"16:41.170","Text":"then you take the original equation that we have and then you add on this expression,"},{"Start":"16:41.170 ","End":"16:47.080","Text":"but another one describing the added action being taken."},{"Start":"16:47.080 ","End":"16:49.270","Text":"Here we said that"},{"Start":"16:49.270 ","End":"16:57.055","Text":"this v_1 rail was the mass being added and the v_2 rail was the mass being extracted."},{"Start":"16:57.055 ","End":"17:01.855","Text":"Then you also have to make sure that you apply this when"},{"Start":"17:01.855 ","End":"17:07.045","Text":"you\u0027re working out your equation for m as a function of time."},{"Start":"17:07.045 ","End":"17:11.890","Text":"You\u0027ll start with your initial mass and then you have to take into"},{"Start":"17:11.890 ","End":"17:17.980","Text":"account this and this for this and this."},{"Start":"17:17.980 ","End":"17:21.370","Text":"Those are the 2 most important points of this lesson."},{"Start":"17:21.370 ","End":"17:24.770","Text":"You just have to remember how to do that."}],"ID":10612},{"Watched":false,"Name":"Varying Mass and Friction","Duration":"16m 48s","ChapterTopicVideoID":9223,"CourseChapterTopicPlaylistID":5427,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.985","Text":"Hello. In this question we\u0027re going to be dealing with"},{"Start":"00:03.985 ","End":"00:08.870","Text":"a varying mass system that also incorporates in friction."},{"Start":"00:08.870 ","End":"00:12.610","Text":"Now, in these types of questions as a trick with how to"},{"Start":"00:12.610 ","End":"00:15.760","Text":"incorporate the friction into our equation,"},{"Start":"00:15.760 ","End":"00:20.660","Text":"and we\u0027re going to see how to do it by answering the following question."},{"Start":"00:20.660 ","End":"00:23.645","Text":"A cart with an initial mass of M_0 is"},{"Start":"00:23.645 ","End":"00:27.520","Text":"traveling on a surface with a coefficient of friction Mu_k."},{"Start":"00:27.520 ","End":"00:33.550","Text":"Because the cart doesn\u0027t have wheels and it\u0027s traveling on a surface over here,"},{"Start":"00:33.550 ","End":"00:35.350","Text":"so there\u0027s a coefficient of friction."},{"Start":"00:35.350 ","End":"00:36.895","Text":"At one end of the cart,"},{"Start":"00:36.895 ","End":"00:40.375","Text":"there\u0027s a hose which ejects water at a rate of Alpha,"},{"Start":"00:40.375 ","End":"00:42.545","Text":"and at a velocity of u_0."},{"Start":"00:42.545 ","End":"00:47.030","Text":"The hose is at an angle theta to the x-axis."},{"Start":"00:47.030 ","End":"00:49.505","Text":"If this is the x-axis,"},{"Start":"00:49.505 ","End":"00:51.500","Text":"this is our angle,"},{"Start":"00:51.500 ","End":"00:52.730","Text":"theta over here,"},{"Start":"00:52.730 ","End":"00:56.885","Text":"and we\u0027re going to say that our x axis is like this."},{"Start":"00:56.885 ","End":"01:00.364","Text":"This direction is the positive direction."},{"Start":"01:00.364 ","End":"01:05.720","Text":"Our hose is at this angle and water is"},{"Start":"01:05.720 ","End":"01:11.285","Text":"exiting the cause at a velocity of u_0 from this hose,"},{"Start":"01:11.285 ","End":"01:13.460","Text":"and at a rate of Alpha."},{"Start":"01:13.460 ","End":"01:16.370","Text":"That\u0027s the important thing to note here."},{"Start":"01:16.370 ","End":"01:19.880","Text":"Our first question is to write an equation of"},{"Start":"01:19.880 ","End":"01:24.295","Text":"motion and our second is to find the velocity as a function of time."},{"Start":"01:24.295 ","End":"01:30.356","Text":"The first thing that we\u0027re going to do is we\u0027re going to answer our first question,"},{"Start":"01:30.356 ","End":"01:34.500","Text":"and for that we\u0027re going to use our beloved equation."},{"Start":"01:34.820 ","End":"01:40.295","Text":"This time, instead of starting with sum of all of the external forces,"},{"Start":"01:40.295 ","End":"01:45.305","Text":"we\u0027re actually going to start with this side of the equation because it\u0027s this side,"},{"Start":"01:45.305 ","End":"01:49.300","Text":"this external force side that is slightly more complicated."},{"Start":"01:49.300 ","End":"01:56.295","Text":"Let\u0027s start first with finding what our mass as a function of time is."},{"Start":"01:56.295 ","End":"02:00.920","Text":"Our mass as a function of t is equal to,"},{"Start":"02:00.920 ","End":"02:04.190","Text":"we know that we\u0027re starting off with M_0,"},{"Start":"02:04.190 ","End":"02:06.734","Text":"our starting mass,"},{"Start":"02:06.734 ","End":"02:12.920","Text":"and then we know that water is leaving the cart at a rate of Alpha,"},{"Start":"02:12.920 ","End":"02:17.440","Text":"so we\u0027ll have negative Alpha per second"},{"Start":"02:17.440 ","End":"02:24.205","Text":"multiplied by t. The next thing we have to look at,"},{"Start":"02:24.205 ","End":"02:26.770","Text":"our dv/dt we\u0027ll get to in a moment,"},{"Start":"02:26.770 ","End":"02:28.975","Text":"is our v_rel,"},{"Start":"02:28.975 ","End":"02:33.940","Text":"our relative velocity with which the water is traveling relative to the carts."},{"Start":"02:33.940 ","End":"02:36.990","Text":"We can say that our v_rel,"},{"Start":"02:36.990 ","End":"02:40.989","Text":"now we\u0027re working on the x-axis remember"},{"Start":"02:40.989 ","End":"02:46.011","Text":"because there\u0027s nothing happening on the y-axis,"},{"Start":"02:46.011 ","End":"02:49.520","Text":"we can say that it equals to,"},{"Start":"02:49.520 ","End":"02:52.065","Text":"so we know how we do this,"},{"Start":"02:52.065 ","End":"02:58.300","Text":"it\u0027s this u_0 multiplied by cosine of the angle."},{"Start":"02:59.960 ","End":"03:03.140","Text":"Cosine theta. Why is it cosine theta?"},{"Start":"03:03.140 ","End":"03:07.970","Text":"Because we want to know what it\u0027s worth on the x-axis which"},{"Start":"03:07.970 ","End":"03:13.360","Text":"is our adjacent side to the theta."},{"Start":"03:13.360 ","End":"03:15.930","Text":"Which means that we need to use cosine,"},{"Start":"03:15.930 ","End":"03:23.298","Text":"and then because we defined that this direction is the positive direction,"},{"Start":"03:23.298 ","End":"03:30.885","Text":"so here we have to put a negative because the hose is a ejecting water backwards."},{"Start":"03:30.885 ","End":"03:33.615","Text":"That\u0027s our v relative."},{"Start":"03:33.615 ","End":"03:36.690","Text":"Then we have to find what dm/dt is."},{"Start":"03:36.690 ","End":"03:44.330","Text":"Our dm/dt is simply equal to Alpha and it\u0027s going to be positive"},{"Start":"03:44.330 ","End":"03:53.065","Text":"because as we\u0027ve seen before that Alpha is positive when we\u0027re losing mass."},{"Start":"03:53.065 ","End":"03:57.994","Text":"Up until now we\u0027ve dealt with the right side of the equation."},{"Start":"03:57.994 ","End":"04:04.715","Text":"Now let\u0027s go over to the more complicated section which is the left side of the equation."},{"Start":"04:04.715 ","End":"04:10.015","Text":"Because we know our coefficient of friction over here is Mu_k,"},{"Start":"04:10.015 ","End":"04:11.840","Text":"then we know that we"},{"Start":"04:11.840 ","End":"04:19.220","Text":"have over"},{"Start":"04:19.220 ","End":"04:24.315","Text":"here our frictional force f_k."},{"Start":"04:24.315 ","End":"04:29.430","Text":"Now we want to find out what f_k is."},{"Start":"04:29.430 ","End":"04:36.285","Text":"We can say that our f_k is equal to Mu_k,"},{"Start":"04:36.285 ","End":"04:40.325","Text":"this is Mu, multiplied by our normal."},{"Start":"04:40.325 ","End":"04:46.230","Text":"Notice that our normal also changes as a function of time because our mass changes."},{"Start":"04:46.230 ","End":"04:52.535","Text":"If you remember, there\u0027s a relationship between the normal force and the mass."},{"Start":"04:52.535 ","End":"04:54.710","Text":"Now a lot of the time,"},{"Start":"04:54.710 ","End":"04:57.115","Text":"students will say that"},{"Start":"04:57.115 ","End":"05:05.630","Text":"that M(t)g is equal to the normal force as a function of time."},{"Start":"05:05.630 ","End":"05:08.420","Text":"But this is incorrect."},{"Start":"05:08.420 ","End":"05:11.900","Text":"This is not right to say because what you\u0027re doing"},{"Start":"05:11.900 ","End":"05:15.770","Text":"here is that you\u0027re saying that the sum of all of"},{"Start":"05:15.770 ","End":"05:24.705","Text":"the forces in the y-direction is equal to the normal minus M_g which is equal to 0,"},{"Start":"05:24.705 ","End":"05:29.450","Text":"and you are saying that because the acceleration in the y-direction is equal to 0."},{"Start":"05:29.450 ","End":"05:34.560","Text":"However, this doesn\u0027t exist here,"},{"Start":"05:34.560 ","End":"05:38.615","Text":"this step, because we\u0027re dealing with varying mass system."},{"Start":"05:38.615 ","End":"05:40.550","Text":"Because the mass isn\u0027t constant,"},{"Start":"05:40.550 ","End":"05:43.500","Text":"you cannot lay this claim."},{"Start":"05:44.180 ","End":"05:51.620","Text":"All you have to remember is that the sum of all the forces equals M_a"},{"Start":"05:51.620 ","End":"06:00.150","Text":"does not apply to changing mass system."},{"Start":"06:00.740 ","End":"06:04.335","Text":"Remember this. Therefore,"},{"Start":"06:04.335 ","End":"06:06.420","Text":"what am I supposed to do?"},{"Start":"06:06.420 ","End":"06:08.055","Text":"As you have noticed,"},{"Start":"06:08.055 ","End":"06:10.065","Text":"over here we were in the gray,"},{"Start":"06:10.065 ","End":"06:17.160","Text":"we were finding out the sum of all of the forces on the x-axis and the x-direction."},{"Start":"06:17.160 ","End":"06:21.980","Text":"Now what we have to do in order to find out what the normal is and"},{"Start":"06:21.980 ","End":"06:26.765","Text":"the force is on the y-axis in order to find out what this f_k is,"},{"Start":"06:26.765 ","End":"06:31.325","Text":"the frictional forces, we have to use this equation,"},{"Start":"06:31.325 ","End":"06:35.520","Text":"our equation here but on the y-axis."},{"Start":"06:36.530 ","End":"06:39.005","Text":"Let\u0027s see how we do this."},{"Start":"06:39.005 ","End":"06:43.340","Text":"We\u0027re going to say that the sum of all of the forces on"},{"Start":"06:43.340 ","End":"06:47.525","Text":"the y-axis is equal to our normal,"},{"Start":"06:47.525 ","End":"06:51.740","Text":"which is dependent on time minus our mass,"},{"Start":"06:51.740 ","End":"06:56.120","Text":"which is varying through time multiplied by gravity,"},{"Start":"06:56.120 ","End":"06:58.010","Text":"which is equal to,"},{"Start":"06:58.010 ","End":"07:02.060","Text":"so this was the left side of the equation, our f external."},{"Start":"07:02.060 ","End":"07:06.180","Text":"Now we have to find the right side of the equation."},{"Start":"07:07.000 ","End":"07:09.819","Text":"Our M(t), let\u0027s wait one second in this,"},{"Start":"07:09.819 ","End":"07:14.630","Text":"our dv/dt means our acceleration in the y-direction."},{"Start":"07:14.630 ","End":"07:19.560","Text":"We know that we have no acceleration in the y-direction,"},{"Start":"07:19.560 ","End":"07:25.385","Text":"so ay=0."},{"Start":"07:25.385 ","End":"07:29.805","Text":"This is the y-direction."},{"Start":"07:29.805 ","End":"07:35.245","Text":"We know that it\u0027s equal to 0 because our cart isn\u0027t moving up or down."},{"Start":"07:35.245 ","End":"07:40.605","Text":"We have a mass varying in time multiplied by 0, so we have 0."},{"Start":"07:40.605 ","End":"07:48.560","Text":"Then we have plus v relative multiplied by dm/dt."},{"Start":"07:48.560 ","End":"07:51.925","Text":"What is v relative?"},{"Start":"07:51.925 ","End":"07:53.500","Text":"Let\u0027s write it over here."},{"Start":"07:53.500 ","End":"08:01.060","Text":"Our v relative in the y-direction is going to be equal to,"},{"Start":"08:01.060 ","End":"08:04.810","Text":"again, like what we did for our v relative in the x-direction,"},{"Start":"08:04.810 ","End":"08:06.580","Text":"so the same in the y-direction."},{"Start":"08:06.580 ","End":"08:08.665","Text":"We have u_0,"},{"Start":"08:08.665 ","End":"08:12.375","Text":"but this time because we want the opposite over here,"},{"Start":"08:12.375 ","End":"08:19.105","Text":"because we want the components and on the y-axis,"},{"Start":"08:19.105 ","End":"08:21.830","Text":"we have the opposite side to the angle."},{"Start":"08:21.830 ","End":"08:25.565","Text":"It\u0027s going to be sine of the angle, so sine theta."},{"Start":"08:25.565 ","End":"08:31.640","Text":"Again, it\u0027s going to be negative because we decided that our y-axis is going upwards,"},{"Start":"08:31.640 ","End":"08:34.530","Text":"that this is the positive direction."},{"Start":"08:34.900 ","End":"08:39.170","Text":"If we do use 0 sine theta,"},{"Start":"08:39.170 ","End":"08:42.755","Text":"we\u0027ll get this in the negative direction pointing downwards,"},{"Start":"08:42.755 ","End":"08:44.450","Text":"so we write negative."},{"Start":"08:44.450 ","End":"08:50.580","Text":"Then we have plus negative u_0 sine of theta."},{"Start":"08:50.580 ","End":"08:59.590","Text":"This was our v_rel multiplied by dm/dt which as we know is alpha."},{"Start":"09:00.050 ","End":"09:02.915","Text":"Now that I have this equation,"},{"Start":"09:02.915 ","End":"09:06.440","Text":"I can just rearrange in order to isolate out this"},{"Start":"09:06.440 ","End":"09:10.750","Text":"N(t) because that\u0027s what I needed to find out my f_k,"},{"Start":"09:10.750 ","End":"09:12.680","Text":"because I know what Mu k is,"},{"Start":"09:12.680 ","End":"09:14.645","Text":"it\u0027s given to me in the question."},{"Start":"09:14.645 ","End":"09:20.405","Text":"Then I will get that my normal force as a function of time is equal to"},{"Start":"09:20.405 ","End":"09:25.530","Text":"my mass as a function of time multiplied by g minus"},{"Start":"09:25.530 ","End":"09:33.360","Text":"u_0 sine of theta multiplied by Alpha."},{"Start":"09:34.260 ","End":"09:36.370","Text":"Now that we have this,"},{"Start":"09:36.370 ","End":"09:39.220","Text":"we can go back to our original equation on"},{"Start":"09:39.220 ","End":"09:46.700","Text":"the x-axis and we can write our final answer for the equation of motion."},{"Start":"09:47.460 ","End":"09:54.595","Text":"What we can do is we can say that the sum of all the forces of the external forces is"},{"Start":"09:54.595 ","End":"10:01.060","Text":"equal to Mu multiplied by K multiplied by our normal force,"},{"Start":"10:01.060 ","End":"10:06.430","Text":"which is multiplied by m(t) times g"},{"Start":"10:06.430 ","End":"10:13.330","Text":"minus Mu_0 sine Theta multiplied by Alpha,"},{"Start":"10:13.330 ","End":"10:20.155","Text":"which will equal m(t) multiplied by"},{"Start":"10:20.155 ","End":"10:27.505","Text":"dv in the x-direction by dt plus V_rel,"},{"Start":"10:27.505 ","End":"10:28.855","Text":"which here we have,"},{"Start":"10:28.855 ","End":"10:31.300","Text":"so plus minus u,"},{"Start":"10:31.300 ","End":"10:38.170","Text":"so it\u0027s just minus u_0 cosine of Theta,"},{"Start":"10:38.170 ","End":"10:43.405","Text":"Mu_0 cosine of Theta multiplied by dm by dt,"},{"Start":"10:43.405 ","End":"10:46.630","Text":"which is multiplied by Alpha."},{"Start":"10:46.630 ","End":"10:52.240","Text":"Then all we have to do is substitute in everywhere we see m(t),"},{"Start":"10:52.240 ","End":"10:54.940","Text":"such as here and here,"},{"Start":"10:54.940 ","End":"10:59.270","Text":"we substitute in this equation."},{"Start":"11:00.660 ","End":"11:05.065","Text":"Then you\u0027ll get this equation and then of course,"},{"Start":"11:05.065 ","End":"11:08.305","Text":"you can multiply out and simplify."},{"Start":"11:08.305 ","End":"11:13.000","Text":"This is the answer to Question 1."},{"Start":"11:13.000 ","End":"11:20.125","Text":"Now, the trick to remember over here is that when you\u0027re trying to work out the normal,"},{"Start":"11:20.125 ","End":"11:21.370","Text":"when you\u0027re dealing with friction,"},{"Start":"11:21.370 ","End":"11:23.283","Text":"you have to work out the normal,"},{"Start":"11:23.283 ","End":"11:27.055","Text":"is that you have to apply this equation,"},{"Start":"11:27.055 ","End":"11:32.425","Text":"also on the y-axis in order to find out the normal,"},{"Start":"11:32.425 ","End":"11:34.915","Text":"because the sum of all the forces equals ma."},{"Start":"11:34.915 ","End":"11:42.955","Text":"This over here does not apply when we\u0027re dealing with varying mass systems."},{"Start":"11:42.955 ","End":"11:46.195","Text":"Now, in order to solve Question number 2,"},{"Start":"11:46.195 ","End":"11:48.580","Text":"find the velocity as a function of time."},{"Start":"11:48.580 ","End":"11:50.170","Text":"What we\u0027re going to have to do,"},{"Start":"11:50.170 ","End":"11:55.720","Text":"we can see that we have here the derivative velocity acceleration,"},{"Start":"11:55.720 ","End":"11:58.585","Text":"so we know that this is a differential equation."},{"Start":"11:58.585 ","End":"12:05.760","Text":"What we want to do is we want to isolate out this dv by dt."},{"Start":"12:05.760 ","End":"12:11.500","Text":"What we\u0027re going to do is that we\u0027re going to restate that all"},{"Start":"12:11.500 ","End":"12:17.418","Text":"of this is equal to mt Just to make it easier to follow what\u0027s happening."},{"Start":"12:17.418 ","End":"12:22.360","Text":"This is also m as a function of t. If you go back a bit in the video,"},{"Start":"12:22.360 ","End":"12:26.140","Text":"you\u0027ll see that that\u0027s the equation that we got for m as a function of time,"},{"Start":"12:26.140 ","End":"12:29.305","Text":"just to make it easier to understand."},{"Start":"12:29.305 ","End":"12:34.270","Text":"Then we\u0027re going to take this expression with the dv by dt to"},{"Start":"12:34.270 ","End":"12:40.120","Text":"one side of the equal side and the other expressions to the other side."},{"Start":"12:40.120 ","End":"12:49.990","Text":"What we\u0027re going to get is we\u0027re going to get dv_x by dt multiplied by m as"},{"Start":"12:49.990 ","End":"12:57.685","Text":"a function of time is equal to"},{"Start":"12:57.685 ","End":"13:05.680","Text":"negative Mu_k multiplied by g multiplied by m"},{"Start":"13:05.680 ","End":"13:13.480","Text":"as a function of time plus Alpha multiplied by u_ 0,"},{"Start":"13:13.480 ","End":"13:20.935","Text":"so I\u0027ve taken this out as a common factor because we can see that we have Alpha and u_0,"},{"Start":"13:20.935 ","End":"13:28.540","Text":"common factors for cosine of Theta and sine Theta over here."},{"Start":"13:28.540 ","End":"13:34.645","Text":"Here we see that our sine Theta is multiplied by Mu_0 Alpha and Mu_k,"},{"Start":"13:34.645 ","End":"13:40.585","Text":"so multiplied by Mu_k sine"},{"Start":"13:40.585 ","End":"13:47.530","Text":"of Theta plus cosine of Theta."},{"Start":"13:47.530 ","End":"13:52.990","Text":"All I\u0027ve done is said that I\u0027m going to solve a differential equation,"},{"Start":"13:52.990 ","End":"13:58.435","Text":"gotten my dv by dt on one side of the equal side and"},{"Start":"13:58.435 ","End":"14:05.425","Text":"rearranged the other expressions in order to make this easier for me to solve."},{"Start":"14:05.425 ","End":"14:08.859","Text":"This is just this rearranged."},{"Start":"14:08.859 ","End":"14:12.020","Text":"You can do it on a piece of paper if it\u0027s unclear."},{"Start":"14:13.110 ","End":"14:16.945","Text":"Now what I\u0027m going to say is, because I\u0027m lazy,"},{"Start":"14:16.945 ","End":"14:22.960","Text":"I\u0027m going to label this whole section as a constant."},{"Start":"14:22.960 ","End":"14:25.330","Text":"Let\u0027s label this c,"},{"Start":"14:25.330 ","End":"14:27.490","Text":"because I don\u0027t want to rewrite this all the time,"},{"Start":"14:27.490 ","End":"14:30.595","Text":"and also it\u0027s not dependent on t,"},{"Start":"14:30.595 ","End":"14:34.330","Text":"this whole expression, which means that it\u0027s a constant."},{"Start":"14:34.330 ","End":"14:40.315","Text":"For the meantime, I\u0027m just gonna write that in order to make this go a bit faster."},{"Start":"14:40.315 ","End":"14:44.230","Text":"Now what I\u0027m going to do is I\u0027m going to divide both sides by"},{"Start":"14:44.230 ","End":"14:49.420","Text":"m(t) and then I\u0027m going to get that my acceleration on the x-axis is"},{"Start":"14:49.420 ","End":"14:52.780","Text":"going to be equal to negative Mu _k"},{"Start":"14:52.780 ","End":"15:00.145","Text":"g plus c divided by m(t),"},{"Start":"15:00.145 ","End":"15:03.490","Text":"because I divided both sides by m(t)."},{"Start":"15:03.490 ","End":"15:07.480","Text":"Now, all that I have to do in order to find the velocity,"},{"Start":"15:07.480 ","End":"15:16.435","Text":"as we know the acceleration is the derivative of velocity,"},{"Start":"15:16.435 ","End":"15:22.030","Text":"so all I have to do is I have to integrate both sides."},{"Start":"15:22.030 ","End":"15:27.550","Text":"Then we\u0027ll just integrate both sides with respect to t"},{"Start":"15:27.550 ","End":"15:32.185","Text":"and then we\u0027ll get that our v as a function of time,"},{"Start":"15:32.185 ","End":"15:36.805","Text":"is equal to the integral from 0 till t of"},{"Start":"15:36.805 ","End":"15:46.730","Text":"negative Mu_k g plus c over M_0 minus Alpha t dt."},{"Start":"15:47.190 ","End":"15:50.140","Text":"Then if we solve this,"},{"Start":"15:50.140 ","End":"15:57.700","Text":"we\u0027ll get negative Mu_k gt negative c"},{"Start":"15:57.700 ","End":"16:06.980","Text":"over Alpha Ln of m_0 minus Alpha t over m_0."},{"Start":"16:07.170 ","End":"16:11.785","Text":"All I\u0027ve done here is integrate."},{"Start":"16:11.785 ","End":"16:15.865","Text":"This you can do, I\u0027m assuming you already know how to do this."},{"Start":"16:15.865 ","End":"16:24.160","Text":"Then this is our final answer for number 2."},{"Start":"16:24.160 ","End":"16:27.130","Text":"Now in this question we were taking into account that are"},{"Start":"16:27.130 ","End":"16:33.115","Text":"starting velocity v at time t=0 was equal to 0."},{"Start":"16:33.115 ","End":"16:34.750","Text":"If this was not the case,"},{"Start":"16:34.750 ","End":"16:40.255","Text":"if our v at time 0 was equal to some constant,"},{"Start":"16:40.255 ","End":"16:45.110","Text":"we would just add over here plus V_0."},{"Start":"16:45.510 ","End":"16:49.220","Text":"That\u0027s the end of our lesson."}],"ID":10613},{"Watched":false,"Name":"Conclusion","Duration":"11m 31s","ChapterTopicVideoID":8108,"CourseChapterTopicPlaylistID":5427,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.045","Text":"Hello. In this lesson,"},{"Start":"00:03.045 ","End":"00:09.330","Text":"we\u0027re going to have a conclusion on the topic of varying mass systems."},{"Start":"00:09.330 ","End":"00:13.440","Text":"Now the first thing that we spoke about,"},{"Start":"00:13.440 ","End":"00:19.649","Text":"was that the sum of all of the forces when dealing with a varying mass system,"},{"Start":"00:19.649 ","End":"00:23.405","Text":"we have to write in its full version, Newton\u0027s second law,"},{"Start":"00:23.405 ","End":"00:28.840","Text":"as dp by dt."},{"Start":"00:29.300 ","End":"00:40.310","Text":"It\u0027s the change in momentum over time and we cannot write that it equals to ma."},{"Start":"00:40.310 ","End":"00:44.210","Text":"The reason we can\u0027t write that it equals to ma is because we\u0027re"},{"Start":"00:44.210 ","End":"00:47.960","Text":"dealing with various mass systems and when you\u0027re writing ma,"},{"Start":"00:47.960 ","End":"00:51.815","Text":"then you\u0027re dealing with a constant mass."},{"Start":"00:51.815 ","End":"00:55.565","Text":"Then we spoke about this equation,"},{"Start":"00:55.565 ","End":"01:00.530","Text":"which says that the sum of all of the external forces is"},{"Start":"01:00.530 ","End":"01:06.605","Text":"equal to mass as a function of time multiplied by the acceleration."},{"Start":"01:06.605 ","End":"01:10.250","Text":"This is the mass as a function of time of the body,"},{"Start":"01:10.250 ","End":"01:11.930","Text":"rocket, or the cart,"},{"Start":"01:11.930 ","End":"01:13.730","Text":"multiplied by its acceleration,"},{"Start":"01:13.730 ","End":"01:17.210","Text":"plus V_relative, we\u0027ll speak about it in a second,"},{"Start":"01:17.210 ","End":"01:24.134","Text":"multiplied by dm by dt and dm by dp"},{"Start":"01:24.134 ","End":"01:33.520","Text":"is the rate of ejection or accretion."},{"Start":"01:34.850 ","End":"01:40.745","Text":"V_relative is the relative velocity of"},{"Start":"01:40.745 ","End":"01:47.110","Text":"the thing being ejected or accreted relative to the body."},{"Start":"01:47.110 ","End":"01:49.340","Text":"Then we have our V over here,"},{"Start":"01:49.340 ","End":"01:53.570","Text":"which is the velocity of the ejected material relative to the ground."},{"Start":"01:53.570 ","End":"01:55.595","Text":"When I say ejected, I, of course,"},{"Start":"01:55.595 ","End":"02:00.325","Text":"am speaking about both ejected and accreted."},{"Start":"02:00.325 ","End":"02:03.800","Text":"Then our vt, so negative vt,"},{"Start":"02:03.800 ","End":"02:09.410","Text":"which is the velocity of the main body relative to the ground,"},{"Start":"02:09.410 ","End":"02:13.570","Text":"so you\u0027re substituting this into here."},{"Start":"02:13.940 ","End":"02:20.165","Text":"Then what we did, is we spoke about the 2 cases where we can use"},{"Start":"02:20.165 ","End":"02:23.300","Text":"this equation over here and we said that there\u0027s"},{"Start":"02:23.300 ","End":"02:26.805","Text":"the case of ejection and the case of accretion."},{"Start":"02:26.805 ","End":"02:32.470","Text":"We said that ejection is when mass is leaving,"},{"Start":"02:32.470 ","End":"02:35.870","Text":"or mass of object,"},{"Start":"02:35.870 ","End":"02:40.020","Text":"the main object, is decreasing."},{"Start":"02:40.020 ","End":"02:44.855","Text":"Then accretion, we said that mass is being added or"},{"Start":"02:44.855 ","End":"02:51.095","Text":"that the mass of the main object is increasing."},{"Start":"02:51.095 ","End":"02:57.958","Text":"Then we said that the 2 main examples for ejection and accretion was, for ejection,"},{"Start":"02:57.958 ","End":"03:04.472","Text":"the main questions you\u0027ll be asked about is speaking about rockets and with accretion,"},{"Start":"03:04.472 ","End":"03:07.925","Text":"the main things that you\u0027ll be asked about is when dealing"},{"Start":"03:07.925 ","End":"03:12.635","Text":"with a cart and water being added."},{"Start":"03:12.635 ","End":"03:20.495","Text":"Then we spoke about how the equation changes if we\u0027re using it for ejection or accretion."},{"Start":"03:20.495 ","End":"03:22.999","Text":"We said that the F external,"},{"Start":"03:22.999 ","End":"03:26.978","Text":"the sum of all the external forces remains,"},{"Start":"03:26.978 ","End":"03:33.900","Text":"we use it in the same way for both the ejection and accretion example."},{"Start":"03:34.390 ","End":"03:39.424","Text":"Then we also said that this expression in our equation,"},{"Start":"03:39.424 ","End":"03:45.350","Text":"our M is a function of t multiplied by the acceleration of the main objects."},{"Start":"03:45.350 ","End":"03:49.792","Text":"Also, it\u0027s the same expression in both cases."},{"Start":"03:49.792 ","End":"03:52.925","Text":"Because our mass is dependent on time,"},{"Start":"03:52.925 ","End":"03:56.480","Text":"we work out the expression for that."},{"Start":"03:56.480 ","End":"04:02.405","Text":"Then our dv by dt also is the same for ejection and accretion."},{"Start":"04:02.405 ","End":"04:06.080","Text":"Then, the next section of our equation is"},{"Start":"04:06.080 ","End":"04:11.045","Text":"our V_relative multiplied by our rate of ejection or accretion."},{"Start":"04:11.045 ","End":"04:17.305","Text":"This is where the formula changes depending on ejection or accretion."},{"Start":"04:17.305 ","End":"04:23.150","Text":"In ejection, our V_relative will usually be given to"},{"Start":"04:23.150 ","End":"04:29.395","Text":"us and it will be V_rel will equal to some U_0."},{"Start":"04:29.395 ","End":"04:33.350","Text":"Now usually, if we\u0027re speaking about a rocket,"},{"Start":"04:33.350 ","End":"04:36.470","Text":"then we\u0027ll say that our rocket is traveling in"},{"Start":"04:36.470 ","End":"04:42.350","Text":"the positive direction and that because if our U_0 is going in this direction,"},{"Start":"04:42.350 ","End":"04:45.665","Text":"so we\u0027ll say that our V_relative is negative U_0."},{"Start":"04:45.665 ","End":"04:49.970","Text":"You have to look at what you\u0027re doing with the signs,"},{"Start":"04:49.970 ","End":"04:51.665","Text":"whether it\u0027s positive or negative."},{"Start":"04:51.665 ","End":"04:54.800","Text":"Sometimes your V_relative,"},{"Start":"04:54.800 ","End":"04:58.070","Text":"your fuel will be emitted from the top of the rocket depending on the question,"},{"Start":"04:58.070 ","End":"05:01.070","Text":"so make sure if it\u0027s positive or negative."},{"Start":"05:01.070 ","End":"05:04.565","Text":"In this example here, it would be negative because"},{"Start":"05:04.565 ","End":"05:10.115","Text":"it\u0027s being emitted in the opposite direction of the direction of travel of the rocket,"},{"Start":"05:10.115 ","End":"05:13.535","Text":"one last thing, when we\u0027re given the V_relative,"},{"Start":"05:13.535 ","End":"05:17.465","Text":"it\u0027s already the velocity of the fuel, for instance,"},{"Start":"05:17.465 ","End":"05:21.273","Text":"already relative to the rocket because the rocket is emitting it,"},{"Start":"05:21.273 ","End":"05:25.445","Text":"so here the V_relative is already relative to the rocket."},{"Start":"05:25.445 ","End":"05:29.255","Text":"If we\u0027re talking about accretion though, for instance,"},{"Start":"05:29.255 ","End":"05:33.335","Text":"then we\u0027ll usually be given the velocity,"},{"Start":"05:33.335 ","End":"05:35.825","Text":"for instance, with the cart,"},{"Start":"05:35.825 ","End":"05:38.280","Text":"with the rain falling in."},{"Start":"05:38.350 ","End":"05:43.672","Text":"We will usually be given the velocity of the rain."},{"Start":"05:43.672 ","End":"05:46.505","Text":"Because the velocity of the rain,"},{"Start":"05:46.505 ","End":"05:49.355","Text":"we know it relative to the ground,"},{"Start":"05:49.355 ","End":"05:52.255","Text":"in order to find our V,"},{"Start":"05:52.255 ","End":"05:56.396","Text":"the velocity of the rain relative to the cart, V_rel,"},{"Start":"05:56.396 ","End":"06:03.025","Text":"we\u0027ll have to do the V of the rain minus the V of the cart."},{"Start":"06:03.025 ","End":"06:07.660","Text":"We\u0027ll have to do V_rain minus the V of the cart."},{"Start":"06:07.660 ","End":"06:11.330","Text":"Now another thing, a little tip that I won\u0027t write here,"},{"Start":"06:11.330 ","End":"06:13.010","Text":"but you have to remember this,"},{"Start":"06:13.010 ","End":"06:16.940","Text":"because it will appear all over mechanics,"},{"Start":"06:16.940 ","End":"06:25.055","Text":"is that you always have to use this equation with regards to the each individual axis."},{"Start":"06:25.055 ","End":"06:30.330","Text":"In the rocket, you\u0027ll be usually dealing with the rocket going up and down the y-axis,"},{"Start":"06:30.330 ","End":"06:34.300","Text":"so you have to make sure that you\u0027re using this equation on the y-axis."},{"Start":"06:34.300 ","End":"06:39.485","Text":"In this example with the accretion and the rain falling."},{"Start":"06:39.485 ","End":"06:43.550","Text":"If the rain is falling vertically downwards perpendicular to the ground,"},{"Start":"06:43.550 ","End":"06:48.260","Text":"then you know that the velocity of the rain in the x-axis=0."},{"Start":"06:48.260 ","End":"06:52.340","Text":"You have to take that into account when using this equation"},{"Start":"06:52.340 ","End":"06:57.320","Text":"according to the x-axis and if the rain is falling at some angle,"},{"Start":"06:57.320 ","End":"06:59.525","Text":"if it\u0027s falling like this,"},{"Start":"06:59.525 ","End":"07:05.240","Text":"then what you have to do is you have to work out the velocity of the rain on the x-axis,"},{"Start":"07:05.240 ","End":"07:09.155","Text":"so you\u0027ll have to multiply it usually by cosine of Theta,"},{"Start":"07:09.155 ","End":"07:10.745","Text":"cosine of the angle,"},{"Start":"07:10.745 ","End":"07:13.760","Text":"in order to work out the values."},{"Start":"07:13.760 ","End":"07:17.910","Text":"Also make sure that you are doing all of that correctly."},{"Start":"07:17.930 ","End":"07:21.980","Text":"Now, when we\u0027re speaking about the dm by dt,"},{"Start":"07:21.980 ","End":"07:25.610","Text":"so the rate of ejection in this case,"},{"Start":"07:25.610 ","End":"07:31.890","Text":"our dm by dt will equal Alpha and this will be positive,"},{"Start":"07:31.890 ","End":"07:37.385","Text":"because this is the rate of ejection of mass."},{"Start":"07:37.385 ","End":"07:41.300","Text":"How much mass is leaving the body per second?"},{"Start":"07:41.300 ","End":"07:47.550","Text":"Then, you can say that your d mass as a function of time,"},{"Start":"07:47.550 ","End":"07:54.530","Text":"so the change in mass as a function of time will equal to negative Alpha,"},{"Start":"07:54.530 ","End":"08:00.500","Text":"because the mass is changing as a function of time and mass is leaving the object,"},{"Start":"08:00.500 ","End":"08:02.315","Text":"which is why it\u0027s a negative."},{"Start":"08:02.315 ","End":"08:04.115","Text":"Then, through this,"},{"Start":"08:04.115 ","End":"08:05.825","Text":"if you integrate both sides,"},{"Start":"08:05.825 ","End":"08:09.185","Text":"you can get that your mass as a function of time is"},{"Start":"08:09.185 ","End":"08:12.830","Text":"equal to sum initial mass, whatever it may be,"},{"Start":"08:12.830 ","End":"08:19.500","Text":"negative Alpha t, because Alpha is the amount of"},{"Start":"08:19.500 ","End":"08:23.300","Text":"mass lost per second and then you multiply it by t in order"},{"Start":"08:23.300 ","End":"08:28.110","Text":"to find the total amount lost in that timeframe."},{"Start":"08:28.490 ","End":"08:31.335","Text":"Then, our dm by dt,"},{"Start":"08:31.335 ","End":"08:33.640","Text":"when dealing with accretion,"},{"Start":"08:33.640 ","End":"08:41.400","Text":"is going to be dm by dt will equal negative Alpha."},{"Start":"08:42.250 ","End":"08:48.430","Text":"This equation has been derived when taking into account the rocket example,"},{"Start":"08:48.430 ","End":"08:50.215","Text":"an example of ejection."},{"Start":"08:50.215 ","End":"08:51.730","Text":"When we\u0027re dealing with accretion,"},{"Start":"08:51.730 ","End":"08:52.860","Text":"because it\u0027s the opposite,"},{"Start":"08:52.860 ","End":"08:55.150","Text":"so we have a minus."},{"Start":"08:55.700 ","End":"09:02.575","Text":"This is the rate that mass is being added to our body."},{"Start":"09:02.575 ","End":"09:07.315","Text":"Then we can say, that our rate of change of the mass"},{"Start":"09:07.315 ","End":"09:12.130","Text":"as a function of time is equal to positive Alpha."},{"Start":"09:12.130 ","End":"09:18.260","Text":"Now, why is this? Because mass is being added to the body per second."},{"Start":"09:18.260 ","End":"09:19.925","Text":"Every second that goes by,"},{"Start":"09:19.925 ","End":"09:24.360","Text":"more and more rain is filling up in this tank."},{"Start":"09:24.360 ","End":"09:29.210","Text":"Then, we\u0027ll get that our equation for our mass as a function of time will"},{"Start":"09:29.210 ","End":"09:34.385","Text":"equal to our initial mass plus Alpha t,"},{"Start":"09:34.385 ","End":"09:36.635","Text":"because it\u0027s the opposite of here."},{"Start":"09:36.635 ","End":"09:40.130","Text":"It\u0027s plus Alpha t because as time goes by,"},{"Start":"09:40.130 ","End":"09:45.900","Text":"our total mass is increasing because we\u0027re dealing with accretion."},{"Start":"09:45.980 ","End":"09:49.850","Text":"Then the last thing that we saw,"},{"Start":"09:49.850 ","End":"09:52.925","Text":"is when dealing with the case of accretion,"},{"Start":"09:52.925 ","End":"09:57.885","Text":"but when we also have friction going together with it."},{"Start":"09:57.885 ","End":"10:03.635","Text":"Then we said that in order to work out what our frictional force was,"},{"Start":"10:03.635 ","End":"10:10.085","Text":"because we know that our frictional force equals some constant multiplied by the normal,"},{"Start":"10:10.085 ","End":"10:13.460","Text":"which we know is also varying in time because the mass is varying in time,"},{"Start":"10:13.460 ","End":"10:18.140","Text":"so then we have to say that the sum of all of the forces this time"},{"Start":"10:18.140 ","End":"10:23.685","Text":"on the y-axis equals N minus mg,"},{"Start":"10:23.685 ","End":"10:29.190","Text":"which is equal to this equation,"},{"Start":"10:29.190 ","End":"10:32.430","Text":"but on the y-axis."},{"Start":"10:32.430 ","End":"10:35.987","Text":"This is the equation on the y-axis,"},{"Start":"10:35.987 ","End":"10:39.467","Text":"so we have N minus Mg equals Mt,"},{"Start":"10:39.467 ","End":"10:41.614","Text":"so mass as a function of time,"},{"Start":"10:41.614 ","End":"10:46.750","Text":"multiplied by the change in velocity over time in the y-direction."},{"Start":"10:46.750 ","End":"10:48.275","Text":"In the example with a car,"},{"Start":"10:48.275 ","End":"10:51.065","Text":"because the car it wasn\u0027t moving in the y-direction,"},{"Start":"10:51.065 ","End":"10:54.980","Text":"so we could have said that this Mt,"},{"Start":"10:54.980 ","End":"10:58.020","Text":"dvy by dt=0,"},{"Start":"10:58.020 ","End":"11:03.225","Text":"plus our V_rel multiplied by dm by dt."},{"Start":"11:03.225 ","End":"11:06.214","Text":"It\u0027s important, if you\u0027re dealing with friction,"},{"Start":"11:06.214 ","End":"11:09.680","Text":"that you also take into account this and remember to"},{"Start":"11:09.680 ","End":"11:14.455","Text":"write down all of your workings out in the y-direction."},{"Start":"11:14.455 ","End":"11:17.615","Text":"That\u0027s the end of this lesson,"},{"Start":"11:17.615 ","End":"11:21.290","Text":"which was just a conclusion on the varying mass systems."},{"Start":"11:21.290 ","End":"11:24.290","Text":"If you want to write down this table because it will help you remember"},{"Start":"11:24.290 ","End":"11:28.010","Text":"the little rules when using this equation, then please do."},{"Start":"11:28.010 ","End":"11:31.740","Text":"Now we\u0027re going to go over to some examples."}],"ID":8224}],"Thumbnail":null,"ID":5427},{"Name":"Exercises","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Rain into a Cart","Duration":"10m 31s","ChapterTopicVideoID":9225,"CourseChapterTopicPlaylistID":5429,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.350 ","End":"00:03.180","Text":"Hello. In this question,"},{"Start":"00:03.180 ","End":"00:06.510","Text":"we\u0027re being told that there\u0027s a cart, this is the cart,"},{"Start":"00:06.510 ","End":"00:09.690","Text":"the square of mass M_0,"},{"Start":"00:09.690 ","End":"00:16.575","Text":"which is traveling at an initial velocity of v_0."},{"Start":"00:16.575 ","End":"00:19.320","Text":"Let\u0027s say that v zero\u0027s in this direction."},{"Start":"00:19.320 ","End":"00:24.900","Text":"At t=0, rain falling perpendicularly to the ground."},{"Start":"00:24.900 ","End":"00:27.945","Text":"The rain is going in this direction,"},{"Start":"00:27.945 ","End":"00:32.025","Text":"90 degrees to the ground."},{"Start":"00:32.025 ","End":"00:35.820","Text":"Starts filling the cart at a rate of q,"},{"Start":"00:35.820 ","End":"00:41.925","Text":"so dm by dt is equal to q."},{"Start":"00:41.925 ","End":"00:46.410","Text":"Then we\u0027re being asked in question number 1."},{"Start":"00:46.690 ","End":"00:52.940","Text":"What is the velocity of the cart as a function of time?"},{"Start":"00:52.940 ","End":"00:57.860","Text":"What are we going to use is we\u0027re going to use this equation over here,"},{"Start":"00:57.860 ","End":"01:02.090","Text":"which is the sum of all external forces equals our mass,"},{"Start":"01:02.090 ","End":"01:03.560","Text":"which changes in time,"},{"Start":"01:03.560 ","End":"01:09.790","Text":"multiplied by dv by dt plus dm by dt multiplied by our u,"},{"Start":"01:09.790 ","End":"01:13.220","Text":"which is v relative times negative 1."},{"Start":"01:13.220 ","End":"01:15.980","Text":"Now, why are we multiplying by negative 1?"},{"Start":"01:15.980 ","End":"01:21.105","Text":"Because this is a case of gaining mass."},{"Start":"01:21.105 ","End":"01:22.590","Text":"Our body is getting mass,"},{"Start":"01:22.590 ","End":"01:24.005","Text":"so it\u0027s a case of accretion,"},{"Start":"01:24.005 ","End":"01:28.000","Text":"which means that we multiply by negative 1,"},{"Start":"01:28.000 ","End":"01:31.170","Text":"the expression of u dm by dt."},{"Start":"01:31.170 ","End":"01:37.010","Text":"Let\u0027s start. The some of our external forces is going to be equal to"},{"Start":"01:37.010 ","End":"01:43.935","Text":"0 because there\u0027s no friction in this question."},{"Start":"01:43.935 ","End":"01:47.735","Text":"Then we\u0027ll say that\u0027s the sum of all external forces,"},{"Start":"01:47.735 ","End":"01:51.440","Text":"which equals m multiplied by dv by dt."},{"Start":"01:51.440 ","End":"01:57.650","Text":"Our m as a function of time is going to be equal to our initial mass,"},{"Start":"01:57.650 ","End":"02:04.370","Text":"which is M_0 plus dm by dt multiplied by t,"},{"Start":"02:04.370 ","End":"02:11.380","Text":"so our dm by dt is q multiplied by t. Now,"},{"Start":"02:11.380 ","End":"02:12.770","Text":"why do we know that it\u0027s plus?"},{"Start":"02:12.770 ","End":"02:16.920","Text":"Because our mass is gaining."},{"Start":"02:17.780 ","End":"02:22.290","Text":"Then I multiply this by dv by dt,"},{"Start":"02:22.290 ","End":"02:24.150","Text":"which I don\u0027t know what this equals to,"},{"Start":"02:24.150 ","End":"02:28.745","Text":"and we will rearrange and solve to find what this equals to later."},{"Start":"02:28.745 ","End":"02:33.870","Text":"Then I add plus my dm by dt, which is q."},{"Start":"02:35.570 ","End":"02:39.550","Text":"Then I multiply by my u,"},{"Start":"02:39.550 ","End":"02:41.150","Text":"which is my v relative."},{"Start":"02:41.150 ","End":"02:45.320","Text":"Now, my v relative is the velocity of"},{"Start":"02:45.320 ","End":"02:50.900","Text":"the rain in the x-axis minus the velocity of the cart in the x-axis."},{"Start":"02:50.900 ","End":"02:53.735","Text":"Now because the rain is falling straight downwards,"},{"Start":"02:53.735 ","End":"02:57.220","Text":"so the velocity of the rain on the x-axis is 0."},{"Start":"02:57.220 ","End":"03:01.460","Text":"Then negative the velocity of the cart and the x-axis, which we don\u0027t know,"},{"Start":"03:01.460 ","End":"03:06.740","Text":"so we\u0027ll just call it v. Then of course we have to multiply by negative"},{"Start":"03:06.740 ","End":"03:13.105","Text":"1 because this is a case of us gaining mass and its accretion."},{"Start":"03:13.105 ","End":"03:18.750","Text":"Now what I want to do is I want to isolate out my dv by dt,"},{"Start":"03:18.750 ","End":"03:23.125","Text":"in order to solve this differential equation."},{"Start":"03:23.125 ","End":"03:29.400","Text":"What I can do is I can say that negative"},{"Start":"03:29.400 ","End":"03:37.920","Text":"Qv=m_0 plus qt multiplied"},{"Start":"03:37.920 ","End":"03:40.840","Text":"by dv by dt."},{"Start":"03:40.840 ","End":"03:45.800","Text":"Then I\u0027ll multiply both sides by dt."},{"Start":"03:45.800 ","End":"03:46.430","Text":"I get"},{"Start":"03:46.430 ","End":"03:47.375","Text":"negative"},{"Start":"03:47.375 ","End":"03:56.880","Text":"qvdt=(M_0+qt)dv."},{"Start":"03:56.880 ","End":"04:02.090","Text":"Then what I want is that on the side where I have my dt,"},{"Start":"04:02.090 ","End":"04:05.090","Text":"I\u0027ll only have an expression with t in it."},{"Start":"04:05.090 ","End":"04:06.605","Text":"On my side of the dv,"},{"Start":"04:06.605 ","End":"04:09.905","Text":"I\u0027ll only have an expression with v in it."},{"Start":"04:09.905 ","End":"04:13.145","Text":"So dt has to match with my t variables,"},{"Start":"04:13.145 ","End":"04:15.575","Text":"and dv has to match with my v variables."},{"Start":"04:15.575 ","End":"04:19.725","Text":"I\u0027m going to cross-divide and cross-multiply."},{"Start":"04:19.725 ","End":"04:27.485","Text":"I\u0027m going to divide both sides by qv and divide both sides by M_0 plus qt."},{"Start":"04:27.485 ","End":"04:29.285","Text":"Then, in the end,"},{"Start":"04:29.285 ","End":"04:33.210","Text":"I\u0027m going to get negative 1 over"},{"Start":"04:33.210 ","End":"04:40.160","Text":"M_0 qt plus d t,"},{"Start":"04:40.160 ","End":"04:45.830","Text":"which will equal to 1/qv,"},{"Start":"04:45.830 ","End":"04:56.305","Text":"d v. Now what I\u0027ll do is I\u0027ll add in my integration signs on both sides."},{"Start":"04:56.305 ","End":"04:58.905","Text":"Now let\u0027s fill in our balance."},{"Start":"04:58.905 ","End":"05:02.910","Text":"In our t, we\u0027re starting from t=0 until"},{"Start":"05:02.910 ","End":"05:07.655","Text":"some time t. Our velocity is starting from initial velocity,"},{"Start":"05:07.655 ","End":"05:12.720","Text":"which is v_0 until some final velocity."},{"Start":"05:12.720 ","End":"05:14.615","Text":"Then if I integrate,"},{"Start":"05:14.615 ","End":"05:17.105","Text":"I\u0027ll see that we will get"},{"Start":"05:17.105 ","End":"05:25.490","Text":"negative 1/q multiplied by Ln of M_0,"},{"Start":"05:25.490 ","End":"05:30.140","Text":"plus q t divided by M_0,"},{"Start":"05:30.140 ","End":"05:32.510","Text":"which will equal to"},{"Start":"05:32.510 ","End":"05:42.090","Text":"1/q Ln"},{"Start":"05:42.090 ","End":"05:43.215","Text":"of v/v_0."},{"Start":"05:43.215 ","End":"05:45.725","Text":"If you don\u0027t know how I did this integration,"},{"Start":"05:45.725 ","End":"05:51.860","Text":"then go back to our lesson about differential equations,"},{"Start":"05:51.860 ","End":"05:55.025","Text":"and also go over integrals."},{"Start":"05:55.025 ","End":"06:00.110","Text":"Now what I want to do is I want to isolate out"},{"Start":"06:00.110 ","End":"06:07.630","Text":"this v. Now let\u0027s take a look at how I\u0027m going to do that."},{"Start":"06:08.780 ","End":"06:14.885","Text":"Now we can see that we can cancel out this 1/q and this 1/q."},{"Start":"06:14.885 ","End":"06:24.585","Text":"Then what we can do is we can move this negative 1 over to this over here."},{"Start":"06:24.585 ","End":"06:29.720","Text":"Instead of having negative 1 multiplied by Ln of this,"},{"Start":"06:29.720 ","End":"06:36.950","Text":"we can just say that the inside the brackets is to the power of negative 1."},{"Start":"06:36.950 ","End":"06:43.740","Text":"Remember, any number multiplied by Ln is equal to Ln to the power of that number."},{"Start":"06:43.880 ","End":"06:53.250","Text":"Then we can raise both sides by e. Then we just get rid of the Lns."},{"Start":"06:53.250 ","End":"06:56.795","Text":"Then we\u0027ll have because this is to the power of negative 1."},{"Start":"06:56.795 ","End":"06:59.300","Text":"We\u0027ll just flip this fraction over."},{"Start":"06:59.300 ","End":"07:07.560","Text":"We\u0027ll get M_0 divided by M_0 plus qt=v/v_0."},{"Start":"07:08.420 ","End":"07:12.905","Text":"Then all we have to do is just isolate out this v,"},{"Start":"07:12.905 ","End":"07:21.680","Text":"and then we\u0027ll get that our v=M_0v_0 divided by M_0 plus qt."},{"Start":"07:21.680 ","End":"07:25.555","Text":"This is our final answer to question number 1."},{"Start":"07:25.555 ","End":"07:32.070","Text":"Now our second question says that when the cart reaches a mass of m_f,"},{"Start":"07:32.070 ","End":"07:36.450","Text":"final mass of m_f, the rain stops."},{"Start":"07:36.450 ","End":"07:38.570","Text":"Then when the rain stops,"},{"Start":"07:38.570 ","End":"07:41.540","Text":"no mass is being added to the cart,"},{"Start":"07:41.540 ","End":"07:47.615","Text":"which means that the velocity will then be equal to v final and it will be constant."},{"Start":"07:47.615 ","End":"07:52.750","Text":"Then we will be asked what is the velocity of the cart after the rain stops?"},{"Start":"07:52.750 ","End":"07:59.145","Text":"We\u0027re trying to find out what our v_f is equal to."},{"Start":"07:59.145 ","End":"08:02.655","Text":"In order to find out what our v_f equals to,"},{"Start":"08:02.655 ","End":"08:06.915","Text":"we have to find out what our t final is?"},{"Start":"08:06.915 ","End":"08:10.380","Text":"At what time this happens."},{"Start":"08:10.380 ","End":"08:20.240","Text":"Then we can substitute this into our equation to find out what our mass is."},{"Start":"08:20.240 ","End":"08:28.085","Text":"Remember we have our equation for mass as a function of time is equal to M_0 plus qt."},{"Start":"08:28.085 ","End":"08:32.785","Text":"We substitute this t into this equation."},{"Start":"08:32.785 ","End":"08:41.010","Text":"Then we substitute this mass into here."},{"Start":"08:41.010 ","End":"08:48.450","Text":"Then from this, we will get our v final."},{"Start":"08:48.450 ","End":"08:54.270","Text":"We have to find out what our t final is by substituting it in here."},{"Start":"08:54.470 ","End":"09:03.970","Text":"We can say we want our mass and we want this to equal to m_f."},{"Start":"09:04.400 ","End":"09:06.915","Text":"We just plug this in."},{"Start":"09:06.915 ","End":"09:15.370","Text":"We say that m_f is equal to M_0 plus qt final."},{"Start":"09:15.370 ","End":"09:19.220","Text":"Now, all we have to do is isolate out the t final."},{"Start":"09:19.220 ","End":"09:22.620","Text":"We\u0027ll get that our t final is equal to"},{"Start":"09:22.620 ","End":"09:30.940","Text":"our M_f minus M_0 divided by q."},{"Start":"09:30.940 ","End":"09:33.185","Text":"This is our t. Now,"},{"Start":"09:33.185 ","End":"09:37.385","Text":"all we have to do is we have to substitute this into"},{"Start":"09:37.385 ","End":"09:42.580","Text":"our equation for our v. Then we will get that v"},{"Start":"09:42.580 ","End":"09:50.745","Text":"final is equal to M_0 multiplied by V_0 divided by M_0 plus"},{"Start":"09:50.745 ","End":"10:00.410","Text":"q multiplied by our t. Now our t is m final minus M_0 divided by q."},{"Start":"10:00.410 ","End":"10:03.545","Text":"Then our q\u0027s can cross out."},{"Start":"10:03.545 ","End":"10:06.995","Text":"Then we\u0027ll also see that our M_0 minus M_0."},{"Start":"10:06.995 ","End":"10:08.540","Text":"This can cross out."},{"Start":"10:08.540 ","End":"10:13.530","Text":"Then we\u0027ll get that our v final is equal to"},{"Start":"10:13.530 ","End":"10:20.200","Text":"m_0 v_0 divided by m final."},{"Start":"10:20.770 ","End":"10:24.990","Text":"Then this is our final answer."},{"Start":"10:24.990 ","End":"10:29.530","Text":"This is the velocity of the cart after the rain stops."},{"Start":"10:29.530 ","End":"10:32.720","Text":"That\u0027s the end of this question."}],"ID":10614},{"Watched":false,"Name":"Rain at an Angle","Duration":"24m 5s","ChapterTopicVideoID":9226,"CourseChapterTopicPlaylistID":5429,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.385","Text":"Hello. In this lesson,"},{"Start":"00:02.385 ","End":"00:07.890","Text":"we\u0027re being asked about 2 different carts that we can see over"},{"Start":"00:07.890 ","End":"00:10.905","Text":"here and we\u0027re being asked to"},{"Start":"00:10.905 ","End":"00:15.315","Text":"write an equation for the velocity of each Cart As a function of time."},{"Start":"00:15.315 ","End":"00:17.520","Text":"Let\u0027s see, we have Cart A,"},{"Start":"00:17.520 ","End":"00:21.390","Text":"which is this 1 over here and we\u0027re being told that Cart A has"},{"Start":"00:21.390 ","End":"00:28.830","Text":"an initial mass M_0 and an initial velocity of V_0 going in the right direction."},{"Start":"00:28.830 ","End":"00:32.985","Text":"They\u0027re traveling in this direction to the right."},{"Start":"00:32.985 ","End":"00:37.680","Text":"Rain falls in the Cart At an angle of Alpha so we can see"},{"Start":"00:37.680 ","End":"00:41.640","Text":"that over here and a velocity U."},{"Start":"00:41.640 ","End":"00:46.665","Text":"The rain is falling in at a velocity of U,"},{"Start":"00:46.665 ","End":"00:50.220","Text":"such that the cart fills up at a rate of Q."},{"Start":"00:50.220 ","End":"00:57.735","Text":"Then we can already write that our dm by dt is equal to q."},{"Start":"00:57.735 ","End":"01:01.530","Text":"Now, Cart B has the same starting mass."},{"Start":"01:01.530 ","End":"01:04.110","Text":"The starting mass is M_0 so"},{"Start":"01:04.110 ","End":"01:11.240","Text":"the same starting mass M_0 as A and the same initial velocity as A."},{"Start":"01:12.620 ","End":"01:15.820","Text":"We can write that in here."},{"Start":"01:15.890 ","End":"01:19.780","Text":"Cart B is already filled with rainwater."},{"Start":"01:19.780 ","End":"01:25.134","Text":"There was a hole in the Cart And water is let out at the same rate as water enters."},{"Start":"01:25.134 ","End":"01:29.990","Text":"Write an equation for the velocity of each Cart As a function of time."},{"Start":"01:29.990 ","End":"01:33.210","Text":"Let\u0027s see how we do this."},{"Start":"01:33.210 ","End":"01:37.075","Text":"Now, of course, we have our equation over here,"},{"Start":"01:37.075 ","End":"01:43.280","Text":"so let\u0027s start off with working out the equation for A."},{"Start":"01:44.600 ","End":"01:48.640","Text":"We have to write the sum of all of our external forces,"},{"Start":"01:48.640 ","End":"01:52.600","Text":"now because we\u0027re not told that there is friction or anything like that,"},{"Start":"01:52.600 ","End":"01:56.890","Text":"so we can say that the sum of the external forces is equal to 0."},{"Start":"01:56.890 ","End":"02:00.355","Text":"That equals our mass."},{"Start":"02:00.355 ","End":"02:05.370","Text":"Now, our equation for mass obviously is as a function of time."},{"Start":"02:05.450 ","End":"02:09.350","Text":"How do we do that? We have our initial mass,"},{"Start":"02:09.350 ","End":"02:13.775","Text":"which is M_0 plus our rate of change times time."},{"Start":"02:13.775 ","End":"02:18.920","Text":"That\u0027s plus q multiplied by t. This is our mass as"},{"Start":"02:18.920 ","End":"02:24.495","Text":"a function of time multiplied by our dv by dt,"},{"Start":"02:24.495 ","End":"02:27.800","Text":"which we right now don\u0027t know what this is."},{"Start":"02:27.800 ","End":"02:31.245","Text":"Plus dm by dt,"},{"Start":"02:31.245 ","End":"02:35.700","Text":"which we know is q multiplied by our u."},{"Start":"02:35.700 ","End":"02:36.795","Text":"Now, I\u0027m reminding you,"},{"Start":"02:36.795 ","End":"02:42.140","Text":"u is the relative velocity so we have to work"},{"Start":"02:42.140 ","End":"02:48.430","Text":"out the relative velocity of the rain and 2 of the carts."},{"Start":"02:48.430 ","End":"02:51.980","Text":"How do we do that? We know that the velocity of the rain is u,"},{"Start":"02:51.980 ","End":"02:56.015","Text":"but it\u0027s going in a diagonal direction so we have to find"},{"Start":"02:56.015 ","End":"03:00.485","Text":"out what the velocity is going in the x direction."},{"Start":"03:00.485 ","End":"03:03.110","Text":"We\u0027re going to say that the x direction"},{"Start":"03:03.110 ","End":"03:06.710","Text":"going in this direction is the positive direction."},{"Start":"03:06.710 ","End":"03:10.040","Text":"We\u0027re going to do because we want"},{"Start":"03:10.040 ","End":"03:15.785","Text":"the adjacent angle so it\u0027s going to be u cosine of Alpha."},{"Start":"03:15.785 ","End":"03:20.030","Text":"Then it\u0027s going to be a negative because you can see that because"},{"Start":"03:20.030 ","End":"03:24.360","Text":"the u is going like this so this is the x component."},{"Start":"03:24.360 ","End":"03:30.155","Text":"It\u0027s in the opposite direction to our V_0 so it\u0027s negative u cosine of Alpha,"},{"Start":"03:30.155 ","End":"03:35.660","Text":"and then negative the velocity of the large object which"},{"Start":"03:35.660 ","End":"03:41.370","Text":"is negative V. Now this V,"},{"Start":"03:41.370 ","End":"03:45.300","Text":"We don\u0027t know we\u0027re yet to figure it out."},{"Start":"03:45.300 ","End":"03:50.645","Text":"Then because the rain is filling up the cart,"},{"Start":"03:50.645 ","End":"03:55.620","Text":"so we know that we have to multiply this by negative 1."},{"Start":"03:56.810 ","End":"03:59.615","Text":"Now, we can just solve this as"},{"Start":"03:59.615 ","End":"04:05.315","Text":"a differential equation and notice that it\u0027s because we only have 2 unknowns."},{"Start":"04:05.315 ","End":"04:08.985","Text":"We have an unknown with v and then"},{"Start":"04:08.985 ","End":"04:14.180","Text":"unknown with t. That means we can do a differential equation."},{"Start":"04:14.180 ","End":"04:20.555","Text":"What I\u0027m going to do is I\u0027m going to move this expression q multiplied by"},{"Start":"04:20.555 ","End":"04:23.780","Text":"negative u cosine Alpha minus v multiplied"},{"Start":"04:23.780 ","End":"04:27.894","Text":"by negative 1 over to the other side of the equation."},{"Start":"04:27.894 ","End":"04:30.155","Text":"I\u0027m going to then get,"},{"Start":"04:30.155 ","End":"04:33.980","Text":"by canceling out the minuses and sorting it out,"},{"Start":"04:33.980 ","End":"04:40.515","Text":"we\u0027re going to get a final thing of negative q multiplied by"},{"Start":"04:40.515 ","End":"04:48.770","Text":"u cosine of Alpha plus v and then the 1 can cancel out,"},{"Start":"04:48.770 ","End":"04:53.520","Text":"which is going to equal M_0"},{"Start":"04:53.520 ","End":"04:59.980","Text":"plus qt multiplied by dv by dt."},{"Start":"05:01.700 ","End":"05:05.600","Text":"Now, what I want to do is I want to separate"},{"Start":"05:05.600 ","End":"05:09.140","Text":"out so that on the side where I have my dv,"},{"Start":"05:09.140 ","End":"05:15.710","Text":"I only have expressions with the variable V and that on the side where I have my dt,"},{"Start":"05:15.710 ","End":"05:18.410","Text":"I only have expressions with the variable"},{"Start":"05:18.410 ","End":"05:22.190","Text":"t. What I\u0027m going to do is I\u0027m going to multiply"},{"Start":"05:22.190 ","End":"05:32.000","Text":"both sides by dt and then I\u0027m going to divide both sides by M_0 plus qt."},{"Start":"05:32.000 ","End":"05:42.515","Text":"Then I\u0027m going to get an expression of dt divided by M_0 plus qt,"},{"Start":"05:42.515 ","End":"05:48.980","Text":"which is going to equal to negative dv divided by"},{"Start":"05:48.980 ","End":"05:54.060","Text":"qu cosine of Alpha"},{"Start":"05:54.060 ","End":"06:00.305","Text":"plus v. If this negative comes from this negative q,"},{"Start":"06:00.305 ","End":"06:01.940","Text":"then just put it up here."},{"Start":"06:01.940 ","End":"06:04.115","Text":"It\u0027s the same thing."},{"Start":"06:04.115 ","End":"06:06.215","Text":"If you don\u0027t see how I did this,"},{"Start":"06:06.215 ","End":"06:10.625","Text":"I just did some multiplication and division simple algebra,"},{"Start":"06:10.625 ","End":"06:13.165","Text":"do it on a piece of paper."},{"Start":"06:13.165 ","End":"06:20.000","Text":"Now, all I have to do is add in my integration signs and add in my bounds."},{"Start":"06:20.000 ","End":"06:27.080","Text":"I\u0027m starting from my t is equal 0 up until some t. Here,"},{"Start":"06:27.080 ","End":"06:32.150","Text":"I\u0027ll add dashes to show you that this t that I\u0027m integrating isn\u0027t the"},{"Start":"06:32.150 ","End":"06:37.520","Text":"same as my t and my bounds and then in my velocity,"},{"Start":"06:37.520 ","End":"06:41.090","Text":"I\u0027m starting from V_0 because that\u0027s my starting velocity"},{"Start":"06:41.090 ","End":"06:45.665","Text":"up until my v as a function of t,"},{"Start":"06:45.665 ","End":"06:49.620","Text":"which relates to this t over here."},{"Start":"06:50.060 ","End":"06:55.210","Text":"Now, we\u0027re going to integrate these expressions,"},{"Start":"06:55.210 ","End":"06:56.970","Text":"so as we can see,"},{"Start":"06:56.970 ","End":"07:05.570","Text":"this is going to be 1/q because we\u0027re multiplying by the"},{"Start":"07:05.570 ","End":"07:15.420","Text":"constant that multiplies this t ln(M_0) plus qt."},{"Start":"07:15.420 ","End":"07:21.379","Text":"We\u0027re substituting in this t divided by M_0."},{"Start":"07:21.379 ","End":"07:23.560","Text":"How did I do this?"},{"Start":"07:23.560 ","End":"07:28.885","Text":"Because when you see that it\u0027s q multiplied by some t,"},{"Start":"07:28.885 ","End":"07:31.600","Text":"then you know that it\u0027s going to be ln."},{"Start":"07:31.600 ","End":"07:33.760","Text":"Why did I do this?"},{"Start":"07:33.760 ","End":"07:38.620","Text":"Because the integral of this is ln divided by the"},{"Start":"07:38.620 ","End":"07:46.210","Text":"constant that multiplies the t. Then when I substitute in the t,"},{"Start":"07:46.210 ","End":"07:48.700","Text":"I\u0027ll have ln of m_0 plus qt,"},{"Start":"07:48.700 ","End":"07:50.815","Text":"and then I\u0027ll substitute in the 0,"},{"Start":"07:50.815 ","End":"07:54.715","Text":"and then I\u0027ll have negative ln of m_0."},{"Start":"07:54.715 ","End":"07:57.480","Text":"Then with logs and lns,"},{"Start":"07:57.480 ","End":"08:00.135","Text":"you can just rewrite it in this format."},{"Start":"08:00.135 ","End":"08:04.710","Text":"If you don\u0027t know that, then please go over your integrating skills."},{"Start":"08:04.710 ","End":"08:09.800","Text":"Then this will equal similar thing over here, negative."},{"Start":"08:09.800 ","End":"08:12.430","Text":"Now, this q we can take out of the integral,"},{"Start":"08:12.430 ","End":"08:15.010","Text":"negative 1 over q."},{"Start":"08:15.010 ","End":"08:24.880","Text":"Then again, the same thing it\u0027s going to be ln of u cosine of Alpha plus v as"},{"Start":"08:24.880 ","End":"08:35.455","Text":"a function of t divided by u cosine of Alpha plus v_0."},{"Start":"08:35.455 ","End":"08:38.590","Text":"The same as on this side."},{"Start":"08:38.590 ","End":"08:41.335","Text":"Then what we can do,"},{"Start":"08:41.335 ","End":"08:43.750","Text":"we see that on both sides we have 1 over q,"},{"Start":"08:43.750 ","End":"08:45.505","Text":"we can cross them both out."},{"Start":"08:45.505 ","End":"08:47.170","Text":"Then this negative,"},{"Start":"08:47.170 ","End":"08:53.830","Text":"we can cross out and write here to the power of negative 1."},{"Start":"08:53.830 ","End":"08:57.070","Text":"Then that will say that the denominator will equal"},{"Start":"08:57.070 ","End":"09:00.610","Text":"the numerator and the numerator will go down and equal the denominator."},{"Start":"09:00.610 ","End":"09:03.085","Text":"You\u0027re just flipping around this fraction,"},{"Start":"09:03.085 ","End":"09:05.275","Text":"when it\u0027s to the power of negative 1."},{"Start":"09:05.275 ","End":"09:07.525","Text":"Then you can write them,"},{"Start":"09:07.525 ","End":"09:12.460","Text":"you can raise both by e and that will get rid of the ln."},{"Start":"09:12.460 ","End":"09:23.080","Text":"Then you\u0027ll get that m_0 plus qt divided by m_0 is equal to this fraction flipped over,"},{"Start":"09:23.080 ","End":"09:29.755","Text":"which will be u cosine of Alpha plus v_0 divided by"},{"Start":"09:29.755 ","End":"09:37.975","Text":"u cosine of Alpha plus v as a function of t. Now,"},{"Start":"09:37.975 ","End":"09:39.790","Text":"because we\u0027re trying to write an equation for"},{"Start":"09:39.790 ","End":"09:42.130","Text":"the velocity of each Cart As a function of time,"},{"Start":"09:42.130 ","End":"09:45.670","Text":"we\u0027re trying to isolate out this."},{"Start":"09:45.670 ","End":"09:50.140","Text":"Then all we have to do is we"},{"Start":"09:50.140 ","End":"09:56.920","Text":"can take this numerator and divide this whole thing."},{"Start":"09:56.920 ","End":"10:03.055","Text":"We\u0027ll have m_0 multiplied by u cosine of Alpha"},{"Start":"10:03.055 ","End":"10:10.045","Text":"plus v_0 divided by m_0 plus qt,"},{"Start":"10:10.045 ","End":"10:15.850","Text":"which will equal u cosine of Alpha plus v as"},{"Start":"10:15.850 ","End":"10:21.970","Text":"a function of t. Then I\u0027ll just minus u cosine of Alpha."},{"Start":"10:21.970 ","End":"10:29.545","Text":"Then all I have to do is write negative u cosine of Alpha is equal to v(t)."},{"Start":"10:29.545 ","End":"10:36.970","Text":"Then this is our final answer."},{"Start":"10:36.970 ","End":"10:40.960","Text":"Now, I know that I did quite a few steps in 1"},{"Start":"10:40.960 ","End":"10:44.560","Text":"go over here and it might be a little bit complicated,"},{"Start":"10:44.560 ","End":"10:47.575","Text":"but really, it\u0027s just simple algebra."},{"Start":"10:47.575 ","End":"10:52.960","Text":"If you didn\u0027t exactly follow what I did and you don\u0027t understand the steps I took,"},{"Start":"10:52.960 ","End":"10:58.300","Text":"please do this on a piece of paper and you\u0027ll see that you\u0027ll get this,"},{"Start":"10:58.300 ","End":"11:00.865","Text":"and that is it."},{"Start":"11:00.865 ","End":"11:04.520","Text":"Now, let\u0027s move over to Cart B."},{"Start":"11:05.730 ","End":"11:10.675","Text":"Now, we\u0027re taking a look at Cart B."},{"Start":"11:10.675 ","End":"11:14.200","Text":"Again, we\u0027re going to use our formula and again,"},{"Start":"11:14.200 ","End":"11:16.750","Text":"because there\u0027s no friction or any external forces,"},{"Start":"11:16.750 ","End":"11:22.280","Text":"we know that the sum of all external forces is equal to 0."},{"Start":"11:22.290 ","End":"11:27.055","Text":"Now, we\u0027re going to write our mass as a function of time."},{"Start":"11:27.055 ","End":"11:34.930","Text":"Now, because we\u0027re told that the rain drips in at the same rates as water is led out,"},{"Start":"11:34.930 ","End":"11:38.229","Text":"we know that our mass is unchanging."},{"Start":"11:38.229 ","End":"11:40.780","Text":"It\u0027s just going to be m_0."},{"Start":"11:40.780 ","End":"11:47.570","Text":"The mass remains constant and then multiply it by dv by dt."},{"Start":"11:49.200 ","End":"11:52.285","Text":"Then we\u0027re going to speak about here."},{"Start":"11:52.285 ","End":"11:54.805","Text":"Our dm by dt."},{"Start":"11:54.805 ","End":"11:56.605","Text":"I didn\u0027t write it earlier,"},{"Start":"11:56.605 ","End":"12:03.410","Text":"but our dm by dt is equal to q."},{"Start":"12:03.510 ","End":"12:11.185","Text":"We know that our dm by dt is equal to q multiplied by the relative velocity."},{"Start":"12:11.185 ","End":"12:18.655","Text":"Now, notice, we have the velocity of the rain going in like this,"},{"Start":"12:18.655 ","End":"12:21.820","Text":"and we have the velocity of the rain going out."},{"Start":"12:21.820 ","End":"12:24.880","Text":"Let\u0027s first speak about the rain going in."},{"Start":"12:24.880 ","End":"12:27.085","Text":"The velocity of the rain going in,"},{"Start":"12:27.085 ","End":"12:28.555","Text":"we have our relative velocity,"},{"Start":"12:28.555 ","End":"12:32.785","Text":"the rain going in relative to the velocity of the cart."},{"Start":"12:32.785 ","End":"12:35.170","Text":"Now the velocity of the cart is v_0."},{"Start":"12:35.170 ","End":"12:40.960","Text":"Our u is going to be the velocity of the rain,"},{"Start":"12:40.960 ","End":"12:42.985","Text":"which has the smaller velocity,"},{"Start":"12:42.985 ","End":"12:46.870","Text":"which is going to be exactly the same as in Cart A."},{"Start":"12:46.870 ","End":"12:51.580","Text":"It\u0027s going to be u cosine Alpha because it\u0027s"},{"Start":"12:51.580 ","End":"12:56.635","Text":"going in at an angle of Alpha and we only want the x-direction."},{"Start":"12:56.635 ","End":"13:01.735","Text":"Notice that the x-direction is going in the left direction."},{"Start":"13:01.735 ","End":"13:04.480","Text":"We said that in the x-direction,"},{"Start":"13:04.480 ","End":"13:06.040","Text":"the right direction is the positive,"},{"Start":"13:06.040 ","End":"13:08.860","Text":"the negative just like in Cart A."},{"Start":"13:08.860 ","End":"13:13.570","Text":"Then negative, the velocity of the cart,"},{"Start":"13:13.570 ","End":"13:18.385","Text":"negative v, v as a function of t,"},{"Start":"13:18.385 ","End":"13:21.080","Text":"which we don\u0027t know."},{"Start":"13:21.150 ","End":"13:27.280","Text":"Now we\u0027re going to speak about the water coming out of the cart."},{"Start":"13:27.280 ","End":"13:29.920","Text":"Our dm by dt is going to be q,"},{"Start":"13:29.920 ","End":"13:31.735","Text":"because we know that it\u0027s the same."},{"Start":"13:31.735 ","End":"13:33.700","Text":"We can say plus q."},{"Start":"13:33.700 ","End":"13:35.920","Text":"Now, we\u0027re dealing with our u,"},{"Start":"13:35.920 ","End":"13:42.055","Text":"which is the relative velocity of the water exiting relative to the cart."},{"Start":"13:42.055 ","End":"13:47.335","Text":"Now, when water exits a moving body,"},{"Start":"13:47.335 ","End":"13:53.305","Text":"the water that is exiting will have the exact same velocity as the moving body."},{"Start":"13:53.305 ","End":"13:59.530","Text":"We\u0027ll know that it will be v minus v."},{"Start":"13:59.530 ","End":"14:07.780","Text":"Because the relative velocity of the water exiting the cart will be 0 to the cart."},{"Start":"14:07.780 ","End":"14:15.910","Text":"Then if this is 0, then this whole section will cross out."},{"Start":"14:15.910 ","End":"14:22.840","Text":"Then I can just rub this out and then we can say that this is the equation."},{"Start":"14:22.840 ","End":"14:28.030","Text":"Oh, sorry, I forgot to multiply by negative 1 because it\u0027s the water coming in."},{"Start":"14:28.030 ","End":"14:29.965","Text":"It\u0027s multiplied by negative 1."},{"Start":"14:29.965 ","End":"14:32.200","Text":"This is our final equation."},{"Start":"14:32.200 ","End":"14:35.590","Text":"Just because the relative velocity of the water coming"},{"Start":"14:35.590 ","End":"14:40.760","Text":"out is equal to 0 relative to the cart."},{"Start":"14:40.860 ","End":"14:46.745","Text":"Now, we\u0027re going to deal with the exact same as what we did and Cart A,"},{"Start":"14:46.745 ","End":"14:50.570","Text":"and we\u0027re going to separate out our components."},{"Start":"14:50.570 ","End":"14:53.975","Text":"What we\u0027re going to do is we\u0027re going to take"},{"Start":"14:53.975 ","End":"14:59.625","Text":"this expression with the q and move it to the other side of the equals sign."},{"Start":"14:59.625 ","End":"15:03.670","Text":"We can say that this is all positives."},{"Start":"15:03.670 ","End":"15:06.190","Text":"Then when we move it here, we\u0027ll get negative q,"},{"Start":"15:06.190 ","End":"15:14.240","Text":"u cosine Alpha plus v. Because negative 1 multiplied by,"},{"Start":"15:14.240 ","End":"15:18.860","Text":"or negative numbers will equal a positive and then we move it to the other side,"},{"Start":"15:18.860 ","End":"15:24.840","Text":"it\u0027s negative, which equals m_0, dv by dt."},{"Start":"15:25.140 ","End":"15:31.175","Text":"Again, what I want to do is I want to have on the side where I have my dv,"},{"Start":"15:31.175 ","End":"15:36.230","Text":"I want only variables with v and on the side where I have my dt,"},{"Start":"15:36.230 ","End":"15:41.210","Text":"I want to have only variables with t. Now,"},{"Start":"15:41.210 ","End":"15:50.250","Text":"we can see that I can divide both sides by m_0 multiply both sides by dt."},{"Start":"15:50.320 ","End":"15:53.960","Text":"Then I\u0027ll have negative q,"},{"Start":"15:53.960 ","End":"15:58.315","Text":"u cosine of Alpha plus"},{"Start":"15:58.315 ","End":"16:04.755","Text":"v divided by m_0 dt will equal dv."},{"Start":"16:04.755 ","End":"16:07.520","Text":"Now, as we can see, over here,"},{"Start":"16:07.520 ","End":"16:12.755","Text":"we have a v. We don\u0027t want it to be on the side of the equal sign, we want it to be here."},{"Start":"16:12.755 ","End":"16:17.585","Text":"Then I can just divide both sides by u cosine Alpha plus v,"},{"Start":"16:17.585 ","End":"16:24.850","Text":"and then I\u0027ll get negative q divided by m_0 dt is equal to dv"},{"Start":"16:24.850 ","End":"16:32.580","Text":"divided by u cosine of Alpha plus v. Perfect."},{"Start":"16:32.580 ","End":"16:34.705","Text":"This is looking good, and now,"},{"Start":"16:34.705 ","End":"16:40.420","Text":"all I have to do is add in my integration signs."},{"Start":"16:40.420 ","End":"16:42.859","Text":"Then just integrate."},{"Start":"16:42.859 ","End":"16:46.205","Text":"First I have to add in my bounds, I forgot to do that."},{"Start":"16:46.205 ","End":"16:50.205","Text":"My t is from t equals 0 until t."},{"Start":"16:50.205 ","End":"16:58.540","Text":"My v is from v_0 until v as a function of t. Now,"},{"Start":"16:58.540 ","End":"17:05.875","Text":"I can say that I have negative q divided by M_0 multiplied by t,"},{"Start":"17:05.875 ","End":"17:08.350","Text":"which equals, now here,"},{"Start":"17:08.350 ","End":"17:10.165","Text":"I\u0027m going to have a ln."},{"Start":"17:10.165 ","End":"17:15.205","Text":"So here I\u0027ll have ln of u,"},{"Start":"17:15.205 ","End":"17:20.440","Text":"cosine of Alpha plus v as a function of t"},{"Start":"17:20.440 ","End":"17:27.440","Text":"divided by u cosine of Alpha plus v_0."},{"Start":"17:27.570 ","End":"17:32.200","Text":"Remember, it\u0027s the exact same thing as in Cart A."},{"Start":"17:32.200 ","End":"17:36.460","Text":"Now, what I want to do is I want to isolate out this v as a function of"},{"Start":"17:36.460 ","End":"17:38.860","Text":"t. Because we\u0027re being asked to write"},{"Start":"17:38.860 ","End":"17:42.190","Text":"an equation for the velocity of each cart as a function of time."},{"Start":"17:42.190 ","End":"17:47.140","Text":"So what I\u0027ll do is I\u0027ll raise each side by e,"},{"Start":"17:47.140 ","End":"17:51.385","Text":"and then the e and the ln will cancel each other out."},{"Start":"17:51.385 ","End":"17:57.145","Text":"Then I\u0027ll multiply both sides by this u cosine of Alpha plus v_0."},{"Start":"17:57.145 ","End":"18:03.325","Text":"Then I\u0027ll minus this u cosine of Alpha in order to isolate this out."},{"Start":"18:03.325 ","End":"18:08.205","Text":"So in the end, what I\u0027ll get is u cosine of Alpha"},{"Start":"18:08.205 ","End":"18:13.425","Text":"plus v_0 multiplied by e to the negative q of"},{"Start":"18:13.425 ","End":"18:19.325","Text":"t divided by M_0 minus u cosine"},{"Start":"18:19.325 ","End":"18:25.045","Text":"of Alpha will equal to v as a function of t. Again,"},{"Start":"18:25.045 ","End":"18:28.255","Text":"I just did a few steps of basic algebra."},{"Start":"18:28.255 ","End":"18:31.090","Text":"If you didn\u0027t understand it or anything else,"},{"Start":"18:31.090 ","End":"18:34.585","Text":"then please do it on a piece of paper alone to go over it."},{"Start":"18:34.585 ","End":"18:39.865","Text":"Then this is our final answer."},{"Start":"18:39.865 ","End":"18:43.825","Text":"Now what we\u0027re going to do is we\u0027re going to check our answers and go over"},{"Start":"18:43.825 ","End":"18:48.650","Text":"a few more details of why these calculations are correct."},{"Start":"18:48.840 ","End":"18:52.555","Text":"First, we\u0027re going to speak about Cart B."},{"Start":"18:52.555 ","End":"18:55.449","Text":"A common question is,"},{"Start":"18:55.449 ","End":"18:58.525","Text":"or why you might not understand why this works,"},{"Start":"18:58.525 ","End":"19:02.455","Text":"this whole equation, because the mass is unchanging."},{"Start":"19:02.455 ","End":"19:05.680","Text":"That\u0027s the difference between Cart A and Cart B. Cart A,"},{"Start":"19:05.680 ","End":"19:07.675","Text":"there\u0027s changing mass,"},{"Start":"19:07.675 ","End":"19:09.865","Text":"and Cart B, the mass remains the same."},{"Start":"19:09.865 ","End":"19:14.544","Text":"Why does the momentum change if the mass is constant?"},{"Start":"19:14.544 ","End":"19:19.960","Text":"That\u0027s because we can see that the rain is falling at an angle of Alpha,"},{"Start":"19:19.960 ","End":"19:26.740","Text":"which means the velocity in the x-direction is causing the momentum to change."},{"Start":"19:26.740 ","End":"19:31.405","Text":"Because remember, momentum depends on velocity and also on mass."},{"Start":"19:31.405 ","End":"19:33.715","Text":"Because the mass is constant,"},{"Start":"19:33.715 ","End":"19:36.700","Text":"but the velocity is not constant,"},{"Start":"19:36.700 ","End":"19:39.160","Text":"then we have our change in momentum,"},{"Start":"19:39.160 ","End":"19:42.085","Text":"and that\u0027s why we can use this equation."},{"Start":"19:42.085 ","End":"19:43.945","Text":"That\u0027s number 1,"},{"Start":"19:43.945 ","End":"19:45.520","Text":"and number 2, now,"},{"Start":"19:45.520 ","End":"19:49.750","Text":"let\u0027s check that these equations actually make sense."},{"Start":"19:49.750 ","End":"19:52.570","Text":"So let\u0027s take a look at Cart A."},{"Start":"19:52.570 ","End":"19:59.365","Text":"Let\u0027s see at time t equals 0, what happens."},{"Start":"19:59.365 ","End":"20:02.095","Text":"When we substitute zero here,"},{"Start":"20:02.095 ","End":"20:08.515","Text":"we\u0027ll get M_0 multiplied by u cosine of Alpha plus v_0 divided by M_0."},{"Start":"20:08.515 ","End":"20:12.010","Text":"Then both of our M_0\u0027s will cross out."},{"Start":"20:12.010 ","End":"20:16.645","Text":"Then we\u0027ll have u cosine of Alpha minus u cosine of Alpha."},{"Start":"20:16.645 ","End":"20:21.655","Text":"Then we\u0027ll just get that v at time t will equal v_0,"},{"Start":"20:21.655 ","End":"20:28.610","Text":"which is correct because we have in our question that our starting velocity is v_0."},{"Start":"20:28.830 ","End":"20:35.390","Text":"Then let\u0027s see what happens when t is equal to infinity."},{"Start":"20:36.120 ","End":"20:39.850","Text":"We\u0027ll see that when t is equal to infinity,"},{"Start":"20:39.850 ","End":"20:42.550","Text":"the denominator over here will be very,"},{"Start":"20:42.550 ","End":"20:46.945","Text":"very large, which means that this whole expression,"},{"Start":"20:46.945 ","End":"20:50.200","Text":"this fraction, will be equal to 0."},{"Start":"20:50.200 ","End":"20:54.640","Text":"Then we\u0027ll get that our velocity at t equals infinity,"},{"Start":"20:54.640 ","End":"20:59.305","Text":"will be equal to negative u cosine of Alpha."},{"Start":"20:59.305 ","End":"21:01.930","Text":"Let\u0027s talk about why this makes sense."},{"Start":"21:01.930 ","End":"21:06.835","Text":"Our initial velocity is v_0 and it\u0027s going in the x direction,"},{"Start":"21:06.835 ","End":"21:10.975","Text":"in this direction, because that\u0027s the direction we said was the positive direction."},{"Start":"21:10.975 ","End":"21:14.660","Text":"Now, as the time progresses,"},{"Start":"21:14.660 ","End":"21:17.880","Text":"because the rain is falling it at an angle,"},{"Start":"21:17.880 ","End":"21:20.115","Text":"it has an x-component,"},{"Start":"21:20.115 ","End":"21:21.825","Text":"and the x-component is going in"},{"Start":"21:21.825 ","End":"21:26.730","Text":"the opposite direction to our initial direction of travel."},{"Start":"21:26.730 ","End":"21:31.375","Text":"As time goes by, eventually,"},{"Start":"21:31.375 ","End":"21:40.765","Text":"our velocity in the positive direction will slow down and eventually reach a stop."},{"Start":"21:40.765 ","End":"21:46.225","Text":"Then the cart will start moving in the opposite direction because"},{"Start":"21:46.225 ","End":"21:52.545","Text":"then it will just be the force of the rain pushing it in the left direction,"},{"Start":"21:52.545 ","End":"21:54.480","Text":"from the force of the rain."},{"Start":"21:54.480 ","End":"21:57.375","Text":"So that\u0027s why that works."},{"Start":"21:57.375 ","End":"22:01.215","Text":"Then let\u0027s take a look at Cart B."},{"Start":"22:01.215 ","End":"22:05.285","Text":"Now, in Cart B, we can see when t equals 0,"},{"Start":"22:05.285 ","End":"22:08.065","Text":"then this expression will equal 1."},{"Start":"22:08.065 ","End":"22:12.865","Text":"Then we\u0027ll have u cosine of Alpha minus u cosine of Alpha plus v_0."},{"Start":"22:12.865 ","End":"22:16.405","Text":"Again, we\u0027ll have that our initial velocity is equal to v_0,"},{"Start":"22:16.405 ","End":"22:18.700","Text":"time t equals 0,"},{"Start":"22:18.700 ","End":"22:22.000","Text":"which is correct, that\u0027s what we\u0027re being told in the question."},{"Start":"22:22.000 ","End":"22:25.225","Text":"Then, at t equals infinity,"},{"Start":"22:25.225 ","End":"22:29.290","Text":"we\u0027ll see that this whole expression will be equal to 0."},{"Start":"22:29.290 ","End":"22:31.405","Text":"Then we\u0027ll get a t equals infinity."},{"Start":"22:31.405 ","End":"22:33.730","Text":"Again, the cart will be moving in"},{"Start":"22:33.730 ","End":"22:38.620","Text":"the left direction at a velocity of negative u cosine of Alpha."},{"Start":"22:38.620 ","End":"22:42.163","Text":"Which again makes sense because the rain is pushing the cart"},{"Start":"22:42.163 ","End":"22:46.630","Text":"backwards at a velocity of u cosine of Alpha,"},{"Start":"22:46.630 ","End":"22:51.460","Text":"which means that it\u0027s decelerating the cart slowly each"},{"Start":"22:51.460 ","End":"22:57.295","Text":"second by u cosine of Alpha until eventually the cart will decelerate,"},{"Start":"22:57.295 ","End":"22:59.560","Text":"come to a stop, and then start moving at"},{"Start":"22:59.560 ","End":"23:03.445","Text":"u cosine of Alpha backwards in the left direction."},{"Start":"23:03.445 ","End":"23:06.160","Text":"You can also see this in the equations."},{"Start":"23:06.160 ","End":"23:08.980","Text":"You can see that this,"},{"Start":"23:08.980 ","End":"23:14.125","Text":"the rain is decelerating."},{"Start":"23:14.125 ","End":"23:17.875","Text":"You have negative u cosine of Alpha and the same over here."},{"Start":"23:17.875 ","End":"23:22.420","Text":"This is from the rain and it\u0027s causing this deceleration to"},{"Start":"23:22.420 ","End":"23:28.219","Text":"an eventual stop and then moving it in the opposite direction."},{"Start":"23:29.040 ","End":"23:31.810","Text":"That is the end of this lesson."},{"Start":"23:31.810 ","End":"23:34.960","Text":"There\u0027s quite a few very important things that we\u0027ve gone through,"},{"Start":"23:34.960 ","End":"23:41.950","Text":"such as the change in momentum that happens even if the mass is unchanging."},{"Start":"23:41.950 ","End":"23:45.940","Text":"How to check your answers by substituting in t equals 0"},{"Start":"23:45.940 ","End":"23:49.930","Text":"and t equals infinity to check that your answers make sense."},{"Start":"23:49.930 ","End":"23:53.170","Text":"Sometimes, it won\u0027t find if you made a mistake,"},{"Start":"23:53.170 ","End":"23:57.054","Text":"but a lot of the time it will find a mistake"},{"Start":"23:57.054 ","End":"24:01.975","Text":"and how to use these 2 equations in these 2 cases."},{"Start":"24:01.975 ","End":"24:06.230","Text":"So that\u0027s it. Let\u0027s go on to other questions."}],"ID":10615},{"Watched":false,"Name":"Sand on a Conveyor Belt","Duration":"6m 7s","ChapterTopicVideoID":9227,"CourseChapterTopicPlaylistID":5429,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.410 ","End":"00:03.465","Text":"Hello. In this question,"},{"Start":"00:03.465 ","End":"00:08.965","Text":"we\u0027re given a conveyor belt which is moving at an initial velocity V_0,"},{"Start":"00:08.965 ","End":"00:13.950","Text":"and there\u0027s some funnel with sand falling onto"},{"Start":"00:13.950 ","End":"00:20.680","Text":"the conveyor belt at a constant rate of dm by dt, which equals At."},{"Start":"00:20.870 ","End":"00:23.880","Text":"Now, there\u0027s some force which is pulling"},{"Start":"00:23.880 ","End":"00:26.140","Text":"the conveyor belt and we don\u0027t know what this force is."},{"Start":"00:26.140 ","End":"00:28.875","Text":"Our first question is, what force F,"},{"Start":"00:28.875 ","End":"00:33.850","Text":"would we need in order for the conveyor belt to travel at a constant velocity."},{"Start":"00:34.400 ","End":"00:38.105","Text":"Let\u0027s see how we\u0027re going to solve this question,"},{"Start":"00:38.105 ","End":"00:42.365","Text":"so because we\u0027re dealing with sand falling onto a conveyor belt,"},{"Start":"00:42.365 ","End":"00:44.790","Text":"it\u0027s a case of accretion."},{"Start":"00:44.790 ","End":"00:48.785","Text":"It\u0027s a similar question to when rain was falling into the cut."},{"Start":"00:48.785 ","End":"00:53.780","Text":"What we\u0027re going to do, we\u0027re going to use our equation of sum of"},{"Start":"00:53.780 ","End":"01:02.405","Text":"all the external forces is equal to mass as a function of time multiplied by dv by dt,"},{"Start":"01:02.405 ","End":"01:06.140","Text":"the acceleration plus V_relative,"},{"Start":"01:06.140 ","End":"01:11.265","Text":"which here is U multiplied by dm by dt,"},{"Start":"01:11.265 ","End":"01:14.560","Text":"our rate of accretion."},{"Start":"01:14.690 ","End":"01:19.179","Text":"Now because this is a case of accretion and not ejection,"},{"Start":"01:19.179 ","End":"01:22.520","Text":"U or V_relative isn\u0027t given to us."},{"Start":"01:22.520 ","End":"01:24.545","Text":"We have to figure out what it is."},{"Start":"01:24.545 ","End":"01:28.729","Text":"I\u0027ll change colors."},{"Start":"01:28.729 ","End":"01:30.500","Text":"We\u0027re going to say that,"},{"Start":"01:30.500 ","End":"01:39.810","Text":"U is equal to the velocity of the sand relative to the ground,"},{"Start":"01:39.810 ","End":"01:45.530","Text":"so V_sand minus the velocity of the conveyor belt."},{"Start":"01:45.530 ","End":"01:48.965","Text":"I\u0027ll just write velocity of the belt,"},{"Start":"01:48.965 ","End":"01:56.530","Text":"because we can see the sand is falling along the y-axis in a straight line."},{"Start":"01:56.540 ","End":"02:00.360","Text":"We\u0027re dealing right now with the x-axis,"},{"Start":"02:00.360 ","End":"02:07.170","Text":"so we can say that the velocity of the sand relative to the x-axis is going to equal 0,"},{"Start":"02:07.170 ","End":"02:11.150","Text":"and the velocity of the belt relative to the x-axis,"},{"Start":"02:11.150 ","End":"02:17.300","Text":"we can see here is V_0, so negative V_0."},{"Start":"02:17.300 ","End":"02:22.235","Text":"Now a little note with why I wrote V_0 is because what we\u0027re being asked is,"},{"Start":"02:22.235 ","End":"02:26.935","Text":"what force would we need in order for the conveyor belt to travel at a constant velocity."},{"Start":"02:26.935 ","End":"02:29.000","Text":"If our initial velocity is V_0,"},{"Start":"02:29.000 ","End":"02:34.400","Text":"in order to keep our conveyor belt traveling at a constant velocity V_0."},{"Start":"02:34.400 ","End":"02:40.850","Text":"We\u0027re going to write here that velocity of the belt is just equal to V_0."},{"Start":"02:40.850 ","End":"02:44.970","Text":"That represents a constant velocity."},{"Start":"02:45.800 ","End":"02:49.765","Text":"Next we\u0027re going to look at our dm by dt."},{"Start":"02:49.765 ","End":"02:55.460","Text":"Now we know that because the sand is falling onto the conveyor belt,"},{"Start":"02:55.460 ","End":"02:58.250","Text":"so mass is being added onto the conveyor belt."},{"Start":"02:58.250 ","End":"03:02.765","Text":"We know that our dm by dt has to be a negative number,"},{"Start":"03:02.765 ","End":"03:07.735","Text":"so all we have to do is just add a negative over here when we put it into the equation."},{"Start":"03:07.735 ","End":"03:09.630","Text":"Let\u0027s write out this equation,"},{"Start":"03:09.630 ","End":"03:13.370","Text":"so the sum of all of our external forces is just going to be equal"},{"Start":"03:13.370 ","End":"03:18.185","Text":"to F. Because we\u0027re being asked this F is,"},{"Start":"03:18.185 ","End":"03:19.475","Text":"this external force,"},{"Start":"03:19.475 ","End":"03:22.970","Text":"that will keep the conveyor belt moving at a constant velocity, V_0,"},{"Start":"03:22.970 ","End":"03:29.885","Text":"so sum of F external is just equal to F. Then we have equals to"},{"Start":"03:29.885 ","End":"03:37.190","Text":"mass as a function of time multiplied by dv by dt."},{"Start":"03:37.190 ","End":"03:40.820","Text":"Now, dv by dt is going to equal 0 because we\u0027re specifically"},{"Start":"03:40.820 ","End":"03:44.720","Text":"told that we want this at a constant velocity,"},{"Start":"03:44.720 ","End":"03:47.480","Text":"which means that acceleration is 0."},{"Start":"03:47.480 ","End":"03:53.040","Text":"M(t) multiplied by dv by dt is just going to be equal to 0,"},{"Start":"03:53.040 ","End":"03:55.485","Text":"and then we have to add plus,"},{"Start":"03:55.485 ","End":"03:57.165","Text":"and then our value for U,"},{"Start":"03:57.165 ","End":"03:59.195","Text":"now our value for you is this,"},{"Start":"03:59.195 ","End":"04:01.675","Text":"which is negative V_0."},{"Start":"04:01.675 ","End":"04:05.710","Text":"Then we\u0027re multiplying by dm by dt,"},{"Start":"04:05.710 ","End":"04:10.465","Text":"which is going to be negative because it\u0027s accretion At,"},{"Start":"04:10.465 ","End":"04:13.345","Text":"so At and negative because it\u0027s accretion."},{"Start":"04:13.345 ","End":"04:21.650","Text":"Then what we\u0027re going to get is that our force is equal to V_0 multiplied by At,"},{"Start":"04:21.650 ","End":"04:24.910","Text":"because a negative times a negative is a positive."},{"Start":"04:24.910 ","End":"04:28.040","Text":"This is the answer for our Question number 1,"},{"Start":"04:28.040 ","End":"04:31.650","Text":"and now let\u0027s go on to Question number 2."},{"Start":"04:31.910 ","End":"04:39.185","Text":"Now the next question that we\u0027re being asked is how much energy is invested per second?"},{"Start":"04:39.185 ","End":"04:45.350","Text":"That means, basically the change of energy per second,"},{"Start":"04:45.350 ","End":"04:47.510","Text":"so dE by dt,"},{"Start":"04:47.510 ","End":"04:50.405","Text":"which is equal to the power,"},{"Start":"04:50.405 ","End":"04:59.360","Text":"which is equal to the force multiplied by the velocity.product."},{"Start":"04:59.360 ","End":"05:01.295","Text":"Now, in our case,"},{"Start":"05:01.295 ","End":"05:04.940","Text":"because our velocity is just in 1 direction,"},{"Start":"05:04.940 ","End":"05:07.640","Text":"just on the x-direction."},{"Start":"05:07.640 ","End":"05:10.730","Text":"Therefore is our force from Question number 1,"},{"Start":"05:10.730 ","End":"05:16.325","Text":"we can see that this.product between these vector quantities is just going to be"},{"Start":"05:16.325 ","End":"05:23.985","Text":"multiplying by the size of each equation of each variable over here."},{"Start":"05:23.985 ","End":"05:29.445","Text":"Our F is just going to be V_0 multiplied by"},{"Start":"05:29.445 ","End":"05:36.390","Text":"At and then our V is just V_0 multiplied by V_0."},{"Start":"05:36.390 ","End":"05:40.980","Text":"Therefore, our final answer of dE by dt,"},{"Start":"05:40.980 ","End":"05:44.870","Text":"how much energy is invested per second,"},{"Start":"05:44.870 ","End":"05:46.580","Text":"which is also equal to the power,"},{"Start":"05:46.580 ","End":"05:51.930","Text":"is going to be V_0^2 multiplied by At."},{"Start":"05:53.990 ","End":"05:57.005","Text":"That is the end of our questions."},{"Start":"05:57.005 ","End":"06:03.230","Text":"I\u0027ve put the important equations that we\u0027ve used in these turquoise boxes."},{"Start":"06:03.230 ","End":"06:07.949","Text":"The final answers have been squared in the red boxes."}],"ID":10616},{"Watched":false,"Name":"Balloon","Duration":"21m 47s","ChapterTopicVideoID":9228,"CourseChapterTopicPlaylistID":5429,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.160","Text":"Hello. In this lesson,"},{"Start":"00:02.160 ","End":"00:04.445","Text":"we\u0027re speaking of a balloon of mass M,"},{"Start":"00:04.445 ","End":"00:06.140","Text":"which is filled with gas."},{"Start":"00:06.140 ","End":"00:09.945","Text":"3/4 of the mass of the balloon is the mass of the gas."},{"Start":"00:09.945 ","End":"00:12.780","Text":"The balloon is released from rest and gas exits"},{"Start":"00:12.780 ","End":"00:16.990","Text":"the balloon at a velocity of u relative to the balloon."},{"Start":"00:17.120 ","End":"00:26.235","Text":"Here we have a balloon and gas is leaving the balloon at a velocity of u."},{"Start":"00:26.235 ","End":"00:30.795","Text":"The balloon accelerates upwards along a straight line at an acceleration"},{"Start":"00:30.795 ","End":"00:37.253","Text":"of 1/2g so acceleration equals 1/2g,"},{"Start":"00:37.253 ","End":"00:39.030","Text":"and then our first question is,"},{"Start":"00:39.030 ","End":"00:41.895","Text":"at what rate is the gas emitted?"},{"Start":"00:41.895 ","End":"00:44.685","Text":"Now usually, we\u0027re given this,"},{"Start":"00:44.685 ","End":"00:48.060","Text":"so we\u0027re trying to find our dm by dt."},{"Start":"00:48.060 ","End":"00:49.693","Text":"This is usually given,"},{"Start":"00:49.693 ","End":"00:51.850","Text":"and they tell us that it\u0027s Alpha."},{"Start":"00:51.850 ","End":"00:53.795","Text":"However, in this question,"},{"Start":"00:53.795 ","End":"00:55.850","Text":"we\u0027re not given our dm by dt,"},{"Start":"00:55.850 ","End":"01:00.800","Text":"which means that we have to find it given this information."},{"Start":"01:00.800 ","End":"01:02.755","Text":"Let\u0027s see how we do that."},{"Start":"01:02.755 ","End":"01:05.460","Text":"We have this equation over here,"},{"Start":"01:05.460 ","End":"01:08.325","Text":"so let\u0027s see how we use this."},{"Start":"01:08.325 ","End":"01:12.990","Text":"First of all, we know that dv is in the y-axis."},{"Start":"01:12.990 ","End":"01:16.940","Text":"Because we\u0027re going to say that we\u0027re only moving in the y-axis and this is"},{"Start":"01:16.940 ","End":"01:22.590","Text":"the positive direction by dt is equal to."},{"Start":"01:22.590 ","End":"01:26.070","Text":"This is our acceleration in the y-axis, which is 1/2g."},{"Start":"01:26.070 ","End":"01:32.745","Text":"We can write that down over here and then let\u0027s speak about the next section."},{"Start":"01:32.745 ","End":"01:34.505","Text":"Our M is a function of time,"},{"Start":"01:34.505 ","End":"01:35.855","Text":"we\u0027ll get there in a minute."},{"Start":"01:35.855 ","End":"01:38.045","Text":"Now let\u0027s just speak about our dm by dt,"},{"Start":"01:38.045 ","End":"01:40.940","Text":"which is what we actually want to find."},{"Start":"01:40.940 ","End":"01:49.175","Text":"Our dm by dt is a positive number because we\u0027re speaking about emission."},{"Start":"01:49.175 ","End":"01:52.520","Text":"We know that our dm by dt is going to be positive because that\u0027s"},{"Start":"01:52.520 ","End":"01:57.035","Text":"how we defined it when we derive the equation."},{"Start":"01:57.035 ","End":"02:06.630","Text":"However, it\u0027s equal to the negative dm(t) by dt."},{"Start":"02:06.880 ","End":"02:10.085","Text":"Let\u0027s just speak about this for 1 second."},{"Start":"02:10.085 ","End":"02:14.420","Text":"The mass exiting is going to be positive because that\u0027s how we defined it."},{"Start":"02:14.420 ","End":"02:16.955","Text":"This means that the mass,"},{"Start":"02:16.955 ","End":"02:18.665","Text":"which is changing over time,"},{"Start":"02:18.665 ","End":"02:20.090","Text":"is going to become less and less,"},{"Start":"02:20.090 ","End":"02:21.515","Text":"and less it\u0027s decreasing,"},{"Start":"02:21.515 ","End":"02:25.230","Text":"so we have to put a minus in front of it."},{"Start":"02:25.840 ","End":"02:28.715","Text":"Now, the next thing is our u,"},{"Start":"02:28.715 ","End":"02:30.275","Text":"which is our v relative."},{"Start":"02:30.275 ","End":"02:38.325","Text":"Here is just u so we have that Vrel is equal to u."},{"Start":"02:38.325 ","End":"02:41.195","Text":"Actually, to make this slightly clear,"},{"Start":"02:41.195 ","End":"02:45.420","Text":"let\u0027s say u_0, and this is u_0."},{"Start":"02:47.360 ","End":"02:51.980","Text":"Now we can put everything together into our equation."},{"Start":"02:51.980 ","End":"02:56.945","Text":"The sum of our external forces in the y-direction is going to be"},{"Start":"02:56.945 ","End":"03:03.830","Text":"negative m as a function of t multiplied by g. Why is this?"},{"Start":"03:03.830 ","End":"03:08.315","Text":"Because we have gravity working downwards and as we know gravity,"},{"Start":"03:08.315 ","End":"03:10.670","Text":"the force acting downwards will be"},{"Start":"03:10.670 ","End":"03:14.885","Text":"mass times gravity and the mass is changing as a function of time,"},{"Start":"03:14.885 ","End":"03:18.965","Text":"then because we said that the upwards direction was the positive direction,"},{"Start":"03:18.965 ","End":"03:23.095","Text":"so the downwards direction is the negative direction,"},{"Start":"03:23.095 ","End":"03:24.875","Text":"so we have a negative here."},{"Start":"03:24.875 ","End":"03:33.695","Text":"Then that equals m as a function of time multiplied by acceleration,"},{"Start":"03:33.695 ","End":"03:35.855","Text":"dv by dt,"},{"Start":"03:35.855 ","End":"03:43.035","Text":"which is multiplied by g over 2 and then we have negative our Vrel,"},{"Start":"03:43.035 ","End":"03:48.405","Text":"which is u_0 multiplied by dm by dt,"},{"Start":"03:48.405 ","End":"03:54.410","Text":"which is negative dm over dt."},{"Start":"03:54.410 ","End":"04:00.530","Text":"Now what I\u0027m left with is a differential equation because the only unknown that I have is"},{"Start":"04:00.530 ","End":"04:03.710","Text":"my M as a function of t. Now what"},{"Start":"04:03.710 ","End":"04:07.745","Text":"I\u0027m going to do is I\u0027m going to rearrange this equation,"},{"Start":"04:07.745 ","End":"04:10.115","Text":"and then I\u0027m going to solve it."},{"Start":"04:10.115 ","End":"04:12.215","Text":"Let\u0027s see how we do this."},{"Start":"04:12.215 ","End":"04:19.030","Text":"We can have negative 3/2g multiplied by m(t)."},{"Start":"04:19.030 ","End":"04:24.338","Text":"I just took this to the other side,"},{"Start":"04:24.338 ","End":"04:30.815","Text":"and then that is going to be equal to these becoming positives,"},{"Start":"04:30.815 ","End":"04:39.070","Text":"so it\u0027s going to be u_0 multiplied by dm as a function of t divided by dt."},{"Start":"04:39.070 ","End":"04:45.665","Text":"Now what I can do is rearrange to get all of my variables,"},{"Start":"04:45.665 ","End":"04:51.065","Text":"my empty M as a function of t variables on the side where I have my dm by dt,"},{"Start":"04:51.065 ","End":"04:55.285","Text":"and all the other variables on the side of my dt."},{"Start":"04:55.285 ","End":"05:06.195","Text":"If I divide both sides by mt and u_0 and multiply both sides by dt, let\u0027s see."},{"Start":"05:06.195 ","End":"05:14.975","Text":"If I multiply both sides by dt divided by u_0m as a function of t,"},{"Start":"05:14.975 ","End":"05:17.255","Text":"then what I will get is"},{"Start":"05:17.255 ","End":"05:25.710","Text":"negative 3/2g over u_0 multiplied by dt,"},{"Start":"05:25.710 ","End":"05:32.580","Text":"which will equal to dm as a function of"},{"Start":"05:32.580 ","End":"05:36.615","Text":"t/m as a function of"},{"Start":"05:36.615 ","End":"05:44.630","Text":"t. Now all I have to do is put in my bounds for the integration."},{"Start":"05:44.630 ","End":"05:50.530","Text":"My t is going to be starting from t=0 until t equals the sum of t"},{"Start":"05:50.530 ","End":"05:56.450","Text":"and my dm is going to be starting from my starting mass,"},{"Start":"05:56.450 ","End":"06:02.210","Text":"which is big M because that is what was given to us in the question, let\u0027s take a look."},{"Start":"06:02.210 ","End":"06:03.955","Text":"A balloon of mass m,"},{"Start":"06:03.955 ","End":"06:06.940","Text":"so our starting mass is m,"},{"Start":"06:06.940 ","End":"06:13.410","Text":"and it will end at the mass as a function of t over there."},{"Start":"06:13.510 ","End":"06:17.345","Text":"All that\u0027s left is that we integrate."},{"Start":"06:17.345 ","End":"06:23.705","Text":"Now, this is a relatively simple integration because here on the side of dt,"},{"Start":"06:23.705 ","End":"06:25.715","Text":"we don\u0027t have any variable of t,"},{"Start":"06:25.715 ","End":"06:32.100","Text":"so it just becomes negative 3/2 multiplied by g,"},{"Start":"06:32.100 ","End":"06:38.030","Text":"over u_0 multiplied by t. Because when we substitute in our bounds,"},{"Start":"06:38.030 ","End":"06:40.205","Text":"the 0 crosses out,"},{"Start":"06:40.205 ","End":"06:49.090","Text":"and then this will equal to ln of m(t)/m."},{"Start":"06:49.090 ","End":"06:55.925","Text":"How did I get this? Because when we have this integral over here,"},{"Start":"06:55.925 ","End":"07:02.510","Text":"becomes ln of m(t),"},{"Start":"07:02.510 ","End":"07:06.670","Text":"and then it\u0027s between m and m(t)."},{"Start":"07:06.670 ","End":"07:08.765","Text":"Then when we substitute this n,"},{"Start":"07:08.765 ","End":"07:16.925","Text":"it becomes ln of m(t) negative ln"},{"Start":"07:16.925 ","End":"07:20.990","Text":"of m. We can"},{"Start":"07:20.990 ","End":"07:25.535","Text":"do circular brackets over here because we\u0027re dealing with only positive numbers."},{"Start":"07:25.535 ","End":"07:36.195","Text":"Then a line of something minus ln of another thing will become ln of m(t) divided by m,"},{"Start":"07:36.195 ","End":"07:38.415","Text":"so that\u0027s over this here."},{"Start":"07:38.415 ","End":"07:40.275","Text":"Now, as we know,"},{"Start":"07:40.275 ","End":"07:44.955","Text":"I want to get my dm by dt variable."},{"Start":"07:44.955 ","End":"07:50.020","Text":"What I\u0027m going to start is by isolating out my mt."},{"Start":"07:50.020 ","End":"07:55.985","Text":"Then what I can do is I can raise both sides by e,"},{"Start":"07:55.985 ","End":"07:59.320","Text":"then my e and my ln will cross out,"},{"Start":"07:59.320 ","End":"08:02.345","Text":"and then I can just multiply both sides by m,"},{"Start":"08:02.345 ","End":"08:10.415","Text":"and then I\u0027ll get that my m(t) is equal to negative m multiplied by"},{"Start":"08:10.415 ","End":"08:18.800","Text":"e to the power of negative 3g/2u_0 multiplied by t. Sorry,"},{"Start":"08:18.800 ","End":"08:19.940","Text":"without this negative here,"},{"Start":"08:19.940 ","End":"08:21.680","Text":"this is a positive."},{"Start":"08:21.680 ","End":"08:25.610","Text":"Now what I want to get as my dm by dt."},{"Start":"08:25.610 ","End":"08:28.880","Text":"What is my dm by dt?"},{"Start":"08:28.880 ","End":"08:36.660","Text":"Now my dm by dt as written over here,"},{"Start":"08:36.660 ","End":"08:45.540","Text":"by dt is equal to my negative dm as a function of t divided by dt."},{"Start":"08:45.540 ","End":"08:50.260","Text":"This is just what\u0027s written over here."},{"Start":"08:50.560 ","End":"08:55.895","Text":"That means I need to take the derivative and multiply it by negative 1,"},{"Start":"08:55.895 ","End":"09:01.100","Text":"of this m(t) in order to get my dm by dt."},{"Start":"09:01.100 ","End":"09:08.900","Text":"Therefore, my dm by dt is equal to the negative derivative,"},{"Start":"09:08.900 ","End":"09:17.570","Text":"which is going to be negative 3g/2u_0 multiplied by m,"},{"Start":"09:17.570 ","End":"09:22.645","Text":"multiplied by e to the negative 3g/2u_0"},{"Start":"09:22.645 ","End":"09:30.255","Text":"multiplied by t. That\u0027s it for question number 1."},{"Start":"09:30.255 ","End":"09:33.145","Text":"Now, for question number 2,"},{"Start":"09:33.145 ","End":"09:36.245","Text":"what is the maximum height the balloon will reach?"},{"Start":"09:36.245 ","End":"09:38.450","Text":"Let\u0027s see how we do that."},{"Start":"09:38.450 ","End":"09:41.330","Text":"Now, this section of the question is"},{"Start":"09:41.330 ","End":"09:44.075","Text":"a little bit tricky and you really have to think about it."},{"Start":"09:44.075 ","End":"09:48.125","Text":"What is happening over here is that we have our balloon,"},{"Start":"09:48.125 ","End":"09:50.945","Text":"which starts over here,"},{"Start":"09:50.945 ","End":"09:52.760","Text":"and gases coming out."},{"Start":"09:52.760 ","End":"09:57.290","Text":"Then the gas will finish coming out at"},{"Start":"09:57.290 ","End":"10:03.710","Text":"around this height and then even after the gas is finished coming out,"},{"Start":"10:03.710 ","End":"10:08.620","Text":"the balloon will still carry on traveling upwards a bit more."},{"Start":"10:08.620 ","End":"10:14.805","Text":"This is just going to be thought of as like a vertical throw."},{"Start":"10:14.805 ","End":"10:18.170","Text":"As if someone\u0027s just throwing the balloon upwards."},{"Start":"10:18.170 ","End":"10:20.210","Text":"The section over here,"},{"Start":"10:20.210 ","End":"10:26.500","Text":"which we\u0027re going to call dy number 2."},{"Start":"10:26.500 ","End":"10:29.745","Text":"Here, there will be a constant acceleration,"},{"Start":"10:29.745 ","End":"10:33.950","Text":"and here over dy number 1"},{"Start":"10:33.950 ","End":"10:40.020","Text":"so the acceleration is constantly changing because the mass is changing."},{"Start":"10:40.960 ","End":"10:46.680","Text":"Let\u0027s see how we do this mathematically and before I say"},{"Start":"10:46.680 ","End":"10:54.110","Text":"that this height will be the maximum height, h_max."},{"Start":"10:54.110 ","End":"10:58.285","Text":"In my first section, my Delta y_1."},{"Start":"10:58.285 ","End":"10:59.665","Text":"We can say that,"},{"Start":"10:59.665 ","End":"11:04.623","Text":"the velocity as a function of time is equal to our starting velocity,"},{"Start":"11:04.623 ","End":"11:10.720","Text":"and it starts from rest so it\u0027s 0 plus acceleration multiplied by t. Here,"},{"Start":"11:10.720 ","End":"11:19.225","Text":"acceleration is g over 2.5g multiplied by t. This is our velocity in our first section."},{"Start":"11:19.225 ","End":"11:21.460","Text":"Let\u0027s make this a bit clearer."},{"Start":"11:21.460 ","End":"11:23.770","Text":"This is that."},{"Start":"11:23.770 ","End":"11:26.800","Text":"Now, from the previous question,"},{"Start":"11:26.800 ","End":"11:34.210","Text":"we got that mass as a function of time was equal to m_e,"},{"Start":"11:34.210 ","End":"11:42.550","Text":"multiply to the power of negative 3g over 2u_0 multiplied by t. Now,"},{"Start":"11:42.550 ","End":"11:45.040","Text":"what we want to do with this equation is we want to find"},{"Start":"11:45.040 ","End":"11:49.330","Text":"the time at which the balloon becomes empty,"},{"Start":"11:49.330 ","End":"11:52.705","Text":"at which there\u0027s no more gas in the balloon."},{"Start":"11:52.705 ","End":"11:55.990","Text":"We\u0027re going to call that time t_1,"},{"Start":"11:55.990 ","End":"11:58.105","Text":"and that\u0027s over here."},{"Start":"11:58.105 ","End":"12:00.790","Text":"We know that the balloon,"},{"Start":"12:00.790 ","End":"12:02.200","Text":"because we\u0027re told in the question,"},{"Start":"12:02.200 ","End":"12:05.470","Text":"3-quarters of the mass of the balloon is the mass of the gas,"},{"Start":"12:05.470 ","End":"12:08.485","Text":"which means that 1-quarter of m,"},{"Start":"12:08.485 ","End":"12:10.210","Text":"is the mass of the balloon."},{"Start":"12:10.210 ","End":"12:12.280","Text":"We want to get at t_1."},{"Start":"12:12.280 ","End":"12:13.510","Text":"Let\u0027s substitute in here,"},{"Start":"12:13.510 ","End":"12:19.525","Text":"t_1 has to equal to m over 4 because that will be the mass of the balloon,"},{"Start":"12:19.525 ","End":"12:21.820","Text":"which means that there\u0027s no more gas in the balloon."},{"Start":"12:21.820 ","End":"12:27.085","Text":"Now, what we want to do is we want to isolate out our t_1."},{"Start":"12:27.085 ","End":"12:30.340","Text":"We can see that on both sides of the equation we have an m,"},{"Start":"12:30.340 ","End":"12:32.020","Text":"so we can cross them out."},{"Start":"12:32.020 ","End":"12:35.575","Text":"Then we can Lan both sides."},{"Start":"12:35.575 ","End":"12:40.000","Text":"On this side we\u0027ll get rid of the e and here we\u0027ll have Lan of a quarter."},{"Start":"12:40.000 ","End":"12:42.400","Text":"Now, as a side note,"},{"Start":"12:42.400 ","End":"12:48.160","Text":"Lan of 1 over 4 is equal to if through the laws of Lan is"},{"Start":"12:48.160 ","End":"12:54.770","Text":"equal to Lan of 1 minus Lan of 4."},{"Start":"12:55.170 ","End":"12:58.120","Text":"Here, when we\u0027ve gotten rid of the e,"},{"Start":"12:58.120 ","End":"13:04.810","Text":"we\u0027ll have negative 3g over 2u_0 multiplied by"},{"Start":"13:04.810 ","End":"13:12.175","Text":"t1 is equal to Lan of 1 minus Lan of 4,"},{"Start":"13:12.175 ","End":"13:13.825","Text":"which is what we saw over here."},{"Start":"13:13.825 ","End":"13:15.415","Text":"Now, if we multiply,"},{"Start":"13:15.415 ","End":"13:18.400","Text":"if we want to get rid of this negative,"},{"Start":"13:18.400 ","End":"13:20.320","Text":"so I\u0027ll multiply both sides by negative 1,"},{"Start":"13:20.320 ","End":"13:22.045","Text":"which means that this will become a positive,"},{"Start":"13:22.045 ","End":"13:24.340","Text":"this a negative and this a positive."},{"Start":"13:24.340 ","End":"13:33.040","Text":"Then this will equal to the side Lan of 4 minus Lan of 1,"},{"Start":"13:33.040 ","End":"13:37.180","Text":"which will equal to Lan of 4 over 1,"},{"Start":"13:37.180 ","End":"13:41.035","Text":"which is equal to Lan of 4."},{"Start":"13:41.035 ","End":"13:46.615","Text":"Then we want to isolate out our t_1 so I\u0027ve t_1 is equal to"},{"Start":"13:46.615 ","End":"13:54.260","Text":"2u_0 over 3g multiplied by Lan of 4."},{"Start":"13:54.510 ","End":"13:58.915","Text":"Now, what we\u0027re going to do is with this information of our t_1,"},{"Start":"13:58.915 ","End":"14:01.900","Text":"we\u0027re going to find the velocity that the balloon"},{"Start":"14:01.900 ","End":"14:04.960","Text":"is traveling in at the moment when all the gas,"},{"Start":"14:04.960 ","End":"14:07.540","Text":"when the last of the gas has left the balloon."},{"Start":"14:07.540 ","End":"14:10.690","Text":"We\u0027re going to substitute that into this equation for"},{"Start":"14:10.690 ","End":"14:15.610","Text":"our v1t is equal to 0 plus g over 2t."},{"Start":"14:15.610 ","End":"14:21.160","Text":"Then we\u0027re going to have that our velocity at this point,"},{"Start":"14:21.160 ","End":"14:24.224","Text":"at time t_1,"},{"Start":"14:24.224 ","End":"14:26.665","Text":"when this is what t_1 is,"},{"Start":"14:26.665 ","End":"14:31.465","Text":"is going to be equal to g over 2 multiplied by our t,"},{"Start":"14:31.465 ","End":"14:38.740","Text":"which is multiplied by 2u_0 over 3g Lan of 4."},{"Start":"14:38.740 ","End":"14:41.020","Text":"Then we can cross out this,"},{"Start":"14:41.020 ","End":"14:46.285","Text":"these 2s, and then we\u0027re left with that velocity."},{"Start":"14:46.285 ","End":"14:50.649","Text":"Now, if we go back to our equations of motion,"},{"Start":"14:50.649 ","End":"14:55.030","Text":"we know that there\u0027s an equation for our position so because we\u0027re on the y-axis,"},{"Start":"14:55.030 ","End":"15:00.820","Text":"so our position is y as a function of time and this is equal to"},{"Start":"15:00.820 ","End":"15:08.050","Text":"our initial velocity multiplied by t plus 1.5A_t^2."},{"Start":"15:08.050 ","End":"15:11.110","Text":"Because our initial velocity is 0,"},{"Start":"15:11.110 ","End":"15:18.820","Text":"this is 0 so then we can say that our position as a function of time,"},{"Start":"15:18.820 ","End":"15:25.725","Text":"when all of the gas has left is equal to 1.5A_t."},{"Start":"15:25.725 ","End":"15:32.580","Text":"We know that our acceleration is g over 2 so that is going to be equal to"},{"Start":"15:32.580 ","End":"15:40.720","Text":"g over 4 because our acceleration is g over 2 multiplied by a 1.5 multiplied by our t,"},{"Start":"15:40.720 ","End":"15:42.490","Text":"which is our t_1^2."},{"Start":"15:42.490 ","End":"15:52.015","Text":"Multiplied by 2u_0 over 3g Lan of 4^2."},{"Start":"15:52.015 ","End":"15:54.040","Text":"Now, what we have here, well,"},{"Start":"15:54.040 ","End":"15:59.360","Text":"we works out our y of t is our height over here."},{"Start":"15:59.820 ","End":"16:03.925","Text":"This is going to be y of t_1,"},{"Start":"16:03.925 ","End":"16:11.360","Text":"so our height is equal to our y of t_1, over here."},{"Start":"16:11.360 ","End":"16:15.330","Text":"Now, we\u0027ve finished with all the information"},{"Start":"16:15.330 ","End":"16:18.885","Text":"that we have for section when our mass is changing,"},{"Start":"16:18.885 ","End":"16:22.935","Text":"when gases being emitted from the balloon and we found out,"},{"Start":"16:22.935 ","End":"16:25.960","Text":"our height when the balloon is empty."},{"Start":"16:25.960 ","End":"16:28.810","Text":"Now, we have to find out this extra heights,"},{"Start":"16:28.810 ","End":"16:30.205","Text":"this increase in height,"},{"Start":"16:30.205 ","End":"16:33.430","Text":"which occurs when the balloon is now empty,"},{"Start":"16:33.430 ","End":"16:35.890","Text":"but because it\u0027s still has a velocity,"},{"Start":"16:35.890 ","End":"16:38.095","Text":"it\u0027s still moving upwards."},{"Start":"16:38.095 ","End":"16:42.370","Text":"Now again, going back to our kinematic equations, if you remember,"},{"Start":"16:42.370 ","End":"16:49.945","Text":"there\u0027s an equation that goes v^2 equals u^2 plus 2a_s,"},{"Start":"16:49.945 ","End":"16:56.690","Text":"where s is our position so our s is our y."},{"Start":"16:57.510 ","End":"17:00.850","Text":"Let\u0027s work out now the upwards motion of"},{"Start":"17:00.850 ","End":"17:05.660","Text":"the empty balloon until the maximum height is reached."},{"Start":"17:06.330 ","End":"17:11.065","Text":"We\u0027re going to have our V^2, which is our final velocity."},{"Start":"17:11.065 ","End":"17:12.775","Text":"We don\u0027t know that yet."},{"Start":"17:12.775 ","End":"17:18.130","Text":"Which is going to equal our u ^2, our initial velocity,"},{"Start":"17:18.130 ","End":"17:20.275","Text":"which is this over here,"},{"Start":"17:20.275 ","End":"17:24.190","Text":"our v_1(t_1), which is the velocity of the empty balloon."},{"Start":"17:24.190 ","End":"17:26.155","Text":"The moment after it empty."},{"Start":"17:26.155 ","End":"17:29.080","Text":"That\u0027s our starting velocity."},{"Start":"17:29.080 ","End":"17:32.920","Text":"The velocity at this point over here is"},{"Start":"17:32.920 ","End":"17:38.300","Text":"our starting velocity for our 2nd section for our delta y_2."},{"Start":"17:39.450 ","End":"17:42.580","Text":"We could also cross all of our gs over here."},{"Start":"17:42.580 ","End":"17:45.970","Text":"It\u0027s going to be equal to u_0 over"},{"Start":"17:45.970 ","End":"17:53.290","Text":"3 Lan of 4 and this is squared plus 2a_s."},{"Start":"17:53.290 ","End":"17:55.435","Text":"Plus 2, now note,"},{"Start":"17:55.435 ","End":"17:59.980","Text":"i acceleration is now going to be caused by gravity,"},{"Start":"17:59.980 ","End":"18:07.285","Text":"which is pulling the balloon down because there\u0027s no more emission from the balloon,"},{"Start":"18:07.285 ","End":"18:12.160","Text":"which means that the acceleration is the force of gravity pulling it down"},{"Start":"18:12.160 ","End":"18:17.335","Text":"so a is negative g multiplied by s,"},{"Start":"18:17.335 ","End":"18:27.415","Text":"which is our y position but at delta y of 2 so multiplied by delta y_2."},{"Start":"18:27.415 ","End":"18:28.855","Text":"Now, notice that,"},{"Start":"18:28.855 ","End":"18:35.170","Text":"our v final is going to be our velocity when max height is reached."},{"Start":"18:35.170 ","End":"18:38.080","Text":"If this is our velocity when our maximum height is reached,"},{"Start":"18:38.080 ","End":"18:40.450","Text":"it means that it is 0 because remember,"},{"Start":"18:40.450 ","End":"18:45.805","Text":"our balloon reaches the maximum height and then it falls back down."},{"Start":"18:45.805 ","End":"18:47.695","Text":"It stops, and then falls back down."},{"Start":"18:47.695 ","End":"18:49.300","Text":"That means that,"},{"Start":"18:49.300 ","End":"18:55.300","Text":"this over here is 0 and then we can move"},{"Start":"18:55.300 ","End":"19:01.360","Text":"this expression of our 2_g delta y_2 to one side,"},{"Start":"19:01.360 ","End":"19:04.330","Text":"and then we can isolate out this."},{"Start":"19:04.330 ","End":"19:06.055","Text":"Then what will get?"},{"Start":"19:06.055 ","End":"19:09.370","Text":"When we move this expression to the other side of the equal side,"},{"Start":"19:09.370 ","End":"19:17.125","Text":"it will become a positive number so we\u0027ll have that 2_g multiplied by delta y_2"},{"Start":"19:17.125 ","End":"19:26.905","Text":"is equal to u_0 divided by 3 multiplied by Lan 4^2."},{"Start":"19:26.905 ","End":"19:29.740","Text":"Then that will mean that,"},{"Start":"19:29.740 ","End":"19:36.955","Text":"our delta y_2 is equal to 1 over 2_g."},{"Start":"19:36.955 ","End":"19:39.010","Text":"We divide both sides by 2_g,"},{"Start":"19:39.010 ","End":"19:44.720","Text":"multiplied by u_0 over 3 Lan of 4^2."},{"Start":"19:45.750 ","End":"19:48.850","Text":"Therefore a maximum height,"},{"Start":"19:48.850 ","End":"19:53.185","Text":"our max, which equals max height,"},{"Start":"19:53.185 ","End":"20:01.720","Text":"is equal to this delta y_2 so delta y_2 plus delta y_1,"},{"Start":"20:01.720 ","End":"20:05.920","Text":"which is equal to 1 over 2_g multiplied by"},{"Start":"20:05.920 ","End":"20:12.355","Text":"u_0 over 3 Lan of 4^2 plus our delta y_1,"},{"Start":"20:12.355 ","End":"20:14.470","Text":"which is going to be this,"},{"Start":"20:14.470 ","End":"20:19.420","Text":"which is g over 4 multiplied by 2_u_0 over"},{"Start":"20:19.420 ","End":"20:26.360","Text":"3_g multiplied by Lan 4 and this is also squared."},{"Start":"20:27.300 ","End":"20:31.870","Text":"Now, you can rearrange this expression in order to make"},{"Start":"20:31.870 ","End":"20:35.860","Text":"it slightly more readable and to look nicer,"},{"Start":"20:35.860 ","End":"20:39.310","Text":"but this is the final answer."},{"Start":"20:39.310 ","End":"20:43.600","Text":"We found now our maximum height so let\u0027s go over"},{"Start":"20:43.600 ","End":"20:48.280","Text":"the steps really quickly of how to find our maximum height."},{"Start":"20:48.280 ","End":"20:50.980","Text":"All we\u0027ve actually done is we\u0027ve found"},{"Start":"20:50.980 ","End":"20:54.580","Text":"the velocity of the balloon whilst the gas is being emitted."},{"Start":"20:54.580 ","End":"20:58.330","Text":"Then we found the time at which the balloon has emitted all of"},{"Start":"20:58.330 ","End":"21:02.815","Text":"the gas so the time at which the balloon is empty,"},{"Start":"21:02.815 ","End":"21:05.410","Text":"then we found the velocity of the balloon,"},{"Start":"21:05.410 ","End":"21:07.945","Text":"the moment or the gas was emitted."},{"Start":"21:07.945 ","End":"21:11.920","Text":"We substitute it in this time that we found into our equation of"},{"Start":"21:11.920 ","End":"21:15.640","Text":"velocity that we found at the beginning to get this"},{"Start":"21:15.640 ","End":"21:20.380","Text":"and then we figured out the position of the balloon at the moment all of the gas was"},{"Start":"21:20.380 ","End":"21:25.240","Text":"emitted so the position corresponding to this velocity at this time,"},{"Start":"21:25.240 ","End":"21:28.450","Text":"and then our second section was to find"},{"Start":"21:28.450 ","End":"21:31.870","Text":"the position of the empty balloon when the maximum height was reached."},{"Start":"21:31.870 ","End":"21:35.860","Text":"We did that by knowing that our final velocity at the maximum height,"},{"Start":"21:35.860 ","End":"21:42.295","Text":"is going to be equal to 0 and knowing that I acceleration is now only due to gravity,"},{"Start":"21:42.295 ","End":"21:48.500","Text":"and then rearranging the equation. That is it."}],"ID":10617},{"Watched":false,"Name":"Funnel and Scales","Duration":"24m 35s","ChapterTopicVideoID":9229,"CourseChapterTopicPlaylistID":5429,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.320 ","End":"00:03.000","Text":"Hello. In this question,"},{"Start":"00:03.000 ","End":"00:08.790","Text":"we\u0027re again going to be dealing with the problem of the type of a varying mass system."},{"Start":"00:08.790 ","End":"00:13.065","Text":"Over here, we\u0027re given a funnel with sand falling through it."},{"Start":"00:13.065 ","End":"00:17.535","Text":"Now, the area of the opening of the funnel is equal to A."},{"Start":"00:17.535 ","End":"00:22.005","Text":"The velocity with which the sand is falling is equal to v_0,"},{"Start":"00:22.005 ","End":"00:26.685","Text":"and the height of the funnel from the scale is equal to h. Now of course,"},{"Start":"00:26.685 ","End":"00:27.840","Text":"as the sand falls,"},{"Start":"00:27.840 ","End":"00:31.840","Text":"we can see the mass increasing on the scale."},{"Start":"00:32.330 ","End":"00:35.475","Text":"One last piece of information,"},{"Start":"00:35.475 ","End":"00:41.200","Text":"the density with which the sand is falling is Rho."},{"Start":"00:41.210 ","End":"00:49.240","Text":"Now, our first question is asking us how much sand exits through the funnel per second?"},{"Start":"00:49.400 ","End":"00:53.535","Text":"How can we answer a question like this?"},{"Start":"00:53.535 ","End":"00:59.820","Text":"We\u0027re being asked dm by dt is equal to what?"},{"Start":"00:59.820 ","End":"01:04.590","Text":"What we can do is we can take some cylinder,"},{"Start":"01:04.590 ","End":"01:08.070","Text":"so the cylinder, and it\u0027s full."},{"Start":"01:08.070 ","End":"01:13.395","Text":"We can say that the length of the cylinder is,"},{"Start":"01:13.395 ","End":"01:16.350","Text":"let\u0027s just call it dx."},{"Start":"01:16.350 ","End":"01:19.815","Text":"Now, we can say that our dm,"},{"Start":"01:19.815 ","End":"01:21.120","Text":"now, what is our dm?"},{"Start":"01:21.120 ","End":"01:23.860","Text":"It\u0027s the mass of this cylinder."},{"Start":"01:23.960 ","End":"01:29.625","Text":"Here, mass of the cylinder is equal to dm,"},{"Start":"01:29.625 ","End":"01:33.885","Text":"a small section of mass of the total amount of sand."},{"Start":"01:33.885 ","End":"01:38.745","Text":"Now, our mass is obviously equal to the density of the sand,"},{"Start":"01:38.745 ","End":"01:43.960","Text":"which is Rho, multiplied by the volume, which is dv."},{"Start":"01:44.210 ","End":"01:46.530","Text":"Now we know what Rho is."},{"Start":"01:46.530 ","End":"01:50.895","Text":"It was given to us in the question and our next question is, what is dv?"},{"Start":"01:50.895 ","End":"01:52.950","Text":"We can say that our dv,"},{"Start":"01:52.950 ","End":"01:54.915","Text":"our change in volume,"},{"Start":"01:54.915 ","End":"01:57.855","Text":"is equal to the area,"},{"Start":"01:57.855 ","End":"02:00.030","Text":"which we know is A,"},{"Start":"02:00.030 ","End":"02:02.580","Text":"multiplied by the length,"},{"Start":"02:02.580 ","End":"02:05.530","Text":"which we know is dx."},{"Start":"02:07.310 ","End":"02:09.750","Text":"Now, we know what our dv is,"},{"Start":"02:09.750 ","End":"02:11.325","Text":"so we can go back to our dm,"},{"Start":"02:11.325 ","End":"02:12.885","Text":"and substitute that in."},{"Start":"02:12.885 ","End":"02:15.887","Text":"We get that our mass,"},{"Start":"02:15.887 ","End":"02:21.820","Text":"our dm is equal to Rho multiplied by A, multiplied by dx."},{"Start":"02:22.790 ","End":"02:27.490","Text":"Now, in order to find our dm by dt,"},{"Start":"02:29.180 ","End":"02:34.125","Text":"we can say that it equals to dm,"},{"Start":"02:34.125 ","End":"02:40.210","Text":"which is Rho Adx divided by dt."},{"Start":"02:40.210 ","End":"02:43.370","Text":"Normal division over here."},{"Start":"02:43.370 ","End":"02:46.339","Text":"Now we know what can we write here?"},{"Start":"02:46.339 ","End":"02:51.215","Text":"That our dx by dt just simply velocity,"},{"Start":"02:51.215 ","End":"02:55.455","Text":"the change in x over the change in time is our velocity."},{"Start":"02:55.455 ","End":"02:58.380","Text":"This is just equal to and our velocity is v_0,"},{"Start":"02:58.380 ","End":"03:03.540","Text":"so Rho A multiplied by v_0,"},{"Start":"03:03.540 ","End":"03:07.470","Text":"and that\u0027s it for question number 1."},{"Start":"03:07.470 ","End":"03:10.395","Text":"Now, our 2nd question is asking,"},{"Start":"03:10.395 ","End":"03:13.815","Text":"what is the velocity of the sand when it hits the scales?"},{"Start":"03:13.815 ","End":"03:19.290","Text":"That means that we\u0027re taking the same cylinder over here,"},{"Start":"03:19.290 ","End":"03:22.680","Text":"but we\u0027re asking what its velocity is when it"},{"Start":"03:22.680 ","End":"03:26.640","Text":"will get to right over here when it hits the scales."},{"Start":"03:26.640 ","End":"03:31.900","Text":"What will be its vf, it\u0027s v final?"},{"Start":"03:32.750 ","End":"03:35.625","Text":"Now, there are 2 ways that we can do this,"},{"Start":"03:35.625 ","End":"03:40.620","Text":"either by using a kinematic equations or by using energy conservation."},{"Start":"03:40.620 ","End":"03:45.540","Text":"So the easiest way in this example is to use energy conservation."},{"Start":"03:45.540 ","End":"03:48.130","Text":"Let\u0027s see what this is."},{"Start":"03:48.620 ","End":"03:56.310","Text":"Our initial energy is going to be equal to half multiplied by our mass,"},{"Start":"03:56.310 ","End":"04:01.200","Text":"which is dm multiplied by our initial velocity squared,"},{"Start":"04:01.200 ","End":"04:09.645","Text":"so 1/2 mv^2 for kinetic energy plus equation for potential energy,"},{"Start":"04:09.645 ","End":"04:13.305","Text":"which is going to be mass times gravity times height,"},{"Start":"04:13.305 ","End":"04:15.209","Text":"mgh, so here,"},{"Start":"04:15.209 ","End":"04:21.870","Text":"it\u0027s dmgh, and then we know that it equals to,"},{"Start":"04:21.870 ","End":"04:23.145","Text":"when it hits the scales,"},{"Start":"04:23.145 ","End":"04:25.215","Text":"it has 0 potential energy,"},{"Start":"04:25.215 ","End":"04:27.210","Text":"so it just has kinetic energy."},{"Start":"04:27.210 ","End":"04:30.420","Text":"It\u0027s going to be 1/2 times mass,"},{"Start":"04:30.420 ","End":"04:35.250","Text":"so dm multiplied by our v_final^2."},{"Start":"04:35.250 ","End":"04:40.110","Text":"Now we can see that our dms cross off,"},{"Start":"04:40.110 ","End":"04:44.850","Text":"and then all we have to do is we have to isolate out our v_final."},{"Start":"04:44.850 ","End":"04:47.460","Text":"Then after doing some very simple algebra,"},{"Start":"04:47.460 ","End":"04:51.075","Text":"we\u0027ll get that our v_final is equal to the square root"},{"Start":"04:51.075 ","End":"04:59.628","Text":"of v_0^2 plus 2gh."},{"Start":"04:59.628 ","End":"05:02.140","Text":"That\u0027s our final answer."},{"Start":"05:02.210 ","End":"05:06.335","Text":"Now, in question number 3, we\u0027re being asked,"},{"Start":"05:06.335 ","End":"05:08.135","Text":"whilst the sand is falling,"},{"Start":"05:08.135 ","End":"05:11.420","Text":"when the scales show a weight of W,"},{"Start":"05:11.420 ","End":"05:12.740","Text":"what is the ratio,"},{"Start":"05:12.740 ","End":"05:17.201","Text":"or the relationship between the real weights of the sand,"},{"Start":"05:17.201 ","End":"05:23.625","Text":"of the certain amount of sand on the scales compared to the indicated mass on the scale?"},{"Start":"05:23.625 ","End":"05:27.330","Text":"What they\u0027re asking is the real mass of the sand on"},{"Start":"05:27.330 ","End":"05:31.950","Text":"the scale compared to what the scale will show the mass to be."},{"Start":"05:31.950 ","End":"05:33.990","Text":"Now, there\u0027s 2 ways to do it."},{"Start":"05:33.990 ","End":"05:40.230","Text":"There\u0027s 1 through using the definition of the derivative,"},{"Start":"05:40.230 ","End":"05:44.235","Text":"and the second way is by using our equation."},{"Start":"05:44.235 ","End":"05:46.935","Text":"Now we\u0027re going to deal with the first way,"},{"Start":"05:46.935 ","End":"05:52.450","Text":"which is using the definition of the derivative."},{"Start":"05:53.210 ","End":"05:58.620","Text":"When we\u0027re using the differential version,"},{"Start":"05:58.620 ","End":"06:04.875","Text":"then we\u0027re going to be using Newton\u0027s second law in the differential wave."},{"Start":"06:04.875 ","End":"06:08.880","Text":"What does that mean? It means that the sum of all of"},{"Start":"06:08.880 ","End":"06:16.480","Text":"our external forces is equal to dp by dt."},{"Start":"06:17.150 ","End":"06:23.786","Text":"Now, the formal definition of dp is equal"},{"Start":"06:23.786 ","End":"06:27.758","Text":"p(t) +"},{"Start":"06:27.758 ","End":"06:36.630","Text":"dt-p (t)."},{"Start":"06:36.630 ","End":"06:41.160","Text":"What does this equation for dp actually mean?"},{"Start":"06:41.160 ","End":"06:43.545","Text":"If we look over here at our diagram,"},{"Start":"06:43.545 ","End":"06:52.380","Text":"when the certain amount of sand of mass dm is falling,"},{"Start":"06:52.380 ","End":"06:55.050","Text":"then it has an initial momentum,"},{"Start":"06:55.050 ","End":"07:00.135","Text":"which will be at its momentum at p equals t,"},{"Start":"07:00.135 ","End":"07:02.955","Text":"somewhere along over here."},{"Start":"07:02.955 ","End":"07:06.135","Text":"But when the dm,"},{"Start":"07:06.135 ","End":"07:10.050","Text":"this small mass of sand reaches the scale,"},{"Start":"07:10.050 ","End":"07:12.945","Text":"it stops, which means it has no velocity,"},{"Start":"07:12.945 ","End":"07:15.090","Text":"which means that it has no momentum,"},{"Start":"07:15.090 ","End":"07:22.910","Text":"which means that it\u0027s p at time t plus dt will equal 0."},{"Start":"07:22.910 ","End":"07:30.250","Text":"Again, it has an initial momentum and then it has its final momentum,"},{"Start":"07:30.250 ","End":"07:35.530","Text":"and its final momentum will equal 0 because it stops because it hits the scale."},{"Start":"07:35.530 ","End":"07:41.440","Text":"That means that it has no momentum because it has no velocity."},{"Start":"07:41.440 ","End":"07:47.275","Text":"That means that our dp will equal to 0."},{"Start":"07:47.275 ","End":"07:50.755","Text":"Because this is our final momentum,"},{"Start":"07:50.755 ","End":"07:56.695","Text":"negative pt which will"},{"Start":"07:56.695 ","End":"08:02.020","Text":"equal dm multiplied by v_final,"},{"Start":"08:02.020 ","End":"08:07.255","Text":"which equals the mass times by the final velocity."},{"Start":"08:07.255 ","End":"08:13.719","Text":"Now, we\u0027re going to choose that this direction is the positive direction,"},{"Start":"08:13.719 ","End":"08:17.635","Text":"which means that our v_final is in the positive direction."},{"Start":"08:17.635 ","End":"08:19.255","Text":"This is how we write this."},{"Start":"08:19.255 ","End":"08:25.525","Text":"Now, this equation, this dm multiplied by v_final,"},{"Start":"08:25.525 ","End":"08:34.240","Text":"it comes because we\u0027re taking the momentum just before the sand hits the scale,"},{"Start":"08:34.240 ","End":"08:38.530","Text":"which is our pt."},{"Start":"08:38.530 ","End":"08:42.220","Text":"Then we\u0027re taking the momentum after the sand hits the scale,"},{"Start":"08:42.220 ","End":"08:45.670","Text":"which is our p(t) plus dt."},{"Start":"08:45.670 ","End":"08:48.310","Text":"Our p(t) plus dt after it hit the scale,"},{"Start":"08:48.310 ","End":"08:50.605","Text":"it\u0027s stopped, so at 0."},{"Start":"08:50.605 ","End":"08:53.094","Text":"Just before it hits the scale,"},{"Start":"08:53.094 ","End":"08:55.375","Text":"we just worked out in question 2,"},{"Start":"08:55.375 ","End":"08:57.190","Text":"what are our v_final is?"},{"Start":"08:57.190 ","End":"09:00.920","Text":"It\u0027s the mass times the velocity."},{"Start":"09:01.350 ","End":"09:11.360","Text":"Now we know that weight W symbolizes mass times gravity."},{"Start":"09:11.640 ","End":"09:14.770","Text":"Now, as the sand is falling,"},{"Start":"09:14.770 ","End":"09:21.894","Text":"sand is going to accumulate on the scale and the weight of this,"},{"Start":"09:21.894 ","End":"09:23.965","Text":"we\u0027re going to call,"},{"Start":"09:23.965 ","End":"09:28.490","Text":"it\u0027s pointing in this direction, weight_real."},{"Start":"09:29.130 ","End":"09:31.660","Text":"This is the weight_real."},{"Start":"09:31.660 ","End":"09:37.135","Text":"Then we have the normal force pointing upwards,"},{"Start":"09:37.135 ","End":"09:42.650","Text":"which is actually equal to the weight_shown."},{"Start":"09:43.760 ","End":"09:47.805","Text":"The normal force pointing upwards is equal to the weight_shown,"},{"Start":"09:47.805 ","End":"09:51.330","Text":"and the weight_shown is whatever the scale reads."},{"Start":"09:51.330 ","End":"09:53.475","Text":"It will have numbers over here."},{"Start":"09:53.475 ","End":"09:54.930","Text":"Whatever the scale reads,"},{"Start":"09:54.930 ","End":"09:58.570","Text":"that is the weight shown and that is the normal force."},{"Start":"09:58.680 ","End":"10:01.915","Text":"Now, the next step that we do is"},{"Start":"10:01.915 ","End":"10:07.195","Text":"we just solve this equation by subbing in everything that we found."},{"Start":"10:07.195 ","End":"10:10.465","Text":"We say that the sum of all our external forces,"},{"Start":"10:10.465 ","End":"10:17.900","Text":"which is the W_real"},{"Start":"10:20.070 ","End":"10:23.410","Text":"because it\u0027s in the positive direction because we said"},{"Start":"10:23.410 ","End":"10:26.020","Text":"that the downwards direction is the positive direction,"},{"Start":"10:26.020 ","End":"10:29.920","Text":"negative or normal, because our normal is going upwards,"},{"Start":"10:29.920 ","End":"10:32.530","Text":"which we\u0027ve said is the negative direction,"},{"Start":"10:32.530 ","End":"10:37.880","Text":"negative W_shown on scale."},{"Start":"10:37.950 ","End":"10:42.550","Text":"This is the sum of all of the external forces."},{"Start":"10:42.550 ","End":"10:49.225","Text":"Then we said that it equals over here dp by dt."},{"Start":"10:49.225 ","End":"10:52.705","Text":"Now, our dp, we said what it is,"},{"Start":"10:52.705 ","End":"10:58.975","Text":"it\u0027s negative dm multiplied by v_final,"},{"Start":"10:58.975 ","End":"11:01.195","Text":"then divided by dt."},{"Start":"11:01.195 ","End":"11:03.055","Text":"Now, in question number 1,"},{"Start":"11:03.055 ","End":"11:05.560","Text":"we worked out what dm by dt is."},{"Start":"11:05.560 ","End":"11:11.020","Text":"We said that it equals Rho AV_0."},{"Start":"11:11.020 ","End":"11:13.765","Text":"Then we can say that it equals"},{"Start":"11:13.765 ","End":"11:22.540","Text":"negative Rho AV_0 multiplied by v_f."},{"Start":"11:22.540 ","End":"11:26.065","Text":"Now, all we have to do to answer the question because,"},{"Start":"11:26.065 ","End":"11:27.520","Text":"sorry, I corrected here its weight,"},{"Start":"11:27.520 ","End":"11:31.480","Text":"not mass because we\u0027re being asked what is the ratio or what is the relationship"},{"Start":"11:31.480 ","End":"11:33.730","Text":"between the real weights of the sand on the scale"},{"Start":"11:33.730 ","End":"11:36.565","Text":"compared to the indicated weight on the scale."},{"Start":"11:36.565 ","End":"11:42.925","Text":"All we have to do, our W_shown is a given."},{"Start":"11:42.925 ","End":"11:45.890","Text":"We can just see it and read it off the scale."},{"Start":"11:47.490 ","End":"11:56.305","Text":"If we want to isolate out this W_real and then divide both sides by W_shown,"},{"Start":"11:56.305 ","End":"11:58.990","Text":"then we will get that relationship."},{"Start":"11:58.990 ","End":"12:07.090","Text":"Then what we\u0027ll have is we\u0027ll have w_ real divided by W_shown."},{"Start":"12:07.090 ","End":"12:14.830","Text":"The real rate weight divided by the weight shown is equal to 1 minus"},{"Start":"12:14.830 ","End":"12:25.490","Text":"Rho AV_0V_final divided by W_shown."},{"Start":"12:27.060 ","End":"12:32.080","Text":"All I did was rearrange this equation to isolate out"},{"Start":"12:32.080 ","End":"12:37.130","Text":"my W_real and that is the final answer."},{"Start":"12:37.830 ","End":"12:41.185","Text":"Now what we\u0027re going to do is we\u0027re going to deal"},{"Start":"12:41.185 ","End":"12:44.365","Text":"with the second way of solving this question,"},{"Start":"12:44.365 ","End":"12:48.505","Text":"which is with our equation."},{"Start":"12:48.505 ","End":"12:52.915","Text":"Let\u0027s just write out the equation right now."},{"Start":"12:52.915 ","End":"12:57.385","Text":"It\u0027s the sum of all of the external forces,"},{"Start":"12:57.385 ","End":"13:00.250","Text":"which is equal to our mass,"},{"Start":"13:00.250 ","End":"13:01.960","Text":"which changes in time,"},{"Start":"13:01.960 ","End":"13:12.860","Text":"multiplied by dv by dt plus v_relative multiplied by dm by dt."},{"Start":"13:13.200 ","End":"13:19.990","Text":"This way, this number 2 is the crux of"},{"Start":"13:19.990 ","End":"13:22.869","Text":"the question because we\u0027re dealing with the equation"},{"Start":"13:22.869 ","End":"13:26.650","Text":"which is speaking about a varying mass system."},{"Start":"13:26.650 ","End":"13:31.405","Text":"Let\u0027s see how we can apply this equation to our question."},{"Start":"13:31.405 ","End":"13:35.500","Text":"Now what we have to do is we have to understand if we\u0027re dealing"},{"Start":"13:35.500 ","End":"13:39.220","Text":"with a case of ejection or with the case of accretion."},{"Start":"13:39.220 ","End":"13:42.715","Text":"Let\u0027s go back to our diagram."},{"Start":"13:42.715 ","End":"13:48.265","Text":"Here\u0027s our diagram. Now how can we know which case we are looking at?"},{"Start":"13:48.265 ","End":"13:52.210","Text":"What we can say is that there\u0027s"},{"Start":"13:52.210 ","End":"14:00.625","Text":"some invisible cart over here and then sand is falling into it."},{"Start":"14:00.625 ","End":"14:04.150","Text":"Now it\u0027s really easy to see that all the sand"},{"Start":"14:04.150 ","End":"14:07.480","Text":"accumulating in this cart or accumulating on"},{"Start":"14:07.480 ","End":"14:16.765","Text":"the scale is going to simply be equal to the case of accretion."},{"Start":"14:16.765 ","End":"14:19.690","Text":"Now, our mass of the cart,"},{"Start":"14:19.690 ","End":"14:24.010","Text":"because it\u0027s imaginary, is obviously 0."},{"Start":"14:24.010 ","End":"14:27.655","Text":"We\u0027re dealing with the case of accretion."},{"Start":"14:27.655 ","End":"14:31.075","Text":"Then we can say that this mass,"},{"Start":"14:31.075 ","End":"14:35.170","Text":"which is collecting inside the cart,"},{"Start":"14:35.170 ","End":"14:41.905","Text":"this weight_real is going to be"},{"Start":"14:41.905 ","End":"14:49.190","Text":"smaller than the weight_shown because the weight_shown is a product of the normal."},{"Start":"14:49.190 ","End":"14:54.575","Text":"The normal is being increased because the sand is falling onto the scale,"},{"Start":"14:54.575 ","End":"14:56.420","Text":"which is then increasing the normal,"},{"Start":"14:56.420 ","End":"15:00.180","Text":"which is therefore showing a larger weight."},{"Start":"15:00.180 ","End":"15:04.780","Text":"Now let\u0027s start by writing out the equation."},{"Start":"15:04.780 ","End":"15:09.995","Text":"The sum of all of our external forces is going to be like this."},{"Start":"15:09.995 ","End":"15:13.820","Text":"We\u0027re going to have our real_weight, which is mg."},{"Start":"15:13.820 ","End":"15:18.465","Text":"It\u0027s just going to be W_real."},{"Start":"15:18.465 ","End":"15:23.530","Text":"Because remember,"},{"Start":"15:23.530 ","End":"15:29.304","Text":"the downwards direction is the positive direction minus our normal,"},{"Start":"15:29.304 ","End":"15:32.930","Text":"which is our W_shown."},{"Start":"15:34.410 ","End":"15:41.350","Text":"Its in fact, m as a function of t g minus our normal."},{"Start":"15:41.350 ","End":"15:45.685","Text":"This is the sum of our external forces."},{"Start":"15:45.685 ","End":"15:50.515","Text":"Then we can say that this equals,"},{"Start":"15:50.515 ","End":"15:53.545","Text":"we have our mass as a function of time,"},{"Start":"15:53.545 ","End":"15:57.640","Text":"which will be our W real divided by g. But our dv"},{"Start":"15:57.640 ","End":"16:03.070","Text":"by dt because there\u0027s no movement in the y-direction,"},{"Start":"16:03.070 ","End":"16:07.510","Text":"which is the direction that we\u0027re looking at the y-axis,"},{"Start":"16:07.510 ","End":"16:12.460","Text":"so it equals to 0 because there\u0027s no acceleration."},{"Start":"16:12.460 ","End":"16:14.995","Text":"This whole thing becomes 0."},{"Start":"16:14.995 ","End":"16:19.240","Text":"Then we have to look at our relative velocity."},{"Start":"16:19.240 ","End":"16:24.220","Text":"Now we have to say what our V relative is."},{"Start":"16:24.220 ","End":"16:28.180","Text":"Our V relative is in actual fact, our V final."},{"Start":"16:28.180 ","End":"16:31.780","Text":"Now why can we just say that instead of doing"},{"Start":"16:31.780 ","End":"16:37.555","Text":"the whole process of V relative is equal to V minus VT. Why can we say that?"},{"Start":"16:37.555 ","End":"16:39.610","Text":"Because our invisible cart,"},{"Start":"16:39.610 ","End":"16:42.925","Text":"which doesn\u0027t exist, isn\u0027t moving."},{"Start":"16:42.925 ","End":"16:47.305","Text":"The sand which is falling into it is falling into"},{"Start":"16:47.305 ","End":"16:52.165","Text":"this stationary cart at the same velocity that is falling to the ground,"},{"Start":"16:52.165 ","End":"16:57.160","Text":"because the cart is moving at the same velocity with that the ground is moving in,"},{"Start":"16:57.160 ","End":"16:59.990","Text":"so they\u0027re not moving."},{"Start":"17:00.180 ","End":"17:04.600","Text":"V_rel is just V_final."},{"Start":"17:04.600 ","End":"17:09.220","Text":"Then we multiply it by our dm by dt,"},{"Start":"17:09.220 ","End":"17:12.430","Text":"which I believe we found in our first question,"},{"Start":"17:12.430 ","End":"17:15.760","Text":"which was Rho AV_0."},{"Start":"17:15.760 ","End":"17:18.415","Text":"Let\u0027s multiply this by this."},{"Start":"17:18.415 ","End":"17:22.010","Text":"It\u0027s Rho AV_0."},{"Start":"17:22.410 ","End":"17:30.025","Text":"Then the last final thing is that we then multiply this expression by negative 1."},{"Start":"17:30.025 ","End":"17:33.655","Text":"Because our dm by dt has to be"},{"Start":"17:33.655 ","End":"17:38.560","Text":"a negative number because we\u0027re dealing with the weight increasing."},{"Start":"17:38.560 ","End":"17:41.350","Text":"If the weight is increasing,"},{"Start":"17:41.350 ","End":"17:43.795","Text":"if weight is being added to the system,"},{"Start":"17:43.795 ","End":"17:46.630","Text":"then our dm by dt is a negative number,"},{"Start":"17:46.630 ","End":"17:49.250","Text":"so we multiply by negative 1."},{"Start":"17:49.250 ","End":"17:54.190","Text":"Now, again, like in number 1 over here,"},{"Start":"17:54.190 ","End":"17:59.545","Text":"we\u0027re asked to find the relationship between our W_real and our W_shown."},{"Start":"17:59.545 ","End":"18:04.030","Text":"Now all we have to do is isolate out or W_real and then divide"},{"Start":"18:04.030 ","End":"18:08.515","Text":"by W_shown in order to get the relationship between the 2."},{"Start":"18:08.515 ","End":"18:14.020","Text":"Then after doing some simple and basic algebra will get W_real divided"},{"Start":"18:14.020 ","End":"18:21.550","Text":"by W_shown will be equal to the exact same answer that we got over here."},{"Start":"18:21.550 ","End":"18:23.905","Text":"If you don\u0027t believe me, you can work it out."},{"Start":"18:23.905 ","End":"18:28.900","Text":"It\u0027s going to be 1 minus Rho AV_ 0 V"},{"Start":"18:28.900 ","End":"18:38.620","Text":"f divided by w_shown."},{"Start":"18:38.620 ","End":"18:39.700","Text":"That is the onset."},{"Start":"18:39.700 ","End":"18:44.320","Text":"We can see that we can solve this type of question in 2 different ways,"},{"Start":"18:44.320 ","End":"18:46.585","Text":"and we\u0027ll get the exact same answer."},{"Start":"18:46.585 ","End":"18:51.250","Text":"Now, I\u0027m going to rub out all the workings out for all of the questions and just leave"},{"Start":"18:51.250 ","End":"18:56.990","Text":"the final answer in order to have space to work out the next questions."},{"Start":"18:58.050 ","End":"19:01.375","Text":"Now, we\u0027re looking at question number 4."},{"Start":"19:01.375 ","End":"19:04.675","Text":"When a weight of W is shown on the scale,"},{"Start":"19:04.675 ","End":"19:08.410","Text":"so remember, over here,"},{"Start":"19:08.410 ","End":"19:12.145","Text":"where we see all of the numbers on the scale, a weight of W,"},{"Start":"19:12.145 ","End":"19:16.315","Text":"so the same way as in question number 3 is shown."},{"Start":"19:16.315 ","End":"19:19.900","Text":"At that moment, the final opening is shut."},{"Start":"19:19.900 ","End":"19:25.795","Text":"Here becomes closed and no more sand can go through."},{"Start":"19:25.795 ","End":"19:29.770","Text":"Then we\u0027re being asked after some moments later,"},{"Start":"19:29.770 ","End":"19:32.695","Text":"what weight will the scale measure?"},{"Start":"19:32.695 ","End":"19:37.225","Text":"What will we see over here on the screen?"},{"Start":"19:37.225 ","End":"19:41.740","Text":"What an actual fact is happening is that our final shots and"},{"Start":"19:41.740 ","End":"19:46.675","Text":"then all the sand that has accumulated on the scale from before,"},{"Start":"19:46.675 ","End":"19:51.080","Text":"we will have, so that will be our W_real."},{"Start":"19:54.930 ","End":"20:01.914","Text":"Then all of the sand that was still in the air when the funnel opening was shut will also"},{"Start":"20:01.914 ","End":"20:08.695","Text":"fall to the scale and will also accumulate and be added to our W_real."},{"Start":"20:08.695 ","End":"20:15.860","Text":"Then we\u0027ll have to add our W_added."},{"Start":"20:16.620 ","End":"20:24.565","Text":"Our W_added is the sand that was still falling in the air when the funnel was shut."},{"Start":"20:24.565 ","End":"20:28.885","Text":"So W_real plus our W_added will equal,"},{"Start":"20:28.885 ","End":"20:32.530","Text":"we\u0027ll call it our W_tag."},{"Start":"20:32.530 ","End":"20:35.815","Text":"This is our new W, our total weight"},{"Start":"20:35.815 ","End":"20:40.205","Text":"on the scale a few moments after the funnel will shut."},{"Start":"20:40.205 ","End":"20:44.835","Text":"Now what we actually want to do in order to find out what our W_tag is,"},{"Start":"20:44.835 ","End":"20:47.540","Text":"we need to calculate what our W_added is."},{"Start":"20:47.540 ","End":"20:51.980","Text":"Our W_real, we already know from our question number 3."},{"Start":"20:51.980 ","End":"20:53.710","Text":"We just have to, in this case,"},{"Start":"20:53.710 ","End":"20:55.645","Text":"isolate out our W_real,"},{"Start":"20:55.645 ","End":"21:01.670","Text":"which just means multiplying both sides by W_shown."},{"Start":"21:01.770 ","End":"21:06.715","Text":"In order to calculate what our W_added is,"},{"Start":"21:06.715 ","End":"21:14.335","Text":"so that is the weight of all of the sand in the pillar that still hasn\u0027t hit the scale."},{"Start":"21:14.335 ","End":"21:18.895","Text":"We\u0027re going to say so we need the weight of the pillar,"},{"Start":"21:18.895 ","End":"21:22.480","Text":"which is mass times gravity,"},{"Start":"21:22.480 ","End":"21:25.000","Text":"the mass of the pillar times gravity."},{"Start":"21:25.000 ","End":"21:28.015","Text":"How are we going to find out what the mass of the gravity is?"},{"Start":"21:28.015 ","End":"21:30.760","Text":"It\u0027s the density with which the sand is falling,"},{"Start":"21:30.760 ","End":"21:32.755","Text":"which is our Rho,"},{"Start":"21:32.755 ","End":"21:38.995","Text":"multiplied by the area of the opening,"},{"Start":"21:38.995 ","End":"21:44.980","Text":"and then multiplied by the height or the length of the pillar,"},{"Start":"21:44.980 ","End":"21:46.375","Text":"the height of the pillar,"},{"Start":"21:46.375 ","End":"21:47.770","Text":"multiplied by height,"},{"Start":"21:47.770 ","End":"21:55.315","Text":"and then multiply it by g. Therefore we can say that our W_tag"},{"Start":"21:55.315 ","End":"21:59.110","Text":"so that our weight that the scale we\u0027ll"},{"Start":"21:59.110 ","End":"22:03.160","Text":"measure a few moments after the final hole was shut,"},{"Start":"22:03.160 ","End":"22:07.460","Text":"will be equal to our W_real."},{"Start":"22:07.530 ","End":"22:13.880","Text":"All I did was isolate out the W_real from my onset question 3."},{"Start":"22:14.100 ","End":"22:17.725","Text":"Plus, our W_added,"},{"Start":"22:17.725 ","End":"22:26.540","Text":"which we said over here was Rho times Ah multiplied by g. That is the final answer."},{"Start":"22:27.870 ","End":"22:31.210","Text":"Now, question number 5."},{"Start":"22:31.210 ","End":"22:33.235","Text":"After question number 4,"},{"Start":"22:33.235 ","End":"22:34.570","Text":"the funnel was shut,"},{"Start":"22:34.570 ","End":"22:38.170","Text":"and there\u0027s a new weight of sand,"},{"Start":"22:38.170 ","End":"22:43.165","Text":"which we now know what it is on the scale."},{"Start":"22:43.165 ","End":"22:47.920","Text":"Now, this weight of sand plus the scale altogether as 1 body"},{"Start":"22:47.920 ","End":"22:53.035","Text":"is accelerated upwards and an acceleration of 5 meters per second squared."},{"Start":"22:53.035 ","End":"22:57.520","Text":"What weight will the scale show after some time?"},{"Start":"22:57.520 ","End":"23:02.110","Text":"Now, all we have to do is we have to say that the sum"},{"Start":"23:02.110 ","End":"23:06.340","Text":"of all the forces is equal to mass times acceleration."},{"Start":"23:06.340 ","End":"23:10.600","Text":"When our acceleration is our 5 meters per second."},{"Start":"23:10.600 ","End":"23:20.050","Text":"Sum of all the forces is going to be our normal force minus W_tag,"},{"Start":"23:20.050 ","End":"23:21.805","Text":"which we worked out in question 4,"},{"Start":"23:21.805 ","End":"23:25.435","Text":"which is going to equal a mass times acceleration."},{"Start":"23:25.435 ","End":"23:26.920","Text":"Now, what is our mass?"},{"Start":"23:26.920 ","End":"23:30.820","Text":"Our mass is our W_tag divided by g,"},{"Start":"23:30.820 ","End":"23:36.235","Text":"because remember, weight is mass times gravity."},{"Start":"23:36.235 ","End":"23:42.800","Text":"Then multiply it by our acceleration, which is 5."},{"Start":"23:43.860 ","End":"23:53.200","Text":"Then if we just isolate out our normal force,"},{"Start":"23:53.200 ","End":"23:56.020","Text":"we will get that our normal force,"},{"Start":"23:56.020 ","End":"24:01.150","Text":"which we know is also equal to our W_shown,"},{"Start":"24:01.400 ","End":"24:05.875","Text":"the W_shown on the scale,"},{"Start":"24:05.875 ","End":"24:13.720","Text":"which will be equal to W_tag divided by g times by 5,"},{"Start":"24:13.720 ","End":"24:19.550","Text":"the acceleration plus W_tag."},{"Start":"24:20.570 ","End":"24:24.585","Text":"That is it. This is our final answer."},{"Start":"24:24.585 ","End":"24:32.600","Text":"The scale will show a weight of 5 times W_tag over g plus W_tag."},{"Start":"24:32.600 ","End":"24:36.260","Text":"This is the end of our questions."}],"ID":10618},{"Watched":false,"Name":"Hose Spraying on Person","Duration":"13m 23s","ChapterTopicVideoID":9230,"CourseChapterTopicPlaylistID":5429,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.410 ","End":"00:03.180","Text":"Hello. In this question,"},{"Start":"00:03.180 ","End":"00:06.675","Text":"we\u0027re being told that a hose sprays water on a person."},{"Start":"00:06.675 ","End":"00:12.720","Text":"The cross-sectional area of the hose is A and the density of the water is rho."},{"Start":"00:12.720 ","End":"00:17.445","Text":"The velocity with which the water exits the hose is v_0."},{"Start":"00:17.445 ","End":"00:19.650","Text":"Then in question number 1,"},{"Start":"00:19.650 ","End":"00:23.850","Text":"we\u0027re being asked to find the force acting on the stationary person being sprayed,"},{"Start":"00:23.850 ","End":"00:26.385","Text":"given that no water is sprayed back."},{"Start":"00:26.385 ","End":"00:29.775","Text":"The water just hits the girl."},{"Start":"00:29.775 ","End":"00:32.445","Text":"Then in the second question,"},{"Start":"00:32.445 ","End":"00:35.520","Text":"we\u0027re being asked to find the force acting on the person"},{"Start":"00:35.520 ","End":"00:39.450","Text":"who was running away from the water at a velocity of v,"},{"Start":"00:39.450 ","End":"00:42.015","Text":"which is smaller than v_0."},{"Start":"00:42.015 ","End":"00:44.760","Text":"Let\u0027s start with question number 1."},{"Start":"00:44.760 ","End":"00:46.915","Text":"How do we do this?"},{"Start":"00:46.915 ","End":"00:52.130","Text":"We\u0027re going to solve this via the definition of the derivative,"},{"Start":"00:52.130 ","End":"01:00.080","Text":"which as we know is the sum of all of the forces is equal to the change in momentum."},{"Start":"01:00.080 ","End":"01:04.290","Text":"Dp naught d rho by dt."},{"Start":"01:05.570 ","End":"01:11.180","Text":"How am I going to figure out what my dp by dt is?"},{"Start":"01:11.180 ","End":"01:14.390","Text":"What I\u0027m going to do is I\u0027m going to find my P,"},{"Start":"01:14.390 ","End":"01:21.360","Text":"my momentum at time t. Then I\u0027m going to find my momentum at"},{"Start":"01:21.360 ","End":"01:28.950","Text":"time t plus dt plus another small excrement of time."},{"Start":"01:28.950 ","End":"01:33.230","Text":"Let\u0027s first work out p of time t. What"},{"Start":"01:33.230 ","End":"01:37.985","Text":"I\u0027m going to do is I\u0027m going to mark out a small cylinder."},{"Start":"01:37.985 ","End":"01:41.090","Text":"We\u0027re going to say that at time t,"},{"Start":"01:41.090 ","End":"01:45.880","Text":"this whole cylinder of water is going to hit the girl over here."},{"Start":"01:45.880 ","End":"01:52.130","Text":"Now what we want to do is we want to find the momentum of this small cylinder."},{"Start":"01:52.130 ","End":"01:59.825","Text":"We\u0027re going to say that the mass of this small cylinder is dm can infinitely small mass."},{"Start":"01:59.825 ","End":"02:02.210","Text":"We\u0027re going to say that the momentum is our mass,"},{"Start":"02:02.210 ","End":"02:04.790","Text":"which is dm, multiplied by our velocity,"},{"Start":"02:04.790 ","End":"02:07.055","Text":"which we know is v_0."},{"Start":"02:07.055 ","End":"02:12.200","Text":"Now, when we look at our momentum at t plus dt."},{"Start":"02:12.200 ","End":"02:16.490","Text":"Because in the question we\u0027re being told that no water is sprayed back."},{"Start":"02:16.490 ","End":"02:19.115","Text":"That means that our velocity is 0."},{"Start":"02:19.115 ","End":"02:22.190","Text":"This is simply going to equal 0."},{"Start":"02:22.190 ","End":"02:25.805","Text":"Now, our unknown right now is our dm,"},{"Start":"02:25.805 ","End":"02:28.415","Text":"our mass of this little cylinder."},{"Start":"02:28.415 ","End":"02:32.660","Text":"Let\u0027s see what our dm is equal to."},{"Start":"02:32.660 ","End":"02:36.470","Text":"It\u0027s going to be the density, which is our rho,"},{"Start":"02:36.470 ","End":"02:41.630","Text":"multiplied by the cross-sectional area of the hose, which is A."},{"Start":"02:41.630 ","End":"02:50.145","Text":"Then we\u0027re going to multiply it by dx which is this distance over here."},{"Start":"02:50.145 ","End":"02:56.450","Text":"Dx is the length of the infinitely small cylinder of water."},{"Start":"02:56.450 ","End":"03:01.940","Text":"Now we can say that because of the equation, speed,"},{"Start":"03:01.940 ","End":"03:05.610","Text":"velocity equals distance over time,"},{"Start":"03:05.610 ","End":"03:08.980","Text":"so our distance here is dx."},{"Start":"03:09.320 ","End":"03:16.125","Text":"That\u0027s dx. If we isolate this out, this is dt."},{"Start":"03:16.125 ","End":"03:24.245","Text":"Then we can say that our dx is equal to our velocity multiplied by our dt."},{"Start":"03:24.245 ","End":"03:27.245","Text":"What is our velocity? It\u0027s v_0."},{"Start":"03:27.245 ","End":"03:38.235","Text":"Then this whole thing will equal rho multiplied by A multiplied by v_0 times dt."},{"Start":"03:38.235 ","End":"03:43.010","Text":"Now what we can do is we can substitute this into our equation of"},{"Start":"03:43.010 ","End":"03:47.285","Text":"the sum of all the forces is equal to dp by dt."},{"Start":"03:47.285 ","End":"03:53.780","Text":"We can say the sum of all of the forces is equal to dp by dt,"},{"Start":"03:53.780 ","End":"04:00.130","Text":"which is equal to p at time t plus dt"},{"Start":"04:00.130 ","End":"04:08.704","Text":"minus p at time t. Our p at time t plus dt is equal to 0."},{"Start":"04:08.704 ","End":"04:13.030","Text":"Then we have negative dm."},{"Start":"04:13.030 ","End":"04:15.195","Text":"We have negative p(t)."},{"Start":"04:15.195 ","End":"04:23.220","Text":"Dm, which we worked out as rho Av_0 multiplied by dt multiplied by v_0."},{"Start":"04:23.220 ","End":"04:25.525","Text":"It becomes v_0^2."},{"Start":"04:25.525 ","End":"04:35.625","Text":"Then of course, all of this divided by dt and also this is divided by dt."},{"Start":"04:35.625 ","End":"04:40.440","Text":"Then we can see that these 2dts over here cross out."},{"Start":"04:40.440 ","End":"04:45.480","Text":"Then we\u0027re just left with negative rho Av_0^2."},{"Start":"04:46.120 ","End":"04:49.250","Text":"Now, the final section to the question,"},{"Start":"04:49.250 ","End":"04:51.710","Text":"because I want to know the force acting on"},{"Start":"04:51.710 ","End":"04:55.900","Text":"the stationary person than through Newton\u0027s third law."},{"Start":"04:55.900 ","End":"05:03.485","Text":"I can say that the sum of all of the forces on the person is equal to"},{"Start":"05:03.485 ","End":"05:12.225","Text":"negative the sum of all of the forces acting against the water."},{"Start":"05:12.225 ","End":"05:16.750","Text":"Therefore, the sum of the forces acting on the person,"},{"Start":"05:16.750 ","End":"05:22.610","Text":"p for person purse is equal to negative of this,"},{"Start":"05:22.610 ","End":"05:30.030","Text":"because this is equal to the sum of the forces acting on the water."},{"Start":"05:30.030 ","End":"05:33.520","Text":"Then I can say that it\u0027s equal to rho Av_0^2."},{"Start":"05:34.580 ","End":"05:39.085","Text":"That is the final answer to question number 1."},{"Start":"05:39.085 ","End":"05:43.615","Text":"Now question number 2 is the exact same thing as question number 1."},{"Start":"05:43.615 ","End":"05:47.480","Text":"The only thing that\u0027s changing here is the velocity."},{"Start":"05:47.480 ","End":"05:51.185","Text":"The check-in question to you is that you have to find"},{"Start":"05:51.185 ","End":"05:56.715","Text":"the relative velocity of the water with regards to the girl."},{"Start":"05:56.715 ","End":"06:02.105","Text":"We\u0027re being told in the question that the girl is also running velocity of v,"},{"Start":"06:02.105 ","End":"06:05.970","Text":"which is smaller than the velocity of the water, which is v_0."},{"Start":"06:06.290 ","End":"06:10.385","Text":"Now in order to find the relative velocity,"},{"Start":"06:10.385 ","End":"06:13.520","Text":"the velocity of the water relative to the girl."},{"Start":"06:13.520 ","End":"06:16.220","Text":"Let\u0027s call it v tilde."},{"Start":"06:16.220 ","End":"06:21.080","Text":"Then we have to say that it\u0027s the velocity of the water"},{"Start":"06:21.080 ","End":"06:27.220","Text":"relative to the Earth minus the velocity of the girl relative to the Earth."},{"Start":"06:27.220 ","End":"06:33.155","Text":"This is the trick. Then all we have to do through that is then"},{"Start":"06:33.155 ","End":"06:38.970","Text":"say that we just substitute this into where I v_0 is."},{"Start":"06:38.970 ","End":"06:47.340","Text":"Then we\u0027ll get that the sum of all of the forces on the person is equal to rho A."},{"Start":"06:47.340 ","End":"06:53.940","Text":"Then instead of v_0 squared will have v_0 minus v^2."},{"Start":"06:53.940 ","End":"06:56.970","Text":"That is the onset of question number 2."},{"Start":"06:56.970 ","End":"06:59.585","Text":"That is the simplest way to do it."},{"Start":"06:59.585 ","End":"07:04.730","Text":"However, now I\u0027m going to show a slightly more complicated way to use it just"},{"Start":"07:04.730 ","End":"07:10.130","Text":"in case you wanted to use the equations and understand it a little bit more."},{"Start":"07:10.130 ","End":"07:12.740","Text":"But if the next section confuses you,"},{"Start":"07:12.740 ","End":"07:19.840","Text":"then just ignore it and just use this here as how to solve this question."},{"Start":"07:19.840 ","End":"07:23.885","Text":"What we\u0027re going to do is we\u0027re going to start off by"},{"Start":"07:23.885 ","End":"07:27.920","Text":"setting it up exactly how we did in question number 1."},{"Start":"07:27.920 ","End":"07:33.150","Text":"We\u0027ll have our momentum at time t on the momentum before,"},{"Start":"07:33.150 ","End":"07:37.955","Text":"which will equal to dm tag. Why is it tag?"},{"Start":"07:37.955 ","End":"07:41.180","Text":"Because it\u0027s not the same dm that we had in question number 1."},{"Start":"07:41.180 ","End":"07:42.500","Text":"We\u0027ll see why in a second,"},{"Start":"07:42.500 ","End":"07:47.020","Text":"multiplied by our velocity before, which is v_0."},{"Start":"07:47.570 ","End":"07:51.125","Text":"Then our momentum after,"},{"Start":"07:51.125 ","End":"07:57.785","Text":"so our momentum at t plus dt is going to equal dm tag,"},{"Start":"07:57.785 ","End":"08:00.260","Text":"the same dm over here."},{"Start":"08:00.260 ","End":"08:05.350","Text":"But now it\u0027s going to be multiplied by v. Now,"},{"Start":"08:05.350 ","End":"08:08.870","Text":"a way of looking at it is that if we have"},{"Start":"08:08.870 ","End":"08:16.315","Text":"over here cylinder of water and it\u0027s hitting the girl."},{"Start":"08:16.315 ","End":"08:19.115","Text":"The second that it hits the girl,"},{"Start":"08:19.115 ","End":"08:20.630","Text":"then it will stay, for instance,"},{"Start":"08:20.630 ","End":"08:23.240","Text":"you can think of it as it will stay on her clothes."},{"Start":"08:23.240 ","End":"08:29.045","Text":"Then she is still running at a velocity of v. That means that"},{"Start":"08:29.045 ","End":"08:32.150","Text":"the water which has now absorbed in"},{"Start":"08:32.150 ","End":"08:35.225","Text":"her clothes will be moving at the exact same velocity,"},{"Start":"08:35.225 ","End":"08:39.590","Text":"which is v. That\u0027s a way of looking at it."},{"Start":"08:39.590 ","End":"08:43.730","Text":"Now we have that it\u0027s dm tag multiplied by"},{"Start":"08:43.730 ","End":"08:50.395","Text":"v. Now what we\u0027re going to want to work out is what is dm tag equal to?"},{"Start":"08:50.395 ","End":"08:53.585","Text":"Just like before, it\u0027s going to be equal to the density of the water,"},{"Start":"08:53.585 ","End":"08:58.135","Text":"which is rho, multiplied by the cross-sectional area of the hose, which is A."},{"Start":"08:58.135 ","End":"09:03.815","Text":"Then it\u0027s going to be multiplied by just like before, the velocity."},{"Start":"09:03.815 ","End":"09:05.270","Text":"Now what\u0027s our velocity?"},{"Start":"09:05.270 ","End":"09:07.930","Text":"We have to take the relative velocity."},{"Start":"09:07.930 ","End":"09:10.295","Text":"As we saw over here,"},{"Start":"09:10.295 ","End":"09:16.525","Text":"our relative velocity of the water relative to the girl is equal to v_0,"},{"Start":"09:16.525 ","End":"09:21.170","Text":"the velocity of the water relative to Earth minus v,"},{"Start":"09:21.170 ","End":"09:24.125","Text":"the velocity of the girl relative to Earth."},{"Start":"09:24.125 ","End":"09:29.130","Text":"Then this multiplied by dt."},{"Start":"09:29.870 ","End":"09:38.240","Text":"Let\u0027s see how we got here instead of if we don\u0027t have the previous question to refer to."},{"Start":"09:38.240 ","End":"09:47.180","Text":"Just like before we saw that dm tag is equal to rho A multiplied by dx tag."},{"Start":"09:47.180 ","End":"09:50.585","Text":"Now, what is dx tags equal to?"},{"Start":"09:50.585 ","End":"09:53.200","Text":"Let\u0027s have a look at the diagram."},{"Start":"09:53.200 ","End":"09:56.070","Text":"Let\u0027s take away the girl for 1 second."},{"Start":"09:56.070 ","End":"09:58.320","Text":"Here\u0027s our cylinder."},{"Start":"09:58.320 ","End":"10:03.730","Text":"We have over here that this distance is dx."},{"Start":"10:03.730 ","End":"10:11.195","Text":"Now if we factor in the girl in the time of dt before"},{"Start":"10:11.195 ","End":"10:13.970","Text":"where the girl was stationary and the last drops of"},{"Start":"10:13.970 ","End":"10:18.695","Text":"this water would have had hit the girl now because the girl is moving."},{"Start":"10:18.695 ","End":"10:24.130","Text":"Some of the last drops of the water in the cylinder won\u0027t hit the girl."},{"Start":"10:24.130 ","End":"10:27.545","Text":"We\u0027ll only actually have this amount of the cylinder"},{"Start":"10:27.545 ","End":"10:32.960","Text":"actually hit the girl because in this time dt because she\u0027s moving,"},{"Start":"10:32.960 ","End":"10:41.800","Text":"she\u0027s not stationary so a certain section will not reach the girl in this time."},{"Start":"10:42.050 ","End":"10:46.365","Text":"Here, what will our dx tag be?"},{"Start":"10:46.365 ","End":"10:52.220","Text":"If we\u0027re going to look at the system of the girl,"},{"Start":"10:52.220 ","End":"10:54.545","Text":"and so in the system of the girl,"},{"Start":"10:54.545 ","End":"10:56.750","Text":"the girl is stationary."},{"Start":"10:56.750 ","End":"11:03.140","Text":"Then we\u0027re saying how much water is hitting this girl in this system relative to her."},{"Start":"11:03.140 ","End":"11:07.925","Text":"Then dx tag, I\u0027ll write it over here."},{"Start":"11:07.925 ","End":"11:14.550","Text":"Dx tag is equal to v tag multiplied by our time,"},{"Start":"11:14.550 ","End":"11:17.355","Text":"which is dt and our v tag."},{"Start":"11:17.355 ","End":"11:21.195","Text":"V tag is equal to v tilde,"},{"Start":"11:21.195 ","End":"11:28.335","Text":"so that equals just v_0 minus v multiplied by dt."},{"Start":"11:28.335 ","End":"11:35.130","Text":"Then we\u0027ll just get rho A multiplied by v_0 minus vdt,"},{"Start":"11:35.130 ","End":"11:37.110","Text":"like we saw over here."},{"Start":"11:37.110 ","End":"11:39.780","Text":"Another way of looking at it."},{"Start":"11:39.780 ","End":"11:45.320","Text":"Again, using this inertial system of the girl is that if the girl is"},{"Start":"11:45.320 ","End":"11:51.410","Text":"standing over here with the inertial system,"},{"Start":"11:51.410 ","End":"11:54.890","Text":"so she is again stationary."},{"Start":"11:54.890 ","End":"11:59.120","Text":"That means that we can say that the water hitting her is traveling"},{"Start":"11:59.120 ","End":"12:03.765","Text":"at a velocity of v_0 minus v when it hits her."},{"Start":"12:03.765 ","End":"12:09.725","Text":"Then again, we can just substitute that in and we\u0027ll get the same result."},{"Start":"12:09.725 ","End":"12:14.830","Text":"Those are a few different ways of looking at how to solve this."},{"Start":"12:14.830 ","End":"12:21.065","Text":"Now we can just finish this just like before saying the sum of all of the forces"},{"Start":"12:21.065 ","End":"12:28.200","Text":"acting on the person is equal to dp by dt,"},{"Start":"12:28.200 ","End":"12:38.040","Text":"which is equal to dm tag v multiplied by v_0 minus v divided by dt."},{"Start":"12:38.040 ","End":"12:43.815","Text":"Then we will get that it equals to"},{"Start":"12:43.815 ","End":"12:53.070","Text":"negative rho Av_0 minus v^2 dt divided by dt."},{"Start":"12:53.070 ","End":"12:56.230","Text":"As you can see, these cross off."},{"Start":"12:56.330 ","End":"13:00.904","Text":"This is the force of the water."},{"Start":"13:00.904 ","End":"13:04.010","Text":"Then we can say that the sum of the force on"},{"Start":"13:04.010 ","End":"13:08.255","Text":"the person is equal to negative the sum of the water."},{"Start":"13:08.255 ","End":"13:15.000","Text":"Which means that it equals to rho Av_0 minus v^2,"},{"Start":"13:15.080 ","End":"13:19.150","Text":"just like we saw right at the beginning."},{"Start":"13:19.160 ","End":"13:23.410","Text":"That\u0027s the end of this lesson."}],"ID":10619},{"Watched":false,"Name":"A Cart With Accretion Ejection and Friction","Duration":"22m 8s","ChapterTopicVideoID":9231,"CourseChapterTopicPlaylistID":5429,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.380 ","End":"00:03.060","Text":"Hello. This question is"},{"Start":"00:03.060 ","End":"00:07.485","Text":"a slightly more challenging question on the topic of varying mass systems."},{"Start":"00:07.485 ","End":"00:12.660","Text":"What we have in this diagram is a cart which is being filled with rainwater."},{"Start":"00:12.660 ","End":"00:16.560","Text":"The rainwater is falling at a velocity of u_1 and"},{"Start":"00:16.560 ","End":"00:20.925","Text":"the cart is being filled up at a rate dn by dt,"},{"Start":"00:20.925 ","End":"00:23.100","Text":"which equals Alpha,"},{"Start":"00:23.100 ","End":"00:26.910","Text":"equals dm by dt."},{"Start":"00:26.910 ","End":"00:31.515","Text":"The rain is falling directly downwards along the y-axis,"},{"Start":"00:31.515 ","End":"00:32.940","Text":"it\u0027s important to note that."},{"Start":"00:32.940 ","End":"00:39.700","Text":"The initial mass of the cart is m_0."},{"Start":"00:39.700 ","End":"00:44.975","Text":"It\u0027s moving with a velocity v_0 in the right direction."},{"Start":"00:44.975 ","End":"00:50.390","Text":"Then, aside from rainwater being added at a rate of Alpha,"},{"Start":"00:50.390 ","End":"00:54.695","Text":"there\u0027s some hose which is letting out water from"},{"Start":"00:54.695 ","End":"01:00.970","Text":"the cart at also a rate of Alpha and at a velocity of u_0."},{"Start":"01:00.970 ","End":"01:08.300","Text":"However, this time when the water is exiting it is at an angle of Theta to the x-axis."},{"Start":"01:08.300 ","End":"01:11.030","Text":"There\u0027s also some coefficient of friction,"},{"Start":"01:11.030 ","End":"01:13.100","Text":"which is equal to Mu k,"},{"Start":"01:13.100 ","End":"01:18.955","Text":"which is acting on the bottom of the cart as it moves forwards."},{"Start":"01:18.955 ","End":"01:25.670","Text":"First question, question number 1 is to find the cart\u0027s equation of motion."},{"Start":"01:25.670 ","End":"01:29.450","Text":"Now, this type of question appears all over physics,"},{"Start":"01:29.450 ","End":"01:31.220","Text":"and what is it talking about?"},{"Start":"01:31.220 ","End":"01:33.155","Text":"What does it refer to?"},{"Start":"01:33.155 ","End":"01:42.045","Text":"What an equation of motion actually means is some equation that involves a position x,"},{"Start":"01:42.045 ","End":"01:44.580","Text":"its derivative v,"},{"Start":"01:44.580 ","End":"01:47.355","Text":"and its derivative a,"},{"Start":"01:47.355 ","End":"01:49.200","Text":"or just 2 of them."},{"Start":"01:49.200 ","End":"01:52.725","Text":"Either x and a or x and v or v, and a,"},{"Start":"01:52.725 ","End":"01:56.000","Text":"it doesn\u0027t matter as long as there\u0027s something to do with"},{"Start":"01:56.000 ","End":"02:03.810","Text":"the position and its various derivatives be at velocity or acceleration."},{"Start":"02:03.860 ","End":"02:08.360","Text":"If you get an equation of x as a function of t,"},{"Start":"02:08.360 ","End":"02:18.805","Text":"then its solution will be the solution of the equation of motion."},{"Start":"02:18.805 ","End":"02:22.130","Text":"Now what we\u0027re going to do is we\u0027re going to begin"},{"Start":"02:22.130 ","End":"02:24.995","Text":"to solve the cart\u0027s equation of motion."},{"Start":"02:24.995 ","End":"02:26.390","Text":"Now as we know,"},{"Start":"02:26.390 ","End":"02:28.730","Text":"we\u0027re going to use our equation,"},{"Start":"02:28.730 ","End":"02:33.845","Text":"which is the sum of all of the external forces is equal to our mass"},{"Start":"02:33.845 ","End":"02:39.400","Text":"as a function of time multiplied by dv by dt,"},{"Start":"02:39.400 ","End":"02:45.020","Text":"plus a relative velocity of the water going in or going"},{"Start":"02:45.020 ","End":"02:52.310","Text":"out relative to the cart multiplied by dm by dt,"},{"Start":"02:52.310 ","End":"02:55.474","Text":"which is the rate of ejection or accretion."},{"Start":"02:55.474 ","End":"02:56.990","Text":"Now, because in this case,"},{"Start":"02:56.990 ","End":"03:02.040","Text":"we have a case of accretion and a case of rejection,"},{"Start":"03:02.180 ","End":"03:07.205","Text":"we\u0027re going to have to add another expression onto this equation."},{"Start":"03:07.205 ","End":"03:12.400","Text":"Let\u0027s now take a look at how we are going to do this."},{"Start":"03:12.400 ","End":"03:18.770","Text":"The first thing that we are going to notice is that because we have Mu k over here,"},{"Start":"03:18.770 ","End":"03:24.035","Text":"that means that we have a frictional force acting in this direction."},{"Start":"03:24.035 ","End":"03:32.285","Text":"Now, the frictional force is always equal to Mu k multiplied by our normal force,"},{"Start":"03:32.285 ","End":"03:35.940","Text":"the normal force, we\u0027re going to look a second at how we solve that,"},{"Start":"03:35.940 ","End":"03:38.665","Text":"but let\u0027s leave it at this in the meantime."},{"Start":"03:38.665 ","End":"03:42.525","Text":"Now let\u0027s take a look at the x-axis,"},{"Start":"03:42.525 ","End":"03:45.935","Text":"we\u0027re going to say that the sum of all the forces,"},{"Start":"03:45.935 ","End":"03:48.635","Text":"the external forces on the x-axis,"},{"Start":"03:48.635 ","End":"03:53.645","Text":"is going to be negative Mu_kN, y negative,"},{"Start":"03:53.645 ","End":"03:57.550","Text":"because we can see that the cart is moving in the right direction,"},{"Start":"03:57.550 ","End":"04:02.820","Text":"we\u0027re going to label this the positive x-direction."},{"Start":"04:03.350 ","End":"04:09.335","Text":"We have negative Mu k because our frictional force is acting in the opposite direction,"},{"Start":"04:09.335 ","End":"04:13.930","Text":"multiplied by n, is going to equal."},{"Start":"04:13.930 ","End":"04:15.975","Text":"Now we can look at our equation."},{"Start":"04:15.975 ","End":"04:18.840","Text":"We have mass as a function of time."},{"Start":"04:18.840 ","End":"04:22.790","Text":"Let\u0027s take a look at what our mass as a function of time is,"},{"Start":"04:22.790 ","End":"04:26.750","Text":"we are starting off with a starting mass,"},{"Start":"04:26.750 ","End":"04:30.020","Text":"which was our initial mass, which was m_0."},{"Start":"04:30.020 ","End":"04:36.965","Text":"Then we have plus Alpha t for the rainwater entering into the cart,"},{"Start":"04:36.965 ","End":"04:43.865","Text":"negative Alpha t for the water exiting via the hose in the cart."},{"Start":"04:43.865 ","End":"04:46.575","Text":"Then these values cross off,"},{"Start":"04:46.575 ","End":"04:51.860","Text":"we can see that our m_t is going to be a constant which will be m_0."},{"Start":"04:51.860 ","End":"04:59.655","Text":"Then we can write here m_0 instead of m_t multiplied by dv by dt."},{"Start":"04:59.655 ","End":"05:03.980","Text":"In the meantime, we\u0027re just going to write dv by dt."},{"Start":"05:03.980 ","End":"05:08.885","Text":"Now we have to add in our u relative,"},{"Start":"05:08.885 ","End":"05:11.465","Text":"right now we\u0027re going to be speaking about"},{"Start":"05:11.465 ","End":"05:17.555","Text":"the u relative of the rainwater which is accumulating inside the cart."},{"Start":"05:17.555 ","End":"05:20.560","Text":"Right now we\u0027re looking at accretion."},{"Start":"05:20.560 ","End":"05:26.825","Text":"We need to find out what our relative velocity of the rain falling into the cart is."},{"Start":"05:26.825 ","End":"05:35.420","Text":"We know that the velocity of the rain relative to the ground is equal 2,"},{"Start":"05:35.420 ","End":"05:42.750","Text":"we\u0027re just going to label it as v rel ground,"},{"Start":"05:44.150 ","End":"05:49.880","Text":"and then minus the velocity with which the cart is traveling."},{"Start":"05:49.880 ","End":"05:52.550","Text":"We\u0027re going to label this in the meantime as v_t."},{"Start":"05:52.550 ","End":"05:57.245","Text":"Then we have to multiply it by dm by dt,"},{"Start":"05:57.245 ","End":"05:59.795","Text":"in this case, it\u0027s Alpha."},{"Start":"05:59.795 ","End":"06:04.400","Text":"Then this we have to multiply by negative 1 because we know that"},{"Start":"06:04.400 ","End":"06:09.430","Text":"in accretion our Alpha has to have a negative sign in front of it."},{"Start":"06:09.430 ","End":"06:17.775","Text":"Now, we have to add in our u rel multiplied by dm by dt over here."},{"Start":"06:17.775 ","End":"06:21.165","Text":"But for our ejected water,"},{"Start":"06:21.165 ","End":"06:25.010","Text":"now we\u0027re going to add on our extra value over here to"},{"Start":"06:25.010 ","End":"06:30.050","Text":"represent also the ejection as well as the accretion."},{"Start":"06:30.050 ","End":"06:37.234","Text":"U relative is going to be very easy because every time we have ejection,"},{"Start":"06:37.234 ","End":"06:41.000","Text":"the ejected material, the velocity of it is going"},{"Start":"06:41.000 ","End":"06:45.105","Text":"to be already relative to the main body."},{"Start":"06:45.105 ","End":"06:47.430","Text":"The water ejected through the hose,"},{"Start":"06:47.430 ","End":"06:52.520","Text":"the velocity is relative to the cart because it\u0027s ejected from the cart."},{"Start":"06:52.520 ","End":"06:57.015","Text":"But remember, because we\u0027re dealing with the x-axis,"},{"Start":"06:57.015 ","End":"07:05.390","Text":"what we have to do is we have to say that our u_0 multiplied by cosine of"},{"Start":"07:05.390 ","End":"07:09.830","Text":"Theta is the velocity with which the water is exiting"},{"Start":"07:09.830 ","End":"07:15.095","Text":"the cart with relative to the x-axis."},{"Start":"07:15.095 ","End":"07:17.990","Text":"Because our x-axis is here,"},{"Start":"07:17.990 ","End":"07:20.690","Text":"which is our adjacent side to the angle,"},{"Start":"07:20.690 ","End":"07:23.525","Text":"which means that we use cosine of Theta."},{"Start":"07:23.525 ","End":"07:26.385","Text":"Then we multiply by dm by dt,"},{"Start":"07:26.385 ","End":"07:29.655","Text":"which is just here, it\u0027s our Alpha."},{"Start":"07:29.655 ","End":"07:33.620","Text":"Here we keep the sign of the Alpha"},{"Start":"07:33.620 ","End":"07:37.580","Text":"as a positive because in ejection the Alpha must be positive,"},{"Start":"07:37.580 ","End":"07:40.070","Text":"our dm by dt is positive."},{"Start":"07:40.070 ","End":"07:42.649","Text":"Now, in this question specifically,"},{"Start":"07:42.649 ","End":"07:46.220","Text":"we can say that the velocity of the rain relative to"},{"Start":"07:46.220 ","End":"07:51.250","Text":"the ground in the x-axis is equal to 0."},{"Start":"07:51.250 ","End":"07:58.867","Text":"Because we can see that the rain is just falling down along the y-axis."},{"Start":"07:58.867 ","End":"08:03.850","Text":"Now we need to find out what our normal force is."},{"Start":"08:03.850 ","End":"08:05.860","Text":"Let\u0027s see how we tackle that."},{"Start":"08:05.860 ","End":"08:09.280","Text":"All we have to do is to use this equation,"},{"Start":"08:09.280 ","End":"08:12.130","Text":"but this time on the y-axis,"},{"Start":"08:12.130 ","End":"08:18.350","Text":"so we\u0027re going to label the y-axis in the positive direction going upwards."},{"Start":"08:18.780 ","End":"08:25.000","Text":"Now we\u0027re going to say that the sum of all of the external forces is going to be,"},{"Start":"08:25.000 ","End":"08:27.805","Text":"so we have our normal going upwards here."},{"Start":"08:27.805 ","End":"08:35.295","Text":"It\u0027s going to be our normal minus our mg. Now our mass is constant, so it\u0027s m_0g."},{"Start":"08:35.295 ","End":"08:43.390","Text":"We saw that over here that our mass is that negative m_0g."},{"Start":"08:43.390 ","End":"08:46.690","Text":"That\u0027s the sum of all external forces on the y-axis,"},{"Start":"08:46.690 ","End":"08:53.335","Text":"which is equal to our mass times our dv by dt in the y direction."},{"Start":"08:53.335 ","End":"08:59.725","Text":"Now, because our acceleration and our movement in the y-direction is equal to 0,"},{"Start":"08:59.725 ","End":"09:01.900","Text":"then our dv by dt is equals to 0."},{"Start":"09:01.900 ","End":"09:06.625","Text":"Our mt multiplied by dv by dt will be equal to 0."},{"Start":"09:06.625 ","End":"09:08.860","Text":"We can just write 0 over here."},{"Start":"09:08.860 ","End":"09:11.170","Text":"Then we can say plus."},{"Start":"09:11.170 ","End":"09:14.260","Text":"Now we need our v relative."},{"Start":"09:14.260 ","End":"09:19.930","Text":"First we\u0027re going to deal with the first case of the rain falling into the cart."},{"Start":"09:19.930 ","End":"09:27.550","Text":"Our v relative is going to be the velocity of the rain in relation to the ground."},{"Start":"09:27.550 ","End":"09:30.010","Text":"That is going to be here."},{"Start":"09:30.010 ","End":"09:32.020","Text":"We can see it, it\u0027s over here, so U_1,"},{"Start":"09:32.020 ","End":"09:33.940","Text":"but because it\u0027s going downwards,"},{"Start":"09:33.940 ","End":"09:36.430","Text":"which we\u0027ve said as the negative direction."},{"Start":"09:36.430 ","End":"09:39.865","Text":"We have negative U_1 and then"},{"Start":"09:39.865 ","End":"09:44.875","Text":"negative the velocity that the cart is traveling in the y-direction,"},{"Start":"09:44.875 ","End":"09:47.830","Text":"which as we know, the cart isn\u0027t traveling in the y direction,"},{"Start":"09:47.830 ","End":"09:51.190","Text":"so we can say negative 0."},{"Start":"09:51.190 ","End":"09:55.360","Text":"Then we\u0027re going to once again multiply it by Alpha"},{"Start":"09:55.360 ","End":"10:00.130","Text":"and multiply that by negative 1 because we\u0027re dealing with accretion."},{"Start":"10:00.130 ","End":"10:03.325","Text":"Then for our next case,"},{"Start":"10:03.325 ","End":"10:06.040","Text":"which is our ejection of the water."},{"Start":"10:06.040 ","End":"10:09.475","Text":"It\u0027s going to be the exact same thing as on the x-axis,"},{"Start":"10:09.475 ","End":"10:13.555","Text":"but instead of cosine Theta is going to be sine Theta."},{"Start":"10:13.555 ","End":"10:16.030","Text":"Why is it sine Theta?"},{"Start":"10:16.030 ","End":"10:17.889","Text":"This is a negative."},{"Start":"10:17.889 ","End":"10:22.630","Text":"It\u0027s sine Theta because now we want on the y-axis,"},{"Start":"10:22.630 ","End":"10:24.790","Text":"which is the side opposite to the angle,"},{"Start":"10:24.790 ","End":"10:30.460","Text":"so we use sine and it\u0027s a negative because it\u0027s going to be pointing downwards,"},{"Start":"10:30.460 ","End":"10:33.760","Text":"which is the negative direction."},{"Start":"10:33.760 ","End":"10:42.415","Text":"It\u0027s negative U_0 sine of Theta and then multiplied by the dm by dt, which is Alpha."},{"Start":"10:42.415 ","End":"10:48.260","Text":"Our Alpha remains positive because this is the ejection case."},{"Start":"10:48.540 ","End":"10:53.020","Text":"Now all we have to do is to isolate out the normal."},{"Start":"10:53.020 ","End":"10:58.420","Text":"We just have to move this m_0g to the other side of the equal sign."},{"Start":"10:58.420 ","End":"11:06.190","Text":"We\u0027re going to end up with our normal equaling to m_0g plus k,"},{"Start":"11:06.190 ","End":"11:08.290","Text":"because a negative times a negative."},{"Start":"11:08.290 ","End":"11:19.195","Text":"Plus Alpha U_1 minus Alpha U_0 sine of Theta."},{"Start":"11:19.195 ","End":"11:25.555","Text":"Now notice that because this whole expression for the normal is not dependent on time,"},{"Start":"11:25.555 ","End":"11:28.195","Text":"so we can say that it\u0027s a constant."},{"Start":"11:28.195 ","End":"11:32.350","Text":"In order to save ourselves a lot of time and a lot of effort writing out"},{"Start":"11:32.350 ","End":"11:38.410","Text":"this whole expression each time we want to write out what our normal is like over here."},{"Start":"11:38.410 ","End":"11:41.500","Text":"All we\u0027re going to do is we\u0027re just going to label it as the normal."},{"Start":"11:41.500 ","End":"11:44.020","Text":"Then our final answer,"},{"Start":"11:44.020 ","End":"11:48.140","Text":"we\u0027re going to substitute in this entire expression."},{"Start":"11:48.510 ","End":"11:51.610","Text":"Now that we have everything,"},{"Start":"11:51.610 ","End":"11:54.400","Text":"so we can just rearrange our equation on"},{"Start":"11:54.400 ","End":"11:59.185","Text":"the x-axis in order to get our equation of motion."},{"Start":"11:59.185 ","End":"12:03.940","Text":"What we\u0027re going to have is the sum of all of the forces on the x-axis,"},{"Start":"12:03.940 ","End":"12:08.560","Text":"which we know is negative Mu_k multiplied by n,"},{"Start":"12:08.560 ","End":"12:09.955","Text":"which is a constant,"},{"Start":"12:09.955 ","End":"12:15.475","Text":"which equals m_0 multiplied by dv by dt,"},{"Start":"12:15.475 ","End":"12:18.490","Text":"which is our only unknown in this whole equation."},{"Start":"12:18.490 ","End":"12:27.490","Text":"Plus Alpha times vt minus U_0 Alpha cosine"},{"Start":"12:27.490 ","End":"12:30.590","Text":"of Theta."},{"Start":"12:30.600 ","End":"12:35.245","Text":"Now what we\u0027re left with is a differential equation."},{"Start":"12:35.245 ","End":"12:37.645","Text":"Once we solve it,"},{"Start":"12:37.645 ","End":"12:41.800","Text":"we will get the solution for the equation of motion."},{"Start":"12:41.800 ","End":"12:48.250","Text":"Because we have here velocity and here we have the derivative of the velocity,"},{"Start":"12:48.250 ","End":"12:50.155","Text":"which is the acceleration."},{"Start":"12:50.155 ","End":"12:55.280","Text":"Here we finished our question number 1."},{"Start":"12:55.470 ","End":"12:59.005","Text":"Now we\u0027re going to move on to question number 2,"},{"Start":"12:59.005 ","End":"13:01.150","Text":"which is what is vf of the cart?"},{"Start":"13:01.150 ","End":"13:05.230","Text":"We\u0027re being asked to find the final velocity of the cart."},{"Start":"13:05.230 ","End":"13:12.610","Text":"Let me just remind that if we\u0027re being told that we\u0027re finding the final velocity,"},{"Start":"13:12.610 ","End":"13:17.350","Text":"then we know that because it\u0027s the final velocity it must be constant,"},{"Start":"13:17.350 ","End":"13:23.650","Text":"which therefore will mean that our dv by dt,"},{"Start":"13:23.650 ","End":"13:28.525","Text":"which equals our acceleration, will equal 0."},{"Start":"13:28.525 ","End":"13:33.715","Text":"Then we can just substitute that into our equation over here,"},{"Start":"13:33.715 ","End":"13:35.980","Text":"where we have dv by dt,"},{"Start":"13:35.980 ","End":"13:38.110","Text":"this expression will become 0."},{"Start":"13:38.110 ","End":"13:44.620","Text":"We\u0027ll have then negative Mu_k multiplied by n,"},{"Start":"13:44.620 ","End":"13:48.550","Text":"which will equal to Alpha vt,"},{"Start":"13:48.550 ","End":"13:53.800","Text":"which here will be alpha v final v at time t, our final t,"},{"Start":"13:53.800 ","End":"14:03.280","Text":"minus U_0 Alpha cosine"},{"Start":"14:03.280 ","End":"14:05.035","Text":"of Theta."},{"Start":"14:05.035 ","End":"14:10.390","Text":"Then all we have to do because we\u0027re being asked to find what our v_f is,"},{"Start":"14:10.390 ","End":"14:13.990","Text":"to just isolate out our v final."},{"Start":"14:13.990 ","End":"14:19.615","Text":"Then what we\u0027ll get is that a v final is equal to"},{"Start":"14:19.615 ","End":"14:24.590","Text":"U_0 Alpha"},{"Start":"14:26.250 ","End":"14:30.490","Text":"cosine of"},{"Start":"14:30.490 ","End":"14:39.325","Text":"Theta minus Mu_k n divided by 1 over Alpha."},{"Start":"14:39.325 ","End":"14:43.760","Text":"Then we have it. That\u0027s our final answer for question number 2."},{"Start":"14:44.160 ","End":"14:46.765","Text":"Now in question number 3,"},{"Start":"14:46.765 ","End":"14:52.315","Text":"we\u0027re being asked what is the velocity of the cart as a function of time?"},{"Start":"14:52.315 ","End":"14:56.470","Text":"Now, the way that we solve it is by looking at"},{"Start":"14:56.470 ","End":"15:01.885","Text":"our equation from number 1 and solving it like a differential equation."},{"Start":"15:01.885 ","End":"15:05.043","Text":"Let\u0027s see how we do that."},{"Start":"15:05.043 ","End":"15:07.170","Text":"We have this equation,"},{"Start":"15:07.170 ","End":"15:09.495","Text":"which is our answer to question number 1."},{"Start":"15:09.495 ","End":"15:15.690","Text":"Now we have to rearrange it in order to get derivative over here,"},{"Start":"15:15.690 ","End":"15:18.690","Text":"dv by dt on 1 side of the equal side and"},{"Start":"15:18.690 ","End":"15:22.440","Text":"all the other terms of the equation on the other side."},{"Start":"15:22.440 ","End":"15:25.395","Text":"We\u0027re just going to do some simple algebra."},{"Start":"15:25.395 ","End":"15:32.085","Text":"What we\u0027ll get is M_0 dv by dt is equal to"},{"Start":"15:32.085 ","End":"15:39.045","Text":"u_0 Alpha cosine of Theta minus"},{"Start":"15:39.045 ","End":"15:49.020","Text":"Mu_k N minus Alpha v(t)."},{"Start":"15:49.020 ","End":"15:50.625","Text":"Why did I set it up this way?"},{"Start":"15:50.625 ","End":"15:54.045","Text":"Because the expression with the dv by dt,"},{"Start":"15:54.045 ","End":"15:56.895","Text":"I need to have on 1 side because that\u0027s what"},{"Start":"15:56.895 ","End":"15:59.790","Text":"I need to do when solving a differential equation."},{"Start":"15:59.790 ","End":"16:08.670","Text":"My term for u_0 Alpha cosine Theta minus Mu_k N is a constant."},{"Start":"16:08.670 ","End":"16:12.540","Text":"This is constant."},{"Start":"16:12.540 ","End":"16:22.270","Text":"Then my Alpha multiplied by v(t) is my variable that I\u0027m going to be integrating by."},{"Start":"16:22.720 ","End":"16:26.180","Text":"Again for the sake of saving time,"},{"Start":"16:26.180 ","End":"16:28.400","Text":"I\u0027m just going to say that because this is constant,"},{"Start":"16:28.400 ","End":"16:31.580","Text":"we\u0027re just going to label it C. Then we don\u0027t have to write"},{"Start":"16:31.580 ","End":"16:35.725","Text":"down this equation and carry it over every time."},{"Start":"16:35.725 ","End":"16:42.750","Text":"Now what we\u0027re going to do is we\u0027re going to multiply both sides by dt."},{"Start":"16:42.750 ","End":"16:50.430","Text":"We\u0027re going to be left with M_0 dv is equal to C minus"},{"Start":"16:50.430 ","End":"16:59.470","Text":"Alpha v(t) and then all of this multiplied by dt."},{"Start":"17:00.830 ","End":"17:04.830","Text":"Now what I want to do is I want to manipulate"},{"Start":"17:04.830 ","End":"17:09.180","Text":"this equation in order to get what is inside this orange box,"},{"Start":"17:09.180 ","End":"17:12.840","Text":"which has some function with variable"},{"Start":"17:12.840 ","End":"17:19.005","Text":"(v)dv equals some function where the variable (t)dt."},{"Start":"17:19.005 ","End":"17:21.735","Text":"What I\u0027m going to do because my v is over here,"},{"Start":"17:21.735 ","End":"17:27.945","Text":"I\u0027m just going to divide both sides by C minus Alpha v(t)."},{"Start":"17:27.945 ","End":"17:29.790","Text":"What I\u0027m going to get in the end"},{"Start":"17:29.790 ","End":"17:38.549","Text":"is C minus Alpha"},{"Start":"17:38.549 ","End":"17:42.900","Text":"v(t) multiplied by dv,"},{"Start":"17:42.900 ","End":"17:46.335","Text":"which will equal just dt."},{"Start":"17:46.335 ","End":"17:52.170","Text":"Then all I\u0027m going to do is add my integration signs to both sides."},{"Start":"17:52.170 ","End":"17:54.300","Text":"Then also, you can do"},{"Start":"17:54.300 ","End":"17:58.875","Text":"this integration as a definite integral or as an indefinite integral."},{"Start":"17:58.875 ","End":"18:02.355","Text":"But I always like to do it as a definite integral."},{"Start":"18:02.355 ","End":"18:06.825","Text":"t is going from time of 0 up until my time of t."},{"Start":"18:06.825 ","End":"18:12.120","Text":"My velocity is going from my velocity at t equals 0,"},{"Start":"18:12.120 ","End":"18:15.210","Text":"which we can see in the diagram."},{"Start":"18:15.210 ","End":"18:19.480","Text":"Over here, is v_0, my initial velocity."},{"Start":"18:19.790 ","End":"18:25.035","Text":"We\u0027re going from v_0 until my velocity at time"},{"Start":"18:25.035 ","End":"18:32.265","Text":"t. Notice my velocity at time t isn\u0027t the same as in my bounds,"},{"Start":"18:32.265 ","End":"18:37.755","Text":"isn\u0027t the same as my velocity at time t that I\u0027m integrating on."},{"Start":"18:37.755 ","End":"18:43.690","Text":"I\u0027m integrating on this until this bound of v(t), but it doesn\u0027t matter."},{"Start":"18:44.720 ","End":"18:55.635","Text":"My integral of dt will just simply equal t. Then if I integrate this side,"},{"Start":"18:55.635 ","End":"18:57.900","Text":"we\u0027re going to have M_0,"},{"Start":"18:57.900 ","End":"19:04.560","Text":"which is a constant divided by the derivative of the denominator,"},{"Start":"19:04.560 ","End":"19:08.430","Text":"which is just negative Alpha multiplied by"},{"Start":"19:08.430 ","End":"19:15.690","Text":"ln of C minus Alpha v(t),"},{"Start":"19:15.690 ","End":"19:21.660","Text":"then with the bounds between v_0 and v(t),"},{"Start":"19:21.660 ","End":"19:30.915","Text":"which is just going to be equal to negative M_0 over Alpha"},{"Start":"19:30.915 ","End":"19:35.310","Text":"multiplied by ln of C minus"},{"Start":"19:35.310 ","End":"19:42.090","Text":"Alpha v(t) divided by C minus v_0,"},{"Start":"19:42.090 ","End":"19:44.895","Text":"which will equal t. Now,"},{"Start":"19:44.895 ","End":"19:51.375","Text":"why do I have here C minus Alpha v(t) and here C just minus v_0."},{"Start":"19:51.375 ","End":"19:59.800","Text":"Why is that? Because my Alpha is my rate of accretion or my rate of ejection."},{"Start":"20:04.340 ","End":"20:06.885","Text":"At t equals 0,"},{"Start":"20:06.885 ","End":"20:08.460","Text":"no time has elapsed,"},{"Start":"20:08.460 ","End":"20:14.190","Text":"which means that the rates just doesn\u0027t exist yet because no time has"},{"Start":"20:14.190 ","End":"20:17.505","Text":"elapsed for any mass to have been added"},{"Start":"20:17.505 ","End":"20:21.150","Text":"to my cards or to have been subtracted from my card."},{"Start":"20:21.150 ","End":"20:23.890","Text":"I don\u0027t have to take it into account."},{"Start":"20:24.500 ","End":"20:26.745","Text":"That is for that."},{"Start":"20:26.745 ","End":"20:28.995","Text":"Now I have this expression."},{"Start":"20:28.995 ","End":"20:32.250","Text":"I\u0027m just going to rearrange it a little bit."},{"Start":"20:32.250 ","End":"20:42.600","Text":"I\u0027m going to say that my ln of C minus Alpha v(t) divided by C minus v_0 is going"},{"Start":"20:42.600 ","End":"20:49.260","Text":"to be equal to negative Alpha over M_0 multiplied by t. I"},{"Start":"20:49.260 ","End":"20:57.960","Text":"just divided both sides by negative 0 divided by Alpha."},{"Start":"20:57.960 ","End":"21:06.885","Text":"Then I\u0027m going to raise both sides of the expression by e. That will get rid of my ln."},{"Start":"21:06.885 ","End":"21:14.370","Text":"Then I\u0027m gonna have seen negative Alpha v(t) divided by C negative v_0"},{"Start":"21:14.370 ","End":"21:22.785","Text":"is equal to e to the negative Alpha over M_0 multiplied by t. Now,"},{"Start":"21:22.785 ","End":"21:30.520","Text":"because I\u0027m being asked for the velocity as a function of time,"},{"Start":"21:30.530 ","End":"21:37.185","Text":"all I\u0027m going to do is I\u0027m just going to isolate out my v(t) over here."},{"Start":"21:37.185 ","End":"21:39.900","Text":"Through simple algebra, we\u0027ll see that we\u0027ll"},{"Start":"21:39.900 ","End":"21:47.415","Text":"get our final answer for our velocity as a function of time is equal to"},{"Start":"21:47.415 ","End":"21:55.515","Text":"negative Alpha times e to the power of negative Alpha over M_0 multiplied by t. Then that"},{"Start":"21:55.515 ","End":"22:05.970","Text":"multiplied by C minus v minus C. That\u0027s the end of this question."},{"Start":"22:05.970 ","End":"22:09.490","Text":"There will be more questions to come."}],"ID":10620},{"Watched":false,"Name":"Elastic Collision - Water Sprayed On Cart","Duration":"26m 50s","ChapterTopicVideoID":9232,"CourseChapterTopicPlaylistID":5429,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.290 ","End":"00:03.165","Text":"Hello. In this exercise,"},{"Start":"00:03.165 ","End":"00:05.715","Text":"again, we\u0027re dealing with a change in mass."},{"Start":"00:05.715 ","End":"00:08.408","Text":"We\u0027re being told that we have a cart,"},{"Start":"00:08.408 ","End":"00:11.910","Text":"and its initial mass is M. Then we have"},{"Start":"00:11.910 ","End":"00:16.680","Text":"a hose that water is coming out of it at a velocity of v0."},{"Start":"00:16.680 ","End":"00:19.725","Text":"The rate at which water is coming out is"},{"Start":"00:19.725 ","End":"00:26.370","Text":"M.. Now we\u0027re told that there\u0027s no friction acting on the cards because of those wheels,"},{"Start":"00:26.370 ","End":"00:30.795","Text":"and we\u0027re told that there\u0027s elastic collision relative to the cart."},{"Start":"00:30.795 ","End":"00:36.960","Text":"Now, we can imagine as if there\u0027s a man standing in the cart,"},{"Start":"00:36.960 ","End":"00:45.760","Text":"and he\u0027s in the system which is the cart because we\u0027re looking at relative to the cart."},{"Start":"00:45.760 ","End":"00:49.235","Text":"Then as far as this man is concerned,"},{"Start":"00:49.235 ","End":"00:50.780","Text":"the cart isn\u0027t moving."},{"Start":"00:50.780 ","End":"00:58.165","Text":"He\u0027s stationary, but he just sees water hitting the cart and elastic collision occurring."},{"Start":"00:58.165 ","End":"01:02.180","Text":"What we\u0027re being asked in the question is,"},{"Start":"01:02.180 ","End":"01:07.205","Text":"what is the velocity of the cart as a function of time?"},{"Start":"01:07.205 ","End":"01:11.185","Text":"This is what we want to find out."},{"Start":"01:11.185 ","End":"01:15.665","Text":"How are we going to answer this is by using our equation."},{"Start":"01:15.665 ","End":"01:21.245","Text":"The sum of the external forces is equal to the mass of the cart,"},{"Start":"01:21.245 ","End":"01:23.650","Text":"which is changing over time,"},{"Start":"01:23.650 ","End":"01:28.970","Text":"multiplied by dv by dt of the cart of the large object,"},{"Start":"01:28.970 ","End":"01:30.350","Text":"it\u0027s always of the main object,"},{"Start":"01:30.350 ","End":"01:33.305","Text":"so cart in this as well."},{"Start":"01:33.305 ","End":"01:35.420","Text":"Then v relative,"},{"Start":"01:35.420 ","End":"01:37.415","Text":"which is our relative velocity,"},{"Start":"01:37.415 ","End":"01:43.820","Text":"and multiply by dm by dt."},{"Start":"01:43.820 ","End":"01:48.975","Text":"Remember that this m is different to"},{"Start":"01:48.975 ","End":"01:54.425","Text":"this m over here because it\u0027s the rates that the water is hitting it."},{"Start":"01:54.425 ","End":"02:00.550","Text":"In this section, we\u0027re speaking about the water."},{"Start":"02:01.820 ","End":"02:09.110","Text":"How do we apply the notion of water going into the cart in this question?"},{"Start":"02:09.110 ","End":"02:11.630","Text":"Because as we can see, the water is hitting the cart and"},{"Start":"02:11.630 ","End":"02:15.530","Text":"then bouncing back, being sprayed back."},{"Start":"02:15.530 ","End":"02:17.015","Text":"How do we do this?"},{"Start":"02:17.015 ","End":"02:22.730","Text":"If we imagine that we\u0027re extending the cart over here."},{"Start":"02:22.730 ","End":"02:26.045","Text":"Then we can see that the water is going into the cart,"},{"Start":"02:26.045 ","End":"02:27.920","Text":"and then going back out."},{"Start":"02:27.920 ","End":"02:30.680","Text":"That\u0027s how we\u0027re going to look at this."},{"Start":"02:30.680 ","End":"02:33.979","Text":"In that case we have to add onto this expression."},{"Start":"02:33.979 ","End":"02:35.060","Text":"We have our V"},{"Start":"02:35.060 ","End":"02:42.065","Text":"relative relative going in"},{"Start":"02:42.065 ","End":"02:48.030","Text":"to our extended cart multiplied by dm by dt."},{"Start":"02:48.030 ","End":"02:52.670","Text":"Then we\u0027re going to add on,"},{"Start":"02:52.670 ","End":"02:56.990","Text":"so we\u0027re going to say plus V rel,"},{"Start":"02:56.990 ","End":"03:00.260","Text":"but this time going out."},{"Start":"03:00.260 ","End":"03:06.570","Text":"Multiplied by dm by dt going out."},{"Start":"03:08.750 ","End":"03:12.395","Text":"We\u0027re looking at it right now,"},{"Start":"03:12.395 ","End":"03:14.780","Text":"is if we have a cart,"},{"Start":"03:14.780 ","End":"03:16.100","Text":"with rain falling in,"},{"Start":"03:16.100 ","End":"03:21.725","Text":"and with some hole with the water falling back out."},{"Start":"03:21.725 ","End":"03:28.110","Text":"This is the exact situation as what we have here with our cart."},{"Start":"03:29.210 ","End":"03:34.340","Text":"Now let\u0027s speak about how we\u0027re going to plug this into the equation."},{"Start":"03:34.340 ","End":"03:39.015","Text":"Sum of external forces is equal to 0."},{"Start":"03:39.015 ","End":"03:41.720","Text":"Because we said that there\u0027s no friction acting on the cart,"},{"Start":"03:41.720 ","End":"03:44.300","Text":"which means that there\u0027s no forces acting,"},{"Start":"03:44.300 ","End":"03:47.645","Text":"and of course we\u0027re only looking in the x direction."},{"Start":"03:47.645 ","End":"03:51.170","Text":"The sum of all of the forces in the x-direction is equal to 0,"},{"Start":"03:51.170 ","End":"03:55.235","Text":"which is equal to our mass as a function of time,"},{"Start":"03:55.235 ","End":"03:57.560","Text":"we\u0027ll get to that soon,"},{"Start":"03:57.560 ","End":"04:01.930","Text":"multiplied by dv by dt."},{"Start":"04:01.930 ","End":"04:07.650","Text":"Then we have to now speak about V relative in."},{"Start":"04:07.650 ","End":"04:10.505","Text":"Let\u0027s see how we do that."},{"Start":"04:10.505 ","End":"04:13.020","Text":"Scroll a bit to the side."},{"Start":"04:13.150 ","End":"04:23.270","Text":"V rel going in is going to be equal to the velocity of the water relative to Earth,"},{"Start":"04:23.270 ","End":"04:30.065","Text":"which is V_0 minus the velocity of the cart."},{"Start":"04:30.065 ","End":"04:35.600","Text":"The velocity of the cart is going to be V as a function of t,"},{"Start":"04:35.600 ","End":"04:40.910","Text":"so that\u0027s our v relative in."},{"Start":"04:40.910 ","End":"04:45.830","Text":"Then we can say that V relative out."},{"Start":"04:45.830 ","End":"04:49.610","Text":"The water, we\u0027re looking,"},{"Start":"04:49.610 ","End":"04:53.615","Text":"this is our v in this direction,"},{"Start":"04:53.615 ","End":"04:56.645","Text":"and then V out is in this direction."},{"Start":"04:56.645 ","End":"05:04.520","Text":"Now because this is all compared to a viewer which is standing over here."},{"Start":"05:04.520 ","End":"05:08.315","Text":"We were in the system of the cart."},{"Start":"05:08.315 ","End":"05:12.185","Text":"If we\u0027re looking really in the system of the cart,"},{"Start":"05:12.185 ","End":"05:16.935","Text":"then we can say that it\u0027s negative V rel in."},{"Start":"05:16.935 ","End":"05:22.685","Text":"Of course, if we were looking in the system of the ground of Earth,"},{"Start":"05:22.685 ","End":"05:26.630","Text":"than it would be slightly different than the V rel out and V rel"},{"Start":"05:26.630 ","End":"05:31.460","Text":"would not be equal and opposite to each other."},{"Start":"05:31.460 ","End":"05:35.510","Text":"They\u0027re also equal and opposite because we know"},{"Start":"05:35.510 ","End":"05:39.485","Text":"that the collision between the water and the cart is elastic collision,"},{"Start":"05:39.485 ","End":"05:41.600","Text":"which means that no energy is lost,"},{"Start":"05:41.600 ","End":"05:45.750","Text":"which means that the velocity is maintained."},{"Start":"05:46.430 ","End":"05:51.450","Text":"Because the relative out is equal to negative v relative in,"},{"Start":"05:51.450 ","End":"05:57.960","Text":"so we can say that it\u0027s equal to negative V_0 plus V(t)."},{"Start":"05:57.960 ","End":"06:04.620","Text":"Now let\u0027s take a look at what M(t) is equal to."},{"Start":"06:04.850 ","End":"06:10.340","Text":"Now M(t), let\u0027s see what this is."},{"Start":"06:10.340 ","End":"06:16.310","Text":"Because we know that our dm by DT in and our DM by DT out are the same,"},{"Start":"06:16.310 ","End":"06:19.790","Text":"so we know that our M(t) is"},{"Start":"06:19.790 ","End":"06:23.690","Text":"going to be just equal to n because the masses are unchanging."},{"Start":"06:23.690 ","End":"06:25.385","Text":"Every mass that is added,"},{"Start":"06:25.385 ","End":"06:30.690","Text":"any dm by dt is then taken out in a dm by dt out."},{"Start":"06:30.690 ","End":"06:34.325","Text":"Another thing here we have to multiply by negative 1."},{"Start":"06:34.325 ","End":"06:38.570","Text":"Remember because whenever mass is being added to the cart,"},{"Start":"06:38.570 ","End":"06:42.050","Text":"because of the way that we derive this equation,"},{"Start":"06:42.050 ","End":"06:46.480","Text":"then we multiply it by negative 1."},{"Start":"06:46.970 ","End":"06:50.240","Text":"Let\u0027s add all of this into our equation."},{"Start":"06:50.240 ","End":"06:54.365","Text":"We already know that M(t) is just"},{"Start":"06:54.365 ","End":"06:59.765","Text":"simply going to be m. We can rub out this as a function of t,"},{"Start":"06:59.765 ","End":"07:03.214","Text":"dv by dt remains unknown."},{"Start":"07:03.214 ","End":"07:06.965","Text":"Then we\u0027re going to have V rel,"},{"Start":"07:06.965 ","End":"07:10.985","Text":"in which we said is V_0 negative v(t),"},{"Start":"07:10.985 ","End":"07:12.830","Text":"but it\u0027s multiplied by negative 1,"},{"Start":"07:12.830 ","End":"07:16.654","Text":"so it\u0027s going to be negative V_0 minus"},{"Start":"07:16.654 ","End":"07:23.465","Text":"V(t) in brackets multiplied by dm by dt."},{"Start":"07:23.465 ","End":"07:27.875","Text":"Then we\u0027re going to have V relative out,"},{"Start":"07:27.875 ","End":"07:31.650","Text":"which is going to be negative V_0 plus v(t)."},{"Start":"07:31.650 ","End":"07:34.130","Text":"If we put the negative over here,"},{"Start":"07:34.130 ","End":"07:42.365","Text":"then we\u0027ll have V_0 negative v(t) multiplied by dm by dt."},{"Start":"07:42.365 ","End":"07:47.060","Text":"Now, all I did here was factored in into these brackets."},{"Start":"07:47.060 ","End":"07:48.500","Text":"But if you see, if you open the brackets,"},{"Start":"07:48.500 ","End":"07:49.805","Text":"you\u0027ll get the exact same thing."},{"Start":"07:49.805 ","End":"07:53.370","Text":"You\u0027ll get negative V_0 plus v(t),"},{"Start":"07:53.370 ","End":"07:55.890","Text":"so it matches up."},{"Start":"07:55.890 ","End":"07:58.955","Text":"This is our equation, and now,"},{"Start":"07:58.955 ","End":"08:03.815","Text":"we can just rearrange it because we see that these 2 values are the exact same thing."},{"Start":"08:03.815 ","End":"08:10.180","Text":"We can just say that it\u0027s negative 2 and then rub that out over here."},{"Start":"08:10.700 ","End":"08:14.920","Text":"Now we have dv by dt,"},{"Start":"08:14.920 ","End":"08:18.860","Text":"which is the derivative of v(t)."},{"Start":"08:18.860 ","End":"08:23.194","Text":"How are we going to solve this with a differential equation?"},{"Start":"08:23.194 ","End":"08:26.150","Text":"Now, just before we do that,"},{"Start":"08:26.150 ","End":"08:29.765","Text":"I just realized we still don\u0027t know what our dm by dt is?"},{"Start":"08:29.765 ","End":"08:32.075","Text":"Now our dm by dt,"},{"Start":"08:32.075 ","End":"08:37.100","Text":"the rate that the water comes in and out of our cart after"},{"Start":"08:37.100 ","End":"08:42.965","Text":"the extension is not the same as the rate at which the water is coming out of the hose."},{"Start":"08:42.965 ","End":"08:46.235","Text":"I know that sounds a little bit strange,"},{"Start":"08:46.235 ","End":"08:49.115","Text":"but I\u0027m going to first explain it mathematically,"},{"Start":"08:49.115 ","End":"08:51.290","Text":"why it is different, and then afterwards,"},{"Start":"08:51.290 ","End":"08:54.290","Text":"I\u0027ll explain it in a more easy to understand way,"},{"Start":"08:54.290 ","End":"08:57.270","Text":"so that we can understand it intuitively."},{"Start":"08:57.740 ","End":"09:02.790","Text":"Right now we\u0027re speaking about why m.,"},{"Start":"09:02.790 ","End":"09:08.885","Text":"which we were given in the question as the rate at which the water is leaving the hose,"},{"Start":"09:08.885 ","End":"09:13.760","Text":"which is equal to dm tag by dt."},{"Start":"09:13.760 ","End":"09:19.370","Text":"Dm tag just represents the water coming out of the hose and we said"},{"Start":"09:19.370 ","End":"09:24.345","Text":"that it is not equal to the dm star,"},{"Start":"09:24.345 ","End":"09:26.890","Text":"let\u0027s call it by dt,"},{"Start":"09:26.920 ","End":"09:32.020","Text":"which is the water coming in and out of the cart."},{"Start":"09:32.020 ","End":"09:35.270","Text":"Let\u0027s take a look over here and draw the cart."},{"Start":"09:35.270 ","End":"09:39.680","Text":"If you remember, this was the extended section of the cart."},{"Start":"09:39.680 ","End":"09:43.190","Text":"Then we can say that here there\u0027s the invisible wall,"},{"Start":"09:43.190 ","End":"09:47.930","Text":"but with which the water enters and exits."},{"Start":"09:47.930 ","End":"09:50.360","Text":"Remember it looks something like this."},{"Start":"09:50.360 ","End":"09:57.280","Text":"Here is the rest of the cart where a person is standing."},{"Start":"09:57.280 ","End":"10:03.635","Text":"What are we going to ask is in an infinitely small time of d,"},{"Start":"10:03.635 ","End":"10:11.970","Text":"what mass of water dm enters the cart."},{"Start":"10:12.490 ","End":"10:14.975","Text":"At any given time,"},{"Start":"10:14.975 ","End":"10:18.290","Text":"some kind of cylinder of water,"},{"Start":"10:18.290 ","End":"10:25.730","Text":"so this is a cylinder of water and its length is going to be Delta x,"},{"Start":"10:25.730 ","End":"10:28.865","Text":"so it enters into the cart."},{"Start":"10:28.865 ","End":"10:36.120","Text":"Then the cross-sectional area of the ends is going to be equal to a."},{"Start":"10:36.370 ","End":"10:41.510","Text":"This a matches up to the cross-sectional area of the hose."},{"Start":"10:41.510 ","End":"10:43.535","Text":"If this is the hose,"},{"Start":"10:43.535 ","End":"10:46.025","Text":"this is also a."},{"Start":"10:46.025 ","End":"10:53.130","Text":"Now dm is going to be equal"},{"Start":"10:53.130 ","End":"11:03.025","Text":"to cross sectional area multiplied by Delta x multiplied by Rho,"},{"Start":"11:03.025 ","End":"11:06.340","Text":"which is the density of the water."},{"Start":"11:06.340 ","End":"11:08.800","Text":"To this over here,"},{"Start":"11:08.800 ","End":"11:12.715","Text":"represents the volume of the water."},{"Start":"11:12.715 ","End":"11:16.870","Text":"Now we have to figure out what our Delta x is,"},{"Start":"11:16.870 ","End":"11:24.110","Text":"so our Delta x is going to be equal to V relative in."},{"Start":"11:24.680 ","End":"11:27.420","Text":"Now, why v relative in?"},{"Start":"11:27.420 ","End":"11:31.830","Text":"Because we\u0027re looking at the guy standing in the cart."},{"Start":"11:31.830 ","End":"11:34.320","Text":"We\u0027re not looking at the velocity with which the water"},{"Start":"11:34.320 ","End":"11:37.290","Text":"exits the hose because that will be relative to the hose."},{"Start":"11:37.290 ","End":"11:38.700","Text":"In the question specifically,"},{"Start":"11:38.700 ","End":"11:40.905","Text":"we\u0027re told that it\u0027s relative to the cart."},{"Start":"11:40.905 ","End":"11:46.440","Text":"We have to say that it\u0027s multiplied by V_rel_in multiplied by"},{"Start":"11:46.440 ","End":"11:53.880","Text":"Delta t because its speed is distance times time and then we just rearrange the formula."},{"Start":"11:53.880 ","End":"11:57.660","Text":"Now our V relative in, as we said,"},{"Start":"11:57.660 ","End":"12:03.105","Text":"was equal to v_0 minus v as a function of t,"},{"Start":"12:03.105 ","End":"12:07.815","Text":"and then multiplied by Delta t, like over here."},{"Start":"12:07.815 ","End":"12:18.240","Text":"Then we can say that our Delta m is equal to Rho A multiplied by v_0 minus v"},{"Start":"12:18.240 ","End":"12:23.715","Text":"as a function of t and then multiplied by"},{"Start":"12:23.715 ","End":"12:31.500","Text":"Delta t. Now what\u0027s left for me to find out is what my Rho A is equal to."},{"Start":"12:31.500 ","End":"12:34.065","Text":"I\u0027m going to find out what my Rho A is equal to,"},{"Start":"12:34.065 ","End":"12:37.080","Text":"I\u0027m going to deal with that exact same calculations just on"},{"Start":"12:37.080 ","End":"12:41.400","Text":"the water exiting the hose. Let\u0027s do this."},{"Start":"12:41.400 ","End":"12:45.420","Text":"Here is my hose and here\u0027s the water exiting it."},{"Start":"12:45.420 ","End":"12:50.610","Text":"We know that the water exiting is exiting at a velocity of v_0."},{"Start":"12:50.610 ","End":"12:53.400","Text":"Then I have, in this water that\u0027s,"},{"Start":"12:53.400 ","End":"13:02.160","Text":"exiting some cylinder again of water which is entering into my invisible wall."},{"Start":"13:02.160 ","End":"13:05.740","Text":"Which is this again, into the cart."},{"Start":"13:06.110 ","End":"13:12.090","Text":"This is also going to be of length Delta x,"},{"Start":"13:12.090 ","End":"13:13.785","Text":"but it\u0027s a different Delta x."},{"Start":"13:13.785 ","End":"13:16.020","Text":"Because this Delta x was the relative,"},{"Start":"13:16.020 ","End":"13:19.090","Text":"so I\u0027ll do a tag over here."},{"Start":"13:22.250 ","End":"13:26.580","Text":"This is the Delta x relative to the hose,"},{"Start":"13:26.580 ","End":"13:27.975","Text":"which is relative to the ground."},{"Start":"13:27.975 ","End":"13:29.865","Text":"That\u0027s why I didn\u0027t do a tag."},{"Start":"13:29.865 ","End":"13:37.950","Text":"Then what I\u0027ll say is speed is distance over time,"},{"Start":"13:37.950 ","End":"13:39.270","Text":"and then we rearrange it,"},{"Start":"13:39.270 ","End":"13:42.270","Text":"so my Delta x is equal to my velocity,"},{"Start":"13:42.270 ","End":"13:46.290","Text":"which is v_0 multiplied by my time,"},{"Start":"13:46.290 ","End":"13:50.505","Text":"my Delta t, my infinitely small amount of time."},{"Start":"13:50.505 ","End":"13:54.389","Text":"Then I can say that through the exact same calculation,"},{"Start":"13:54.389 ","End":"13:58.875","Text":"that my Delta m is equal to Rho A,"},{"Start":"13:58.875 ","End":"14:05.235","Text":"so the density of water multiplied by the cross-sectional area of the hose."},{"Start":"14:05.235 ","End":"14:07.560","Text":"Then multiplied by my dx,"},{"Start":"14:07.560 ","End":"14:13.710","Text":"which is v_0 multiplied by Delta t. Now, as we know,"},{"Start":"14:13.710 ","End":"14:19.020","Text":"because we were told in the question that the water exiting is going at a rate of m.,"},{"Start":"14:19.020 ","End":"14:28.845","Text":"so we can say that our m. is equal to Delta m divided by Delta t,"},{"Start":"14:28.845 ","End":"14:34.770","Text":"which is then equal to Rho A multiplied by v_0."},{"Start":"14:34.770 ","End":"14:42.795","Text":"Therefore, my Rho A is equal to m. divided by v_0."},{"Start":"14:42.795 ","End":"14:46.020","Text":"All I\u0027ve done is rearrange this equation."},{"Start":"14:46.020 ","End":"14:47.550","Text":"That\u0027s my Rho A."},{"Start":"14:47.550 ","End":"14:53.505","Text":"Therefore, I can say that my dm by dt over here,"},{"Start":"14:53.505 ","End":"14:58.095","Text":"my Delta m divided by"},{"Start":"14:58.095 ","End":"15:04.755","Text":"Delta t is equal to m. which is my Rho A,"},{"Start":"15:04.755 ","End":"15:11.760","Text":"multiplied by v_0 minus v as a function of t. All this"},{"Start":"15:11.760 ","End":"15:20.620","Text":"divided by v_0 because I already divided both sides by my Delta T over here."},{"Start":"15:20.750 ","End":"15:31.240","Text":"All of this is equal to my dm star that we saw before, divided by dt."},{"Start":"15:33.200 ","End":"15:36.810","Text":"Now, if we look at this equation,"},{"Start":"15:36.810 ","End":"15:38.265","Text":"if the cost isn\u0027t moving,"},{"Start":"15:38.265 ","End":"15:40.965","Text":"which means that our v(t)=0,"},{"Start":"15:40.965 ","End":"15:43.658","Text":"then we see that our v_0\u0027s cancel out,"},{"Start":"15:43.658 ","End":"15:46.770","Text":"and that we\u0027re really left with m. which"},{"Start":"15:46.770 ","End":"15:51.145","Text":"is the rate at which the water is leaving the hose."},{"Start":"15:51.145 ","End":"15:59.835","Text":"So this is why m. does not equal to dm by dt star,"},{"Start":"15:59.835 ","End":"16:02.625","Text":"where this dm by dt is"},{"Start":"16:02.625 ","End":"16:05.640","Text":"the rate that the water is coming"},{"Start":"16:05.640 ","End":"16:09.480","Text":"in and the rate that the water is coming out of our cart."},{"Start":"16:09.480 ","End":"16:13.290","Text":"At the end of the lesson, I\u0027ll give a slightly more intuitive explanation,"},{"Start":"16:13.290 ","End":"16:16.780","Text":"but this is the mathematical explanation."},{"Start":"16:17.180 ","End":"16:20.745","Text":"Now we can substitute in our dm by dt,"},{"Start":"16:20.745 ","End":"16:24.210","Text":"which is actually our dm star by dt."},{"Start":"16:24.210 ","End":"16:26.730","Text":"Let\u0027s just substitute that in."},{"Start":"16:26.730 ","End":"16:31.110","Text":"If we remembered was v_0 minus v_t."},{"Start":"16:31.110 ","End":"16:38.730","Text":"If we remember our dm star by dt was equal to v_0 minus"},{"Start":"16:38.730 ","End":"16:46.920","Text":"v(t) multiplied by m. divided by v_0."},{"Start":"16:46.920 ","End":"16:50.355","Text":"Here we already have v_0 minus v(t)."},{"Start":"16:50.355 ","End":"16:51.942","Text":"We\u0027re multiplying it by this,"},{"Start":"16:51.942 ","End":"16:58.240","Text":"so that\u0027s squared multiplied by m. divided by v_0."},{"Start":"16:58.610 ","End":"17:01.800","Text":"Now we have everything."},{"Start":"17:01.800 ","End":"17:05.775","Text":"We have our m. which is given in the question."},{"Start":"17:05.775 ","End":"17:10.050","Text":"We have our v_0, which we know in the question,"},{"Start":"17:10.050 ","End":"17:12.105","Text":"our m which is also given in the question."},{"Start":"17:12.105 ","End":"17:19.170","Text":"Now we can solve with our differential equation like I promised you a few minutes ago."},{"Start":"17:19.170 ","End":"17:27.180","Text":"Now we\u0027re just going to rearrange this equation to get all of our values with the v in"},{"Start":"17:27.180 ","End":"17:30.600","Text":"them to one side for our dv and all of"},{"Start":"17:30.600 ","End":"17:35.070","Text":"the values with t\u0027s in them to the other side with our dt."},{"Start":"17:35.070 ","End":"17:40.050","Text":"What we\u0027re going to do is we\u0027re going to have that our m"},{"Start":"17:40.050 ","End":"17:44.910","Text":"multiplied by dv by dt is going to be equal to"},{"Start":"17:44.910 ","End":"17:49.410","Text":"2 v_0 minus v as a function of t^2"},{"Start":"17:49.410 ","End":"17:57.225","Text":"multiplied by m. divided by v_0."},{"Start":"17:57.225 ","End":"18:03.473","Text":"Then what we\u0027re going to do is we\u0027re going to multiply both sides by dt,"},{"Start":"18:03.473 ","End":"18:12.480","Text":"and divide both sides by m. Then we\u0027re going to have that our dv=2"},{"Start":"18:12.480 ","End":"18:22.275","Text":"divided by v_0 multiplied by m. multiplied by m. Sorry,"},{"Start":"18:22.275 ","End":"18:28.500","Text":"this is going to be divided by M. We can have 2m."},{"Start":"18:28.500 ","End":"18:34.560","Text":"divided by v_0M multiplied"},{"Start":"18:34.560 ","End":"18:40.750","Text":"by v_0 minus v(t) squared."},{"Start":"18:40.950 ","End":"18:43.720","Text":"Then dt of course."},{"Start":"18:43.720 ","End":"18:51.010","Text":"Now, all we have to do is get our vs over to the side, and then integrate."},{"Start":"18:51.010 ","End":"18:53.275","Text":"Let\u0027s scroll down a little bit."},{"Start":"18:53.275 ","End":"19:02.410","Text":"Then what we will have is we will have dv divided by v0 minus"},{"Start":"19:02.410 ","End":"19:13.100","Text":"v(t) and all of this squared integral which is equal to 2m dot over v0m."},{"Start":"19:13.440 ","End":"19:20.440","Text":"All of these are constants so we can put them outside of the integral sign multiplied by,"},{"Start":"19:20.440 ","End":"19:23.470","Text":"integral according to dt."},{"Start":"19:23.470 ","End":"19:26.740","Text":"We\u0027re going from time 0 up until time t,"},{"Start":"19:26.740 ","End":"19:30.970","Text":"some t. Which means that our velocity at v at"},{"Start":"19:30.970 ","End":"19:35.080","Text":"time 0 is also 0 because we\u0027re assuming that the cart starts at"},{"Start":"19:35.080 ","End":"19:37.870","Text":"rest until v at"},{"Start":"19:37.870 ","End":"19:46.870","Text":"some time t. Now we can just put some tags along to differentiate between,"},{"Start":"19:46.870 ","End":"19:52.290","Text":"what we\u0027re integrating by and our bounds,"},{"Start":"19:52.290 ","End":"19:54.765","Text":"so also here we can put a tag,"},{"Start":"19:54.765 ","End":"19:58.035","Text":"and then we simply have to integrate."},{"Start":"19:58.035 ","End":"20:01.815","Text":"Let\u0027s integrate the left side in the meantime."},{"Start":"20:01.815 ","End":"20:05.430","Text":"Let\u0027s say that we have some variable which is x,"},{"Start":"20:05.430 ","End":"20:11.035","Text":"which is equal to our denominator which is v0 minus v tag of"},{"Start":"20:11.035 ","End":"20:20.660","Text":"t. Then we can say that our dx is equal to negative dv tag."},{"Start":"20:20.940 ","End":"20:26.680","Text":"Then we can switch over our integral to"},{"Start":"20:26.680 ","End":"20:31.930","Text":"be between our bounds that we\u0027ll soon fill in so we\u0027ll just"},{"Start":"20:31.930 ","End":"20:38.710","Text":"say a and b in the meantime of dx divided"},{"Start":"20:38.710 ","End":"20:42.760","Text":"by x^2 and then it\u0027s a negative over"},{"Start":"20:42.760 ","End":"20:47.740","Text":"here which will equal to the integral of our right side which is easy,"},{"Start":"20:47.740 ","End":"20:53.830","Text":"which is 2m dot divided by v0m multiplied by"},{"Start":"20:53.830 ","End":"21:02.870","Text":"t. Then the integral of this negative dx over x^2 is just 1 over x and then if we,"},{"Start":"21:02.870 ","End":"21:07.420","Text":"substitute in what our x is equal to which is v0 minus v tag of"},{"Start":"21:07.420 ","End":"21:12.430","Text":"t so we\u0027ll get that we have 1 over"},{"Start":"21:12.430 ","End":"21:22.945","Text":"v0 minus v tag of t. This is the integral and then it\u0027s between the bounds of 0 and v of"},{"Start":"21:22.945 ","End":"21:31.700","Text":"t which is equal to 2m dot over v0m multiplied by t."},{"Start":"21:32.790 ","End":"21:40.960","Text":"Then all of this will just equal 1 divided by v0 minus v(t)"},{"Start":"21:40.960 ","End":"21:50.390","Text":"minus 1 over v0 which is equal to 2m dot t divided by v0m."},{"Start":"21:50.970 ","End":"21:55.600","Text":"Now we have to do some algebra because we want to isolate"},{"Start":"21:55.600 ","End":"22:00.355","Text":"out this v of t because that\u0027s what we\u0027re looking for in order to answer the question."},{"Start":"22:00.355 ","End":"22:04.255","Text":"Then the rest in order to isolate this out"},{"Start":"22:04.255 ","End":"22:08.140","Text":"is algebra so just to speed things up you can just work"},{"Start":"22:08.140 ","End":"22:15.880","Text":"this out you\u0027ll see that you\u0027ll get finally at the end that vt is equal to v0"},{"Start":"22:15.880 ","End":"22:21.205","Text":"1 minus capital M divided by"},{"Start":"22:21.205 ","End":"22:24.325","Text":"2m dot t plus"},{"Start":"22:24.325 ","End":"22:29.995","Text":"m. You can just do this on the page and there should be the answer that you get."},{"Start":"22:29.995 ","End":"22:34.420","Text":"Obviously, you might get your answer in a different form and that\u0027s also fine."},{"Start":"22:34.420 ","End":"22:36.670","Text":"All you have to do really is to isolate out"},{"Start":"22:36.670 ","End":"22:42.910","Text":"the vt. Then of course we can check and see that when t is equal to 0,"},{"Start":"22:42.910 ","End":"22:46.120","Text":"so this over here will equal to 0 and then we have"},{"Start":"22:46.120 ","End":"22:49.090","Text":"m divided by m which is equal to 1 and then in"},{"Start":"22:49.090 ","End":"22:52.390","Text":"the brackets we\u0027ll have 1 minus 1 which is equal to 0 and"},{"Start":"22:52.390 ","End":"22:56.870","Text":"then we\u0027ll get that our v at time t=0,"},{"Start":"22:57.180 ","End":"23:01.960","Text":"will be equal to 0 which is what we were expecting,"},{"Start":"23:01.960 ","End":"23:05.515","Text":"it\u0027s what we\u0027re being told in the question that we\u0027re starting from rest."},{"Start":"23:05.515 ","End":"23:09.715","Text":"Then we could check the velocity at t equals infinity,"},{"Start":"23:09.715 ","End":"23:12.820","Text":"after a very long period of time and that means that"},{"Start":"23:12.820 ","End":"23:21.070","Text":"this whole fractional equation will become 0 because the denominator will be 2 infinity."},{"Start":"23:21.070 ","End":"23:29.000","Text":"Then we\u0027ll just get that we have v_t equals infinity will just equal to v0."},{"Start":"23:29.070 ","End":"23:33.145","Text":"If the hose is moving,"},{"Start":"23:33.145 ","End":"23:37.440","Text":"is pushing the cart so that makes sense"},{"Start":"23:37.440 ","End":"23:41.790","Text":"that after a very long time the water coming out of the hose will be at"},{"Start":"23:41.790 ","End":"23:45.720","Text":"the same velocity or rather the cart will be at the same velocity as"},{"Start":"23:45.720 ","End":"23:50.860","Text":"the water coming out of the hose because it\u0027s pushing it."},{"Start":"23:51.570 ","End":"23:58.675","Text":"Now what we have over here is our final answer in red but as I promised you,"},{"Start":"23:58.675 ","End":"24:02.635","Text":"let\u0027s have an explanation of why our dm by"},{"Start":"24:02.635 ","End":"24:06.805","Text":"dt is coming out of the hose and into the cart."},{"Start":"24:06.805 ","End":"24:08.680","Text":"Onto the same as I promised"},{"Start":"24:08.680 ","End":"24:12.985","Text":"the slightly more intuitive version rather than the mathematical."},{"Start":"24:12.985 ","End":"24:19.480","Text":"When we\u0027re starting we have our hose over here which is spraying water out,"},{"Start":"24:19.480 ","End":"24:24.860","Text":"and then the cart is located over here."},{"Start":"24:26.550 ","End":"24:32.870","Text":"The water is coming in and it\u0027s going like this, round and out."},{"Start":"24:33.660 ","End":"24:37.870","Text":"The amount of water that\u0027s in the air,"},{"Start":"24:37.870 ","End":"24:40.345","Text":"ignoring gravity right now."},{"Start":"24:40.345 ","End":"24:46.630","Text":"No water is falling it\u0027s just in the air flying towards the cart is going to be x."},{"Start":"24:46.630 ","End":"24:53.000","Text":"This is the distance between the opening of the hose until the water enters the cart."},{"Start":"24:53.280 ","End":"24:58.000","Text":"Now because the cart is moving because it\u0027s on wheels and the hose is"},{"Start":"24:58.000 ","End":"25:02.365","Text":"spraying water so the force of the water is moving the cart."},{"Start":"25:02.365 ","End":"25:08.920","Text":"A moment later we\u0027re going to have the hose over here,"},{"Start":"25:08.920 ","End":"25:11.230","Text":"excuse my drawings and the water is coming out"},{"Start":"25:11.230 ","End":"25:13.930","Text":"again but because the cart has moved further"},{"Start":"25:13.930 ","End":"25:19.930","Text":"away the entrance to the cart will only be over here."},{"Start":"25:19.930 ","End":"25:26.500","Text":"That means that the amount of water in the air is going to be greater,"},{"Start":"25:26.500 ","End":"25:30.220","Text":"it\u0027s no longer x it\u0027s now x tag."},{"Start":"25:30.220 ","End":"25:32.665","Text":"That means that,"},{"Start":"25:32.665 ","End":"25:35.290","Text":"and the water is still coming in and out,"},{"Start":"25:35.290 ","End":"25:40.240","Text":"that\u0027s fine but there\u0027s more water in the air,"},{"Start":"25:40.240 ","End":"25:43.285","Text":"flowing through the air than in the cart."},{"Start":"25:43.285 ","End":"25:45.505","Text":"Of all the water,"},{"Start":"25:45.505 ","End":"25:47.635","Text":"according to the rate of water,"},{"Start":"25:47.635 ","End":"25:52.555","Text":"the rate with which the water is exiting, the hose."},{"Start":"25:52.555 ","End":"25:59.485","Text":"The rate at which the water is exiting the hose some of the water is going to the air,"},{"Start":"25:59.485 ","End":"26:03.070","Text":"and some of that water is going into the carts."},{"Start":"26:03.070 ","End":"26:04.960","Text":"Of course, it doesn\u0027t make a difference if"},{"Start":"26:04.960 ","End":"26:07.810","Text":"this particle of water will in a few seconds be in"},{"Start":"26:07.810 ","End":"26:10.330","Text":"the carts because then a new particle of water"},{"Start":"26:10.330 ","End":"26:13.880","Text":"coming out of the hose will just replace it."},{"Start":"26:14.370 ","End":"26:18.955","Text":"This happens over time as the cart moves further and further away"},{"Start":"26:18.955 ","End":"26:25.690","Text":"the distance of water in the air grows larger and larger,"},{"Start":"26:25.690 ","End":"26:30.580","Text":"so that\u0027s why the rate coming out and"},{"Start":"26:30.580 ","End":"26:36.490","Text":"the rate of the hose and the rate going in to the cart is not the same."},{"Start":"26:36.490 ","End":"26:38.905","Text":"That\u0027s the end of the lesson."},{"Start":"26:38.905 ","End":"26:42.625","Text":"It was quite a difficult and challenging lesson I hope you understood."},{"Start":"26:42.625 ","End":"26:46.015","Text":"If you don\u0027t watch it again or try a few more examples"},{"Start":"26:46.015 ","End":"26:50.570","Text":"because this example was quite difficult. Thank you."}],"ID":10621},{"Watched":false,"Name":"Chain Falling On Scale","Duration":"15m 22s","ChapterTopicVideoID":9233,"CourseChapterTopicPlaylistID":5429,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.590 ","End":"00:03.330","Text":"Hello. In this question,"},{"Start":"00:03.330 ","End":"00:07.320","Text":"we\u0027re being told that a chain of length L and mass M is held"},{"Start":"00:07.320 ","End":"00:12.840","Text":"vertically above a scale such that the bottom end is just touching the scale."},{"Start":"00:12.840 ","End":"00:16.650","Text":"Over here, the chain is just touching the scale."},{"Start":"00:16.650 ","End":"00:19.125","Text":"The chain is then released from rest."},{"Start":"00:19.125 ","End":"00:22.800","Text":"Then we\u0027re being asked to find the weight shown on the scale."},{"Start":"00:22.800 ","End":"00:26.235","Text":"What will the scale show as a function of x"},{"Start":"00:26.235 ","End":"00:30.825","Text":"when x is the distance which the top of the chain fell?"},{"Start":"00:30.825 ","End":"00:32.910","Text":"As the chain falls,"},{"Start":"00:32.910 ","End":"00:34.560","Text":"we\u0027re measuring this distance,"},{"Start":"00:34.560 ","End":"00:39.390","Text":"the difference between the starting position of the top of the chain"},{"Start":"00:39.390 ","End":"00:45.030","Text":"to the position at time t of the top of the chain,"},{"Start":"00:45.030 ","End":"00:48.715","Text":"and this distance is x."},{"Start":"00:48.715 ","End":"00:50.750","Text":"Now throughout this question,"},{"Start":"00:50.750 ","End":"00:57.435","Text":"we\u0027re assuming that the mass of each chain is the same."},{"Start":"00:57.435 ","End":"01:01.485","Text":"The chain is uniform and that the density is uniform."},{"Start":"01:01.485 ","End":"01:04.220","Text":"Now, what we\u0027re going to do is we\u0027re going to work out"},{"Start":"01:04.220 ","End":"01:07.535","Text":"this question with the help of our equation,"},{"Start":"01:07.535 ","End":"01:09.965","Text":"this equation that we know and love,"},{"Start":"01:09.965 ","End":"01:17.635","Text":"because this will make working the question out a lot more easy and a lot more intuitive."},{"Start":"01:17.635 ","End":"01:22.025","Text":"The way that we can do that is that if we put"},{"Start":"01:22.025 ","End":"01:29.330","Text":"some imaginary lines over here in order to make this look like a cart or a box."},{"Start":"01:29.330 ","End":"01:35.855","Text":"Then we have this imaginary cards where mass is being accumulated."},{"Start":"01:35.855 ","End":"01:40.760","Text":"Now we\u0027re being told to find what the weight will be."},{"Start":"01:40.760 ","End":"01:42.575","Text":"What the scale will show."},{"Start":"01:42.575 ","End":"01:45.085","Text":"For those of you that don\u0027t know,"},{"Start":"01:45.085 ","End":"01:47.765","Text":"the weights that the scale will show is"},{"Start":"01:47.765 ","End":"01:53.075","Text":"the normal force and we\u0027re trying to find it as a function of x."},{"Start":"01:53.075 ","End":"01:57.755","Text":"This is our goal. Now, as we can notice,"},{"Start":"01:57.755 ","End":"02:02.365","Text":"in the x-direction, nothing is happening."},{"Start":"02:02.365 ","End":"02:06.335","Text":"We\u0027re going to look at the system in the y-direction."},{"Start":"02:06.335 ","End":"02:08.825","Text":"Then here, because the chain is falling,"},{"Start":"02:08.825 ","End":"02:10.130","Text":"it\u0027s easier to say that"},{"Start":"02:10.130 ","End":"02:14.390","Text":"the positive y direction will be in the downwards direction, but it doesn\u0027t matter."},{"Start":"02:14.390 ","End":"02:15.920","Text":"You can also say it\u0027s upwards,"},{"Start":"02:15.920 ","End":"02:17.655","Text":"but here it\u0027s just easier."},{"Start":"02:17.655 ","End":"02:20.435","Text":"The first thing we\u0027re going to do is we\u0027re going to work out"},{"Start":"02:20.435 ","End":"02:24.090","Text":"the sum of all of the external forces."},{"Start":"02:24.310 ","End":"02:32.705","Text":"We\u0027ll label the sum of all our external forces is going to be equal to the mass,"},{"Start":"02:32.705 ","End":"02:35.540","Text":"which is changing as a function of time multiplied by"},{"Start":"02:35.540 ","End":"02:40.620","Text":"mg. That\u0027s what\u0027s pulling the chain downwards."},{"Start":"02:40.620 ","End":"02:45.510","Text":"Then it\u0027s going to be minus the weight over here,"},{"Start":"02:45.510 ","End":"02:49.445","Text":"minus the normal force because the weight is just the normal force,"},{"Start":"02:49.445 ","End":"02:55.215","Text":"which is pointing in the opposite direction to mg."},{"Start":"02:55.215 ","End":"02:57.070","Text":"Minus our n,"},{"Start":"02:57.070 ","End":"02:58.880","Text":"which is a function of x,"},{"Start":"02:58.880 ","End":"03:01.250","Text":"which is what we\u0027re trying to find."},{"Start":"03:01.250 ","End":"03:07.805","Text":"Now what we\u0027re going to do before we say that this is equal to the rest of the equation."},{"Start":"03:07.805 ","End":"03:13.430","Text":"As far as we\u0027re going to find what our M as a function of t is, which we don\u0027t know."},{"Start":"03:13.430 ","End":"03:17.885","Text":"Now, this is the difficult part of the question, and actually,"},{"Start":"03:17.885 ","End":"03:21.410","Text":"maybe it\u0027s better if we find our M as a function of"},{"Start":"03:21.410 ","End":"03:26.125","Text":"x because at the end we want our N as a function of x."},{"Start":"03:26.125 ","End":"03:32.645","Text":"My mass as a function of distance is equal to Lambda."},{"Start":"03:32.645 ","End":"03:34.280","Text":"Now, what is Lambda?"},{"Start":"03:34.280 ","End":"03:41.740","Text":"Density per unit length and then multiply it by our length."},{"Start":"03:41.740 ","End":"03:48.900","Text":"Then we\u0027ll say that our mass times 0 is equal to"},{"Start":"03:48.900 ","End":"03:52.085","Text":"0 because we are told that"},{"Start":"03:52.085 ","End":"03:55.385","Text":"it\u0027s held vertically and the bottom end is just touching the scale,"},{"Start":"03:55.385 ","End":"03:58.745","Text":"which means it\u0027s not applying any force to the scale,"},{"Start":"03:58.745 ","End":"04:03.515","Text":"which means that our beginning our starting mass is equal to 0."},{"Start":"04:03.515 ","End":"04:07.580","Text":"Now what we need to do is we need to find out what our Lambda is."},{"Start":"04:07.580 ","End":"04:11.380","Text":"What our density per unit length is."},{"Start":"04:11.380 ","End":"04:15.230","Text":"If we, again, return to what\u0027s given to us in the question,"},{"Start":"04:15.230 ","End":"04:18.805","Text":"we know that for the entire length L,"},{"Start":"04:18.805 ","End":"04:21.890","Text":"the mass of the chain is M. We can say that"},{"Start":"04:21.890 ","End":"04:25.145","Text":"our density per unit length is simply our mass"},{"Start":"04:25.145 ","End":"04:31.375","Text":"divided by our length, which is M/L."},{"Start":"04:31.375 ","End":"04:34.330","Text":"Let me just go back to our M,"},{"Start":"04:34.330 ","End":"04:37.325","Text":"our mass as a function of x."},{"Start":"04:37.325 ","End":"04:41.255","Text":"Now, I did my Lambda multiplied by x,"},{"Start":"04:41.255 ","End":"04:46.895","Text":"because x is my distance traveled. It\u0027s my variable."},{"Start":"04:46.895 ","End":"04:51.965","Text":"Now let\u0027s go through our expressions in our equation,"},{"Start":"04:51.965 ","End":"04:54.785","Text":"and then we\u0027ll eventually substitute them in."},{"Start":"04:54.785 ","End":"04:57.560","Text":"Our mass as a function of x,"},{"Start":"04:57.560 ","End":"04:59.000","Text":"we already have over here,"},{"Start":"04:59.000 ","End":"05:06.500","Text":"and it doesn\u0027t matter if it\u0027s mass as a function of x or mass as a function of t, time."},{"Start":"05:06.500 ","End":"05:09.020","Text":"It doesn\u0027t matter. We can substitute in both."},{"Start":"05:09.020 ","End":"05:11.960","Text":"Then our dv by dt."},{"Start":"05:11.960 ","End":"05:14.720","Text":"Our dv in the y-axis,"},{"Start":"05:14.720 ","End":"05:17.925","Text":"because obviously, we\u0027re not looking at the x-axis."},{"Start":"05:17.925 ","End":"05:23.840","Text":"Our dv in the y-direction by dt is referring to the velocity of the cards."},{"Start":"05:23.840 ","End":"05:25.490","Text":"Now we know that the cards,"},{"Start":"05:25.490 ","End":"05:29.125","Text":"which is in fact the scale, is stationary."},{"Start":"05:29.125 ","End":"05:32.465","Text":"It\u0027s not accelerating, it\u0027s not moving in the y-direction."},{"Start":"05:32.465 ","End":"05:34.550","Text":"We can say that that equals to 0."},{"Start":"05:34.550 ","End":"05:41.395","Text":"That means that this whole section will be equal to 0. That was easy."},{"Start":"05:41.395 ","End":"05:49.455","Text":"Now we\u0027re looking at the second part and we want to find our dm by dt."},{"Start":"05:49.455 ","End":"05:57.885","Text":"As we know, our dm by dt is here going to be equal to dM,"},{"Start":"05:57.885 ","End":"06:00.725","Text":"M, by dt."},{"Start":"06:00.725 ","End":"06:02.215","Text":"Now, this looks a bit stupid."},{"Start":"06:02.215 ","End":"06:03.730","Text":"What is the difference?"},{"Start":"06:03.730 ","End":"06:07.240","Text":"But sometimes we can say that our dm by dt,"},{"Start":"06:07.240 ","End":"06:09.850","Text":"this one is equal to negative,"},{"Start":"06:09.850 ","End":"06:12.940","Text":"so it\u0027s sometimes plus and sometimes minus dm by dt."},{"Start":"06:12.940 ","End":"06:16.390","Text":"Now here, because we have mass accumulating in the cart,"},{"Start":"06:16.390 ","End":"06:20.280","Text":"so we can say that it\u0027s positive dm by dt."},{"Start":"06:20.280 ","End":"06:22.175","Text":"When we\u0027re speaking about the small mass,"},{"Start":"06:22.175 ","End":"06:26.890","Text":"we\u0027re speaking about the emission or the accumulation of mass,"},{"Start":"06:26.890 ","End":"06:29.140","Text":"the rates of change."},{"Start":"06:29.140 ","End":"06:32.005","Text":"When we\u0027re speaking about the capital M we\u0027re speaking about"},{"Start":"06:32.005 ","End":"06:36.540","Text":"the total mass of the system."},{"Start":"06:36.540 ","End":"06:43.850","Text":"Now the problem here is that we found our M to be M as a function of x,"},{"Start":"06:43.850 ","End":"06:50.300","Text":"and then we can\u0027t derive it according to t. What we have to do is we have to"},{"Start":"06:50.300 ","End":"06:53.300","Text":"find our x as a function of"},{"Start":"06:53.300 ","End":"06:58.055","Text":"t and then we can place it and substitute it into this equation."},{"Start":"06:58.055 ","End":"07:01.010","Text":"Then you will see how we can solve this."},{"Start":"07:01.010 ","End":"07:02.990","Text":"Let\u0027s see how we do this."},{"Start":"07:02.990 ","End":"07:09.830","Text":"Now, we know that our acceleration of the chain downwards is equal to g,"},{"Start":"07:09.830 ","End":"07:14.570","Text":"because the chain is released from rest and it\u0027s falling, it\u0027s in free fall."},{"Start":"07:14.570 ","End":"07:16.790","Text":"The acceleration downwards, because we said that"},{"Start":"07:16.790 ","End":"07:19.355","Text":"the downwards direction is the positive direction,"},{"Start":"07:19.355 ","End":"07:22.775","Text":"is equal to g. Then again,"},{"Start":"07:22.775 ","End":"07:25.355","Text":"because the chain is released from rest,"},{"Start":"07:25.355 ","End":"07:28.310","Text":"so we can say that our velocity is then equal"},{"Start":"07:28.310 ","End":"07:31.265","Text":"to just the acceleration multiplied by the time,"},{"Start":"07:31.265 ","End":"07:34.295","Text":"which will equal to gt."},{"Start":"07:34.295 ","End":"07:41.345","Text":"Then, as we know, that x.=v."},{"Start":"07:41.345 ","End":"07:45.020","Text":"We can say that this is equal to x dot."},{"Start":"07:45.020 ","End":"07:49.130","Text":"Then if we integrate both sides,"},{"Start":"07:49.130 ","End":"07:56.070","Text":"then we will get that our x as a function of t is equal to 1/2gt^2."},{"Start":"07:58.020 ","End":"08:04.315","Text":"Now I just substitute this x into my x for my M(x)."},{"Start":"08:04.315 ","End":"08:09.295","Text":"Therefore, my M as a function of"},{"Start":"08:09.295 ","End":"08:14.890","Text":"t is going to be Lambda multiplied by x as a function of t."},{"Start":"08:14.890 ","End":"08:20.800","Text":"Multiply by half gt^2"},{"Start":"08:20.800 ","End":"08:26.650","Text":"and then all I have to do is in order to find my dm by dt."},{"Start":"08:26.650 ","End":"08:36.040","Text":"Therefore my dm by dt will equal to Lambda multiplied by gt."},{"Start":"08:36.040 ","End":"08:40.525","Text":"All I did was to take the derivative."},{"Start":"08:40.525 ","End":"08:43.870","Text":"Now, another way that we could look at it is if"},{"Start":"08:43.870 ","End":"08:48.348","Text":"we can solve this through the chain rule,"},{"Start":"08:48.348 ","End":"08:53.740","Text":"and the chain rule says that our dm by dt is equal"},{"Start":"08:53.740 ","End":"09:01.360","Text":"to dm by dx multiplied by dx by dt."},{"Start":"09:01.360 ","End":"09:03.145","Text":"This is just the chain rule."},{"Start":"09:03.145 ","End":"09:05.800","Text":"Let\u0027s see if we get the same answer."},{"Start":"09:05.800 ","End":"09:16.090","Text":"Our dm by dx will equal to Lambda and then multiplied by our dx by dt so where\u0027s our x?"},{"Start":"09:16.090 ","End":"09:24.705","Text":"It\u0027s over here. Taking the derivative will equal to gt."},{"Start":"09:24.705 ","End":"09:28.830","Text":"Then we can see that we got the exact same answer."},{"Start":"09:28.830 ","End":"09:31.175","Text":"That\u0027s another way that we can do it."},{"Start":"09:31.175 ","End":"09:37.840","Text":"Now, the last thing that we have to speak about is our u and as we know,"},{"Start":"09:37.840 ","End":"09:43.510","Text":"our u is equal to our relative velocity, so V_rel."},{"Start":"09:43.510 ","End":"09:48.445","Text":"Notice that the chain is falling into our cart,"},{"Start":"09:48.445 ","End":"09:50.275","Text":"but the cart is at rest."},{"Start":"09:50.275 ","End":"09:53.950","Text":"We can just take the relative velocity of the chain"},{"Start":"09:53.950 ","End":"09:58.120","Text":"to the cart as the same as the velocity of the chain to the ground."},{"Start":"09:58.120 ","End":"10:00.940","Text":"We\u0027ve already figured this out,"},{"Start":"10:00.940 ","End":"10:09.890","Text":"and we already said that it equals to g multiplied by t. This is also equal to u."},{"Start":"10:10.230 ","End":"10:16.795","Text":"Now let\u0027s continue this our equation and fill in all the blanks."},{"Start":"10:16.795 ","End":"10:23.335","Text":"We have our Lambda multiplied by xg."},{"Start":"10:23.335 ","End":"10:32.200","Text":"That is our M as a function of x minus our normal as a function of x,"},{"Start":"10:32.200 ","End":"10:36.730","Text":"which will equal to we\u0027ve already said that our M dv by dt is equal to"},{"Start":"10:36.730 ","End":"10:41.380","Text":"0 and plus our dm by dt which we worked"},{"Start":"10:41.380 ","End":"10:43.180","Text":"out over here to be"},{"Start":"10:43.180 ","End":"10:51.220","Text":"Lambda gt and then we\u0027re going to multiply it by our relative velocity,"},{"Start":"10:51.220 ","End":"10:55.840","Text":"which we just said is the same velocity that the chain is falling on."},{"Start":"10:55.840 ","End":"11:06.715","Text":"Then we\u0027ll multiply that by gt and then because of the way that we derived the equation,"},{"Start":"11:06.715 ","End":"11:12.190","Text":"we said that it was for emission so losing mass,"},{"Start":"11:12.190 ","End":"11:14.500","Text":"sorry, but now because we\u0027re gaining mass,"},{"Start":"11:14.500 ","End":"11:18.565","Text":"so we have to multiply it by this negative 1."},{"Start":"11:18.565 ","End":"11:22.420","Text":"Whenever you\u0027re gaining mass into your cart,"},{"Start":"11:22.420 ","End":"11:25.315","Text":"you multiply by negative 1."},{"Start":"11:25.315 ","End":"11:29.125","Text":"Now because we want to isolate out our normal,"},{"Start":"11:29.125 ","End":"11:35.990","Text":"so we can just isolate it out and then we\u0027ll have Lambda xg"},{"Start":"11:36.180 ","End":"11:42.055","Text":"plus Lambda g^2t^2 will"},{"Start":"11:42.055 ","End":"11:47.620","Text":"equal to our normal function as a function of x."},{"Start":"11:47.620 ","End":"11:54.760","Text":"This isn\u0027t x, but now because we have our equation with the variable t inside,"},{"Start":"11:54.760 ","End":"11:57.970","Text":"and we don\u0027t want that because we want this as a function of x."},{"Start":"11:57.970 ","End":"12:02.650","Text":"We can go back to our equation over here and we can isolate out the"},{"Start":"12:02.650 ","End":"12:08.530","Text":"t. The t ^2 and say that our t^2 is equal to 2x"},{"Start":"12:08.530 ","End":"12:12.610","Text":"divided by g and substitute this N and then we\u0027ll get"},{"Start":"12:12.610 ","End":"12:20.095","Text":"the Lambda xg plus Lambda g^2 multiplied by t^2,"},{"Start":"12:20.095 ","End":"12:28.923","Text":"which is multiplied by 2x over g and then our gs will cross out,"},{"Start":"12:28.923 ","End":"12:35.170","Text":"and then we\u0027ll see that we have 2x Lambda multiplied by g,"},{"Start":"12:35.170 ","End":"12:39.070","Text":"so then we\u0027ll have 2x Lambda g plus Lambda xg."},{"Start":"12:39.070 ","End":"12:45.730","Text":"That means that 3 Lambda xg is equal to our normal."},{"Start":"12:45.730 ","End":"12:49.060","Text":"Then we can substitute N what we have for our Lambda,"},{"Start":"12:49.060 ","End":"12:55.060","Text":"which is M divided by L. Then we\u0027ll get that our normal as a function"},{"Start":"12:55.060 ","End":"13:04.010","Text":"of x is equal to 3M divided by L multiplied by xg."},{"Start":"13:04.380 ","End":"13:07.570","Text":"This is our final answer."},{"Start":"13:07.570 ","End":"13:10.840","Text":"Why does this make sense? Let\u0027s take a look."},{"Start":"13:10.840 ","End":"13:15.160","Text":"If we substitute N that our x is equal to 0,"},{"Start":"13:15.160 ","End":"13:19.090","Text":"okay, so right at the beginning, our starting."},{"Start":"13:19.090 ","End":"13:23.545","Text":"Then, we\u0027ll see that our normal is equal to 0, which is correct."},{"Start":"13:23.545 ","End":"13:26.515","Text":"That\u0027s what we\u0027re given at the beginning of the question."},{"Start":"13:26.515 ","End":"13:31.645","Text":"Now what happens if our x is equal to L?"},{"Start":"13:31.645 ","End":"13:36.234","Text":"Then we\u0027ll see that our N as a function of x,"},{"Start":"13:36.234 ","End":"13:41.530","Text":"when x equals L will be equal to 3Mg."},{"Start":"13:41.530 ","End":"13:43.735","Text":"Why is this logical?"},{"Start":"13:43.735 ","End":"13:47.200","Text":"We\u0027ll see that this whole equation and all of"},{"Start":"13:47.200 ","End":"13:52.855","Text":"our derivation is only correct while as the chain is still falling,"},{"Start":"13:52.855 ","End":"13:57.730","Text":"and hitting the scale."},{"Start":"13:57.730 ","End":"14:02.515","Text":"A moment after the last piece of the chain has hit the scale,"},{"Start":"14:02.515 ","End":"14:06.115","Text":"then we\u0027ll get to this equation,"},{"Start":"14:06.115 ","End":"14:10.255","Text":"N equals Mg. Why do we have,"},{"Start":"14:10.255 ","End":"14:12.205","Text":"when the chain is still falling,"},{"Start":"14:12.205 ","End":"14:14.905","Text":"that N will equal 3Mg."},{"Start":"14:14.905 ","End":"14:19.195","Text":"Now, this happens because while as the chain is falling,"},{"Start":"14:19.195 ","End":"14:25.045","Text":"the normal force opposing the chain falling has to stop the chain,"},{"Start":"14:25.045 ","End":"14:28.510","Text":"because the chain hits the scale and then it stops."},{"Start":"14:28.510 ","End":"14:30.070","Text":"There\u0027s no more movement up,"},{"Start":"14:30.070 ","End":"14:31.675","Text":"or down, or sideways."},{"Start":"14:31.675 ","End":"14:36.730","Text":"Which means that the normal force aside from having to be equal"},{"Start":"14:36.730 ","End":"14:42.100","Text":"and opposite to the mass of the chain which is already resting on the scale,"},{"Start":"14:42.100 ","End":"14:46.660","Text":"it has to be grazed enough in order to stop in 1"},{"Start":"14:46.660 ","End":"14:52.555","Text":"go in 1 swift movement the the rest of the chain,"},{"Start":"14:52.555 ","End":"14:54.580","Text":"which is still falling."},{"Start":"14:54.580 ","End":"14:59.050","Text":"The force that it needs in order to do that here in this example is"},{"Start":"14:59.050 ","End":"15:06.640","Text":"3 times the normal force when the chain is just at rest on the scale."},{"Start":"15:06.640 ","End":"15:13.675","Text":"I\u0027ll write in a different color that the normal force a moment after x equals L,"},{"Start":"15:13.675 ","End":"15:17.125","Text":"when the whole chain is resting on the scale will equal to"},{"Start":"15:17.125 ","End":"15:23.120","Text":"Mg. That is the end of this lesson."}],"ID":10622},{"Watched":false,"Name":"Lifting A String Off The Table","Duration":"18m 29s","ChapterTopicVideoID":9234,"CourseChapterTopicPlaylistID":5429,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.470 ","End":"00:03.945","Text":"Hello. In this lesson,"},{"Start":"00:03.945 ","End":"00:05.835","Text":"we\u0027re being asked a question."},{"Start":"00:05.835 ","End":"00:10.195","Text":"A string of mass m and of length l is resting on a table."},{"Start":"00:10.195 ","End":"00:13.200","Text":"A force of F begins lifting the string off"},{"Start":"00:13.200 ","End":"00:17.115","Text":"the table from the tip of the string at a velocity of v_0."},{"Start":"00:17.115 ","End":"00:22.155","Text":"The string is being lifted upwards at a velocity of v_0."},{"Start":"00:22.155 ","End":"00:24.705","Text":"Then our first question is,"},{"Start":"00:24.705 ","End":"00:27.270","Text":"what is the force as a function of time?"},{"Start":"00:27.270 ","End":"00:33.750","Text":"Our second question is how much energy is wasted as a function of the distance traveled?"},{"Start":"00:33.750 ","End":"00:41.245","Text":"The first question that we\u0027re going to answer is going to be question number 1."},{"Start":"00:41.245 ","End":"00:44.460","Text":"Let\u0027s see how we do this."},{"Start":"00:45.170 ","End":"00:52.768","Text":"The first thing that we have to do is we have to decide what our large body is,"},{"Start":"00:52.768 ","End":"00:57.520","Text":"and what are small body is in order to know how to apply this equation."},{"Start":"00:57.520 ","End":"01:00.915","Text":"We\u0027re going to say that our large body,"},{"Start":"01:00.915 ","End":"01:03.662","Text":"everything will be relative to it,"},{"Start":"01:03.662 ","End":"01:08.705","Text":"will be the part of the string that has already been lifted off of the table."},{"Start":"01:08.705 ","End":"01:15.845","Text":"Then the string that\u0027s still lying on the table is going to be our small body."},{"Start":"01:15.845 ","End":"01:20.405","Text":"All our calculations are going to be relative to the string up in the air."},{"Start":"01:20.405 ","End":"01:22.445","Text":"Because we already know the velocity of"},{"Start":"01:22.445 ","End":"01:25.820","Text":"it and we\u0027re trying to find the force which is lifting it,"},{"Start":"01:25.820 ","End":"01:29.310","Text":"so we have to use that section of the string."},{"Start":"01:29.630 ","End":"01:34.835","Text":"The first thing we\u0027re going to do is we\u0027re going to find the sum of all of the forces."},{"Start":"01:34.835 ","End":"01:39.920","Text":"First of all, let\u0027s decide our direction of the axis and we\u0027ll"},{"Start":"01:39.920 ","End":"01:45.565","Text":"say that the positive direction is going in the upwards direction."},{"Start":"01:45.565 ","End":"01:51.825","Text":"That means that gravity is acting downwards in the negative direction."},{"Start":"01:51.825 ","End":"01:57.415","Text":"We know that we\u0027re going to have our mass times gravity,"},{"Start":"01:57.415 ","End":"01:59.985","Text":"our mg, which is pulling the string down."},{"Start":"01:59.985 ","End":"02:01.940","Text":"Remember that it\u0027s our mass,"},{"Start":"02:01.940 ","End":"02:06.773","Text":"which is dependent on time multiplied by gravity,"},{"Start":"02:06.773 ","End":"02:11.685","Text":"and that\u0027s going in the negative direction because it\u0027s pulling downwards."},{"Start":"02:11.685 ","End":"02:13.650","Text":"Then we have the force."},{"Start":"02:13.650 ","End":"02:16.610","Text":"We\u0027ve said that the force because it\u0027s in the upwards direction,"},{"Start":"02:16.610 ","End":"02:17.990","Text":"and the upwards direction is positive,"},{"Start":"02:17.990 ","End":"02:20.180","Text":"so we add a positive over here."},{"Start":"02:20.180 ","End":"02:25.460","Text":"Then we have to say that this equals to"},{"Start":"02:25.460 ","End":"02:31.110","Text":"our mass as a function of time multiplied by dv by dt."},{"Start":"02:31.110 ","End":"02:34.430","Text":"Now right now what we\u0027ll notice is because our mass as"},{"Start":"02:34.430 ","End":"02:37.835","Text":"a function of time is being multiplied by dv by dt,"},{"Start":"02:37.835 ","End":"02:40.880","Text":"which is the acceleration of the string."},{"Start":"02:40.880 ","End":"02:45.410","Text":"Now because we\u0027re being told that the string is being lifted at a velocity of v_0,"},{"Start":"02:45.410 ","End":"02:53.345","Text":"there is no acceleration so our dv by dt here in this example is equal to 0."},{"Start":"02:53.345 ","End":"02:57.860","Text":"This whole thing is equal to 0 and false."},{"Start":"02:57.860 ","End":"03:02.660","Text":"Then we just have to say that it\u0027s our v relative."},{"Start":"03:02.660 ","End":"03:04.925","Text":"What is our v relative?"},{"Start":"03:04.925 ","End":"03:09.380","Text":"Relative to the part of the string in the air."},{"Start":"03:09.380 ","End":"03:15.125","Text":"We know that v_rel is equal to the velocity of the smaller body."},{"Start":"03:15.125 ","End":"03:18.110","Text":"Here we said it\u0027s the string which is on the table."},{"Start":"03:18.110 ","End":"03:20.765","Text":"Because it\u0027s stationary, it\u0027s 0,"},{"Start":"03:20.765 ","End":"03:23.990","Text":"negative the velocity of the larger body."},{"Start":"03:23.990 ","End":"03:27.170","Text":"Here we said that the larger body was the string up in"},{"Start":"03:27.170 ","End":"03:30.950","Text":"the air that\u0027s already been lifted and we\u0027re told in the question that is v_0,"},{"Start":"03:30.950 ","End":"03:36.540","Text":"so our v relative is just negative v_0."},{"Start":"03:36.700 ","End":"03:44.730","Text":"We\u0027ll have here negative v_0 multiplied by dm by dt."},{"Start":"03:45.380 ","End":"03:52.550","Text":"Then again, because we\u0027re speaking of the larger body as the string in the air,"},{"Start":"03:52.550 ","End":"03:57.230","Text":"so if I draw some imaginary box over here,"},{"Start":"03:57.230 ","End":"03:58.835","Text":"so as we lift the string,"},{"Start":"03:58.835 ","End":"04:00.050","Text":"more and more massive,"},{"Start":"04:00.050 ","End":"04:03.785","Text":"the string is entering into our invisible box in the air."},{"Start":"04:03.785 ","End":"04:06.650","Text":"Which means, as we know,"},{"Start":"04:06.650 ","End":"04:10.955","Text":"how we derive the equation that if mass is being added,"},{"Start":"04:10.955 ","End":"04:14.075","Text":"then we have to multiply it by negative 1."},{"Start":"04:14.075 ","End":"04:19.235","Text":"The negative 1 and the negative over here cancel out,"},{"Start":"04:19.235 ","End":"04:24.720","Text":"so it\u0027s equal to v_0 multiplied by dm by dt."},{"Start":"04:28.130 ","End":"04:31.910","Text":"Now the next thing that we have to solve is to find"},{"Start":"04:31.910 ","End":"04:35.530","Text":"out what this dm by dt is because it\u0027s not told."},{"Start":"04:35.530 ","End":"04:41.640","Text":"We know that dm by dt is"},{"Start":"04:41.640 ","End":"04:48.220","Text":"equal to negative dM by dt."},{"Start":"04:48.530 ","End":"04:52.410","Text":"Now let\u0027s see what our M as a function of time is,"},{"Start":"04:52.410 ","End":"04:55.580","Text":"and then we\u0027ll take the negative derivative of it."},{"Start":"04:55.580 ","End":"04:59.220","Text":"Then we\u0027ll just substitute that into here."},{"Start":"04:59.630 ","End":"05:07.140","Text":"Therefore, let\u0027s see what our M as a function of t is equal to."},{"Start":"05:07.190 ","End":"05:12.755","Text":"Our mass as a function of time is equal to the distance,"},{"Start":"05:12.755 ","End":"05:14.705","Text":"the length of the string."},{"Start":"05:14.705 ","End":"05:17.870","Text":"But because it\u0027s as a function of time,"},{"Start":"05:17.870 ","End":"05:23.240","Text":"as we know that velocity is equal to distance over time."},{"Start":"05:23.240 ","End":"05:29.495","Text":"We have our v_0 and we\u0027re trying to find this distance which is dependent by time."},{"Start":"05:29.495 ","End":"05:33.230","Text":"We know therefore that our velocity multiplied"},{"Start":"05:33.230 ","End":"05:37.225","Text":"by time is equal to the distance traveled as a function of time."},{"Start":"05:37.225 ","End":"05:40.530","Text":"We know that it\u0027s equal to v_0 multiplied"},{"Start":"05:40.530 ","End":"05:44.640","Text":"by t because we know that our velocity is equal to v_0."},{"Start":"05:44.680 ","End":"05:50.300","Text":"Then we have to multiply it by the density of"},{"Start":"05:50.300 ","End":"05:53.420","Text":"the string and the density of the string is"},{"Start":"05:53.420 ","End":"05:57.065","Text":"just the mass of the string divided by the length."},{"Start":"05:57.065 ","End":"06:03.630","Text":"We know the mass is m and we know the length it\u0027s l. These Ms are different,"},{"Start":"06:03.630 ","End":"06:05.890","Text":"so I\u0027ll do a tag over here."},{"Start":"06:08.420 ","End":"06:12.515","Text":"It\u0027s the mass of the string divided by the length."},{"Start":"06:12.515 ","End":"06:16.460","Text":"Now I have to take the negative derivative of this."},{"Start":"06:16.460 ","End":"06:24.560","Text":"Therefore, negative dM as a function of time divided by dt is going to"},{"Start":"06:24.560 ","End":"06:33.450","Text":"be equal to negative v_0 m over l because this t crosses off when we take the derivative."},{"Start":"06:33.580 ","End":"06:37.845","Text":"Then our answer, so we can go back."},{"Start":"06:37.845 ","End":"06:43.350","Text":"We\u0027ll have negative M as a function of t. We have it over here."},{"Start":"06:43.350 ","End":"06:51.855","Text":"It\u0027s negative v_0 t m over l multiplied by g plus F,"},{"Start":"06:51.855 ","End":"06:55.365","Text":"which is equal to v_0."},{"Start":"06:55.365 ","End":"07:01.770","Text":"Then we\u0027ll substitute in our dm by dt multiplied by negative v_0 m over"},{"Start":"07:01.770 ","End":"07:08.510","Text":"l. Negative v_0^2 m over l. Sorry,"},{"Start":"07:08.510 ","End":"07:09.860","Text":"I made a calculation error here."},{"Start":"07:09.860 ","End":"07:17.420","Text":"It\u0027s meant to be positive v_0^2 m over l. Because we want to put in our dm by dt,"},{"Start":"07:17.420 ","End":"07:22.840","Text":"which is the negative dM by dt."},{"Start":"07:22.840 ","End":"07:24.485","Text":"If this is this,"},{"Start":"07:24.485 ","End":"07:29.780","Text":"then therefore my dm by dt is the negative of this,"},{"Start":"07:29.780 ","End":"07:33.740","Text":"which equals v_0 multiplied by m over l,"},{"Start":"07:33.740 ","End":"07:38.770","Text":"so positive m over l. Therefore, our force,"},{"Start":"07:38.770 ","End":"07:42.210","Text":"which is what we were searching for the whole time,"},{"Start":"07:42.230 ","End":"07:49.034","Text":"is going to be equal to v_0^2 m over l,"},{"Start":"07:49.034 ","End":"07:52.760","Text":"so we\u0027re just isolating out this force,"},{"Start":"07:52.760 ","End":"08:00.710","Text":"plus v_0 t m over l g."},{"Start":"08:00.710 ","End":"08:04.570","Text":"Now what we\u0027re going to do is we\u0027re going to look at the second part,"},{"Start":"08:04.570 ","End":"08:09.175","Text":"which is how much energy is wasted as a function of the distance traveled."},{"Start":"08:09.175 ","End":"08:12.295","Text":"The energy wasted,"},{"Start":"08:12.295 ","End":"08:17.365","Text":"we\u0027re going to label it as Delta E,"},{"Start":"08:17.365 ","End":"08:19.285","Text":"the change in energy."},{"Start":"08:19.285 ","End":"08:23.080","Text":"Of course, it\u0027s in the y-direction because we\u0027re only looking at the y-direction,"},{"Start":"08:23.080 ","End":"08:25.510","Text":"there is no movement in the x-direction."},{"Start":"08:25.510 ","End":"08:33.190","Text":"It\u0027s equal to the energy that the force put in to lifting the string,"},{"Start":"08:33.190 ","End":"08:38.635","Text":"and as we know, that\u0027s the integral of the force."},{"Start":"08:38.635 ","End":"08:46.900","Text":"Then dy, which is the route that the string took."},{"Start":"08:46.900 ","End":"08:52.525","Text":"The route that it took to lift off the table."},{"Start":"08:52.525 ","End":"08:57.805","Text":"Then we\u0027re going to say that this is bounded by its height at 0,"},{"Start":"08:57.805 ","End":"08:59.605","Text":"so when it\u0027s still on the table,"},{"Start":"08:59.605 ","End":"09:01.690","Text":"until its maximum height,"},{"Start":"09:01.690 ","End":"09:05.920","Text":"which is some height in the y-direction, so y."},{"Start":"09:05.920 ","End":"09:11.155","Text":"Then minus the energy that is in the string,"},{"Start":"09:11.155 ","End":"09:15.670","Text":"which is equal to, as we know,"},{"Start":"09:15.670 ","End":"09:22.390","Text":"the potential energy plus the kinetic energy."},{"Start":"09:22.400 ","End":"09:25.440","Text":"Again, let\u0027s go over this."},{"Start":"09:25.440 ","End":"09:27.735","Text":"The energy wasted,"},{"Start":"09:27.735 ","End":"09:31.220","Text":"the Delta energy in the y-direction,"},{"Start":"09:31.220 ","End":"09:37.450","Text":"is equal to the energy put in to the system by the force,"},{"Start":"09:37.450 ","End":"09:46.570","Text":"which is the integral of the way that the string is lifted from the height 0,"},{"Start":"09:46.570 ","End":"09:48.025","Text":"which is on the table,"},{"Start":"09:48.025 ","End":"09:51.190","Text":"until a height y off the table."},{"Start":"09:51.190 ","End":"09:56.365","Text":"Then negative the energy which is now stored in the string,"},{"Start":"09:56.365 ","End":"10:00.685","Text":"which is the sum of the potential energy in the string,"},{"Start":"10:00.685 ","End":"10:03.865","Text":"as well as the kinetic energy in the string."},{"Start":"10:03.865 ","End":"10:08.020","Text":"Now let\u0027s do this calculation."},{"Start":"10:08.020 ","End":"10:10.750","Text":"Now, the first thing that I want to look"},{"Start":"10:10.750 ","End":"10:14.290","Text":"at is I want to look at how we solve this integral."},{"Start":"10:14.290 ","End":"10:20.245","Text":"Now, notice that we\u0027re integrating the force with respect to y."},{"Start":"10:20.245 ","End":"10:24.930","Text":"However, if we look at our answer for question number 1,"},{"Start":"10:24.930 ","End":"10:28.725","Text":"our force has the variable t not y."},{"Start":"10:28.725 ","End":"10:30.165","Text":"We need to change that."},{"Start":"10:30.165 ","End":"10:37.960","Text":"Now we know that the relationship between the position and t, the time,"},{"Start":"10:38.100 ","End":"10:43.480","Text":"y being the position is equal to our velocity,"},{"Start":"10:43.480 ","End":"10:45.730","Text":"which here is v_0 multiplied by"},{"Start":"10:45.730 ","End":"10:50.575","Text":"our t. It\u0027s just our velocity is distance over time equation,"},{"Start":"10:50.575 ","End":"10:55.990","Text":"again, where our distance is y."},{"Start":"10:55.990 ","End":"11:04.795","Text":"Therefore, we can say that our t is just equal to y divided by v_0."},{"Start":"11:04.795 ","End":"11:07.720","Text":"Now in this expression for our force,"},{"Start":"11:07.720 ","End":"11:10.495","Text":"we just have to substitute in wherever we see t,"},{"Start":"11:10.495 ","End":"11:13.100","Text":"this y divided by v_0."},{"Start":"11:13.440 ","End":"11:21.250","Text":"Then we will get therefore that our force is equal"},{"Start":"11:21.250 ","End":"11:29.800","Text":"to v_0^2 m over l plus v_0 multiplied by y divided by v_0."},{"Start":"11:29.800 ","End":"11:34.825","Text":"The v_0s will cancel out leaving just a y multiplied by m"},{"Start":"11:34.825 ","End":"11:40.030","Text":"over l multiplied by g. This is now"},{"Start":"11:40.030 ","End":"11:46.150","Text":"our force when it has the variable y Naught"},{"Start":"11:46.150 ","End":"11:52.360","Text":"t. Now if we\u0027re going to integrate on this,"},{"Start":"11:52.360 ","End":"11:54.715","Text":"so let\u0027s carry this one over here."},{"Start":"11:54.715 ","End":"12:02.935","Text":"We\u0027ll get that our Delta E with respect to y is equal to this integral from 0 until y."},{"Start":"12:02.935 ","End":"12:06.490","Text":"Over here in the first expression of the F,"},{"Start":"12:06.490 ","End":"12:08.920","Text":"we see that there\u0027s no variable y,"},{"Start":"12:08.920 ","End":"12:10.645","Text":"so we just multiply it by y."},{"Start":"12:10.645 ","End":"12:16.675","Text":"We\u0027ll have v_0^2 m over l multiplied by y."},{"Start":"12:16.675 ","End":"12:19.780","Text":"Then in the second expression we have a y,"},{"Start":"12:19.780 ","End":"12:28.555","Text":"so we have y^2 divided by 2 multiplied by our m over l multiplied by g,"},{"Start":"12:28.555 ","End":"12:30.820","Text":"because those are constants."},{"Start":"12:30.820 ","End":"12:34.849","Text":"This is this."},{"Start":"12:35.820 ","End":"12:43.715","Text":"Now what we\u0027re going to do is we\u0027re going to minus the energies."},{"Start":"12:43.715 ","End":"12:48.585","Text":"We have our potential energy now,"},{"Start":"12:48.585 ","End":"12:52.760","Text":"E_p, let\u0027s see this over here."},{"Start":"12:52.760 ","End":"12:55.015","Text":"E_p as we know,"},{"Start":"12:55.015 ","End":"12:59.620","Text":"is equal to mass times gravity times height."},{"Start":"12:59.620 ","End":"13:03.200","Text":"Let\u0027s see what our mass is."},{"Start":"13:03.450 ","End":"13:06.790","Text":"I\u0027m going to put a tag over here to show that it\u0027s not"},{"Start":"13:06.790 ","End":"13:12.040","Text":"the same mass which is given in the question over here,"},{"Start":"13:12.040 ","End":"13:14.050","Text":"so it\u0027s representing mass."},{"Start":"13:14.050 ","End":"13:16.360","Text":"Let\u0027s see what it is. Now, we know that"},{"Start":"13:16.360 ","End":"13:19.945","Text":"the mass because it\u0027s changing as the string is being lifted,"},{"Start":"13:19.945 ","End":"13:26.935","Text":"we know that our mass is equal to the density multiplied by the length of the string."},{"Start":"13:26.935 ","End":"13:30.070","Text":"Again, I\u0027m doing a tag to show you that this length is"},{"Start":"13:30.070 ","End":"13:34.760","Text":"different to this length given in the question over here."},{"Start":"13:35.730 ","End":"13:42.080","Text":"The mass of a string is its density multiplied by the length of the string."},{"Start":"13:42.900 ","End":"13:45.010","Text":"Its position in the air,"},{"Start":"13:45.010 ","End":"13:46.855","Text":"rather, not the length of the string."},{"Start":"13:46.855 ","End":"13:51.535","Text":"But when it\u0027s raised to position y above the table,"},{"Start":"13:51.535 ","End":"13:56.200","Text":"so over here, let\u0027s say this is position y above the table."},{"Start":"13:56.200 ","End":"13:59.600","Text":"What is the mass of the string then?"},{"Start":"14:00.150 ","End":"14:03.820","Text":"The density, as we saw in question number 1,"},{"Start":"14:03.820 ","End":"14:07.660","Text":"is equal to mass divided by length,"},{"Start":"14:07.660 ","End":"14:11.590","Text":"and then our position when it\u0027s in position y in the air,"},{"Start":"14:11.590 ","End":"14:15.130","Text":"so at the upper bound of our integral,"},{"Start":"14:15.130 ","End":"14:18.370","Text":"it\u0027s going to be multiplied by y."},{"Start":"14:18.370 ","End":"14:20.950","Text":"This is our mass."},{"Start":"14:20.950 ","End":"14:22.930","Text":"Now our g, we know it\u0027s g,"},{"Start":"14:22.930 ","End":"14:25.795","Text":"and then let\u0027s see what our h is."},{"Start":"14:25.795 ","End":"14:31.990","Text":"Now, our h is the height that the string has traveled."},{"Start":"14:31.990 ","End":"14:33.370","Text":"Now as we\u0027ve just said,"},{"Start":"14:33.370 ","End":"14:38.350","Text":"the string has been lifted to a height y above the table."},{"Start":"14:38.350 ","End":"14:42.550","Text":"However, we\u0027re dealing not with a point mass,"},{"Start":"14:42.550 ","End":"14:44.050","Text":"this would be a point mass."},{"Start":"14:44.050 ","End":"14:46.555","Text":"We\u0027re dealing with a long gangling mass,"},{"Start":"14:46.555 ","End":"14:51.400","Text":"it\u0027s not a point, which means that we can\u0027t speak of it as a point mass."},{"Start":"14:51.400 ","End":"14:53.995","Text":"We can\u0027t say that it was lifted y into the air."},{"Start":"14:53.995 ","End":"14:56.500","Text":"We have to look at its center of mass."},{"Start":"14:56.500 ","End":"14:58.135","Text":"Where is its center of mass?"},{"Start":"14:58.135 ","End":"15:01.450","Text":"Right in the center of the string. Right about here."},{"Start":"15:01.450 ","End":"15:07.825","Text":"That means that our height is going to be the position in the air divided by 2,"},{"Start":"15:07.825 ","End":"15:11.605","Text":"its center of mass because it\u0027s not a point mass."},{"Start":"15:11.605 ","End":"15:14.830","Text":"Let\u0027s put this back into the equation."},{"Start":"15:14.830 ","End":"15:19.810","Text":"Potential energy is going to be our m,"},{"Start":"15:19.810 ","End":"15:27.955","Text":"which is mass over l times y multiplied by our g,"},{"Start":"15:27.955 ","End":"15:31.240","Text":"multiplied by our height,"},{"Start":"15:31.240 ","End":"15:33.160","Text":"which is the height of our center of mass,"},{"Start":"15:33.160 ","End":"15:35.270","Text":"which is y over 2."},{"Start":"15:35.310 ","End":"15:41.110","Text":"Then we\u0027ll see that our y will be squared."},{"Start":"15:41.110 ","End":"15:47.305","Text":"We can already rearrange this now and put our 2 over here and our y^2 over here."},{"Start":"15:47.305 ","End":"15:54.370","Text":"Now we have to add to this kinetic energy."},{"Start":"15:54.370 ","End":"15:57.640","Text":"What is our kinetic energy?"},{"Start":"15:57.640 ","End":"16:03.400","Text":"As we know, our kinetic energy is equal to 1/2 our mass,"},{"Start":"16:03.400 ","End":"16:06.905","Text":"tag again to differentiate from the mass in the question,"},{"Start":"16:06.905 ","End":"16:11.164","Text":"multiplied by our velocity squared."},{"Start":"16:11.164 ","End":"16:16.060","Text":"Our mass, just like above, it\u0027s the same."},{"Start":"16:16.060 ","End":"16:19.805","Text":"Then our velocity squared is just going to be"},{"Start":"16:19.805 ","End":"16:22.010","Text":"our v_0^2 because in"},{"Start":"16:22.010 ","End":"16:25.985","Text":"the question we\u0027re dealing with a constant velocity and we were told that it\u0027s v_0."},{"Start":"16:25.985 ","End":"16:29.415","Text":"Let\u0027s again add this to the question."},{"Start":"16:29.415 ","End":"16:34.885","Text":"It\u0027s going to be plus 1/2 and then m"},{"Start":"16:34.885 ","End":"16:41.960","Text":"over l multiplied by y multiplied by v_0^2."},{"Start":"16:42.240 ","End":"16:49.940","Text":"This is the energy which is wasted as a function of the distance traveled."},{"Start":"16:49.940 ","End":"16:53.060","Text":"Now, in order to just cancel out"},{"Start":"16:53.060 ","End":"16:59.960","Text":"this long and ugly expression because it actually is quite a short answer in the end,"},{"Start":"16:59.960 ","End":"17:06.005","Text":"we can see that this and this cancel each other out."},{"Start":"17:06.005 ","End":"17:10.085","Text":"Let\u0027s see. We both have y^2 divided by 2,"},{"Start":"17:10.085 ","End":"17:12.965","Text":"y^2 divided by 2, m over l,"},{"Start":"17:12.965 ","End":"17:17.665","Text":"m over l and g. These 2 cancel out."},{"Start":"17:17.665 ","End":"17:19.730","Text":"Then we can look over here."},{"Start":"17:19.730 ","End":"17:21.530","Text":"We have v _0^2,"},{"Start":"17:21.530 ","End":"17:24.215","Text":"v _0^2, m over l,"},{"Start":"17:24.215 ","End":"17:27.575","Text":"m over l and y, y."},{"Start":"17:27.575 ","End":"17:34.585","Text":"This is like 1 times this v_0^2 m over l y negative,"},{"Start":"17:34.585 ","End":"17:37.400","Text":"so a negative and a positive is a negative,"},{"Start":"17:37.400 ","End":"17:40.290","Text":"1/2 m over l y v_0^2."},{"Start":"17:40.290 ","End":"17:47.000","Text":"We\u0027re going to be left with our Delta E in the y-direction is"},{"Start":"17:47.000 ","End":"17:54.740","Text":"going to be simply equal to 1/2 v_0^2 m over l y."},{"Start":"17:55.170 ","End":"18:01.820","Text":"This is our final answer over here."},{"Start":"18:01.860 ","End":"18:06.460","Text":"You see how easy this was."},{"Start":"18:06.460 ","End":"18:11.255","Text":"It\u0027s just important when you\u0027re dealing with questions like this to just really"},{"Start":"18:11.255 ","End":"18:16.790","Text":"follow a set way of working it out because everything is very easy."},{"Start":"18:16.790 ","End":"18:22.130","Text":"You just have to think about the problem and remember the main equations,"},{"Start":"18:22.130 ","End":"18:29.660","Text":"which is this equation and this equation. That\u0027s it."}],"ID":10623},{"Watched":false,"Name":"Rocket Orbiting A Star","Duration":"11m 10s","ChapterTopicVideoID":9235,"CourseChapterTopicPlaylistID":5429,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.140 ","End":"00:02.940","Text":"Hello. In this question,"},{"Start":"00:02.940 ","End":"00:07.560","Text":"we\u0027re being told that a rocket of mass m_0 orbits a star of mass M,"},{"Start":"00:07.560 ","End":"00:10.125","Text":"a distance of r_0 away,"},{"Start":"00:10.125 ","End":"00:14.295","Text":"so this rocket has a mass of m_0,"},{"Start":"00:14.295 ","End":"00:19.440","Text":"and it\u0027s a distance of r_0 away from this star,"},{"Start":"00:19.440 ","End":"00:24.630","Text":"which is in itself mass M. The rocket is"},{"Start":"00:24.630 ","End":"00:30.240","Text":"moving at a velocity of v_0 around the sky,"},{"Start":"00:30.240 ","End":"00:33.490","Text":"so it will change direction all the time."},{"Start":"00:33.950 ","End":"00:36.705","Text":"At time t=0,"},{"Start":"00:36.705 ","End":"00:40.010","Text":"the rocket emits a gas at a relative velocity of u,"},{"Start":"00:40.010 ","End":"00:41.960","Text":"and then the rate of Alpha,"},{"Start":"00:41.960 ","End":"00:47.400","Text":"find the equation of motion of the rocket in the radial direction."},{"Start":"00:48.650 ","End":"00:55.280","Text":"Now, because we know that at some stage a rocket is going to be emitting gas,"},{"Start":"00:55.280 ","End":"01:01.460","Text":"which means that its mass is going to decrease,"},{"Start":"01:01.460 ","End":"01:04.370","Text":"which means that its velocity will increase."},{"Start":"01:04.370 ","End":"01:06.920","Text":"Then if its velocity increases,"},{"Start":"01:06.920 ","End":"01:11.425","Text":"then the radius between it and the star will also increase."},{"Start":"01:11.425 ","End":"01:17.414","Text":"All of what is rested over here is our initial conditions."},{"Start":"01:17.414 ","End":"01:20.300","Text":"What we\u0027re going to say is,"},{"Start":"01:20.300 ","End":"01:22.595","Text":"we\u0027re going to say that the sum of"},{"Start":"01:22.595 ","End":"01:27.560","Text":"our forces in the radial direction it\u0027s going to be equal"},{"Start":"01:27.560 ","End":"01:36.330","Text":"to the mass of the rocket multiplied by the acceleration in the radial direction."},{"Start":"01:37.490 ","End":"01:40.280","Text":"Now, the force in"},{"Start":"01:40.280 ","End":"01:45.680","Text":"the radial direction is going to be governed by the force of attraction,"},{"Start":"01:45.680 ","End":"01:50.545","Text":"so the gravitational attraction between the rocket and the style."},{"Start":"01:50.545 ","End":"01:57.370","Text":"We know that it\u0027s going to be the mass of the rocket."},{"Start":"01:57.370 ","End":"01:59.850","Text":"Well, it\u0027s going to be changing in time,"},{"Start":"01:59.850 ","End":"02:03.155","Text":"so we\u0027ll write just m multiplied by"},{"Start":"02:03.155 ","End":"02:07.475","Text":"the mass of the star divided by the radius between them."},{"Start":"02:07.475 ","End":"02:11.495","Text":"Now I\u0027m going to say that it\u0027s divided by r^2,"},{"Start":"02:11.495 ","End":"02:15.020","Text":"and not r_0^2 squared because r_0^2 is"},{"Start":"02:15.020 ","End":"02:17.555","Text":"the starting radius between them but"},{"Start":"02:17.555 ","End":"02:20.320","Text":"the radius will increase later on in the question we\u0027ll see,"},{"Start":"02:20.320 ","End":"02:27.080","Text":"so I want this equation to be correct for all the radiuses that are possible."},{"Start":"02:27.080 ","End":"02:30.695","Text":"Then I add in a minus over here."},{"Start":"02:30.695 ","End":"02:34.145","Text":"Now why? Because we\u0027re dealing with radial forces."},{"Start":"02:34.145 ","End":"02:39.860","Text":"Whenever we\u0027re dealing with radial forces or radial acceleration or whatever it might be,"},{"Start":"02:39.860 ","End":"02:46.020","Text":"then whenever we have this radial direction,"},{"Start":"02:46.020 ","End":"02:49.310","Text":"if the force is pointing in the inwards direction then it\u0027s a"},{"Start":"02:49.310 ","End":"02:53.960","Text":"negative and if the force is pointing outwards direction then it\u0027s a positive."},{"Start":"02:53.960 ","End":"02:58.520","Text":"Over here because the gravity is pulling the rocket inwards,"},{"Start":"02:58.520 ","End":"03:02.820","Text":"so it\u0027s a negative force."},{"Start":"03:09.860 ","End":"03:15.935","Text":"Our radial acceleration is equal to r double-dot"},{"Start":"03:15.935 ","End":"03:21.890","Text":"negative Omega^2 r. If you don\u0027t remember this,"},{"Start":"03:21.890 ","End":"03:25.490","Text":"please go back to the lessons on radial acceleration,"},{"Start":"03:25.490 ","End":"03:30.580","Text":"centrifugal acceleration, it\u0027s the same thing just different names."},{"Start":"03:30.580 ","End":"03:36.560","Text":"Notice that if in the question they told us that the radius remains constant,"},{"Start":"03:36.560 ","End":"03:39.515","Text":"then this r double-dot would cross out."},{"Start":"03:39.515 ","End":"03:44.210","Text":"However, because we know that the radius is in fact changing,"},{"Start":"03:44.210 ","End":"03:47.365","Text":"so we write it like this."},{"Start":"03:47.365 ","End":"03:52.430","Text":"Now, it\u0027s generally not that comfortable to work with Omega,"},{"Start":"03:52.430 ","End":"03:57.050","Text":"so I know that my tangential velocity in"},{"Start":"03:57.050 ","End":"04:02.875","Text":"the direction of the tangent is equal to Omega r,"},{"Start":"04:02.875 ","End":"04:10.010","Text":"so then I can just rearrange this to say that it\u0027s m multiplied by r"},{"Start":"04:10.010 ","End":"04:14.740","Text":"double-dot negative v tangential"},{"Start":"04:14.740 ","End":"04:22.750","Text":"squared divided by r^2."},{"Start":"04:22.750 ","End":"04:26.480","Text":"Now, what we want to do is we want to connect."},{"Start":"04:26.480 ","End":"04:30.410","Text":"We have going in this direction,"},{"Start":"04:30.410 ","End":"04:34.100","Text":"our v tangential, it\u0027s orbiting."},{"Start":"04:34.100 ","End":"04:42.530","Text":"Now we want to connect this v tangential to our radial forces, so how do we do that?"},{"Start":"04:42.530 ","End":"04:46.640","Text":"We\u0027re going to be using this equation."},{"Start":"04:47.190 ","End":"04:54.565","Text":"The sum of all of the forces in the tangential direction is equal to 0,"},{"Start":"04:54.565 ","End":"04:59.500","Text":"and that equals to our mass as a function of time."},{"Start":"04:59.500 ","End":"05:03.230","Text":"Now we know that our mass as a function of time is going to be m_0,"},{"Start":"05:03.230 ","End":"05:07.410","Text":"and then because we\u0027re emitting gases,"},{"Start":"05:07.410 ","End":"05:11.825","Text":"so it\u0027s going to be negative because we\u0027re reducing the mass,"},{"Start":"05:11.825 ","End":"05:17.210","Text":"the rate that the gases are being emitted multiplied by time."},{"Start":"05:18.590 ","End":"05:23.795","Text":"This is our mass as a function of time and then"},{"Start":"05:23.795 ","End":"05:28.875","Text":"we\u0027re going to multiply that by dv by dt,"},{"Start":"05:28.875 ","End":"05:32.970","Text":"and then we have to add in our v relative."},{"Start":"05:32.970 ","End":"05:34.010","Text":"Now our v relative,"},{"Start":"05:34.010 ","End":"05:37.130","Text":"we\u0027re told in the question, where is it?"},{"Start":"05:37.130 ","End":"05:41.880","Text":"The relative velocity of the gas is u."},{"Start":"05:41.960 ","End":"05:47.215","Text":"However, we\u0027ll notice that the u is going in this direction,"},{"Start":"05:47.215 ","End":"05:50.410","Text":"which is opposite to the direction of travel of the rocket,"},{"Start":"05:50.410 ","End":"05:55.105","Text":"so we\u0027re going to add in a negative because it\u0027s in the opposite direction."},{"Start":"05:55.105 ","End":"05:59.875","Text":"Negative u multiply it by our dm by dt,"},{"Start":"05:59.875 ","End":"06:05.065","Text":"which we\u0027re also being told that it\u0027s our rate which is Alpha."},{"Start":"06:05.065 ","End":"06:08.050","Text":"Now, I\u0027m not going to multiply it by negative 1"},{"Start":"06:08.050 ","End":"06:11.185","Text":"because we\u0027re dealing with emission of gas,"},{"Start":"06:11.185 ","End":"06:14.110","Text":"the mass of the rocket is being reduced which means that"},{"Start":"06:14.110 ","End":"06:17.480","Text":"we don\u0027t multiply by negative 1,"},{"Start":"06:17.480 ","End":"06:24.640","Text":"so this is v_rel and this is our dm by dt."},{"Start":"06:26.750 ","End":"06:31.205","Text":"Now, we can notice that we only have"},{"Start":"06:31.205 ","End":"06:35.330","Text":"our unknowns with regards to v and with regards to t,"},{"Start":"06:35.330 ","End":"06:37.015","Text":"these are variables,"},{"Start":"06:37.015 ","End":"06:43.890","Text":"so the way that we can solve this equation is through a differential equation."},{"Start":"06:43.940 ","End":"06:47.960","Text":"Again, what I want to do is get my dv on the side with"},{"Start":"06:47.960 ","End":"06:53.620","Text":"all my variables and my dt on the side with all my t variables."},{"Start":"06:53.620 ","End":"07:02.840","Text":"All I\u0027m going to do is I\u0027m going to isolate out this and I\u0027m just going"},{"Start":"07:02.840 ","End":"07:11.535","Text":"to rearrange so that I get my u Alpha divided by m_0 minus Alpha t,"},{"Start":"07:11.535 ","End":"07:18.330","Text":"dt is going to simply be equal to dv."},{"Start":"07:18.670 ","End":"07:24.710","Text":"I just rearrange this expression in order to get my dv\u0027s and dt\u0027s,"},{"Start":"07:24.710 ","End":"07:28.730","Text":"and then I\u0027m going to integrate both sides and I\u0027m going to put tags over"},{"Start":"07:28.730 ","End":"07:32.945","Text":"here to show that my bounds are different to what I\u0027m integrating by,"},{"Start":"07:32.945 ","End":"07:42.780","Text":"and I\u0027m integrating from time 0 until time t and from my starting velocity which was v_0"},{"Start":"07:42.780 ","End":"07:48.920","Text":"until my velocity at time t. Notice that"},{"Start":"07:48.920 ","End":"07:55.790","Text":"the integral on the left side of the equation is going to be an integral with ln,"},{"Start":"07:55.790 ","End":"08:04.199","Text":"so we\u0027re going to have u Alpha divided by negative Alpha multiplied"},{"Start":"08:04.199 ","End":"08:13.230","Text":"by ln of m_0 minus Alpha t divided by m_0."},{"Start":"08:13.230 ","End":"08:17.550","Text":"Remember because ln of something minus learn of something"},{"Start":"08:17.550 ","End":"08:22.230","Text":"else is just ln of the something divided by something else,"},{"Start":"08:22.670 ","End":"08:29.705","Text":"and this division is because we\u0027re multiplying by the inner derivative."},{"Start":"08:29.705 ","End":"08:34.625","Text":"If you don\u0027t remember this, then please go over your integrating skills."},{"Start":"08:34.625 ","End":"08:38.740","Text":"Then over here on the side of the v,"},{"Start":"08:38.740 ","End":"08:44.270","Text":"then we will simply have our v as a function of t,"},{"Start":"08:44.270 ","End":"08:46.670","Text":"which is what we\u0027re looking for,"},{"Start":"08:46.670 ","End":"08:50.300","Text":"minus our v_0,"},{"Start":"08:50.300 ","End":"08:53.780","Text":"so this is in fact what we\u0027re searching for,"},{"Start":"08:53.780 ","End":"08:55.270","Text":"what we need to isolate."},{"Start":"08:55.270 ","End":"08:57.070","Text":"Let\u0027s isolate this out."},{"Start":"08:57.070 ","End":"08:58.630","Text":"This and this crosses out,"},{"Start":"08:58.630 ","End":"09:00.760","Text":"so what we will have is"},{"Start":"09:00.760 ","End":"09:05.090","Text":"v_0 minus u ln"},{"Start":"09:05.090 ","End":"09:12.185","Text":"of m_0 minus Alpha t divided by m_0,"},{"Start":"09:12.185 ","End":"09:20.165","Text":"which is equal to v as a function of t. Then"},{"Start":"09:20.165 ","End":"09:23.750","Text":"our final answer is simply going to be substituting"},{"Start":"09:23.750 ","End":"09:28.665","Text":"in this v is a function of t into this,"},{"Start":"09:28.665 ","End":"09:32.485","Text":"so this is also going to be in the tangential direction."},{"Start":"09:32.485 ","End":"09:35.765","Text":"Then we\u0027re going to get that our final answer is equal to"},{"Start":"09:35.765 ","End":"09:40.780","Text":"negative m multiplied by m divided by r^2,"},{"Start":"09:40.780 ","End":"09:48.835","Text":"which is going to be equal to our m multiplied by r double-dot minus and then"},{"Start":"09:48.835 ","End":"09:54.585","Text":"our v_0 minus u ln of"},{"Start":"09:54.585 ","End":"10:02.340","Text":"m_0 minus Alpha t divided by m_0,"},{"Start":"10:02.340 ","End":"10:11.190","Text":"and then this is going to be squared divided by r^2."},{"Start":"10:12.880 ","End":"10:17.275","Text":"This is our final answer."},{"Start":"10:17.275 ","End":"10:22.015","Text":"I hope you understood how we connected the 2 equations."},{"Start":"10:22.015 ","End":"10:29.880","Text":"All we did is we found the sum of the forces in the radial direction then we"},{"Start":"10:29.880 ","End":"10:33.660","Text":"translated that into what we\u0027re given from"},{"Start":"10:33.660 ","End":"10:37.740","Text":"the question because this was just the recipe,"},{"Start":"10:37.740 ","End":"10:40.790","Text":"then we substituted in all of our unknowns,"},{"Start":"10:40.790 ","End":"10:45.080","Text":"and then we saw that we have a variable,"},{"Start":"10:45.080 ","End":"10:49.490","Text":"this v tangential that we didn\u0027t know what it was."},{"Start":"10:49.490 ","End":"10:57.185","Text":"Then what we did was we used this equation to find out what v tangential will equal"},{"Start":"10:57.185 ","End":"11:01.670","Text":"as the gas is emitted from the rockets and then we"},{"Start":"11:01.670 ","End":"11:07.090","Text":"simply substituted in to get to our final answer over here."},{"Start":"11:07.090 ","End":"11:11.020","Text":"That is the end of the lesson."}],"ID":10624}],"Thumbnail":null,"ID":5429}]

[{"ID":5426,"Videos":[8222]},{"ID":5427,"Videos":[10610,10611,10612,10613,8224]},{"ID":5429,"Videos":[10614,10615,10616,10617,10618,10619,10620,10621,10622,10623,10624]}];

[8222];

1

3