What is Momentum and Newtons Second Law
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Introduction to Impulse
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Conserving Momentum and External Forces
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Types of Collisions
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Conservation of Momentum During Short Timed Collisions
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Momentum Conclusion
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[{"Name":"What is Momentum and Newtons Second Law","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"What is Momentum","Duration":"4m 29s","ChapterTopicVideoID":9053,"CourseChapterTopicPlaylistID":5391,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.445","Text":"Hello. In this lecture,"},{"Start":"00:02.445 ","End":"00:06.795","Text":"I want to talk about momentum and its connection to Newton\u0027s second law."},{"Start":"00:06.795 ","End":"00:10.110","Text":"First, I want to start with a mathematical explanation."},{"Start":"00:10.110 ","End":"00:14.700","Text":"Momentum is symbolized by the letter P and it is a vector."},{"Start":"00:14.700 ","End":"00:19.875","Text":"The P vector equals mass times the velocity vector, another vector."},{"Start":"00:19.875 ","End":"00:22.800","Text":"We know that momentum is in the same direction as"},{"Start":"00:22.800 ","End":"00:26.370","Text":"velocity because it\u0027s a multiple of the velocity vector."},{"Start":"00:26.370 ","End":"00:29.520","Text":"That\u0027s in terms of direction and in terms of magnitude,"},{"Start":"00:29.520 ","End":"00:34.075","Text":"we just multiply that velocity vector by mass and that\u0027ll give us the magnitude."},{"Start":"00:34.075 ","End":"00:38.285","Text":"Let\u0027s leave math aside and talk about this in intuitive terms for a second."},{"Start":"00:38.285 ","End":"00:43.235","Text":"Personally, the first time I heard about momentum in the context of a physics course,"},{"Start":"00:43.235 ","End":"00:44.720","Text":"it didn\u0027t make much sense to me."},{"Start":"00:44.720 ","End":"00:48.470","Text":"I didn\u0027t think it was necessary because if it\u0027s just the velocity times the mass,"},{"Start":"00:48.470 ","End":"00:50.374","Text":"why don\u0027t we just use the velocity?"},{"Start":"00:50.374 ","End":"00:52.100","Text":"But now when I think about momentum,"},{"Start":"00:52.100 ","End":"00:55.385","Text":"the first thing that comes to mind is a rhinoceros."},{"Start":"00:55.385 ","End":"00:58.100","Text":"You might ask, why a rhinoceros?"},{"Start":"00:58.100 ","End":"00:59.495","Text":"Well, think of it this way."},{"Start":"00:59.495 ","End":"01:04.340","Text":"Imagine that the rhinoceros is running at 50 kilometers an hour."},{"Start":"01:04.340 ","End":"01:07.775","Text":"Now it\u0027s going to take a lot of force to stop that rhinoceros,"},{"Start":"01:07.775 ","End":"01:11.015","Text":"even though 50 kilometers an hour isn\u0027t all that fast."},{"Start":"01:11.015 ","End":"01:16.520","Text":"If we compare that to say a cyclist that\u0027s riding at 50 kilometers an hour,"},{"Start":"01:16.520 ","End":"01:18.890","Text":"it\u0027s going to take a lot less force to stop"},{"Start":"01:18.890 ","End":"01:21.995","Text":"the cyclist than it will take to stop the rhinoceros."},{"Start":"01:21.995 ","End":"01:24.980","Text":"Obviously, the difference between the 2 isn\u0027t velocity,"},{"Start":"01:24.980 ","End":"01:26.750","Text":"they\u0027re going the exact same speed,"},{"Start":"01:26.750 ","End":"01:28.100","Text":"the differences in the mass,"},{"Start":"01:28.100 ","End":"01:29.660","Text":"and that\u0027s the point of momentum."},{"Start":"01:29.660 ","End":"01:34.890","Text":"Momentum is a force that takes a mass into account when talking about magnitude."},{"Start":"01:35.000 ","End":"01:38.810","Text":"What we can conclude from this is that the rhinoceros has"},{"Start":"01:38.810 ","End":"01:42.095","Text":"a lot more momentum when running at 50 kilometers an hour,"},{"Start":"01:42.095 ","End":"01:43.850","Text":"as opposed to the cyclist riding at"},{"Start":"01:43.850 ","End":"01:47.585","Text":"50 kilometers an hour and that has to do with the rhinoceros\u0027 mass."},{"Start":"01:47.585 ","End":"01:50.015","Text":"If the rhinoceros weighs 10 times as much,"},{"Start":"01:50.015 ","End":"01:54.990","Text":"it will have 10 times the momentum and require 10 times the force to stop it."},{"Start":"01:55.580 ","End":"01:58.970","Text":"Essentially, one intuitive way to look at"},{"Start":"01:58.970 ","End":"02:03.215","Text":"momentum is that as the momentum of an object increases,"},{"Start":"02:03.215 ","End":"02:07.790","Text":"its influence during a collision will increase in kind."},{"Start":"02:07.790 ","End":"02:09.305","Text":"This is an important point."},{"Start":"02:09.305 ","End":"02:11.435","Text":"When we talk about stopping an object,"},{"Start":"02:11.435 ","End":"02:13.475","Text":"that means exerting force on it."},{"Start":"02:13.475 ","End":"02:15.710","Text":"What\u0027s the formula for this force?"},{"Start":"02:15.710 ","End":"02:17.780","Text":"Well, it comes from Newton\u0027s second law."},{"Start":"02:17.780 ","End":"02:22.475","Text":"In fact, the original version of Newton\u0027s second law was written in this format."},{"Start":"02:22.475 ","End":"02:25.990","Text":"The sum of forces equals dP over dt."},{"Start":"02:25.990 ","End":"02:29.105","Text":"What this means is that to change the momentum,"},{"Start":"02:29.105 ","End":"02:32.315","Text":"you have to exert force and in the case of the rhinoceros,"},{"Start":"02:32.315 ","End":"02:35.480","Text":"you have to exert a lot of force because of its enormous mass."},{"Start":"02:35.480 ","End":"02:37.300","Text":"Whereas with the bicyclist,"},{"Start":"02:37.300 ","End":"02:41.360","Text":"you have to exert relatively less force because of its smaller mass."},{"Start":"02:41.360 ","End":"02:44.195","Text":"Another way to understand this is to think about"},{"Start":"02:44.195 ","End":"02:47.405","Text":"a rhinoceros hitting a car or a bicycle hitting a car."},{"Start":"02:47.405 ","End":"02:50.720","Text":"The rhinoceros hitting the car will do a lot of damage because of"},{"Start":"02:50.720 ","End":"02:53.150","Text":"the enormous momentum and the amount of force of"},{"Start":"02:53.150 ","End":"02:56.435","Text":"the car would have to exert in order to stop the rhinoceros."},{"Start":"02:56.435 ","End":"02:59.360","Text":"Whereas because the bicycle has a lot less mass,"},{"Start":"02:59.360 ","End":"03:03.200","Text":"it won\u0027t do much damage at all to the car because of the little amount of momentum"},{"Start":"03:03.200 ","End":"03:07.245","Text":"required or the little amount of force required to stop its momentum."},{"Start":"03:07.245 ","End":"03:11.810","Text":"These are 2 rather important equations and you should save them on your equation sheet."},{"Start":"03:11.810 ","End":"03:14.300","Text":"The other thing to remember here is that P,"},{"Start":"03:14.300 ","End":"03:16.445","Text":"the momentum, is written as a vector."},{"Start":"03:16.445 ","End":"03:19.645","Text":"Really we\u0027re talking about 3 different formula here."},{"Start":"03:19.645 ","End":"03:26.705","Text":"The first is P_x=mV_x, the second,"},{"Start":"03:26.705 ","End":"03:30.710","Text":"P_y=mV_y, and"},{"Start":"03:30.710 ","End":"03:37.910","Text":"lastly, P_z=mV_z."},{"Start":"03:37.910 ","End":"03:39.965","Text":"What this means is that when I\u0027m talking about"},{"Start":"03:39.965 ","End":"03:42.935","Text":"a given object and its momentum, say the rhinoceros,"},{"Start":"03:42.935 ","End":"03:48.110","Text":"I can talk about its momentum in each of the 3 directions or in each of the 3 elements."},{"Start":"03:48.110 ","End":"03:52.460","Text":"This particular rhinoceros seems to have momentum only along the x-axis,"},{"Start":"03:52.460 ","End":"03:55.250","Text":"but not along the y or the z axes."},{"Start":"03:55.250 ","End":"03:57.995","Text":"Of course, the same is true for the sum of forces."},{"Start":"03:57.995 ","End":"04:01.085","Text":"We\u0027re also talking about 3 different formula here,"},{"Start":"04:01.085 ","End":"04:03.260","Text":"one for x, one for y,"},{"Start":"04:03.260 ","End":"04:04.720","Text":"and one for z."},{"Start":"04:04.720 ","End":"04:10.745","Text":"The sum of forces along the x-axis equals dPx over dt."},{"Start":"04:10.745 ","End":"04:17.956","Text":"The sum of forces along the y-axis equals dPy over"},{"Start":"04:17.956 ","End":"04:25.710","Text":"dt and the sum of forces along the z-axis equals dPz dt."}],"ID":9326},{"Watched":false,"Name":"The Connection to Newton\u0026#39;s Second Law","Duration":"1m 28s","ChapterTopicVideoID":9054,"CourseChapterTopicPlaylistID":5391,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.200 ","End":"00:03.360","Text":"The last thing I want to bring up is the connection"},{"Start":"00:03.360 ","End":"00:05.685","Text":"between this version of Newton\u0027s 2nd law,"},{"Start":"00:05.685 ","End":"00:09.465","Text":"the version involving momentum to the more familiar version to us,"},{"Start":"00:09.465 ","End":"00:12.465","Text":"Sigma F equals ma."},{"Start":"00:12.465 ","End":"00:16.830","Text":"I want to show you how we get from this to this and along the way you\u0027ll"},{"Start":"00:16.830 ","End":"00:20.595","Text":"see that the momentum version is a general version that is more correct,"},{"Start":"00:20.595 ","End":"00:25.210","Text":"whereas the ma version only works in certain circumstances."},{"Start":"00:25.490 ","End":"00:34.020","Text":"Let\u0027s write this equation from above Sigma F equals dP,"},{"Start":"00:34.020 ","End":"00:43.520","Text":"dt but we\u0027re going to replace P with its definition from this equation, mv."},{"Start":"00:43.550 ","End":"00:48.065","Text":"We know we have to take a derivative of this in terms of time,"},{"Start":"00:48.065 ","End":"00:51.245","Text":"and we can take m out if it\u0027s constant that is,"},{"Start":"00:51.245 ","End":"00:52.445","Text":"and move it on the side."},{"Start":"00:52.445 ","End":"00:57.690","Text":"We have m times dv, dt."},{"Start":"00:59.200 ","End":"01:02.180","Text":"By now, you should know that dv,"},{"Start":"01:02.180 ","End":"01:06.730","Text":"dt equals acceleration, so we have is ma."},{"Start":"01:06.730 ","End":"01:09.185","Text":"These are really the same equation."},{"Start":"01:09.185 ","End":"01:14.180","Text":"The only difference is that the ma version supposes that the mass is constant,"},{"Start":"01:14.180 ","End":"01:16.190","Text":"and for those who are interested later,"},{"Start":"01:16.190 ","End":"01:20.209","Text":"we\u0027ll talk about what we do in an instance when the mass is not constant."},{"Start":"01:20.209 ","End":"01:23.330","Text":"In that case, the ma formula really is"},{"Start":"01:23.330 ","End":"01:26.750","Text":"inaccurate and we have to use this formula but for now,"},{"Start":"01:26.750 ","End":"01:29.070","Text":"that\u0027s the end of this lecture."}],"ID":9327}],"Thumbnail":null,"ID":5391},{"Name":"Introduction to Impulse","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Impulse a Basic Description","Duration":"4m 11s","ChapterTopicVideoID":9055,"CourseChapterTopicPlaylistID":5392,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.020 ","End":"00:04.050","Text":"Hello. In this lecture I want to talk about impulse."},{"Start":"00:04.050 ","End":"00:09.225","Text":"The way we\u0027ll do that is by analyzing an example where a soccer player kicks a ball."},{"Start":"00:09.225 ","End":"00:11.400","Text":"As soon as the player kicks the ball,"},{"Start":"00:11.400 ","End":"00:14.385","Text":"he or she is exerting force upon that ball,"},{"Start":"00:14.385 ","End":"00:18.210","Text":"and we\u0027ll call that force F. Let\u0027s assume we know the force to be F,"},{"Start":"00:18.210 ","End":"00:20.520","Text":"and then we also know the amount of time involved."},{"Start":"00:20.520 ","End":"00:24.360","Text":"We\u0027ll call that Delta t. The last thing we know is that the ball"},{"Start":"00:24.360 ","End":"00:28.500","Text":"has a mass of m. Now that we know these things,"},{"Start":"00:28.500 ","End":"00:30.690","Text":"we can incorporate Newton\u0027s Second Law,"},{"Start":"00:30.690 ","End":"00:32.505","Text":"and we\u0027ll use the classic formula,"},{"Start":"00:32.505 ","End":"00:34.740","Text":"the one that we learned about in the last lecture,"},{"Start":"00:34.740 ","End":"00:39.755","Text":"and that is the sum of forces equals dP over dt,"},{"Start":"00:39.755 ","End":"00:42.035","Text":"that is momentum over time."},{"Start":"00:42.035 ","End":"00:45.620","Text":"You should also know from the last lecture how to get from this version of"},{"Start":"00:45.620 ","End":"00:50.590","Text":"the formula to the version we\u0027re more familiar with the sum of forces equals ma."},{"Start":"00:50.590 ","End":"00:55.700","Text":"Now what I want to do is multiply both sides of this by dt."},{"Start":"00:55.700 ","End":"00:57.455","Text":"When I do this calculation,"},{"Start":"00:57.455 ","End":"01:00.130","Text":"the result is Sigma F,"},{"Start":"01:00.130 ","End":"01:05.925","Text":"the sum of forces times dt equals dP."},{"Start":"01:05.925 ","End":"01:10.715","Text":"The reason I did this is now on the right side, I have dP,"},{"Start":"01:10.715 ","End":"01:13.610","Text":"my momentum, and on the left side I have the sum of"},{"Start":"01:13.610 ","End":"01:17.090","Text":"forces and dt things having to do with force and time."},{"Start":"01:17.090 ","End":"01:20.360","Text":"So on the left side, I have things having to do with influence,"},{"Start":"01:20.360 ","End":"01:23.375","Text":"having to it the kicker and the force exerted."},{"Start":"01:23.375 ","End":"01:25.550","Text":"Whereas on the right side I have momentum,"},{"Start":"01:25.550 ","End":"01:29.329","Text":"something having to do with the ball itself or the object being influenced."},{"Start":"01:29.329 ","End":"01:31.715","Text":"Another way to think about this is on the left,"},{"Start":"01:31.715 ","End":"01:33.520","Text":"I have things having to do with the kicker,"},{"Start":"01:33.520 ","End":"01:36.655","Text":"and on the right, I have things having to do with the ball."},{"Start":"01:36.655 ","End":"01:40.310","Text":"The next step is to take an integral of both sides."},{"Start":"01:40.310 ","End":"01:41.600","Text":"On the left side,"},{"Start":"01:41.600 ","End":"01:46.625","Text":"I\u0027ll take an integral from t equals 0 to t equals Delta_t."},{"Start":"01:46.625 ","End":"01:48.410","Text":"This will be over time of course,"},{"Start":"01:48.410 ","End":"01:50.128","Text":"because we\u0027re dealing with dt."},{"Start":"01:50.128 ","End":"01:54.080","Text":"On the left, we\u0027ll take an integral based on momentum."},{"Start":"01:54.080 ","End":"01:56.660","Text":"So it\u0027ll be P when t=0,"},{"Start":"01:56.660 ","End":"02:00.424","Text":"and we\u0027ll call this P_i the initial momentum."},{"Start":"02:00.424 ","End":"02:04.320","Text":"On top, let me move this out of the way,"},{"Start":"02:04.460 ","End":"02:13.525","Text":"I\u0027ll get p of Delta t and that we\u0027ll call our final or terminal momentum."},{"Start":"02:13.525 ","End":"02:19.600","Text":"What we have here on the left is the amount of force exerted over the course of time,"},{"Start":"02:19.600 ","End":"02:21.280","Text":"and we have a name for this."},{"Start":"02:21.280 ","End":"02:23.810","Text":"The amount of force that I exerted over time,"},{"Start":"02:23.810 ","End":"02:26.710","Text":"we call that J or impulse."},{"Start":"02:26.710 ","End":"02:30.265","Text":"What this really means is the more force that I exert,"},{"Start":"02:30.265 ","End":"02:31.520","Text":"the larger my impulse,"},{"Start":"02:31.520 ","End":"02:32.635","Text":"and at the same time,"},{"Start":"02:32.635 ","End":"02:34.855","Text":"the more time that I exert that force,"},{"Start":"02:34.855 ","End":"02:38.000","Text":"the larger my impulse is as well."},{"Start":"02:38.070 ","End":"02:40.285","Text":"Now, on the right side,"},{"Start":"02:40.285 ","End":"02:43.915","Text":"our result is basically P_F, our final P,"},{"Start":"02:43.915 ","End":"02:48.470","Text":"minus P_i our initial P. When we have final minus initial,"},{"Start":"02:48.470 ","End":"02:49.775","Text":"we can write that as Delta,"},{"Start":"02:49.775 ","End":"02:58.435","Text":"so Delta P. This J equals Delta P is a formula we\u0027re going to want to remember."},{"Start":"02:58.435 ","End":"03:03.290","Text":"This is the formula, and you should make sure to put it on your formula sheet."},{"Start":"03:03.290 ","End":"03:06.184","Text":"To summarize, basically this is impulse,"},{"Start":"03:06.184 ","End":"03:08.000","Text":"and you can think of it in two ways."},{"Start":"03:08.000 ","End":"03:11.855","Text":"Impulse is either a new way of ordering Newton\u0027s Second Law,"},{"Start":"03:11.855 ","End":"03:15.770","Text":"where you take dt and put it on the left side with the sum of forces,"},{"Start":"03:15.770 ","End":"03:21.080","Text":"and the other way to think about it is the total influence of one object on another."},{"Start":"03:21.080 ","End":"03:24.080","Text":"In our situation, it\u0027s how much force really"},{"Start":"03:24.080 ","End":"03:28.135","Text":"the kicker is exerting on the ball over the course of time."},{"Start":"03:28.135 ","End":"03:32.210","Text":"When it comes down to actually using impulse in a problem,"},{"Start":"03:32.210 ","End":"03:35.705","Text":"what you want to do is find your sum of forces times dt,"},{"Start":"03:35.705 ","End":"03:37.595","Text":"and take an integral of that."},{"Start":"03:37.595 ","End":"03:42.650","Text":"Using that result, you\u0027ll know the change in momentum over the course of time."},{"Start":"03:42.650 ","End":"03:45.545","Text":"Once I know the change in momentum,"},{"Start":"03:45.545 ","End":"03:47.750","Text":"I can use that to find all things,"},{"Start":"03:47.750 ","End":"03:50.915","Text":"including changes in velocity or acceleration."},{"Start":"03:50.915 ","End":"03:54.425","Text":"The other way to use impulse is backwards, so to speak."},{"Start":"03:54.425 ","End":"03:56.375","Text":"First you find your Delta P,"},{"Start":"03:56.375 ","End":"03:57.590","Text":"the change in momentum,"},{"Start":"03:57.590 ","End":"03:59.945","Text":"and from there you can use J."},{"Start":"03:59.945 ","End":"04:04.945","Text":"Once you know J, you can find the amount of force exerted over time."},{"Start":"04:04.945 ","End":"04:07.725","Text":"This ends the basic explanation."},{"Start":"04:07.725 ","End":"04:11.730","Text":"From here I\u0027ll show you a few things you can do with impulse."}],"ID":9328},{"Watched":false,"Name":"Impulse of a Constant Force","Duration":"1m 51s","ChapterTopicVideoID":9056,"CourseChapterTopicPlaylistID":5392,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.595","Text":"Hello. In the last video,"},{"Start":"00:02.595 ","End":"00:04.815","Text":"we discussed what impulse is."},{"Start":"00:04.815 ","End":"00:05.940","Text":"We defined it,"},{"Start":"00:05.940 ","End":"00:08.280","Text":"and we said that the sum of the impulse or"},{"Start":"00:08.280 ","End":"00:12.435","Text":"the impulse of the sum of forces equals the change in momentum."},{"Start":"00:12.435 ","End":"00:21.900","Text":"Just to remind you, we said that momentum or P equals mv and that the change in P,"},{"Start":"00:21.900 ","End":"00:28.870","Text":"the change in momentum equals P final minus P initial."},{"Start":"00:30.170 ","End":"00:34.500","Text":"You should also keep in mind that momentum is a vector and that it"},{"Start":"00:34.500 ","End":"00:38.865","Text":"has the exact same axis as the v, velocity vector."},{"Start":"00:38.865 ","End":"00:43.715","Text":"Let\u0027s suppose for a second that we want to find the impulse of 1 force,"},{"Start":"00:43.715 ","End":"00:47.300","Text":"not the sum of forces like we have up here, just 1 force."},{"Start":"00:47.300 ","End":"00:52.320","Text":"We do is take an integral of just that 1 force times dt."},{"Start":"00:52.900 ","End":"00:57.350","Text":"In this case, this will not equal the change in"},{"Start":"00:57.350 ","End":"01:01.745","Text":"momentum unless this is the only force acting upon our object."},{"Start":"01:01.745 ","End":"01:06.290","Text":"Only the impulse of the sum of forces equals the change in momentum,"},{"Start":"01:06.290 ","End":"01:09.295","Text":"not the impulse of a single force."},{"Start":"01:09.295 ","End":"01:13.385","Text":"You should notice by now that J is also a vector,"},{"Start":"01:13.385 ","End":"01:14.880","Text":"impulse is a vector."},{"Start":"01:14.880 ","End":"01:19.980","Text":"If the force that we\u0027re talking about is a constant force,"},{"Start":"01:21.760 ","End":"01:25.910","Text":"that is a IF,"},{"Start":"01:25.910 ","End":"01:29.375","Text":"then the F, the force,"},{"Start":"01:29.375 ","End":"01:30.990","Text":"can move to the outside of this integral,"},{"Start":"01:30.990 ","End":"01:34.395","Text":"and we have F by the integral of dt."},{"Start":"01:34.395 ","End":"01:39.815","Text":"What we get is the force times the change in time Delta t,"},{"Start":"01:39.815 ","End":"01:43.525","Text":"or the range of time that this force is acting."},{"Start":"01:43.525 ","End":"01:47.900","Text":"The impulse of a single constant force equals that force,"},{"Start":"01:47.900 ","End":"01:51.600","Text":"F multiplied by the change in time."}],"ID":9329},{"Watched":false,"Name":"Example for Calculating Impulse","Duration":"3m 7s","ChapterTopicVideoID":9057,"CourseChapterTopicPlaylistID":5392,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.260 ","End":"00:05.325","Text":"Let\u0027s do a quick example of calculating impulse."},{"Start":"00:05.325 ","End":"00:07.600","Text":"A football player, I\u0027m from America,"},{"Start":"00:07.600 ","End":"00:10.690","Text":"we call it soccer, but for the sake of the international audience,"},{"Start":"00:10.690 ","End":"00:13.360","Text":"a football player kicks a ball with a mass of"},{"Start":"00:13.360 ","End":"00:16.900","Text":"2 kilograms with a constant force of 50 newtons."},{"Start":"00:16.900 ","End":"00:20.605","Text":"The player is touching the ball for 0.2 seconds."},{"Start":"00:20.605 ","End":"00:24.560","Text":"What will be the velocity with which the ball travels?"},{"Start":"00:25.560 ","End":"00:33.980","Text":"Let\u0027s assume that our ball started at rest and that it was kicked with a force of 50N."},{"Start":"00:34.400 ","End":"00:38.080","Text":"In that case, to find the impulse J,"},{"Start":"00:38.080 ","End":"00:44.690","Text":"what we do is take an integral of our 50 Newtons times dt over the time,"},{"Start":"00:44.690 ","End":"00:49.900","Text":"which was from times 0 to the times 0.2 seconds."},{"Start":"00:49.900 ","End":"00:54.515","Text":"Now we can do because our 50 Newtons is a constant force."},{"Start":"00:54.515 ","End":"00:57.410","Text":"We can take it to the outside of our integral and then we just"},{"Start":"00:57.410 ","End":"01:00.980","Text":"take an integral from 0 to 0.2 seconds of dt,"},{"Start":"01:00.980 ","End":"01:05.765","Text":"and our result is 0.2 seconds."},{"Start":"01:05.765 ","End":"01:10.245","Text":"When we multiply 50N times 0.2 seconds,"},{"Start":"01:10.245 ","End":"01:14.515","Text":"our result is 1 Newton second."},{"Start":"01:14.515 ","End":"01:17.749","Text":"If I want to think about this in terms of directions,"},{"Start":"01:17.749 ","End":"01:20.240","Text":"I can assume that this is my x-axis."},{"Start":"01:20.240 ","End":"01:24.985","Text":"Therefore I\u0027m moving along my x-axis as well at the end."},{"Start":"01:24.985 ","End":"01:27.320","Text":"This is in the direction of x."},{"Start":"01:27.320 ","End":"01:31.250","Text":"Now the next step is to take this to find my velocity."},{"Start":"01:31.250 ","End":"01:33.620","Text":"We know that the change in momentum,"},{"Start":"01:33.620 ","End":"01:38.480","Text":"Delta p equals J, the impulse."},{"Start":"01:38.480 ","End":"01:44.410","Text":"Put another way, Delta P is P final minus P initial."},{"Start":"01:44.410 ","End":"01:50.915","Text":"We can rewrite this because each P equals M times V. We can say that P final"},{"Start":"01:50.915 ","End":"01:59.160","Text":"is MV final and P initial is MV initial and these are vectors."},{"Start":"01:59.390 ","End":"02:03.705","Text":"We know that this has to equal 1 Newton times seconds."},{"Start":"02:03.705 ","End":"02:08.580","Text":"We get M times V_f minus V_i."},{"Start":"02:08.580 ","End":"02:15.760","Text":"Again, these are vectors equals 1 Newton second."},{"Start":"02:16.400 ","End":"02:23.675","Text":"We know that the mass equals 2 kilograms and we know that V initial equals 0,"},{"Start":"02:23.675 ","End":"02:25.475","Text":"our ball starts at rest."},{"Start":"02:25.475 ","End":"02:27.125","Text":"When you do the calculation,"},{"Start":"02:27.125 ","End":"02:34.040","Text":"2 kilograms times something minus 0 equals 1 Newton\u0027s second and in the end,"},{"Start":"02:34.040 ","End":"02:40.605","Text":"V final equals 0.5 meters per second."},{"Start":"02:40.605 ","End":"02:47.330","Text":"Because Mks, or meters times kilograms per second is the equivalent of Newton\u0027s seconds."},{"Start":"02:47.330 ","End":"02:50.210","Text":"I know that my unit here will be meters per second."},{"Start":"02:50.210 ","End":"02:53.375","Text":"I know that this is also in the direction of x"},{"Start":"02:53.375 ","End":"02:57.120","Text":"because the velocity goes in the same direction as the momentum."},{"Start":"02:57.120 ","End":"02:59.660","Text":"The momentum goes in the same direction as J,"},{"Start":"02:59.660 ","End":"03:02.755","Text":"the impulse which is in the direction of x."},{"Start":"03:02.755 ","End":"03:07.990","Text":"Now we\u0027ve found our answer and that\u0027s the end of the example."}],"ID":9330},{"Watched":false,"Name":"Sum of the Impulses Equals the Impulse of the Sum of Forces","Duration":"2m 12s","ChapterTopicVideoID":9058,"CourseChapterTopicPlaylistID":5392,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.575","Text":"Hello. In this lecture,"},{"Start":"00:02.575 ","End":"00:04.855","Text":"I want to show you that the sum of impulses,"},{"Start":"00:04.855 ","End":"00:11.430","Text":"that is the sum of each impulse of each force equals the impulse of the sum of forces."},{"Start":"00:11.430 ","End":"00:14.470","Text":"Let\u0027s start with the impulse of the sum of forces."},{"Start":"00:14.470 ","End":"00:19.150","Text":"We know that J sum of forces is Sigma F"},{"Start":"00:19.150 ","End":"00:25.915","Text":"equals the integral of the sum of forces times dt."},{"Start":"00:25.915 ","End":"00:28.285","Text":"Now, if we were to break that down,"},{"Start":"00:28.285 ","End":"00:33.490","Text":"we would get the integral of Force 1 plus Force"},{"Start":"00:33.490 ","End":"00:39.750","Text":"2 plus Force 3 plus Force 4,"},{"Start":"00:39.750 ","End":"00:43.375","Text":"and on and on and on until we reach our last force."},{"Start":"00:43.375 ","End":"00:45.835","Text":"That\u0027s going to be times dt."},{"Start":"00:45.835 ","End":"00:48.865","Text":"Now if we broke this integral down,"},{"Start":"00:48.865 ","End":"00:51.970","Text":"we would get individual integrals for each force."},{"Start":"00:51.970 ","End":"00:58.000","Text":"This equals integral of Force 1 dt plus"},{"Start":"00:58.000 ","End":"01:05.570","Text":"integral of Force 2 dt plus integral of Force 3 dt,"},{"Start":"01:05.570 ","End":"01:09.710","Text":"and on and on and on until we reach our last force."},{"Start":"01:10.200 ","End":"01:13.014","Text":"Now look at each of these integrals,"},{"Start":"01:13.014 ","End":"01:17.335","Text":"and it looks like our impulse of an individual force from a form."},{"Start":"01:17.335 ","End":"01:20.805","Text":"This is J of Force 1,"},{"Start":"01:20.805 ","End":"01:25.545","Text":"and this is J of Force 2,"},{"Start":"01:25.545 ","End":"01:29.640","Text":"and this is J of Force 3,"},{"Start":"01:29.640 ","End":"01:32.710","Text":"and on and on and on until we reach our last force."},{"Start":"01:32.710 ","End":"01:38.370","Text":"We have here is a sum of each of the impulses."},{"Start":"01:38.560 ","End":"01:40.835","Text":"The upshot from this,"},{"Start":"01:40.835 ","End":"01:43.850","Text":"what it really means is that it doesn\u0027t matter whether you"},{"Start":"01:43.850 ","End":"01:47.390","Text":"first do a sum of forces and then take an integral of that."},{"Start":"01:47.390 ","End":"01:50.300","Text":"Or whether you just take the integral of each force"},{"Start":"01:50.300 ","End":"01:53.779","Text":"individually and take the sum of the integrals or the impulses,"},{"Start":"01:53.779 ","End":"01:55.535","Text":"you\u0027ll get the same result."},{"Start":"01:55.535 ","End":"01:58.475","Text":"The sum of the impulses of the impulse of"},{"Start":"01:58.475 ","End":"02:02.435","Text":"each force equals the impulse of the sum of forces,"},{"Start":"02:02.435 ","End":"02:05.345","Text":"and both of those equal Delta p,"},{"Start":"02:05.345 ","End":"02:08.015","Text":"the change in momentum."},{"Start":"02:08.015 ","End":"02:13.380","Text":"I hope this explanation makes sense and now we\u0027ll move on to an example."}],"ID":9331},{"Watched":false,"Name":"Example 2 - Two Forces on an Object","Duration":"5m 12s","ChapterTopicVideoID":9059,"CourseChapterTopicPlaylistID":5392,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.170","Text":"In this example, you\u0027re given an object with a mass of"},{"Start":"00:04.170 ","End":"00:07.890","Text":"3 kilograms and told that 2 forces are acting upon it."},{"Start":"00:07.890 ","End":"00:12.030","Text":"F_1 is acting with a force of 10 Newtons towards the right,"},{"Start":"00:12.030 ","End":"00:18.360","Text":"and F_2 with a force of 50 Newtons at a 30-degree angle with respect to the x-axis."},{"Start":"00:18.360 ","End":"00:23.250","Text":"Both of these forces are acting for a period of 0.5 seconds."},{"Start":"00:23.250 ","End":"00:27.330","Text":"In part a, we\u0027re asked to find the impulse of each force."},{"Start":"00:27.330 ","End":"00:32.880","Text":"For part a, we can start with force 1 and because it\u0027s a constant force,"},{"Start":"00:32.880 ","End":"00:34.515","Text":"we don\u0027t need to use an integral,"},{"Start":"00:34.515 ","End":"00:38.600","Text":"we can take F_1 times the change in time."},{"Start":"00:38.600 ","End":"00:44.985","Text":"That equals 10 newtons times 0.5 seconds."},{"Start":"00:44.985 ","End":"00:51.140","Text":"Our result is 5 Newton seconds in the direction of x."},{"Start":"00:51.140 ","End":"00:52.820","Text":"For force 2,"},{"Start":"00:52.820 ","End":"00:54.695","Text":"we can go through the same procedure."},{"Start":"00:54.695 ","End":"00:56.570","Text":"F_2 is also constant,"},{"Start":"00:56.570 ","End":"00:59.555","Text":"so we can multiply it by the change in time."},{"Start":"00:59.555 ","End":"01:04.190","Text":"We end up with 15 newtons times 0.5 seconds,"},{"Start":"01:04.190 ","End":"01:09.109","Text":"and that equals 7.5N times seconds."},{"Start":"01:09.109 ","End":"01:12.410","Text":"However, the direction here is a little bit different because"},{"Start":"01:12.410 ","End":"01:15.755","Text":"we are going with respect to the x-axis at a 30-degree angle,"},{"Start":"01:15.755 ","End":"01:20.785","Text":"we can consider this 7.5N times seconds as the magnitude."},{"Start":"01:20.785 ","End":"01:23.380","Text":"When we want to find the direction,"},{"Start":"01:23.380 ","End":"01:26.515","Text":"we can break this down just like any other force vector."},{"Start":"01:26.515 ","End":"01:29.320","Text":"We can break it down into x and y elements."},{"Start":"01:29.320 ","End":"01:38.090","Text":"J_2 along the x-axis equals 7.5 times the cosine of 30 degrees,"},{"Start":"01:38.090 ","End":"01:46.370","Text":"that equals 7.5 times the square root of 3/2."},{"Start":"01:48.740 ","End":"01:58.780","Text":"That\u0027s in the direction of x. J2 on the y-axis equals 7.5 times sine 30 degrees."},{"Start":"01:58.780 ","End":"02:03.525","Text":"That equals 7.5 times 1/2,"},{"Start":"02:03.525 ","End":"02:05.265","Text":"and that\u0027s in the direction of y."},{"Start":"02:05.265 ","End":"02:07.015","Text":"In part b,"},{"Start":"02:07.015 ","End":"02:11.255","Text":"when I\u0027m asked to find the total impulse applied to the object,"},{"Start":"02:11.255 ","End":"02:14.255","Text":"what I can do is take the sum of the impulses"},{"Start":"02:14.255 ","End":"02:17.890","Text":"along the x-axis and then the sum along the y-axis."},{"Start":"02:17.890 ","End":"02:28.365","Text":"Along the x-axis, I have 5 plus 7.5 root 3 over 2 newtons seconds."},{"Start":"02:28.365 ","End":"02:36.500","Text":"Along the y-axis, I have 7.5/2."},{"Start":"02:36.500 ","End":"02:40.100","Text":"That\u0027s my whole y-axis."},{"Start":"02:40.100 ","End":"02:42.635","Text":"In terms of finding a single answer,"},{"Start":"02:42.635 ","End":"02:44.720","Text":"what we can do is first add these together,"},{"Start":"02:44.720 ","End":"02:50.295","Text":"so we get 11.5N seconds for x,"},{"Start":"02:50.295 ","End":"02:56.475","Text":"and for y we get 3.75N seconds."},{"Start":"02:56.475 ","End":"02:58.400","Text":"When we want to put them together,"},{"Start":"02:58.400 ","End":"03:02.495","Text":"the magnitude will be the same as what we do with normal force vectors."},{"Start":"03:02.495 ","End":"03:07.755","Text":"We take the square root of x-squared plus y-squared."},{"Start":"03:07.755 ","End":"03:14.715","Text":"11.5 squared plus 3.75 squared,"},{"Start":"03:14.715 ","End":"03:20.155","Text":"and that equals 12.1 Newton seconds."},{"Start":"03:20.155 ","End":"03:24.110","Text":"We could also find the angle taking y divided by x,"},{"Start":"03:24.110 ","End":"03:26.460","Text":"but we\u0027re not going to do that now."},{"Start":"03:27.550 ","End":"03:29.840","Text":"Now in part c,"},{"Start":"03:29.840 ","End":"03:34.160","Text":"I\u0027m asked to find the velocity of the object after the forces have acted upon it."},{"Start":"03:34.160 ","End":"03:38.645","Text":"That means that time equals 0.5 once our forces have done their work."},{"Start":"03:38.645 ","End":"03:41.840","Text":"We know the way we find this is taking momentum,"},{"Start":"03:41.840 ","End":"03:45.380","Text":"the change in momentum equals the impulse."},{"Start":"03:45.380 ","End":"03:47.450","Text":"But because we\u0027re talking about velocity,"},{"Start":"03:47.450 ","End":"03:49.030","Text":"we also need to account for direction."},{"Start":"03:49.030 ","End":"03:52.070","Text":"We can just break this down into our different elements."},{"Start":"03:52.070 ","End":"03:56.930","Text":"Let\u0027s say Delta P_x equals J_x,"},{"Start":"03:56.930 ","End":"04:00.330","Text":"and Delta P_y equals J_y."},{"Start":"04:01.850 ","End":"04:04.500","Text":"Starting with the x-axis,"},{"Start":"04:04.500 ","End":"04:11.330","Text":"we know that P equals m times v. M times V_x final minus V_x initial,"},{"Start":"04:11.330 ","End":"04:18.325","Text":"which is 0, has to equal 11.5 Newton seconds."},{"Start":"04:18.325 ","End":"04:21.590","Text":"Because we know that the mass equals 3 kilograms,"},{"Start":"04:21.590 ","End":"04:23.510","Text":"we can say that V_x F,"},{"Start":"04:23.510 ","End":"04:25.625","Text":"our final velocity for x,"},{"Start":"04:25.625 ","End":"04:30.990","Text":"equals 11.5 divided by 3 meters per second."},{"Start":"04:30.990 ","End":"04:33.455","Text":"The same way with our y element,"},{"Start":"04:33.455 ","End":"04:35.070","Text":"that V_y F,"},{"Start":"04:35.070 ","End":"04:37.130","Text":"our final y velocity,"},{"Start":"04:37.130 ","End":"04:42.930","Text":"equals 3.75 divided by 3 meters per second."},{"Start":"04:43.520 ","End":"04:45.735","Text":"These are 2 answers."},{"Start":"04:45.735 ","End":"04:50.060","Text":"The idea here is we can find the impulse of each force separately and"},{"Start":"04:50.060 ","End":"04:55.510","Text":"combine the impulse of each force to find the total impulse applied to the object."},{"Start":"04:55.510 ","End":"04:58.820","Text":"Then we know once we have our total impulse,"},{"Start":"04:58.820 ","End":"05:01.670","Text":"we can set that equal to the change in momentum."},{"Start":"05:01.670 ","End":"05:06.605","Text":"From there, we can find our velocity or our final velocity as it may be."},{"Start":"05:06.605 ","End":"05:12.240","Text":"If we need to, we can work along the x and the y axis separately like we did here."}],"ID":9332},{"Watched":false,"Name":"Impulse of an Average Force","Duration":"3m 10s","ChapterTopicVideoID":9060,"CourseChapterTopicPlaylistID":5392,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.445","Text":"Hello. In this video,"},{"Start":"00:02.445 ","End":"00:04.754","Text":"I want to illustrate for you how our impulse,"},{"Start":"00:04.754 ","End":"00:08.490","Text":"which we\u0027ve previously defined as the integral of your force,"},{"Start":"00:08.490 ","End":"00:17.175","Text":"F times dt can also be found using the average force times Delta t, the change in time."},{"Start":"00:17.175 ","End":"00:22.605","Text":"What this means is that if you have a non-constant force or even a constant force,"},{"Start":"00:22.605 ","End":"00:25.180","Text":"you can find Delta f,"},{"Start":"00:26.090 ","End":"00:29.940","Text":"you can find your average force and multiply that by"},{"Start":"00:29.940 ","End":"00:33.585","Text":"Delta t to find your answer instead of having to do an integral."},{"Start":"00:33.585 ","End":"00:35.880","Text":"For some of you, this may be something you\u0027re already"},{"Start":"00:35.880 ","End":"00:38.370","Text":"familiar with or that you already understand,"},{"Start":"00:38.370 ","End":"00:39.660","Text":"or that you feel comfortable with."},{"Start":"00:39.660 ","End":"00:42.000","Text":"In that case, you can move on to the next video."},{"Start":"00:42.000 ","End":"00:45.125","Text":"For anyone for whom this is a new concept,"},{"Start":"00:45.125 ","End":"00:48.595","Text":"I\u0027d encourage you to watch the explanation that follows."},{"Start":"00:48.595 ","End":"00:52.460","Text":"First, let\u0027s talk about how we define an average force."},{"Start":"00:52.460 ","End":"00:56.270","Text":"An average force equals, by definition,"},{"Start":"00:56.270 ","End":"00:58.175","Text":"the integral of your force,"},{"Start":"00:58.175 ","End":"01:03.440","Text":"F times dt divided by your total change in time."},{"Start":"01:03.440 ","End":"01:06.515","Text":"In a case when you\u0027re going from 0 to Delta t,"},{"Start":"01:06.515 ","End":"01:08.000","Text":"it will be your upper limit,"},{"Start":"01:08.000 ","End":"01:12.440","Text":"delta t. Now let\u0027s multiply both sides"},{"Start":"01:12.440 ","End":"01:17.255","Text":"by delta t. What you get is average force times Delta t,"},{"Start":"01:17.255 ","End":"01:18.755","Text":"which is what we have up here,"},{"Start":"01:18.755 ","End":"01:23.570","Text":"equals an integral of f, your force dt."},{"Start":"01:23.570 ","End":"01:27.560","Text":"I assume this is familiar to you because it\u0027s exactly what\u0027s listed above."},{"Start":"01:27.560 ","End":"01:30.455","Text":"In fact, this equals impulse."},{"Start":"01:30.455 ","End":"01:32.270","Text":"If this is still unclear,"},{"Start":"01:32.270 ","End":"01:34.370","Text":"let me illustrate through an example."},{"Start":"01:34.370 ","End":"01:38.030","Text":"Let\u0027s assume we have a force F and that it exerts"},{"Start":"01:38.030 ","End":"01:41.555","Text":"1 Newton of force for a time period of 2 seconds,"},{"Start":"01:41.555 ","End":"01:44.675","Text":"Delta T. After 2 seconds,"},{"Start":"01:44.675 ","End":"01:48.140","Text":"the force gets stronger and it exerts 3 newtons of"},{"Start":"01:48.140 ","End":"01:52.505","Text":"force over the course of the following 2 seconds."},{"Start":"01:52.505 ","End":"01:56.195","Text":"Intuitively, we can see that our average force,"},{"Start":"01:56.195 ","End":"02:03.365","Text":"f average equals 2 Newtons and that it\u0027s over a time period of 4 seconds."},{"Start":"02:03.365 ","End":"02:07.895","Text":"The first thing we can do is traditionally compute"},{"Start":"02:07.895 ","End":"02:12.070","Text":"the impulse of each of these fours periods, then add them together."},{"Start":"02:12.070 ","End":"02:17.540","Text":"J, the impulse equals 1 n times 2 seconds,"},{"Start":"02:17.540 ","End":"02:19.130","Text":"which is our F1,"},{"Start":"02:19.130 ","End":"02:25.290","Text":"so to speak, plus 3 newtons times 2 seconds,"},{"Start":"02:25.290 ","End":"02:28.695","Text":"which is our F2, again, so to speak."},{"Start":"02:28.695 ","End":"02:30.645","Text":"That equals 2 plus 6,"},{"Start":"02:30.645 ","End":"02:34.280","Text":"which is 8 Newtons times seconds."},{"Start":"02:34.280 ","End":"02:36.095","Text":"Now using our new formula,"},{"Start":"02:36.095 ","End":"02:43.635","Text":"we can take our f average and say that J equals f average times Delta t."},{"Start":"02:43.635 ","End":"02:52.115","Text":"Our result here is 2 Newtons times 4 seconds or 8 Newtons times seconds."},{"Start":"02:52.115 ","End":"02:54.350","Text":"The exact same result."},{"Start":"02:54.350 ","End":"02:57.965","Text":"The takeaway here is that you can now use"},{"Start":"02:57.965 ","End":"03:00.740","Text":"the average force times Delta t instead of having to"},{"Start":"03:00.740 ","End":"03:03.595","Text":"take an integral and it can help you in certain situations."},{"Start":"03:03.595 ","End":"03:06.710","Text":"In fact, the next video is an example that shows you"},{"Start":"03:06.710 ","End":"03:11.400","Text":"a certain situation where taking the average force can make things easier."}],"ID":9333},{"Watched":false,"Name":"Impulse of an Average Force Example","Duration":"3m 52s","ChapterTopicVideoID":9061,"CourseChapterTopicPlaylistID":5392,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.530 ","End":"00:05.835","Text":"In this problem, we\u0027re given a ball with a mass of 1 kilogram."},{"Start":"00:05.835 ","End":"00:09.495","Text":"We\u0027re told that it\u0027s thrown against a wall with the velocity of"},{"Start":"00:09.495 ","End":"00:13.590","Text":"2 meters per second and that it bounces back."},{"Start":"00:13.590 ","End":"00:17.565","Text":"Imagine this is the ball bouncing back with an identical velocity."},{"Start":"00:17.565 ","End":"00:20.985","Text":"That\u0027s the velocity of 2 meters per second."},{"Start":"00:20.985 ","End":"00:24.845","Text":"Now, of course, the velocities here are identical in terms of magnitude,"},{"Start":"00:24.845 ","End":"00:27.160","Text":"but not in terms of direction."},{"Start":"00:27.160 ","End":"00:33.635","Text":"In part a, we\u0027re asked to find the impulse J that was applied to the ball."},{"Start":"00:33.635 ","End":"00:38.960","Text":"Now the impulse equals the integral of the force F times dt."},{"Start":"00:38.960 ","End":"00:40.670","Text":"However, in this problem,"},{"Start":"00:40.670 ","End":"00:43.110","Text":"we\u0027re not given any information about the force,"},{"Start":"00:43.110 ","End":"00:44.810","Text":"so we can\u0027t really use this method."},{"Start":"00:44.810 ","End":"00:47.405","Text":"We have to find something else that equals the impulse."},{"Start":"00:47.405 ","End":"00:49.745","Text":"Luckily, we have Delta P,"},{"Start":"00:49.745 ","End":"00:51.185","Text":"the change of momentum."},{"Start":"00:51.185 ","End":"00:54.335","Text":"The change of momentum we know also equals the impulse."},{"Start":"00:54.335 ","End":"01:04.200","Text":"The change of momentum Delta P can be broken down into P_F minus P_i."},{"Start":"01:04.330 ","End":"01:11.280","Text":"Or further into MV_F minus MV_i."},{"Start":"01:11.280 ","End":"01:12.600","Text":"Before we assign our v,"},{"Start":"01:12.600 ","End":"01:14.675","Text":"so we have to choose our x-axis."},{"Start":"01:14.675 ","End":"01:17.965","Text":"Let\u0027s say that everything going to the right is positive,"},{"Start":"01:17.965 ","End":"01:19.825","Text":"so that\u0027s the direction of our x-axis."},{"Start":"01:19.825 ","End":"01:21.845","Text":"Everything to the left is negative."},{"Start":"01:21.845 ","End":"01:23.655","Text":"If that\u0027s the case,"},{"Start":"01:23.655 ","End":"01:33.055","Text":"then MV_F minus V_i equals 1 kilogram or mass we\u0027ll write that in a second,"},{"Start":"01:33.055 ","End":"01:38.075","Text":"times negative 2 minus 2."},{"Start":"01:38.075 ","End":"01:40.660","Text":"Now you\u0027ll see here this isn\u0027t going to equal 0."},{"Start":"01:40.660 ","End":"01:43.870","Text":"You may have thought so above because we have identical velocities,"},{"Start":"01:43.870 ","End":"01:48.040","Text":"but in fact, we\u0027re not going to get an impulse of 0."},{"Start":"01:48.040 ","End":"01:49.915","Text":"When we plug this in,"},{"Start":"01:49.915 ","End":"01:58.485","Text":"we get 1 kilogram times negative 4 meters per second,"},{"Start":"01:58.485 ","End":"02:03.130","Text":"and that equals negative 4 newtons seconds,"},{"Start":"02:03.130 ","End":"02:06.309","Text":"and that equals J or impulse."},{"Start":"02:06.309 ","End":"02:08.290","Text":"That answers part a."},{"Start":"02:08.290 ","End":"02:12.220","Text":"You\u0027ll notice that negative 4 is our coefficient here."},{"Start":"02:12.220 ","End":"02:15.970","Text":"That means that our impulse is being applied in a negative direction,"},{"Start":"02:15.970 ","End":"02:18.580","Text":"which actually makes sense because if you"},{"Start":"02:18.580 ","End":"02:23.140","Text":"consider that the normal force is what\u0027s doing this impulse,"},{"Start":"02:23.140 ","End":"02:24.865","Text":"what\u0027s exerting the impulse,"},{"Start":"02:24.865 ","End":"02:26.395","Text":"that\u0027s the normal force."},{"Start":"02:26.395 ","End":"02:30.350","Text":"Then it would make sense that your force is acting backwards or in a negative direction."},{"Start":"02:30.350 ","End":"02:32.255","Text":"This actually answers part b."},{"Start":"02:32.255 ","End":"02:33.290","Text":"When we see this negative,"},{"Start":"02:33.290 ","End":"02:36.110","Text":"it implies something and we can answer that part b,"},{"Start":"02:36.110 ","End":"02:38.485","Text":"who exerted the impulse described above?"},{"Start":"02:38.485 ","End":"02:40.435","Text":"It\u0027s the normal force."},{"Start":"02:40.435 ","End":"02:44.150","Text":"Let\u0027s move on to part c. In part c,"},{"Start":"02:44.150 ","End":"02:47.450","Text":"we\u0027re asked to calculate the average normal force that the wall"},{"Start":"02:47.450 ","End":"02:51.215","Text":"exerted if the period of exertion was 0.2 seconds."},{"Start":"02:51.215 ","End":"02:54.305","Text":"We\u0027re trying to find the average normal force."},{"Start":"02:54.305 ","End":"02:56.765","Text":"We can say about the average normal force,"},{"Start":"02:56.765 ","End":"03:00.350","Text":"because the normal force is responsible for the impulse above,"},{"Start":"03:00.350 ","End":"03:06.870","Text":"that means that the average normal force times Delta t equals J, our impulse."},{"Start":"03:06.870 ","End":"03:12.680","Text":"We know that our impulse is negative 4N times seconds."},{"Start":"03:12.680 ","End":"03:16.280","Text":"We know that Delta t was 0.2,"},{"Start":"03:16.280 ","End":"03:17.815","Text":"that\u0027s given to us,"},{"Start":"03:17.815 ","End":"03:21.920","Text":"so that\u0027s multiplied by the average normal force."},{"Start":"03:21.920 ","End":"03:25.445","Text":"We find that the average normal force equals"},{"Start":"03:25.445 ","End":"03:32.900","Text":"negative 4N seconds over 0.2 seconds."},{"Start":"03:32.900 ","End":"03:37.220","Text":"Our answer is that the average normal force is"},{"Start":"03:37.220 ","End":"03:42.720","Text":"negative 20 newtons in the direction of x,"},{"Start":"03:43.580 ","End":"03:46.745","Text":"so our answer is negative 20N."},{"Start":"03:46.745 ","End":"03:49.145","Text":"We know that it\u0027s negative along the x-axis,"},{"Start":"03:49.145 ","End":"03:52.830","Text":"and that answer is all 3 parts of our example."}],"ID":9334},{"Watched":false,"Name":"Impuse of a non-Constant Force","Duration":"3m 5s","ChapterTopicVideoID":9062,"CourseChapterTopicPlaylistID":5392,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.050 ","End":"00:03.630","Text":"In this lecture, I want to talk about how we"},{"Start":"00:03.630 ","End":"00:06.705","Text":"calculate the impulse of a non-constant force."},{"Start":"00:06.705 ","End":"00:10.740","Text":"Up until now, we haven\u0027t really talked about a case with a non-constant force."},{"Start":"00:10.740 ","End":"00:15.645","Text":"But let\u0027s assume you have an object and it\u0027s moving towards the right along the x-axis,"},{"Start":"00:15.645 ","End":"00:19.130","Text":"and the force is moving along at a non-constant speed,"},{"Start":"00:19.130 ","End":"00:25.140","Text":"let\u0027s say it\u0027s moving at 3t along the x-axis."},{"Start":"00:25.430 ","End":"00:29.270","Text":"In this situation, to calculate the impulse,"},{"Start":"00:29.270 ","End":"00:34.070","Text":"what you\u0027ll do is just take an integral of this function, so 3tdt."},{"Start":"00:34.070 ","End":"00:38.525","Text":"Let\u0027s say that this function is active for 2 seconds."},{"Start":"00:38.525 ","End":"00:41.000","Text":"Delta t equals 2 seconds."},{"Start":"00:41.000 ","End":"00:45.755","Text":"In that case, our limits would be from t equals 0 to t equals 2."},{"Start":"00:45.755 ","End":"00:51.210","Text":"This would equal 6 times t squared over 2."},{"Start":"00:51.210 ","End":"00:54.510","Text":"That would be 2 squared over 2."},{"Start":"00:54.510 ","End":"00:59.350","Text":"Our answer would be 12 newton seconds."},{"Start":"00:59.900 ","End":"01:02.690","Text":"Essentially if we\u0027re given a function,"},{"Start":"01:02.690 ","End":"01:07.225","Text":"we can just take an integral of that function with our allotted amount of time."},{"Start":"01:07.225 ","End":"01:09.605","Text":"Now there\u0027s a second method that can be useful,"},{"Start":"01:09.605 ","End":"01:13.430","Text":"especially if you\u0027re giving your function in the form of a graph."},{"Start":"01:13.430 ","End":"01:20.630","Text":"Essentially what you can do is take the impulse to equal the area of the graph."},{"Start":"01:20.630 ","End":"01:23.735","Text":"In this case, it\u0027d be the graph of f as a function of"},{"Start":"01:23.735 ","End":"01:28.200","Text":"t. You take the area of this, the shaded-in portion."},{"Start":"01:28.340 ","End":"01:30.890","Text":"In the example on the left,"},{"Start":"01:30.890 ","End":"01:36.245","Text":"if I have a force that it seems is equal to 2 for the first second,"},{"Start":"01:36.245 ","End":"01:38.195","Text":"and then from seconds 1-2,"},{"Start":"01:38.195 ","End":"01:41.270","Text":"it rises linearly from 2-3 newtons."},{"Start":"01:41.270 ","End":"01:44.935","Text":"Then what I need to do is find the shaded in area here."},{"Start":"01:44.935 ","End":"01:49.415","Text":"The best way for me to do this is to break this down into 2 different shapes."},{"Start":"01:49.415 ","End":"01:53.765","Text":"One shape representing the first part of the function, this part here,"},{"Start":"01:53.765 ","End":"02:00.045","Text":"and that will be shape 1 equals 2 Newtons times 1 second."},{"Start":"02:00.045 ","End":"02:02.390","Text":"Shape 2 is a trapezoid."},{"Start":"02:02.390 ","End":"02:04.055","Text":"I can use my trapezoid formula."},{"Start":"02:04.055 ","End":"02:05.450","Text":"I have 2 bases."},{"Start":"02:05.450 ","End":"02:08.630","Text":"My first base has a length of 2,"},{"Start":"02:08.630 ","End":"02:12.245","Text":"and my second base has a length of 3."},{"Start":"02:12.245 ","End":"02:14.675","Text":"I\u0027m going to multiply that by the height,"},{"Start":"02:14.675 ","End":"02:17.330","Text":"which is 1, and divide that by 2."},{"Start":"02:17.330 ","End":"02:20.240","Text":"That\u0027s your trapezoid formula if you\u0027re not familiar."},{"Start":"02:20.240 ","End":"02:24.080","Text":"What you end up with here is 2 plus 3 is 5,"},{"Start":"02:24.080 ","End":"02:28.485","Text":"5 times 1 divided by 2 equals 5 halves,"},{"Start":"02:28.485 ","End":"02:30.375","Text":"and above you get 2."},{"Start":"02:30.375 ","End":"02:32.175","Text":"My total is,"},{"Start":"02:32.175 ","End":"02:35.955","Text":"5 halves can also be thought of as 2.5,"},{"Start":"02:35.955 ","End":"02:43.350","Text":"so my total is 4.5 newton seconds."},{"Start":"02:43.350 ","End":"02:46.565","Text":"The takeaway here is just what we did."},{"Start":"02:46.565 ","End":"02:47.700","Text":"If we have a function,"},{"Start":"02:47.700 ","End":"02:52.145","Text":"we can just take the integral of the function and see how long we are using the function,"},{"Start":"02:52.145 ","End":"02:53.780","Text":"usually in terms of seconds."},{"Start":"02:53.780 ","End":"02:55.025","Text":"If we have a graph,"},{"Start":"02:55.025 ","End":"02:58.010","Text":"what we need to do is calculate the area of the graph."},{"Start":"02:58.010 ","End":"02:59.870","Text":"We can do that by breaking it down into"},{"Start":"02:59.870 ","End":"03:02.795","Text":"different shapes if necessary and adding in our units."},{"Start":"03:02.795 ","End":"03:06.690","Text":"Again, I assumed here we\u0027re using seconds generally, we will."}],"ID":9335},{"Watched":false,"Name":"Summary of Impulse","Duration":"3m 4s","ChapterTopicVideoID":9063,"CourseChapterTopicPlaylistID":5392,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.290 ","End":"00:05.460","Text":"Let\u0027s summarize what we\u0027ve talked about so far with regards to impulse."},{"Start":"00:05.460 ","End":"00:09.675","Text":"First of all, we talked about the classic version of Newton\u0027s 2nd law,"},{"Start":"00:09.675 ","End":"00:15.375","Text":"which was new to us, that the sum of forces equals dp over dt."},{"Start":"00:15.375 ","End":"00:19.275","Text":"From there, we defined impulse, J."},{"Start":"00:19.275 ","End":"00:25.245","Text":"We said that impulse equals the integral of the force F times dt."},{"Start":"00:25.245 ","End":"00:32.145","Text":"That\u0027s over some period of time Delta t. We said that if we have a constant force,"},{"Start":"00:32.145 ","End":"00:38.280","Text":"we can say that the impulse is also equal to the force times the change in time,"},{"Start":"00:38.280 ","End":"00:43.340","Text":"Delta t. That\u0027s only if we have a constant force."},{"Start":"00:43.340 ","End":"00:46.765","Text":"However, in all cases,"},{"Start":"00:46.765 ","End":"00:55.580","Text":"we can say that the average force times Delta t,"},{"Start":"00:55.580 ","End":"00:57.560","Text":"also equals the impulse."},{"Start":"00:57.560 ","End":"01:01.040","Text":"Also keep in mind that the impulse is a vector,"},{"Start":"01:01.040 ","End":"01:05.435","Text":"meaning that the impulse is applied in the same direction that the force is exerted."},{"Start":"01:05.435 ","End":"01:07.805","Text":"We also talked about the total impulse,"},{"Start":"01:07.805 ","End":"01:12.695","Text":"that\u0027s J total, that is applied to a given object."},{"Start":"01:12.695 ","End":"01:18.080","Text":"That equals the integral of the sum of forces times dt,"},{"Start":"01:18.080 ","End":"01:21.980","Text":"again over the time Delta t. We can also calculate"},{"Start":"01:21.980 ","End":"01:26.305","Text":"it by adding together the impulse of each force."},{"Start":"01:26.305 ","End":"01:31.110","Text":"We can say that J_1 plus J_2, etc."},{"Start":"01:31.110 ","End":"01:33.185","Text":"Until we reach our final impulse,"},{"Start":"01:33.185 ","End":"01:37.630","Text":"or the sum of all impulses equals the total impulse."},{"Start":"01:37.630 ","End":"01:39.170","Text":"We can find this 2 ways,"},{"Start":"01:39.170 ","End":"01:42.080","Text":"either by summing the forces and taking an integral or"},{"Start":"01:42.080 ","End":"01:45.560","Text":"finding each impulse individually and summing them together."},{"Start":"01:45.560 ","End":"01:49.670","Text":"Now, your total impulse is equal to Delta p,"},{"Start":"01:49.670 ","End":"01:53.390","Text":"the change momentum, and all of these things are vectors."},{"Start":"01:53.390 ","End":"01:56.615","Text":"We know that Delta p, your change momentum,"},{"Start":"01:56.615 ","End":"02:02.980","Text":"is also equal to P final minus P initial."},{"Start":"02:03.320 ","End":"02:06.020","Text":"Add 1 more thing here."},{"Start":"02:06.020 ","End":"02:08.914","Text":"If we start at rest or start with 0 momentum,"},{"Start":"02:08.914 ","End":"02:10.520","Text":"that means that pf,"},{"Start":"02:10.520 ","End":"02:14.400","Text":"our final momentum equals our total impulse."},{"Start":"02:14.400 ","End":"02:16.415","Text":"That\u0027s another way to think about momentum."},{"Start":"02:16.415 ","End":"02:21.515","Text":"Momentum equals the total impulse that accumulates in a given object,"},{"Start":"02:21.515 ","End":"02:24.875","Text":"or you can say that total impulse equals"},{"Start":"02:24.875 ","End":"02:29.030","Text":"the amount of momentum that accumulates in a given object."},{"Start":"02:29.030 ","End":"02:33.500","Text":"Given the circumstance that we start at rest or start with 0 momentum."},{"Start":"02:33.500 ","End":"02:38.255","Text":"One final note is if we\u0027re given a non-constant force,"},{"Start":"02:38.255 ","End":"02:43.265","Text":"then we can graph it out and the area underneath the graph,"},{"Start":"02:43.265 ","End":"02:49.585","Text":"if we shade it in, that should equal our impulse, J."},{"Start":"02:49.585 ","End":"02:53.555","Text":"This graph has to be force over time"},{"Start":"02:53.555 ","End":"02:58.340","Text":"t. These are the important things to remember about impulse."},{"Start":"02:58.340 ","End":"03:04.620","Text":"From here we\u0027ll move on to a lecture about the conservation of momentum in collisions."}],"ID":9336}],"Thumbnail":null,"ID":5392},{"Name":"Conserving Momentum and External Forces","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"The Law of Conservation of Momentum and External Forces","Duration":"8m 33s","ChapterTopicVideoID":9064,"CourseChapterTopicPlaylistID":5393,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/9064.jpeg","UploadDate":"2021-11-19T08:08:26.1870000","DurationForVideoObject":"PT8M33S","Description":null,"MetaTitle":"The Law of Conservation of Momentum and External Forces: Video + Workbook | Proprep","MetaDescription":"Momentum and Impulse - Conserving Momentum and External Forces. 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/momentum-and-impulse/conserving-momentum-and-external-forces/vid9337","VideoComments":[],"Subtitles":[{"Start":"00:00.170 ","End":"00:03.360","Text":"In this lecture, I want to talk about the law of"},{"Start":"00:03.360 ","End":"00:06.135","Text":"conservation of momentum and external forces."},{"Start":"00:06.135 ","End":"00:10.050","Text":"This is a particularly important lecture as the concepts we talk about"},{"Start":"00:10.050 ","End":"00:14.895","Text":"here we\u0027ll use a lot to solve many different exercises in the future."},{"Start":"00:14.895 ","End":"00:17.325","Text":"Take a look at this system we have here."},{"Start":"00:17.325 ","End":"00:18.735","Text":"We have 2 objects,"},{"Start":"00:18.735 ","End":"00:22.080","Text":"we can think of each one as a billiard ball or a snicker ball,"},{"Start":"00:22.080 ","End":"00:23.640","Text":"and they\u0027re headed towards each other,"},{"Start":"00:23.640 ","End":"00:29.200","Text":"one with a velocity V_1 and the other with the velocity V_2."},{"Start":"00:29.870 ","End":"00:34.455","Text":"First let\u0027s define the total momentum in the system,"},{"Start":"00:34.455 ","End":"00:39.060","Text":"P_T or p total as equal to P_1,"},{"Start":"00:39.060 ","End":"00:41.595","Text":"the momentum of the first ball,"},{"Start":"00:41.595 ","End":"00:46.630","Text":"plus P_2, the momentum of the second ball."},{"Start":"00:47.120 ","End":"00:51.909","Text":"Let\u0027s look at this system when the 2 balls collide."},{"Start":"00:52.000 ","End":"00:54.544","Text":"At the time of collision,"},{"Start":"00:54.544 ","End":"01:00.295","Text":"the first ball is going to exert a force F_1 onto the second ball."},{"Start":"01:00.295 ","End":"01:06.260","Text":"The second ball is going to exert a force F_2 right back onto ball 1."},{"Start":"01:06.260 ","End":"01:09.320","Text":"We know, based on Newton\u0027s third law,"},{"Start":"01:09.320 ","End":"01:14.275","Text":"that F_1 equals negative F_2,"},{"Start":"01:14.275 ","End":"01:16.085","Text":"meaning that they\u0027re equal and opposite."},{"Start":"01:16.085 ","End":"01:18.815","Text":"They have the same magnitude in opposite directions."},{"Start":"01:18.815 ","End":"01:24.640","Text":"F_1 and F_2 have the same magnitude and point in opposite directions."},{"Start":"01:24.640 ","End":"01:28.505","Text":"Let\u0027s suppose that the time of the collision is Delta t,"},{"Start":"01:28.505 ","End":"01:30.050","Text":"so that each of these forces,"},{"Start":"01:30.050 ","End":"01:34.290","Text":"F_1 and F_2 are acting for Delta t time."},{"Start":"01:34.370 ","End":"01:38.158","Text":"Now let\u0027s look at the object on the right."},{"Start":"01:38.158 ","End":"01:39.619","Text":"These are both the vectors."},{"Start":"01:39.619 ","End":"01:42.590","Text":"We know that the force is affecting the velocity of"},{"Start":"01:42.590 ","End":"01:45.590","Text":"the ball on the right because Newton\u0027s second law tells us"},{"Start":"01:45.590 ","End":"01:48.414","Text":"that force equals mass times acceleration"},{"Start":"01:48.414 ","End":"01:52.130","Text":"and acceleration is directly related to velocity."},{"Start":"01:52.130 ","End":"01:56.180","Text":"In other terms, we could say that the force 1 is applying"},{"Start":"01:56.180 ","End":"02:00.170","Text":"an impulse to the object on the right which is affecting its velocity."},{"Start":"02:00.170 ","End":"02:04.810","Text":"As a result, I want to find the impulse which we\u0027ll call J_1."},{"Start":"02:04.810 ","End":"02:08.705","Text":"We\u0027ll call this J_1 because it\u0027s in accordance with F_1."},{"Start":"02:08.705 ","End":"02:15.380","Text":"We know that J_1 has to equal the integral of F_1,"},{"Start":"02:15.380 ","End":"02:18.615","Text":"dt over the time,"},{"Start":"02:18.615 ","End":"02:24.860","Text":"Delta t. We also know that this equals Delta P_1,"},{"Start":"02:24.860 ","End":"02:27.690","Text":"the change in momentum 1."},{"Start":"02:28.820 ","End":"02:35.555","Text":"Using my impulse, I can calculate the change in momentum or the change in velocity,"},{"Start":"02:35.555 ","End":"02:38.485","Text":"phenomena focus on the change in momentum."},{"Start":"02:38.485 ","End":"02:42.305","Text":"Now let\u0027s look at the object on the left, the second ball."},{"Start":"02:42.305 ","End":"02:43.940","Text":"We can do the same calculation."},{"Start":"02:43.940 ","End":"02:53.320","Text":"We know that J_2 equals the integral of F_2 dt,"},{"Start":"02:53.320 ","End":"02:55.655","Text":"again, over the time period,"},{"Start":"02:55.655 ","End":"03:00.005","Text":"Delta t. We know that that also equals negative"},{"Start":"03:00.005 ","End":"03:06.900","Text":"integral of F_1 dt over the same time period."},{"Start":"03:07.420 ","End":"03:10.700","Text":"We know that because of the formula above."},{"Start":"03:10.700 ","End":"03:15.350","Text":"We know that if it equals negative F_1 dt integral,"},{"Start":"03:15.350 ","End":"03:19.595","Text":"there also equals negative Delta P_1."},{"Start":"03:19.595 ","End":"03:26.275","Text":"At the same time, we know that J_2 has to equal Delta P_2."},{"Start":"03:26.275 ","End":"03:30.440","Text":"What we get from this is we know that the change in momentum of"},{"Start":"03:30.440 ","End":"03:35.785","Text":"the second object equals the negative change in momentum of the first object."},{"Start":"03:35.785 ","End":"03:39.955","Text":"This P_T, we can also call it P_T_i, initial,"},{"Start":"03:39.955 ","End":"03:45.035","Text":"so to speak, is the total momentum before the collision."},{"Start":"03:45.035 ","End":"03:47.720","Text":"We\u0027re also going to talk about a second P_T."},{"Start":"03:47.720 ","End":"03:52.900","Text":"This is going to be P_T final or the P_T_F, the collision."},{"Start":"03:52.900 ","End":"03:56.315","Text":"If we calculate this second total momentum,"},{"Start":"03:56.315 ","End":"04:02.170","Text":"the total momentum equals P_1 plus the change in P_1."},{"Start":"04:02.170 ","End":"04:04.710","Text":"This is a vector, of course."},{"Start":"04:04.710 ","End":"04:11.385","Text":"That plus P_2 plus the change in P_2."},{"Start":"04:11.385 ","End":"04:16.850","Text":"Now we know that the change in P_2 equals negative change in P_1."},{"Start":"04:16.850 ","End":"04:19.020","Text":"Let\u0027s switch these out."},{"Start":"04:19.340 ","End":"04:28.420","Text":"You\u0027ll see that negative Delta P_1 and positive Delta P_1 fall out."},{"Start":"04:28.420 ","End":"04:31.680","Text":"What you\u0027re left with is that P_TF,"},{"Start":"04:31.680 ","End":"04:36.310","Text":"your total momentum after the collision equals P_T_i,"},{"Start":"04:36.310 ","End":"04:39.225","Text":"your total momentum before the collision."},{"Start":"04:39.225 ","End":"04:44.975","Text":"Really what happened here is that P_1 gained some momentum and P_2 lost some momentum,"},{"Start":"04:44.975 ","End":"04:48.745","Text":"but the total momentum in the system remained the same."},{"Start":"04:48.745 ","End":"04:53.930","Text":"Keep in mind, each individual object\u0027s momentum has change."},{"Start":"04:53.930 ","End":"04:57.170","Text":"P_1 has had a change in momentum."},{"Start":"04:57.170 ","End":"05:02.415","Text":"This is the total momentum after the collision."},{"Start":"05:02.415 ","End":"05:04.170","Text":"The same is true for P_2,"},{"Start":"05:04.170 ","End":"05:05.585","Text":"the object on the left."},{"Start":"05:05.585 ","End":"05:08.930","Text":"This is the total momentum after the collision,"},{"Start":"05:08.930 ","End":"05:10.085","Text":"meaning that it\u0027s changed."},{"Start":"05:10.085 ","End":"05:13.040","Text":"There is a change for each individual object,"},{"Start":"05:13.040 ","End":"05:18.780","Text":"but the system as a whole maintains or conserves the same amount of momentum."},{"Start":"05:18.780 ","End":"05:23.040","Text":"When won\u0027t we experience this conservation of momentum?"},{"Start":"05:23.040 ","End":"05:26.059","Text":"That\u0027s only in a case when we have another object."},{"Start":"05:26.059 ","End":"05:29.060","Text":"Let\u0027s say a third object here that"},{"Start":"05:29.060 ","End":"05:33.995","Text":"exercises or exerts a certain amount of force on 1 object or the other, maybe both."},{"Start":"05:33.995 ","End":"05:37.040","Text":"In this case, let\u0027s say it exerts some force,"},{"Start":"05:37.040 ","End":"05:38.600","Text":"F_3 we\u0027ll call this,"},{"Start":"05:38.600 ","End":"05:40.160","Text":"on the object on the left."},{"Start":"05:40.160 ","End":"05:43.145","Text":"What will happen is because of this force,"},{"Start":"05:43.145 ","End":"05:48.770","Text":"the momentum and the acceleration and the velocity of the object on the left will"},{"Start":"05:48.770 ","End":"05:51.380","Text":"change without having any relationship or"},{"Start":"05:51.380 ","End":"05:54.710","Text":"change in the object on the right in terms of its velocity,"},{"Start":"05:54.710 ","End":"05:57.330","Text":"acceleration, or momentum."},{"Start":"05:57.940 ","End":"06:03.575","Text":"On the right, I will write the circumstances when we have the conservation of momentum,"},{"Start":"06:03.575 ","End":"06:05.120","Text":"and it is as follows."},{"Start":"06:05.120 ","End":"06:08.810","Text":"Conservation of momentum within a system occurs"},{"Start":"06:08.810 ","End":"06:12.890","Text":"when the sum of external forces acting on the system equals 0."},{"Start":"06:12.890 ","End":"06:17.945","Text":"That means that either you have no external forces or the sum of those forces is 0."},{"Start":"06:17.945 ","End":"06:22.750","Text":"Say that F_3\u0027s force was offset by another force called F_4."},{"Start":"06:22.750 ","End":"06:26.900","Text":"External forces, those forces that can screw up conservation of"},{"Start":"06:26.900 ","End":"06:32.180","Text":"momentum are considered forces which are caused by an object outside of the system."},{"Start":"06:32.180 ","End":"06:36.965","Text":"In this case, F_3 is caused by something that is not one of these 2 billiard balls."},{"Start":"06:36.965 ","End":"06:40.205","Text":"Now this is ultimately a somewhat arbitrary decision."},{"Start":"06:40.205 ","End":"06:44.900","Text":"It\u0027s only because I chose to have P_1 and P_2 or only the momentum of"},{"Start":"06:44.900 ","End":"06:47.030","Text":"the first 2 objects as part of my system"},{"Start":"06:47.030 ","End":"06:49.970","Text":"beforehand that this is not considered part of my system."},{"Start":"06:49.970 ","End":"06:53.525","Text":"I could have from the start also considered P_3,"},{"Start":"06:53.525 ","End":"06:56.620","Text":"the momentum of this third mysterious object."},{"Start":"06:56.620 ","End":"07:00.665","Text":"It would be part of my system and I would then have a conservation of momentum."},{"Start":"07:00.665 ","End":"07:01.970","Text":"Of course, in this case,"},{"Start":"07:01.970 ","End":"07:03.515","Text":"my momentum would be different."},{"Start":"07:03.515 ","End":"07:05.435","Text":"I\u0027m talking about a larger system now,"},{"Start":"07:05.435 ","End":"07:07.025","Text":"a system with more elements."},{"Start":"07:07.025 ","End":"07:08.660","Text":"I would have a different momentum,"},{"Start":"07:08.660 ","End":"07:10.970","Text":"but I would have conservation of momentum."},{"Start":"07:10.970 ","End":"07:13.175","Text":"Now you\u0027re probably asking yourself,"},{"Start":"07:13.175 ","End":"07:16.070","Text":"why not just always take any object and include it"},{"Start":"07:16.070 ","End":"07:19.415","Text":"in your systems that you always have conservation of momentum?"},{"Start":"07:19.415 ","End":"07:21.185","Text":"Well, the answer is sometimes,"},{"Start":"07:21.185 ","End":"07:24.409","Text":"it\u0027s not effective because I can\u0027t perhaps calculate"},{"Start":"07:24.409 ","End":"07:29.345","Text":"the momentum of a particular object so it doesn\u0027t work within my system."},{"Start":"07:29.345 ","End":"07:32.435","Text":"By the way, this doesn\u0027t only work with 2 objects,"},{"Start":"07:32.435 ","End":"07:34.310","Text":"a system can have multiple objects."},{"Start":"07:34.310 ","End":"07:37.010","Text":"I could have the same situation with 3 balls."},{"Start":"07:37.010 ","End":"07:40.480","Text":"As long as I can take a sum of my momentum from the beginning,"},{"Start":"07:40.480 ","End":"07:43.550","Text":"no matter which ball is giving which some of its momentum,"},{"Start":"07:43.550 ","End":"07:47.165","Text":"there\u0027ll be conservation within the system. Let\u0027s summarize."},{"Start":"07:47.165 ","End":"07:50.870","Text":"I can make a system or choose a system of any number of objects."},{"Start":"07:50.870 ","End":"07:53.644","Text":"As long as I take a sum of the momentum,"},{"Start":"07:53.644 ","End":"07:57.988","Text":"meaning the sum of the momentum of each object before the collision,"},{"Start":"07:57.988 ","End":"07:59.795","Text":"then I know that after the collision,"},{"Start":"07:59.795 ","End":"08:02.720","Text":"all of the momentum will be preserved within the system."},{"Start":"08:02.720 ","End":"08:05.135","Text":"Momentum may shift between objects,"},{"Start":"08:05.135 ","End":"08:09.035","Text":"but as long as I know my total momentum before the collision,"},{"Start":"08:09.035 ","End":"08:12.380","Text":"I also know my total momentum after the collision."},{"Start":"08:12.380 ","End":"08:17.278","Text":"The only time this is not the case is when I have an external force,"},{"Start":"08:17.278 ","End":"08:21.905","Text":"an external object acting upon the system using external force."},{"Start":"08:21.905 ","End":"08:26.190","Text":"When that is the case, I do not have conservation of momentum."},{"Start":"08:26.330 ","End":"08:29.190","Text":"This law is rather important."},{"Start":"08:29.190 ","End":"08:33.570","Text":"You may want to circle it in red and add it to your formula sheet."}],"ID":9337},{"Watched":false,"Name":"The Formula for Conservation of Energy and How to Approach Exercises","Duration":"3m 51s","ChapterTopicVideoID":9065,"CourseChapterTopicPlaylistID":5393,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.110 ","End":"00:02.160","Text":"In the last lecture,"},{"Start":"00:02.160 ","End":"00:04.755","Text":"we talked about the conservation of momentum."},{"Start":"00:04.755 ","End":"00:08.430","Text":"I stated that conservation of momentum within a system"},{"Start":"00:08.430 ","End":"00:12.360","Text":"occurs when the sum of external forces acting on the system is"},{"Start":"00:12.360 ","End":"00:15.570","Text":"zero and we then defined external forces as"},{"Start":"00:15.570 ","End":"00:19.785","Text":"those forces which are caused by objects outside of our closed system,"},{"Start":"00:19.785 ","End":"00:21.645","Text":"the system mentioned above."},{"Start":"00:21.645 ","End":"00:26.610","Text":"In this lecture, I want to talk about the formula for conservation of momentum and"},{"Start":"00:26.610 ","End":"00:31.575","Text":"how we use conservation of momentum in exercises and how we approach those exercises."},{"Start":"00:31.575 ","End":"00:34.410","Text":"I want to start with how we approach the exercises."},{"Start":"00:34.410 ","End":"00:38.820","Text":"The first question I\u0027m going to ask myself when looking at one of these exercises is the"},{"Start":"00:38.820 ","End":"00:46.115","Text":"following: does the sum of external forces acting upon my system equals zero?"},{"Start":"00:46.115 ","End":"00:48.905","Text":"If the answer to this question is yes,"},{"Start":"00:48.905 ","End":"00:52.100","Text":"the sum of external forces does equal zero,"},{"Start":"00:52.100 ","End":"00:55.775","Text":"that means that we have the conservation of momentum,"},{"Start":"00:55.775 ","End":"00:59.015","Text":"P. If the answer to this question is no,"},{"Start":"00:59.015 ","End":"01:01.745","Text":"it means that we don\u0027t have the conservation of momentum,"},{"Start":"01:01.745 ","End":"01:03.585","Text":"but we can possibly use it later,"},{"Start":"01:03.585 ","End":"01:04.635","Text":"so we\u0027ll get back to that."},{"Start":"01:04.635 ","End":"01:08.444","Text":"For now let\u0027s start with situation where the answer to our question is yes,"},{"Start":"01:08.444 ","End":"01:12.125","Text":"in fact, the sum of external forces does equal zero."},{"Start":"01:12.125 ","End":"01:16.415","Text":"The way that we define that situation is when P_Ti,"},{"Start":"01:16.415 ","End":"01:18.904","Text":"the initial total momentum,"},{"Start":"01:18.904 ","End":"01:24.005","Text":"equals P_Tf, the final total momentum."},{"Start":"01:24.005 ","End":"01:27.575","Text":"In another way we can write this is P total PT,"},{"Start":"01:27.575 ","End":"01:32.900","Text":"our total momentum equals m_1 times v_1,"},{"Start":"01:32.900 ","End":"01:36.185","Text":"the mass of our first object times its velocity,"},{"Start":"01:36.185 ","End":"01:40.775","Text":"plus the mass of our second object times velocity of that object,"},{"Start":"01:40.775 ","End":"01:42.290","Text":"and so on and so forth."},{"Start":"01:42.290 ","End":"01:44.060","Text":"For the sake of this problem,"},{"Start":"01:44.060 ","End":"01:45.755","Text":"we\u0027ll start with 2 objects."},{"Start":"01:45.755 ","End":"01:49.280","Text":"Generally speaking, collisions happen between only 2 objects anyways,"},{"Start":"01:49.280 ","End":"01:52.290","Text":"but also make the math a little simpler here."},{"Start":"01:52.580 ","End":"01:57.560","Text":"Let\u0027s assume we\u0027re talking about a problem with the collision of 2 objects."},{"Start":"01:57.560 ","End":"02:00.260","Text":"In that case, we can think of this side here as"},{"Start":"02:00.260 ","End":"02:03.575","Text":"our initial momentum and this will equal our final momentum,"},{"Start":"02:03.575 ","End":"02:07.530","Text":"which will be m_1u_1 plus"},{"Start":"02:07.530 ","End":"02:11.520","Text":"m_2u_2 and you were"},{"Start":"02:11.520 ","End":"02:15.800","Text":"going to use this symbol u to symbolize our velocity after the collision,"},{"Start":"02:15.800 ","End":"02:19.040","Text":"whereas v is the velocity before the collision."},{"Start":"02:19.040 ","End":"02:22.249","Text":"Now we can talk about a very important distinction."},{"Start":"02:22.249 ","End":"02:27.845","Text":"We can say that the conservation of momentum can happen on one axis but not on the other."},{"Start":"02:27.845 ","End":"02:29.780","Text":"Or it can happen on the other axis,"},{"Start":"02:29.780 ","End":"02:32.000","Text":"but not the first and how do we know that?"},{"Start":"02:32.000 ","End":"02:34.894","Text":"Well, we can say that the sum of forces,"},{"Start":"02:34.894 ","End":"02:37.295","Text":"external forces ext,"},{"Start":"02:37.295 ","End":"02:39.980","Text":"along the x-axis equals 0."},{"Start":"02:39.980 ","End":"02:42.980","Text":"If we broke down our forces into x and y components."},{"Start":"02:42.980 ","End":"02:46.940","Text":"In that case, we could apply the above equation to the x-axis,"},{"Start":"02:46.940 ","End":"03:00.080","Text":"saying m_1v_1x plus m_2v_2x equals m_1u_1x"},{"Start":"03:00.080 ","End":"03:04.990","Text":"plus m_2u_2x."},{"Start":"03:04.990 ","End":"03:08.240","Text":"This can be true for only the x axis."},{"Start":"03:08.240 ","End":"03:11.075","Text":"It can also be true for only the y-axis,"},{"Start":"03:11.075 ","End":"03:13.385","Text":"or it could be true for both axis."},{"Start":"03:13.385 ","End":"03:16.325","Text":"We could have a situation where, for example,"},{"Start":"03:16.325 ","End":"03:19.735","Text":"the forces along the x-axis equals 0."},{"Start":"03:19.735 ","End":"03:23.115","Text":"Then this equation is correct or true,"},{"Start":"03:23.115 ","End":"03:26.900","Text":"but on the y-axis there is not conservation of momentum"},{"Start":"03:26.900 ","End":"03:30.680","Text":"because the external forces along the y-axis do not equal zero,"},{"Start":"03:30.680 ","End":"03:33.559","Text":"in which case, this would not be an applicable equation"},{"Start":"03:33.559 ","End":"03:36.685","Text":"and of course the same is true for the z-axis as well."},{"Start":"03:36.685 ","End":"03:41.030","Text":"I\u0027m going to write this in here in red because it\u0027s important to remember,"},{"Start":"03:41.030 ","End":"03:47.060","Text":"conservation of momentum can occur along each or any axis independently."},{"Start":"03:47.060 ","End":"03:52.140","Text":"This should go in your formula sheet and now we\u0027ll move on to the next topic."}],"ID":9338},{"Watched":false,"Name":"The Law of Conservation of Momentum from a Different Perspective","Duration":"3m 34s","ChapterTopicVideoID":9066,"CourseChapterTopicPlaylistID":5393,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.810","Text":"I know that we\u0027ve talked a lot about the law of conservation of momentum,"},{"Start":"00:03.810 ","End":"00:06.570","Text":"but I want to give 1 more short explanation and"},{"Start":"00:06.570 ","End":"00:09.450","Text":"it will be short to show you a different perspective."},{"Start":"00:09.450 ","End":"00:12.405","Text":"Let\u0027s start with Newton\u0027s 2nd law of motion."},{"Start":"00:12.405 ","End":"00:15.270","Text":"Sigma F, the sum of forces,"},{"Start":"00:15.270 ","End":"00:20.280","Text":"equals dp over dt."},{"Start":"00:20.280 ","End":"00:22.860","Text":"Remember, this is the new version of the 2nd law of"},{"Start":"00:22.860 ","End":"00:27.010","Text":"motion that we talked about when first starting with momentum."},{"Start":"00:27.170 ","End":"00:30.270","Text":"Because we\u0027re talking about the sum of forces,"},{"Start":"00:30.270 ","End":"00:35.910","Text":"we know that this law works for the total momentum, that is P_T."},{"Start":"00:35.910 ","End":"00:40.760","Text":"We know that P_T equals the sum of all of our different momenta."},{"Start":"00:40.760 ","End":"00:49.055","Text":"P_1 plus P_2 plus P_3 and so on and so forth until you reach your final object."},{"Start":"00:49.055 ","End":"00:55.505","Text":"We sum up or add up the momenta of each of these objects, and that gives us P_T."},{"Start":"00:55.505 ","End":"00:59.180","Text":"Similarly, when we calculate the sum of forces,"},{"Start":"00:59.180 ","End":"01:03.075","Text":"we need to add up each force that is acting upon each object."},{"Start":"01:03.075 ","End":"01:04.325","Text":"For the first object,"},{"Start":"01:04.325 ","End":"01:07.610","Text":"we might have Forces 1 and 2 and we need to add those."},{"Start":"01:07.610 ","End":"01:09.080","Text":"For a 2nd or 3rd object,"},{"Start":"01:09.080 ","End":"01:11.330","Text":"we might have 1 or many forces that we need to"},{"Start":"01:11.330 ","End":"01:13.910","Text":"add to that and in the same way that we have P_1,"},{"Start":"01:13.910 ","End":"01:18.815","Text":"we would have F_1 or P_2 F_2 and so on and so forth."},{"Start":"01:18.815 ","End":"01:25.894","Text":"The question is, what happens when the sum of forces equals 0?"},{"Start":"01:25.894 ","End":"01:28.325","Text":"When the sum of forces equals 0,"},{"Start":"01:28.325 ","End":"01:31.130","Text":"that means that dp dt equals 0."},{"Start":"01:31.130 ","End":"01:35.000","Text":"When we take the derivative of something over time and that equals 0,"},{"Start":"01:35.000 ","End":"01:37.175","Text":"that means that our variable itself,"},{"Start":"01:37.175 ","End":"01:41.740","Text":"in this case P_T or P is constant."},{"Start":"01:41.740 ","End":"01:43.850","Text":"If P_T is constant,"},{"Start":"01:43.850 ","End":"01:48.120","Text":"that means that we have conservation of momentum."},{"Start":"01:52.120 ","End":"01:55.130","Text":"Now let\u0027s break this down a little further."},{"Start":"01:55.130 ","End":"01:58.295","Text":"We have internal forces and external forces."},{"Start":"01:58.295 ","End":"02:03.095","Text":"Internal forces are the forces that are exerted by our objects themselves."},{"Start":"02:03.095 ","End":"02:04.909","Text":"If, for example,"},{"Start":"02:04.909 ","End":"02:09.198","Text":"Object 1 is to exert some force on Object 2,"},{"Start":"02:09.198 ","End":"02:13.820","Text":"then Object 2 would have to exert an equal and opposite force on Object 1."},{"Start":"02:13.820 ","End":"02:15.860","Text":"We know that from Newton\u0027s 3rd law."},{"Start":"02:15.860 ","End":"02:24.120","Text":"F_1 2 equals negative F_2 1 equal and opposite."},{"Start":"02:24.530 ","End":"02:29.015","Text":"Therefore, if I take the sum of the internal forces,"},{"Start":"02:29.015 ","End":"02:31.700","Text":"those are the forces where 1 is acting on"},{"Start":"02:31.700 ","End":"02:35.490","Text":"2 and 2 is acting on 3 and 3 is acting on 1, etc."},{"Start":"02:35.490 ","End":"02:40.745","Text":"The sum of those internal forces is 0,"},{"Start":"02:40.745 ","End":"02:42.065","Text":"and that\u0027s always true."},{"Start":"02:42.065 ","End":"02:44.960","Text":"The sum of my internal forces is 0."},{"Start":"02:44.960 ","End":"02:46.595","Text":"Because this is true,"},{"Start":"02:46.595 ","End":"02:49.700","Text":"we don\u0027t have to look at the internal forces every time"},{"Start":"02:49.700 ","End":"02:53.150","Text":"we look at an equation when talking about conservation of momentum,"},{"Start":"02:53.150 ","End":"02:58.805","Text":"all we need to do is look at the external forces and if our external forces,"},{"Start":"02:58.805 ","End":"03:02.870","Text":"that is, if our external forces equals 0,"},{"Start":"03:02.870 ","End":"03:09.530","Text":"that means that our sum of total forces also equals 0 and if that is true,"},{"Start":"03:09.530 ","End":"03:14.730","Text":"then we have conservation of P, conservation of momentum."},{"Start":"03:14.780 ","End":"03:18.590","Text":"Therefore, we can look at the law of conservation of"},{"Start":"03:18.590 ","End":"03:22.325","Text":"momentum from this perspective and say,"},{"Start":"03:22.325 ","End":"03:25.985","Text":"if the sum of external forces equals 0,"},{"Start":"03:25.985 ","End":"03:29.885","Text":"then the total momentum P_T is conserved and again,"},{"Start":"03:29.885 ","End":"03:35.700","Text":"external forces are forces which are caused by an object outside of the system."}],"ID":9339},{"Watched":false,"Name":"Example of Conservation of Momentum","Duration":"8m 20s","ChapterTopicVideoID":9067,"CourseChapterTopicPlaylistID":5393,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.020","Text":"In this example, we\u0027re told that 2 balls have a collision."},{"Start":"00:04.020 ","End":"00:06.105","Text":"The first ball, m_1,"},{"Start":"00:06.105 ","End":"00:10.350","Text":"is given with a mass of 3 kilograms and a velocity of"},{"Start":"00:10.350 ","End":"00:15.490","Text":"3 meters per second and it\u0027s going in the direction of the x-axis."},{"Start":"00:16.430 ","End":"00:19.005","Text":"The second ball, m_2,"},{"Start":"00:19.005 ","End":"00:22.680","Text":"is given a mass of 2 kilograms and is at rest."},{"Start":"00:22.680 ","End":"00:26.155","Text":"This is the diagram of what it looks like before the collision."},{"Start":"00:26.155 ","End":"00:31.430","Text":"After the collision, we see that m_2 goes at an upward angle of"},{"Start":"00:31.430 ","End":"00:37.285","Text":"30 degrees with respect to the x-axis at a velocity of V_2."},{"Start":"00:37.285 ","End":"00:41.280","Text":"We\u0027re told that m_1, the first ball,"},{"Start":"00:41.280 ","End":"00:49.165","Text":"goes at a downwards angle of 45 degrees with respect to the x-axis at a velocity of V_1."},{"Start":"00:49.165 ","End":"00:52.160","Text":"We\u0027re given the mass of both balls and"},{"Start":"00:52.160 ","End":"00:56.165","Text":"the initial velocity of really both balls for m_2 it\u0027s 0,"},{"Start":"00:56.165 ","End":"00:58.505","Text":"for m_1 it\u0027s 3 meters per second."},{"Start":"00:58.505 ","End":"01:02.630","Text":"We\u0027re also told the angles at which they are going after the collision."},{"Start":"01:02.630 ","End":"01:09.690","Text":"What we need to find is the value of V_1 and V_2."},{"Start":"01:10.490 ","End":"01:14.420","Text":"Now I need to approach the problem and the best way to start,"},{"Start":"01:14.420 ","End":"01:17.369","Text":"is to ask if there is conservation of momentum."},{"Start":"01:17.369 ","End":"01:21.230","Text":"The best way to do that is to look at the sum of external forces."},{"Start":"01:21.230 ","End":"01:24.215","Text":"If the sum of external forces equals 0,"},{"Start":"01:24.215 ","End":"01:26.330","Text":"that means that we have conservation of"},{"Start":"01:26.330 ","End":"01:29.965","Text":"momentum and then I can use my law of conservation of momentum."},{"Start":"01:29.965 ","End":"01:34.745","Text":"Let\u0027s ask that the thing to look at is total momentum here, P_T."},{"Start":"01:34.745 ","End":"01:36.545","Text":"P_T in our case,"},{"Start":"01:36.545 ","End":"01:41.350","Text":"equals P_1 plus P_2."},{"Start":"01:41.350 ","End":"01:43.505","Text":"If you look at our problem,"},{"Start":"01:43.505 ","End":"01:45.500","Text":"they\u0027re only seem to be internal forces."},{"Start":"01:45.500 ","End":"01:48.935","Text":"It\u0027s m_1 colliding with m_2."},{"Start":"01:48.935 ","End":"01:51.680","Text":"There\u0027s no mg to worry about,"},{"Start":"01:51.680 ","End":"01:53.975","Text":"there\u0027s no other object coming into the problem."},{"Start":"01:53.975 ","End":"01:57.380","Text":"From here it seems that we only have internal forces,"},{"Start":"01:57.380 ","End":"01:59.465","Text":"meaning there are no external forces."},{"Start":"01:59.465 ","End":"02:04.895","Text":"In that case, P_T is constant and when P_T is constant,"},{"Start":"02:04.895 ","End":"02:07.850","Text":"it means that we have conservation of momentum."},{"Start":"02:07.850 ","End":"02:10.490","Text":"Now that I know that I have conservation of"},{"Start":"02:10.490 ","End":"02:13.970","Text":"momentum because I\u0027m only dealing with internal forces,"},{"Start":"02:13.970 ","End":"02:16.475","Text":"I can write out my equation."},{"Start":"02:16.475 ","End":"02:23.425","Text":"m_1v_1 plus m_2v_2 equals m_1u_1 plus m_2u_2."},{"Start":"02:23.425 ","End":"02:27.760","Text":"It\u0027s basically a breakdown of P initial and P final."},{"Start":"02:27.760 ","End":"02:30.985","Text":"What I can do is take this by each axis."},{"Start":"02:30.985 ","End":"02:33.055","Text":"Let\u0027s first do the x-axis,"},{"Start":"02:33.055 ","End":"02:35.910","Text":"P_x and that equals,"},{"Start":"02:35.910 ","End":"02:38.460","Text":"well, our mass 1 is 3 kilograms,"},{"Start":"02:38.460 ","End":"02:40.550","Text":"we\u0027ll bring in our units later if we need to,"},{"Start":"02:40.550 ","End":"02:47.929","Text":"and velocity 1 is 3 meters per second plus the mass of the second object is 2 kilograms,"},{"Start":"02:47.929 ","End":"02:49.280","Text":"but the velocity is 0,"},{"Start":"02:49.280 ","End":"02:51.050","Text":"so this will zero out."},{"Start":"02:51.050 ","End":"02:54.650","Text":"m_1u_1 is again 3 kilograms."},{"Start":"02:54.650 ","End":"02:56.300","Text":"I don\u0027t know what u_1 is."},{"Start":"02:56.300 ","End":"02:59.550","Text":"We are calling it v_1 here, so v_1,"},{"Start":"02:59.550 ","End":"03:01.725","Text":"and we have to remember that angle too,"},{"Start":"03:01.725 ","End":"03:07.860","Text":"times the cosine of 45 degrees."},{"Start":"03:07.860 ","End":"03:12.285","Text":"For m_2, it\u0027s 2 times v_2,"},{"Start":"03:12.285 ","End":"03:14.640","Text":"another thing we don\u0027t know, another variable,"},{"Start":"03:14.640 ","End":"03:18.460","Text":"and cosine of 30 degrees."},{"Start":"03:18.740 ","End":"03:21.530","Text":"Now let\u0027s do this on the y-axis too,"},{"Start":"03:21.530 ","End":"03:24.995","Text":"and I\u0027ll actually move this back so I have a little more room in case."},{"Start":"03:24.995 ","End":"03:30.515","Text":"P_y equals 0 plus 0."},{"Start":"03:30.515 ","End":"03:32.180","Text":"That\u0027s the initial velocity."},{"Start":"03:32.180 ","End":"03:35.480","Text":"There is no velocity along the y-axis,"},{"Start":"03:35.480 ","End":"03:40.230","Text":"so both of the momenta zero out and the final momentum,"},{"Start":"03:40.230 ","End":"03:42.795","Text":"for the first object m_1,"},{"Start":"03:42.795 ","End":"03:45.870","Text":"is going to be 3, the mass,"},{"Start":"03:45.870 ","End":"03:52.980","Text":"times v_1, our unknown velocity, sine 45 degrees."},{"Start":"03:52.980 ","End":"03:55.190","Text":"This has to be negative because we\u0027re going in"},{"Start":"03:55.190 ","End":"03:57.815","Text":"a negative direction with respect to our y-axis."},{"Start":"03:57.815 ","End":"03:59.405","Text":"If this is our y-axis,"},{"Start":"03:59.405 ","End":"04:00.770","Text":"x is positive to the right."},{"Start":"04:00.770 ","End":"04:05.055","Text":"y is positive to the top of our screen so,"},{"Start":"04:05.055 ","End":"04:09.760","Text":"this is in a negative direction so this is negative."},{"Start":"04:10.580 ","End":"04:13.080","Text":"Add to that m_2,"},{"Start":"04:13.080 ","End":"04:22.680","Text":"which is 2 kilograms times v_2 sine 30 degrees and this is positive."},{"Start":"04:22.680 ","End":"04:26.200","Text":"From this point, you just need to plug in the values of"},{"Start":"04:26.200 ","End":"04:29.905","Text":"each cosine and sine and you can solve the problem."},{"Start":"04:29.905 ","End":"04:39.065","Text":"I\u0027ll write the solution out here for you and v_1 equals 0.98 meters per second,"},{"Start":"04:39.065 ","End":"04:49.730","Text":"and v_2 equals 2.078 meters per second."},{"Start":"04:49.730 ","End":"04:52.675","Text":"With that, let\u0027s move on to part b."},{"Start":"04:52.675 ","End":"04:57.485","Text":"In part b, we\u0027re asked to find the impulse applied to each object."},{"Start":"04:57.485 ","End":"05:02.775","Text":"We need to find J_1 and we need to find J_2. Let\u0027s start with J_1."},{"Start":"05:02.775 ","End":"05:06.769","Text":"J_1 equals, we know based on our formulas,"},{"Start":"05:06.769 ","End":"05:11.465","Text":"the integral of the force times dt."},{"Start":"05:11.465 ","End":"05:13.010","Text":"However, in this problem,"},{"Start":"05:13.010 ","End":"05:15.200","Text":"we\u0027re not given any data about the force"},{"Start":"05:15.200 ","End":"05:17.585","Text":"so we\u0027ll go into another version of this formula."},{"Start":"05:17.585 ","End":"05:20.565","Text":"The impulse J equals Delta P,"},{"Start":"05:20.565 ","End":"05:22.050","Text":"the change in momentum."},{"Start":"05:22.050 ","End":"05:24.990","Text":"Of course this is all for object 1."},{"Start":"05:24.990 ","End":"05:30.385","Text":"If that\u0027s the case, we can break this down based on the x-axis and the y-axis."},{"Start":"05:30.385 ","End":"05:32.700","Text":"Let\u0027s start with the x axis,"},{"Start":"05:32.700 ","End":"05:35.770","Text":"J_1_x equals Delta P_1_x."},{"Start":"05:38.510 ","End":"05:48.990","Text":"The same is true for y. J_1_y equals Delta P_1_y."},{"Start":"05:48.990 ","End":"05:53.305","Text":"Starting with J_1_x, we can plug in"},{"Start":"05:53.305 ","End":"05:58.435","Text":"our solution for v_1 to the equations above and find our answer."},{"Start":"05:58.435 ","End":"06:04.410","Text":"J_1_ x equals the final momentum minus the initial momentum."},{"Start":"06:04.410 ","End":"06:06.614","Text":"P final is 3"},{"Start":"06:06.614 ","End":"06:16.455","Text":"times 0.98. and that is multiplied by the cosine of 45 degrees,"},{"Start":"06:16.455 ","End":"06:20.730","Text":"which is 1 over root 2."},{"Start":"06:20.730 ","End":"06:23.635","Text":"Subtract from that the initial momentum,"},{"Start":"06:23.635 ","End":"06:26.230","Text":"which is 3 times 3 or 9."},{"Start":"06:26.230 ","End":"06:35.630","Text":"Our answer for J_1_x is negative 6.921 Newton seconds."},{"Start":"06:35.820 ","End":"06:39.170","Text":"Now, moving on to J_1_y,"},{"Start":"06:39.710 ","End":"06:42.590","Text":"we can do a similar calculation."},{"Start":"06:42.590 ","End":"06:47.110","Text":"The final momentum is negative 3,"},{"Start":"06:47.110 ","End":"06:51.170","Text":"remember the negative sign from here, times v_1,"},{"Start":"06:51.170 ","End":"06:57.240","Text":"which is 0.98, times the sine of 45 degrees which"},{"Start":"06:57.240 ","End":"07:03.970","Text":"is also 1 over root 2 and minus the initial momentum, which is 0."},{"Start":"07:03.970 ","End":"07:13.510","Text":"Our answer is negative 2.079 Newton seconds."},{"Start":"07:13.510 ","End":"07:15.400","Text":"In terms of J_2,"},{"Start":"07:15.400 ","End":"07:17.285","Text":"the impulse for the second object,"},{"Start":"07:17.285 ","End":"07:18.965","Text":"we can do the same procedure."},{"Start":"07:18.965 ","End":"07:21.185","Text":"However, we can also take a little shortcut."},{"Start":"07:21.185 ","End":"07:23.590","Text":"Because we have conservation of momentum,"},{"Start":"07:23.590 ","End":"07:26.960","Text":"we know that the change in momentum for object 1 is"},{"Start":"07:26.960 ","End":"07:30.740","Text":"equal and opposite to the change in momentum for object 2."},{"Start":"07:30.740 ","End":"07:39.020","Text":"That means that J_1_x"},{"Start":"07:39.020 ","End":"07:48.880","Text":"equals negative J_2_x and that J_1_y equals negative J_2_y."},{"Start":"07:49.670 ","End":"07:54.380","Text":"Because we have conservation of momentum along each of the axis,"},{"Start":"07:54.380 ","End":"07:56.225","Text":"we know this to be true."},{"Start":"07:56.225 ","End":"08:01.145","Text":"You can also do the calculations and double-check that this is correct,"},{"Start":"08:01.145 ","End":"08:05.270","Text":"but we should know based on the conservation of momentum that this is true."},{"Start":"08:05.270 ","End":"08:12.770","Text":"The other thing you can say is that the change in the total momentum in the system is 0,"},{"Start":"08:12.770 ","End":"08:17.090","Text":"and therefore the total impulse equals 0 so this should be true."},{"Start":"08:17.090 ","End":"08:20.100","Text":"With that, we end our example."}],"ID":9340},{"Watched":false,"Name":"What Happens When Momentum Is Not Conserved","Duration":"3m 24s","ChapterTopicVideoID":9068,"CourseChapterTopicPlaylistID":5393,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.110 ","End":"00:02.250","Text":"In an earlier lecture,"},{"Start":"00:02.250 ","End":"00:07.455","Text":"we talked about how to approach the questions using the following idea."},{"Start":"00:07.455 ","End":"00:12.555","Text":"We asked is the sum of external forces equal to 0?"},{"Start":"00:12.555 ","End":"00:16.200","Text":"If yes, then we have a conservation of momentum."},{"Start":"00:16.200 ","End":"00:19.675","Text":"If no, we\u0027ll figure out how to deal with it later."},{"Start":"00:19.675 ","End":"00:22.440","Text":"In this lecture we\u0027re going to address these questions"},{"Start":"00:22.440 ","End":"00:25.335","Text":"whether is not conservation of momentum."},{"Start":"00:25.335 ","End":"00:28.590","Text":"In these questions, we\u0027re going to not be able to"},{"Start":"00:28.590 ","End":"00:32.865","Text":"use the conservation of momentum formula as we\u0027ve used it so far."},{"Start":"00:32.865 ","End":"00:35.790","Text":"We can\u0027t say that P_i=P_F."},{"Start":"00:35.790 ","End":"00:37.980","Text":"That is not true."},{"Start":"00:37.980 ","End":"00:43.575","Text":"But what we can say is that the change in total momentum"},{"Start":"00:43.575 ","End":"00:50.840","Text":"P_T equals the impulse of the external forces J_external."},{"Start":"00:50.840 ","End":"00:54.500","Text":"We\u0027ve actually already done this in a sense, if you recall,"},{"Start":"00:54.500 ","End":"00:57.860","Text":"we had 1 problem where we talked about a ball being thrown"},{"Start":"00:57.860 ","End":"01:01.985","Text":"against the wall and the calculation was in the end,"},{"Start":"01:01.985 ","End":"01:10.595","Text":"that the change in momentum Delta P was equal to J_N the impulse of the normal force."},{"Start":"01:10.595 ","End":"01:13.562","Text":"Why was that? Because the wall wasn\u0027t part of our system."},{"Start":"01:13.562 ","End":"01:15.860","Text":"Our closed system only included the ball."},{"Start":"01:15.860 ","End":"01:18.650","Text":"The walls force, or the force that the wall"},{"Start":"01:18.650 ","End":"01:22.540","Text":"exerted was outside of our system and therefore external."},{"Start":"01:22.540 ","End":"01:25.745","Text":"J_N equals the external impulse."},{"Start":"01:25.745 ","End":"01:29.990","Text":"We can apply the same concept to other situations as well."},{"Start":"01:29.990 ","End":"01:32.740","Text":"Let\u0027s say we have a collision between 2 balls."},{"Start":"01:32.740 ","End":"01:36.515","Text":"They\u0027re going to collide and at some point during this interaction,"},{"Start":"01:36.515 ","End":"01:40.085","Text":"a mysterious force F exerts force on"},{"Start":"01:40.085 ","End":"01:44.590","Text":"1 or both of my objects and it does that for the time Delta t,"},{"Start":"01:44.590 ","End":"01:46.340","Text":"F is a vector here."},{"Start":"01:46.340 ","End":"01:51.320","Text":"In that case, instead of using the equation or formula for the conservation of momentum,"},{"Start":"01:51.320 ","End":"01:59.910","Text":"I can instead say that the change in momentum Delta P=P_F minus P_i also"},{"Start":"01:59.910 ","End":"02:09.140","Text":"vectors and that equals the impulse J of this mysterious force F. Of course,"},{"Start":"02:09.140 ","End":"02:10.820","Text":"using my other formula,"},{"Start":"02:10.820 ","End":"02:13.310","Text":"I know that J_F,"},{"Start":"02:13.310 ","End":"02:18.654","Text":"or the impulse of the mysterious force equals an integral of that same force,"},{"Start":"02:18.654 ","End":"02:25.925","Text":"dt during the time period Delta t. Basically to summarize here,"},{"Start":"02:25.925 ","End":"02:29.045","Text":"when we do have a conservation of momentum,"},{"Start":"02:29.045 ","End":"02:31.805","Text":"that means that we can say P_i=P_F,"},{"Start":"02:31.805 ","End":"02:34.445","Text":"your initial momentum equals your final momentum."},{"Start":"02:34.445 ","End":"02:36.125","Text":"But if that is not the case,"},{"Start":"02:36.125 ","End":"02:37.595","Text":"then you can take Delta P,"},{"Start":"02:37.595 ","End":"02:38.750","Text":"the change in momentum,"},{"Start":"02:38.750 ","End":"02:40.730","Text":"or P_F minus P_i,"},{"Start":"02:40.730 ","End":"02:45.145","Text":"and set that equal to the impulse of the external forces."},{"Start":"02:45.145 ","End":"02:48.380","Text":"If you take an integral of the external force over"},{"Start":"02:48.380 ","End":"02:51.880","Text":"the time Delta t with which it\u0027s acting you can find your answer."},{"Start":"02:51.880 ","End":"02:59.015","Text":"Or another way to think about this is P_i plus J_external that is,"},{"Start":"02:59.015 ","End":"03:03.650","Text":"your initial momentum plus the impulse of the external force"},{"Start":"03:03.650 ","End":"03:08.875","Text":"or forces equals P_F, your final momentum."},{"Start":"03:08.875 ","End":"03:12.560","Text":"Maybe this is worthwhile to put on your formula sheet."},{"Start":"03:12.560 ","End":"03:17.960","Text":"Delta P_T=J_external, or the change"},{"Start":"03:17.960 ","End":"03:24.179","Text":"in total momentum equals the impulse of the external force or forces."}],"ID":9341}],"Thumbnail":null,"ID":5393},{"Name":"Types of Collisions","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Types of Collisions","Duration":"3m 52s","ChapterTopicVideoID":9069,"CourseChapterTopicPlaylistID":5394,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.020 ","End":"00:04.605","Text":"In this lecture, I want to talk about types of collisions."},{"Start":"00:04.605 ","End":"00:08.940","Text":"What we\u0027re first going to do is break this down into 2 basic types of collisions."},{"Start":"00:08.940 ","End":"00:11.685","Text":"The first is an elastic collision,"},{"Start":"00:11.685 ","End":"00:14.145","Text":"and the second is inelastic."},{"Start":"00:14.145 ","End":"00:17.640","Text":"An elastic collision is a collision that has"},{"Start":"00:17.640 ","End":"00:21.990","Text":"both conservation of momentum and conservation of energy."},{"Start":"00:21.990 ","End":"00:24.690","Text":"Every collision has conservation of momentum,"},{"Start":"00:24.690 ","End":"00:29.595","Text":"but elastic collisions are unique because they also have conservation of energy."},{"Start":"00:29.595 ","End":"00:34.815","Text":"So if you have a problem where it\u0027s written that you\u0027re looking at an elastic collision,"},{"Start":"00:34.815 ","End":"00:37.440","Text":"it means you automatically can understand that"},{"Start":"00:37.440 ","End":"00:40.470","Text":"you\u0027re looking at a problem with conservation of energy."},{"Start":"00:40.470 ","End":"00:43.005","Text":"As opposed to elastic collisions,"},{"Start":"00:43.005 ","End":"00:47.915","Text":"inelastic collisions are collisions that do not have conservation of energy."},{"Start":"00:47.915 ","End":"00:50.765","Text":"They only have conservation of momentum."},{"Start":"00:50.765 ","End":"00:53.320","Text":"So this is the only thing that separates them."},{"Start":"00:53.320 ","End":"00:56.630","Text":"Elastic collisions have conservation of energy;"},{"Start":"00:56.630 ","End":"00:59.135","Text":"inelastic collisions do not."},{"Start":"00:59.135 ","End":"01:01.865","Text":"1 more thing I\u0027d like to bring up here"},{"Start":"01:01.865 ","End":"01:05.015","Text":"is there\u0027s something called a perfectly elastic collision."},{"Start":"01:05.015 ","End":"01:07.580","Text":"If you continue studying physics at some point,"},{"Start":"01:07.580 ","End":"01:09.140","Text":"this will have a different meaning."},{"Start":"01:09.140 ","End":"01:10.565","Text":"However, for now,"},{"Start":"01:10.565 ","End":"01:12.035","Text":"you can think about that,"},{"Start":"01:12.035 ","End":"01:13.640","Text":"a perfectly elastic collision,"},{"Start":"01:13.640 ","End":"01:18.979","Text":"that is, as identical to inelastic collision for all intents and purposes."},{"Start":"01:18.979 ","End":"01:20.750","Text":"Now, the next thing,"},{"Start":"01:20.750 ","End":"01:22.355","Text":"once you understand the difference between"},{"Start":"01:22.355 ","End":"01:27.340","Text":"elastic and inelastic collisions is to look at some special cases."},{"Start":"01:27.340 ","End":"01:29.765","Text":"In terms of inelastic collisions,"},{"Start":"01:29.765 ","End":"01:31.580","Text":"we have 2 special cases."},{"Start":"01:31.580 ","End":"01:34.355","Text":"The first is called a plastic collision."},{"Start":"01:34.355 ","End":"01:39.725","Text":"Now, some students will confuse between plastic and elastic."},{"Start":"01:39.725 ","End":"01:41.694","Text":"However, they\u0027re in fact opposites."},{"Start":"01:41.694 ","End":"01:44.635","Text":"Plastic collisions are always inelastic."},{"Start":"01:44.635 ","End":"01:47.930","Text":"1 easy way to remember is to think about plaster."},{"Start":"01:47.930 ","End":"01:49.820","Text":"When you throw plaster at something,"},{"Start":"01:49.820 ","End":"01:51.173","Text":"it will usually stick,"},{"Start":"01:51.173 ","End":"01:54.005","Text":"and that\u0027s what\u0027s happening in a plastic collision."},{"Start":"01:54.005 ","End":"01:57.635","Text":"You have 2 objects that move together after the collision."},{"Start":"01:57.635 ","End":"02:00.020","Text":"For example, a projectile, say,"},{"Start":"02:00.020 ","End":"02:02.480","Text":"a bullet that\u0027s hitting a tree and gets stuck in"},{"Start":"02:02.480 ","End":"02:06.595","Text":"the tree trunk or 2 balls that collide and stick to each other."},{"Start":"02:06.595 ","End":"02:09.500","Text":"As soon as the 2 objects collide,"},{"Start":"02:09.500 ","End":"02:12.610","Text":"they stick together, that is a plastic collision."},{"Start":"02:12.610 ","End":"02:15.620","Text":"You can take away from this that in plastic collisions or"},{"Start":"02:15.620 ","End":"02:18.920","Text":"if a problem tells you you\u0027re dealing with a plastic collision,"},{"Start":"02:18.920 ","End":"02:21.640","Text":"you do not have conservation of energy."},{"Start":"02:21.640 ","End":"02:25.790","Text":"The opposite of a plastic collision is recoil."},{"Start":"02:25.790 ","End":"02:27.695","Text":"When you have a collision with recoil,"},{"Start":"02:27.695 ","End":"02:31.160","Text":"it means you have objects that move together before the collision,"},{"Start":"02:31.160 ","End":"02:33.295","Text":"but after the collision, they separate."},{"Start":"02:33.295 ","End":"02:38.345","Text":"2 examples of that are when a bullet is shot from a gun and an explosion,"},{"Start":"02:38.345 ","End":"02:40.273","Text":"and we\u0027ll talk about why later."},{"Start":"02:40.273 ","End":"02:41.765","Text":"But in terms of recoil,"},{"Start":"02:41.765 ","End":"02:45.684","Text":"what you need to know is that you\u0027re not dealing with conservation of energy."},{"Start":"02:45.684 ","End":"02:48.110","Text":"In a case of recoil or a plastic collision,"},{"Start":"02:48.110 ","End":"02:50.605","Text":"there is not conservation of energy."},{"Start":"02:50.605 ","End":"02:53.505","Text":"In terms of the elastic special cases,"},{"Start":"02:53.505 ","End":"02:54.830","Text":"we\u0027ll deal with them later."},{"Start":"02:54.830 ","End":"02:57.724","Text":"For now, just know that when you have head-on collisions,"},{"Start":"02:57.724 ","End":"03:01.985","Text":"or particularly head-on collision between 2 objects with equal masses,"},{"Start":"03:01.985 ","End":"03:04.390","Text":"something unique happens, and additionally,"},{"Start":"03:04.390 ","End":"03:08.330","Text":"when you have a non head-on collision between 2 objects of equal masses,"},{"Start":"03:08.330 ","End":"03:11.060","Text":"when 1 object is at rest, fun things happen."},{"Start":"03:11.060 ","End":"03:12.485","Text":"We\u0027re not going to deal with that now."},{"Start":"03:12.485 ","End":"03:14.555","Text":"We\u0027ll look at these soon."},{"Start":"03:14.555 ","End":"03:17.630","Text":"For now, the big takeaway from this lecture is"},{"Start":"03:17.630 ","End":"03:21.230","Text":"the difference between inelastic and elastic collisions."},{"Start":"03:21.230 ","End":"03:25.974","Text":"Elastic collisions have conservation of momentum and conservation of energy."},{"Start":"03:25.974 ","End":"03:29.826","Text":"Inelastic collisions only conserve momentum,"},{"Start":"03:29.826 ","End":"03:33.200","Text":"and there are 2 special cases of inelastic collisions,"},{"Start":"03:33.200 ","End":"03:36.475","Text":"which are plastic and recoil collisions,"},{"Start":"03:36.475 ","End":"03:41.360","Text":"and they have to do with objects sticking to each other after and before the collision."},{"Start":"03:41.360 ","End":"03:42.883","Text":"Now I want to make it clear here."},{"Start":"03:42.883 ","End":"03:44.189","Text":"This is a lot of information."},{"Start":"03:44.189 ","End":"03:46.850","Text":"You don\u0027t have to worry about memorizing it right now."},{"Start":"03:46.850 ","End":"03:48.770","Text":"But if you can get down the basic concepts,"},{"Start":"03:48.770 ","End":"03:50.230","Text":"it can be rather helpful."},{"Start":"03:50.230 ","End":"03:53.520","Text":"With that, let\u0027s move on to our next lecture."}],"ID":9342},{"Watched":false,"Name":"Plastic and Recoil Collisions","Duration":"6m 27s","ChapterTopicVideoID":9070,"CourseChapterTopicPlaylistID":5394,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.280","Text":"I want to talk a little bit more about plastic and recoil collisions."},{"Start":"00:05.280 ","End":"00:07.665","Text":"If you recall, in the last lecture,"},{"Start":"00:07.665 ","End":"00:11.850","Text":"we said that plastic collisions are when objects move together after"},{"Start":"00:11.850 ","End":"00:16.695","Text":"collisions and recoil is when objects move together before collisions."},{"Start":"00:16.695 ","End":"00:18.945","Text":"Moving backwards a little bit,"},{"Start":"00:18.945 ","End":"00:27.850","Text":"you remember that our conservation of momentum equation is m_1 v_1 plus"},{"Start":"00:27.850 ","End":"00:32.479","Text":"m_2 v_2 equals"},{"Start":"00:32.479 ","End":"00:41.070","Text":"m_1 u_1 plus m_2 u_2."},{"Start":"00:41.320 ","End":"00:44.405","Text":"In the case of a plastic collision,"},{"Start":"00:44.405 ","End":"00:46.520","Text":"we know that after the collision happens"},{"Start":"00:46.520 ","End":"00:49.085","Text":"when we\u0027re talking on the u side or the right side,"},{"Start":"00:49.085 ","End":"00:54.990","Text":"u_1 equals u_2 because the 2 objects are moving together."},{"Start":"00:56.090 ","End":"00:58.145","Text":"If this is the case,"},{"Start":"00:58.145 ","End":"01:04.475","Text":"then we can rewrite this equation as m_1 v_1 plus"},{"Start":"01:04.475 ","End":"01:14.775","Text":"m_2 v_2 equals m_1 plus m_2 times u."},{"Start":"01:14.775 ","End":"01:20.245","Text":"This u is the velocity of the 2 objects moving together."},{"Start":"01:20.245 ","End":"01:23.435","Text":"Normally in the case of a plastic collision,"},{"Start":"01:23.435 ","End":"01:26.300","Text":"we\u0027ll jump straight to this form of the equation."},{"Start":"01:26.300 ","End":"01:29.840","Text":"But if you understand how the conservation of momentum works,"},{"Start":"01:29.840 ","End":"01:31.555","Text":"you\u0027ll get here anyways."},{"Start":"01:31.555 ","End":"01:34.745","Text":"This is something you can put on your formula sheet."},{"Start":"01:34.745 ","End":"01:39.185","Text":"But I\u0027ll put it in a pink box because it\u0027s not necessary to put it on your formula sheet."},{"Start":"01:39.185 ","End":"01:42.210","Text":"But if you\u0027d like to, you\u0027re more than welcome to."},{"Start":"01:42.310 ","End":"01:44.735","Text":"Now, in the case of recoil,"},{"Start":"01:44.735 ","End":"01:47.015","Text":"you have the exact opposite thing happening."},{"Start":"01:47.015 ","End":"01:52.145","Text":"You have 2 objects that are moving together at the beginning and at the end,"},{"Start":"01:52.145 ","End":"01:54.260","Text":"they separate after your collision."},{"Start":"01:54.260 ","End":"01:57.080","Text":"I\u0027ll also put that formula here in"},{"Start":"01:57.080 ","End":"02:01.020","Text":"pink so you can put it in your formula sheet if you\u0027d like."},{"Start":"02:01.580 ","End":"02:04.410","Text":"Now that we have these 2 formulas,"},{"Start":"02:04.410 ","End":"02:06.180","Text":"let\u0027s use them in an example."},{"Start":"02:06.180 ","End":"02:09.330","Text":"In our example, we have a rifle."},{"Start":"02:09.410 ","End":"02:11.625","Text":"The mass of the rifle,"},{"Start":"02:11.625 ","End":"02:13.185","Text":"we\u0027ll call this m_1,"},{"Start":"02:13.185 ","End":"02:16.035","Text":"is given as 2 kilograms."},{"Start":"02:16.035 ","End":"02:18.480","Text":"This rifle shoots a bullet,"},{"Start":"02:18.480 ","End":"02:23.885","Text":"and the bullet is going in the direction of the x-axis,"},{"Start":"02:23.885 ","End":"02:27.340","Text":"and that is going to be m_2,"},{"Start":"02:27.340 ","End":"02:33.365","Text":"and the mass of the bullet will be 0.05 kilograms, that\u0027s 50 grams."},{"Start":"02:33.365 ","End":"02:37.445","Text":"We know the velocity of this bullet once it\u0027s shot,"},{"Start":"02:37.445 ","End":"02:39.350","Text":"this is the velocity after the collision,"},{"Start":"02:39.350 ","End":"02:47.030","Text":"so it\u0027s u_2, and u_2 equals 500 meters per second."},{"Start":"02:47.030 ","End":"02:53.105","Text":"Now in our system, we know v equals 0 because the rifle starts at rest."},{"Start":"02:53.105 ","End":"02:56.090","Text":"What we need to find is u_1,"},{"Start":"02:56.090 ","End":"02:59.335","Text":"which is the recoil of the gun."},{"Start":"02:59.335 ","End":"03:01.310","Text":"If we find u_1,"},{"Start":"03:01.310 ","End":"03:02.705","Text":"we\u0027ve solved our problem."},{"Start":"03:02.705 ","End":"03:06.470","Text":"At this point, it should be pretty obvious that we\u0027re dealing with a recoil collision,"},{"Start":"03:06.470 ","End":"03:09.080","Text":"so we can use our recoil equation."},{"Start":"03:09.080 ","End":"03:13.820","Text":"First m_1 plus m_2 times v. Well, if v equals 0,"},{"Start":"03:13.820 ","End":"03:17.390","Text":"then the whole left side equals 0,"},{"Start":"03:17.390 ","End":"03:21.350","Text":"so 0 equals m_1 times u_1."},{"Start":"03:21.350 ","End":"03:23.825","Text":"We know m_1 is 2 kilograms,"},{"Start":"03:23.825 ","End":"03:29.115","Text":"u_1 is our unknown variable, so that\u0027s u_1."},{"Start":"03:29.115 ","End":"03:31.820","Text":"Now, we need to add to that m_2,"},{"Start":"03:31.820 ","End":"03:37.610","Text":"which we know to be 0.05 kilograms or 50 grams, and u_2,"},{"Start":"03:37.610 ","End":"03:44.410","Text":"which we also know that\u0027s also a vector to be 500 meters per second."},{"Start":"03:44.410 ","End":"03:46.590","Text":"Now, we can solve for u_1."},{"Start":"03:46.590 ","End":"03:54.670","Text":"It turns out that u_1 equals about 12.5 meters per second."},{"Start":"03:54.740 ","End":"03:59.270","Text":"Now, let\u0027s continue this problem and assume that this bullet travels and that we"},{"Start":"03:59.270 ","End":"04:03.020","Text":"have a log here at the end of our firing range."},{"Start":"04:03.020 ","End":"04:06.445","Text":"The bullet hits the log and gets stuck in the log."},{"Start":"04:06.445 ","End":"04:10.415","Text":"Now, we\u0027ve gone from a recoil collision to a plastic collision."},{"Start":"04:10.415 ","End":"04:13.475","Text":"Let\u0027s assume that the bullet doesn\u0027t lose any speed,"},{"Start":"04:13.475 ","End":"04:16.460","Text":"that it\u0027s still traveling at 500 meters per second."},{"Start":"04:16.460 ","End":"04:19.280","Text":"Let\u0027s also assume that m_3,"},{"Start":"04:19.280 ","End":"04:20.890","Text":"the mass of the log,"},{"Start":"04:20.890 ","End":"04:22.875","Text":"also equals 2 kilograms,"},{"Start":"04:22.875 ","End":"04:25.095","Text":"just like with the gun."},{"Start":"04:25.095 ","End":"04:27.065","Text":"Then the question is,"},{"Start":"04:27.065 ","End":"04:28.595","Text":"with this plastic collision,"},{"Start":"04:28.595 ","End":"04:32.750","Text":"what is the velocity of this tree stump or"},{"Start":"04:32.750 ","End":"04:37.160","Text":"this log with the bullet inside after the collision."},{"Start":"04:37.160 ","End":"04:41.264","Text":"We\u0027ll call that u star, u asterisk."},{"Start":"04:41.264 ","End":"04:45.665","Text":"The way we\u0027re going to find that is using our plastic collisions equation."},{"Start":"04:45.665 ","End":"04:49.995","Text":"What we know is m_1 plus v_1."},{"Start":"04:49.995 ","End":"04:53.190","Text":"Well, m_1, we\u0027ll call that actually m_2."},{"Start":"04:53.190 ","End":"05:02.170","Text":"That\u0027s 0.05 kilograms times 500 meters per second."},{"Start":"05:02.170 ","End":"05:04.375","Text":"We know that v_2,"},{"Start":"05:04.375 ","End":"05:06.845","Text":"the velocity of our log is 0,"},{"Start":"05:06.845 ","End":"05:10.800","Text":"so m_2 times v_2 equals 0."},{"Start":"05:11.990 ","End":"05:14.640","Text":"Now, to find u asterisk,"},{"Start":"05:14.640 ","End":"05:16.499","Text":"we need to add our 2 masses,"},{"Start":"05:16.499 ","End":"05:21.995","Text":"0.05 kilograms plus 2 kilograms,"},{"Start":"05:21.995 ","End":"05:27.030","Text":"and multiply that by u asterisk or u star."},{"Start":"05:27.030 ","End":"05:31.280","Text":"Let\u0027s just assume for the sake of simplicity that the bullet has no weight."},{"Start":"05:31.280 ","End":"05:33.080","Text":"It\u0027s more or less the same thing here."},{"Start":"05:33.080 ","End":"05:36.485","Text":"Instead of multiplying 2.05 kilograms by u,"},{"Start":"05:36.485 ","End":"05:38.860","Text":"we\u0027ll multiply 2 kilograms by u."},{"Start":"05:38.860 ","End":"05:40.730","Text":"Low and behold, our answer,"},{"Start":"05:40.730 ","End":"05:48.890","Text":"u star equals 12.5 meters per second. Now, why is that?"},{"Start":"05:48.890 ","End":"05:51.980","Text":"Well, we have the same mass for our log as we have for"},{"Start":"05:51.980 ","End":"05:55.265","Text":"our rifle and the same velocity for the bullet in both equations."},{"Start":"05:55.265 ","End":"06:00.570","Text":"It would make sense that u star or u asterisk equals u_1."},{"Start":"06:01.370 ","End":"06:05.970","Text":"Now, we have our solutions for the example."},{"Start":"06:06.220 ","End":"06:12.770","Text":"Just keep in mind that recoil and plastic collisions are normal collisions."},{"Start":"06:12.770 ","End":"06:14.945","Text":"Let\u0027s make this a vector."},{"Start":"06:14.945 ","End":"06:17.945","Text":"We can talk about them in these normal terms."},{"Start":"06:17.945 ","End":"06:20.870","Text":"The big takeaway are these 2 equations which are"},{"Start":"06:20.870 ","End":"06:25.070","Text":"seemingly opposite from each other and more or less the same thing."},{"Start":"06:25.070 ","End":"06:28.470","Text":"With that, let\u0027s move on to our next lecture."}],"ID":9343},{"Watched":false,"Name":"Elastic Collisions","Duration":"5m 34s","ChapterTopicVideoID":9071,"CourseChapterTopicPlaylistID":5394,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.320","Text":"In this lecture, I want to talk about elastic collisions."},{"Start":"00:04.320 ","End":"00:06.345","Text":"When you have an elastic collision,"},{"Start":"00:06.345 ","End":"00:11.655","Text":"it means that you have both conservation of momentum and conservation of energy."},{"Start":"00:11.655 ","End":"00:14.592","Text":"Here we have our conservation of momentum formula,"},{"Start":"00:14.592 ","End":"00:15.760","Text":"you should know it by now."},{"Start":"00:15.760 ","End":"00:18.870","Text":"Of course this is written in vector form,"},{"Start":"00:18.870 ","End":"00:23.415","Text":"but can also be written in terms of each axis or each element on its own."},{"Start":"00:23.415 ","End":"00:26.925","Text":"Here we have the formula for the conservation of energy."},{"Start":"00:26.925 ","End":"00:30.675","Text":"Now this is the formula for the conservation of kinetic energy."},{"Start":"00:30.675 ","End":"00:33.285","Text":"There\u0027s kinetic energy and potential energy."},{"Start":"00:33.285 ","End":"00:35.400","Text":"But for our purposes here,"},{"Start":"00:35.400 ","End":"00:39.375","Text":"we\u0027re only going to focus on kinetic energy because functionally,"},{"Start":"00:39.375 ","End":"00:42.350","Text":"potential energy won\u0027t change enough during"},{"Start":"00:42.350 ","End":"00:46.620","Text":"the time periods we\u0027re talking about to matter for our purposes."},{"Start":"00:47.050 ","End":"00:51.590","Text":"This is our equation of conservation of kinetic energy."},{"Start":"00:51.590 ","End":"00:54.145","Text":"You could even write that in if you\u0027d like."},{"Start":"00:54.145 ","End":"00:56.085","Text":"But it\u0027s what we\u0027ll be using."},{"Start":"00:56.085 ","End":"00:58.970","Text":"If you notice this is only one formula,"},{"Start":"00:58.970 ","End":"01:00.905","Text":"unlike conservation of momentum,"},{"Start":"01:00.905 ","End":"01:04.715","Text":"where you can break it down based on x and y and possibly z."},{"Start":"01:04.715 ","End":"01:08.240","Text":"Conservation of energy or energy itself has no direction."},{"Start":"01:08.240 ","End":"01:10.239","Text":"v_1, for example,"},{"Start":"01:10.239 ","End":"01:17.290","Text":"includes both v_x and v_y and possibly v_z if we\u0027re talking about 3 dimensions."},{"Start":"01:17.420 ","End":"01:22.655","Text":"The takeaway from this first part is that when we know we have an elastic collision,"},{"Start":"01:22.655 ","End":"01:27.590","Text":"it means we also have conservation of energy and we can use this formula as well."},{"Start":"01:27.590 ","End":"01:31.445","Text":"Generally speaking, it means we\u0027ll have one more unknown variable,"},{"Start":"01:31.445 ","End":"01:35.840","Text":"and from there we\u0027ll use this formula to help us solve for that variable as well."},{"Start":"01:35.840 ","End":"01:37.630","Text":"Let\u0027s move on to an example."},{"Start":"01:37.630 ","End":"01:41.110","Text":"We\u0027re going to recycle an example from conservation of momentum,"},{"Start":"01:41.110 ","End":"01:43.130","Text":"and we\u0027re going to change the question a little bit."},{"Start":"01:43.130 ","End":"01:46.580","Text":"First of all, we\u0027ve now defined this as an elastic collision."},{"Start":"01:46.580 ","End":"01:49.355","Text":"That means that we can use conservation of energy."},{"Start":"01:49.355 ","End":"01:52.895","Text":"The second thing is we no longer have the given mass,"},{"Start":"01:52.895 ","End":"01:55.520","Text":"m_2, the mass of the second ball."},{"Start":"01:55.520 ","End":"01:58.610","Text":"But we\u0027re still talking about a collision and now we need to"},{"Start":"01:58.610 ","End":"02:02.975","Text":"find u_2 and u_1 as well as m_2."},{"Start":"02:02.975 ","End":"02:08.520","Text":"That\u0027s the velocity of each ball after the collision and the mass m_2."},{"Start":"02:09.080 ","End":"02:13.490","Text":"Let\u0027s start by writing out our conservation of momentum equation,"},{"Start":"02:13.490 ","End":"02:16.300","Text":"and we can do that along each axis separately."},{"Start":"02:16.300 ","End":"02:19.625","Text":"The momentum along the x-axis, using this equation,"},{"Start":"02:19.625 ","End":"02:24.725","Text":"m_1 is 3 kilograms and v_1 is 3 meters per second,"},{"Start":"02:24.725 ","End":"02:27.635","Text":"and m_2 times v_2 is 0,"},{"Start":"02:27.635 ","End":"02:29.120","Text":"because velocity 2,"},{"Start":"02:29.120 ","End":"02:30.580","Text":"v_2 is 0,"},{"Start":"02:30.580 ","End":"02:32.685","Text":"and that equals m_1,"},{"Start":"02:32.685 ","End":"02:35.660","Text":"3 kilograms times u_1,"},{"Start":"02:35.660 ","End":"02:40.960","Text":"some unknown variable, times the angle with respect to the x-axis."},{"Start":"02:40.960 ","End":"02:45.125","Text":"We\u0027re going to take cosine of 45 degrees,"},{"Start":"02:45.125 ","End":"02:48.390","Text":"plus m_2, another unknown,"},{"Start":"02:48.390 ","End":"02:51.755","Text":"times u_2, our third unknown variable,"},{"Start":"02:51.755 ","End":"02:54.755","Text":"times their angle with respect to the x-axis,"},{"Start":"02:54.755 ","End":"02:58.045","Text":"cosine of 30 degrees."},{"Start":"02:58.045 ","End":"03:00.595","Text":"Now along the y-axis,"},{"Start":"03:00.595 ","End":"03:03.900","Text":"P_y is initially 0."},{"Start":"03:03.900 ","End":"03:06.800","Text":"Because there\u0027s no movement at the beginning along the y-axis,"},{"Start":"03:06.800 ","End":"03:08.930","Text":"everything is along the x-axis."},{"Start":"03:08.930 ","End":"03:17.660","Text":"That equals m_2 u_2 sine 30,"},{"Start":"03:17.660 ","End":"03:19.570","Text":"30 degrees that is,"},{"Start":"03:19.570 ","End":"03:24.695","Text":"minus 3, because this is in the negative direction."},{"Start":"03:24.695 ","End":"03:29.560","Text":"It\u0027s going downwards. u_1,"},{"Start":"03:29.570 ","End":"03:31.905","Text":"3 is our mass,"},{"Start":"03:31.905 ","End":"03:33.600","Text":"u_1 is our velocity,"},{"Start":"03:33.600 ","End":"03:38.680","Text":"times the sine of 45 degrees."},{"Start":"03:38.800 ","End":"03:42.560","Text":"Now let\u0027s write out our kinetic energy formula."},{"Start":"03:42.560 ","End":"03:46.130","Text":"Energy is first,"},{"Start":"03:46.130 ","End":"03:52.475","Text":"1/2 of m_1 times v_1^2 is 1/2 of 3"},{"Start":"03:52.475 ","End":"03:59.085","Text":"times 3^2 plus 1/2 of m_2 times v_2^2 is 0."},{"Start":"03:59.085 ","End":"04:02.085","Text":"Because once again, v_2 is 0,"},{"Start":"04:02.085 ","End":"04:08.770","Text":"equals 1/2 of 3 times"},{"Start":"04:09.740 ","End":"04:19.440","Text":"u_1^2 plus 1/2 of m_2u_2^2."},{"Start":"04:19.440 ","End":"04:22.985","Text":"From here I have my 3 unknown variables,"},{"Start":"04:22.985 ","End":"04:28.005","Text":"u_1, m_2 and u_2."},{"Start":"04:28.005 ","End":"04:32.670","Text":"They appear throughout, m_2, u_2, u_1."},{"Start":"04:32.670 ","End":"04:35.670","Text":"We have u_1 here, m_2 and u_2."},{"Start":"04:35.670 ","End":"04:39.230","Text":"Using some algebra, I can solve the problem."},{"Start":"04:39.230 ","End":"04:42.785","Text":"I\u0027m not going to waste your time in the algebra because we\u0027re here to learn physics."},{"Start":"04:42.785 ","End":"04:46.990","Text":"But I encourage you to do the math on your own just to get some practice,"},{"Start":"04:46.990 ","End":"04:56.820","Text":"and your solutions are as follows, m_2=0.223 kilograms."},{"Start":"04:56.820 ","End":"05:03.890","Text":"u_2, that is the final velocity or the post collision velocity of the second object,"},{"Start":"05:03.890 ","End":"05:12.250","Text":"the second ball is 29.5 meters per second, and u_1,"},{"Start":"05:12.250 ","End":"05:22.790","Text":"the velocity of our first ball after the collision is 1.55 meters per second."},{"Start":"05:22.790 ","End":"05:28.300","Text":"Just to summarize, using 3 equations we\u0027re able to solve for our 3 variables,"},{"Start":"05:28.300 ","End":"05:31.775","Text":"and that is the basic idea with elastic collisions."},{"Start":"05:31.775 ","End":"05:34.740","Text":"Now let\u0027s move on to our next subject."}],"ID":9344},{"Watched":false,"Name":"Head on Elastic Collisions","Duration":"5m 8s","ChapterTopicVideoID":9072,"CourseChapterTopicPlaylistID":5394,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.020 ","End":"00:04.275","Text":"In this lecture, I want to talk about a specific subject."},{"Start":"00:04.275 ","End":"00:08.955","Text":"This is one of our special cases of elastic collisions and it\u0027s a head-on collision."},{"Start":"00:08.955 ","End":"00:13.455","Text":"What a head-on collision means is that our first ball or ball on the left,"},{"Start":"00:13.455 ","End":"00:15.120","Text":"you can call this number 1,"},{"Start":"00:15.120 ","End":"00:18.450","Text":"is traveling towards our second ball, ball number 2."},{"Start":"00:18.450 ","End":"00:21.135","Text":"This is all along one straight line."},{"Start":"00:21.135 ","End":"00:22.980","Text":"Imagine that this is the x-axis,"},{"Start":"00:22.980 ","End":"00:26.085","Text":"and it\u0027s only traveling along that x-axis."},{"Start":"00:26.085 ","End":"00:28.290","Text":"It bumps into the second ball,"},{"Start":"00:28.290 ","End":"00:30.885","Text":"collides with it along that same axis,"},{"Start":"00:30.885 ","End":"00:33.810","Text":"and continues along the same axis."},{"Start":"00:33.810 ","End":"00:38.670","Text":"This is all one-dimensional along one line or one axis."},{"Start":"00:38.670 ","End":"00:40.205","Text":"To be totally clear,"},{"Start":"00:40.205 ","End":"00:46.340","Text":"this case only works when neither object deviates at all from the straight line."},{"Start":"00:46.340 ","End":"00:52.225","Text":"In this case, we can replace our energy formula with the following."},{"Start":"00:52.225 ","End":"01:00.180","Text":"v_1 plus u_1 equals v_2 plus u_2."},{"Start":"01:00.180 ","End":"01:05.240","Text":"You should put this in your formula sheet because it can save you a lot of time,"},{"Start":"01:05.240 ","End":"01:07.205","Text":"a lot of busy work on a test,"},{"Start":"01:07.205 ","End":"01:09.455","Text":"or in solving an exercise."},{"Start":"01:09.455 ","End":"01:12.995","Text":"One more thing to note is that we don\u0027t need to worry about vectors,"},{"Start":"01:12.995 ","End":"01:14.690","Text":"not only because we\u0027re talking about energy,"},{"Start":"01:14.690 ","End":"01:17.500","Text":"but also because this is all on one axis,"},{"Start":"01:17.500 ","End":"01:22.070","Text":"This is only the case when we\u0027re talking about one axis, one-dimensional motion."},{"Start":"01:22.070 ","End":"01:24.980","Text":"The second ball or the first ball can\u0027t deviate with"},{"Start":"01:24.980 ","End":"01:28.890","Text":"any degrees up or downwards in this case."},{"Start":"01:29.800 ","End":"01:32.555","Text":"For this formula to be true,"},{"Start":"01:32.555 ","End":"01:34.895","Text":"we don\u0027t need the masses to be equal,"},{"Start":"01:34.895 ","End":"01:37.580","Text":"we\u0027re only talking about v_1 and u_1,"},{"Start":"01:37.580 ","End":"01:39.155","Text":"and v_2 and u_2."},{"Start":"01:39.155 ","End":"01:43.595","Text":"We\u0027ll talk about the situation when the masses are equal in a second."},{"Start":"01:43.595 ","End":"01:47.285","Text":"You can also write this formula another way,"},{"Start":"01:47.285 ","End":"01:52.895","Text":"which is basically taking v_2 over to the left side and u_1 over to the right side."},{"Start":"01:52.895 ","End":"01:54.185","Text":"It looks like this,"},{"Start":"01:54.185 ","End":"02:01.680","Text":"v_1 minus v_2 equals negative u_1 minus u_2."},{"Start":"02:01.680 ","End":"02:04.145","Text":"What this describes to us is"},{"Start":"02:04.145 ","End":"02:09.065","Text":"that the relative velocity switches directions after the collision."},{"Start":"02:09.065 ","End":"02:13.520","Text":"I want to show you how we get to this equation mathematically"},{"Start":"02:13.520 ","End":"02:18.290","Text":"by using our conservation of momentum and conservation of energy equations."},{"Start":"02:18.290 ","End":"02:20.554","Text":"It doesn\u0027t take very long to demonstrate."},{"Start":"02:20.554 ","End":"02:22.775","Text":"But if you are really in a rush,"},{"Start":"02:22.775 ","End":"02:25.440","Text":"you can feel free to move forward."},{"Start":"02:26.590 ","End":"02:30.230","Text":"Let\u0027s start by taking both equations and"},{"Start":"02:30.230 ","End":"02:33.710","Text":"moving everything having to do with our first object to the left side,"},{"Start":"02:33.710 ","End":"02:36.800","Text":"and everything having to do with the second object to the right side."},{"Start":"02:36.800 ","End":"02:40.700","Text":"Our conservation of momentum equation will look like this,"},{"Start":"02:40.700 ","End":"02:47.385","Text":"m_1 v_1 minus m_1 u_1"},{"Start":"02:47.385 ","End":"02:55.375","Text":"equals m_2 u_2 minus m_2 v_2."},{"Start":"02:55.375 ","End":"03:00.005","Text":"The reason that I\u0027m not using vectors here is we\u0027re talking all on one axis,"},{"Start":"03:00.005 ","End":"03:02.320","Text":"so we don\u0027t need to worry about direction."},{"Start":"03:02.320 ","End":"03:05.765","Text":"We can move on to our conservation of energy formula"},{"Start":"03:05.765 ","End":"03:09.520","Text":"and remove our one-halves and read it as follows,"},{"Start":"03:09.520 ","End":"03:18.365","Text":"m_1 v_1^2 minus m_1 u_1^2,"},{"Start":"03:18.365 ","End":"03:28.620","Text":"equals m_2 u_2^2 minus m_2 v_2^2."},{"Start":"03:28.620 ","End":"03:32.075","Text":"Now we can take m_1 on our left side of each equation,"},{"Start":"03:32.075 ","End":"03:34.730","Text":"and m_2 on the right side of each equation,"},{"Start":"03:34.730 ","End":"03:37.070","Text":"and move it outside of some parentheses."},{"Start":"03:37.070 ","End":"03:39.210","Text":"It looked like this."},{"Start":"03:39.290 ","End":"03:43.325","Text":"Now the next step is to take our second equation,"},{"Start":"03:43.325 ","End":"03:45.290","Text":"the conservation of energy equation,"},{"Start":"03:45.290 ","End":"03:50.330","Text":"and make it our numerator and divide it by our conservation of momentum equation."},{"Start":"03:50.330 ","End":"03:54.379","Text":"The first thing is all of our m\u0027s will fall out. They\u0027ll simplify."},{"Start":"03:54.379 ","End":"03:57.850","Text":"What we\u0027re left with is our conservation of energy,"},{"Start":"03:57.850 ","End":"04:03.510","Text":"v^2 minus u^2 or u^2 minus v^2 over r, conservation of momentum."},{"Start":"04:03.510 ","End":"04:14.190","Text":"We can rewrite v_1^2 minus u_1^2 as v_1 minus u_1 times v_1 plus u_1."},{"Start":"04:14.190 ","End":"04:19.785","Text":"That\u0027ll be over v_1 minus u_1,"},{"Start":"04:19.785 ","End":"04:27.689","Text":"and that equals u_2 minus v_2 times"},{"Start":"04:27.689 ","End":"04:32.070","Text":"u_2 plus"},{"Start":"04:32.070 ","End":"04:38.310","Text":"v_2 over u_2 minus v_2."},{"Start":"04:38.310 ","End":"04:43.650","Text":"Again, this is the same as u_2^2 minus v_2 squared."},{"Start":"04:43.650 ","End":"04:47.900","Text":"You might already be able to see that these terms will simplify."},{"Start":"04:47.900 ","End":"04:54.870","Text":"What you\u0027re left with is v_1 plus u_1 equals v_2 plus u_2."},{"Start":"04:55.460 ","End":"04:59.120","Text":"That explains the math behind what\u0027s going on here."},{"Start":"04:59.120 ","End":"05:02.420","Text":"That ends our lecture on head-on elastic collisions."},{"Start":"05:02.420 ","End":"05:04.820","Text":"From here I suggest that you try to"},{"Start":"05:04.820 ","End":"05:08.730","Text":"work out the equation on your own and see the result that you get."}],"ID":9345},{"Watched":false,"Name":"Head on Elastic Collisions with Equal Mass","Duration":"2m 47s","ChapterTopicVideoID":9073,"CourseChapterTopicPlaylistID":5394,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.950","Text":"In the last lecture,"},{"Start":"00:01.950 ","End":"00:04.710","Text":"we talked about head-on elastic collisions."},{"Start":"00:04.710 ","End":"00:08.835","Text":"In this lecture, I want to talk about an even more specific example,"},{"Start":"00:08.835 ","End":"00:12.165","Text":"a head-on elastic collision with equal mass."},{"Start":"00:12.165 ","End":"00:15.540","Text":"That means that the mass of Ball 1 equals the mass of"},{"Start":"00:15.540 ","End":"00:19.500","Text":"Ball 2 and they have a head-on collision like we described before."},{"Start":"00:19.500 ","End":"00:22.575","Text":"We already know that instead of the energy equation,"},{"Start":"00:22.575 ","End":"00:25.260","Text":"we can use our equation in the red box."},{"Start":"00:25.260 ","End":"00:28.230","Text":"As a result, if our masses are equal,"},{"Start":"00:28.230 ","End":"00:32.085","Text":"so if m_1 equals m_2,"},{"Start":"00:32.085 ","End":"00:34.260","Text":"then that means that our conservation of"},{"Start":"00:34.260 ","End":"00:37.365","Text":"momentum equation doesn\u0027t need to use mass at all."},{"Start":"00:37.365 ","End":"00:41.885","Text":"What you\u0027ll see as a result is that the first ball,"},{"Start":"00:41.885 ","End":"00:45.590","Text":"Ball 1, will move until it collides with Ball 2."},{"Start":"00:45.590 ","End":"00:47.750","Text":"Ball 1 will stop in place,"},{"Start":"00:47.750 ","End":"00:50.960","Text":"and Ball 2 will move in the exact same direction that"},{"Start":"00:50.960 ","End":"00:55.050","Text":"Ball 1 was moving with the exact same velocity."},{"Start":"00:56.360 ","End":"00:59.505","Text":"Let\u0027s see how this works mathematically."},{"Start":"00:59.505 ","End":"01:02.630","Text":"First of all, we can simplify our conservation"},{"Start":"01:02.630 ","End":"01:05.615","Text":"of momentum equation because every mass is the same"},{"Start":"01:05.615 ","End":"01:14.370","Text":"to v_1 plus v_2 equals u_1 plus u_2."},{"Start":"01:14.370 ","End":"01:17.735","Text":"We can also, instead of our conservation of energy equation,"},{"Start":"01:17.735 ","End":"01:19.130","Text":"use this equation,"},{"Start":"01:19.130 ","End":"01:26.655","Text":"v_1 plus u_1 equals v_2 plus u_2."},{"Start":"01:26.655 ","End":"01:28.830","Text":"As the problem is set up,"},{"Start":"01:28.830 ","End":"01:32.920","Text":"v_2 equals 0, our second object starts at rest."},{"Start":"01:32.920 ","End":"01:37.070","Text":"In that case, we get the following equations v_1 equals u_1"},{"Start":"01:37.070 ","End":"01:42.110","Text":"plus u_2 and v_1 plus u_1 equals u_2."},{"Start":"01:42.110 ","End":"01:46.105","Text":"What we end up getting is if we add these 2 equations together,"},{"Start":"01:46.105 ","End":"01:57.130","Text":"2 times v_1 plus u_1 equals u_1 plus 2 times u_2."},{"Start":"01:57.130 ","End":"01:59.525","Text":"As you might be able to see already,"},{"Start":"01:59.525 ","End":"02:03.410","Text":"the u_1 is dropped out and what we\u0027re left with is 2 v and we"},{"Start":"02:03.410 ","End":"02:08.115","Text":"can drop the 2s as well, v_1 equals u_2."},{"Start":"02:08.115 ","End":"02:10.760","Text":"In that sense, the velocity of mass,"},{"Start":"02:10.760 ","End":"02:13.430","Text":"1 of the first object before it hits"},{"Start":"02:13.430 ","End":"02:18.820","Text":"the second object is the same as the velocity of the second object after it is hit."},{"Start":"02:18.820 ","End":"02:22.680","Text":"Our end result is that u_2 equals"},{"Start":"02:22.680 ","End":"02:26.405","Text":"v_1 and if you were to plug this back into our first equation,"},{"Start":"02:26.405 ","End":"02:32.225","Text":"you\u0027d also find that u_1 equals 0 just like v_2 and this is really important."},{"Start":"02:32.225 ","End":"02:37.985","Text":"What it confirms for us is that the first ball will hit the second ball stop in place,"},{"Start":"02:37.985 ","End":"02:39.620","Text":"and the second ball will continue with"},{"Start":"02:39.620 ","End":"02:42.905","Text":"the exact same velocity in the exact same direction."},{"Start":"02:42.905 ","End":"02:44.495","Text":"It\u0027s important to remember this."},{"Start":"02:44.495 ","End":"02:47.610","Text":"It can save you a lot of time in the future."}],"ID":9346},{"Watched":false,"Name":"Non-Head On Collisions with Equal Mass","Duration":"4m 12s","ChapterTopicVideoID":9074,"CourseChapterTopicPlaylistID":5394,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.350 ","End":"00:04.935","Text":"This is the last example with these 2 balls."},{"Start":"00:04.935 ","End":"00:07.350","Text":"I\u0027m sure at this point you\u0027re a little sick of them,"},{"Start":"00:07.350 ","End":"00:10.140","Text":"but this is really useful stuff and I assure you it\u0027s"},{"Start":"00:10.140 ","End":"00:13.455","Text":"worthwhile to know what\u0027s going on by heart if necessary."},{"Start":"00:13.455 ","End":"00:16.290","Text":"This last example is another special case."},{"Start":"00:16.290 ","End":"00:20.310","Text":"It is a non-head-on elastic collision with equal mass."},{"Start":"00:20.310 ","End":"00:22.080","Text":"That means that these 2 balls,"},{"Start":"00:22.080 ","End":"00:24.570","Text":"Ball 1 and Ball 2 have equal mass,"},{"Start":"00:24.570 ","End":"00:28.585","Text":"and Ball 2 starts at rest but Ball 1 hits Ball 2"},{"Start":"00:28.585 ","End":"00:34.085","Text":"not head-on rather maybe a little bit below are a little bit above the center."},{"Start":"00:34.085 ","End":"00:36.305","Text":"When this is the case,"},{"Start":"00:36.305 ","End":"00:39.360","Text":"what happens is after the collision,"},{"Start":"00:39.370 ","End":"00:43.540","Text":"each ball will go in a separate direction."},{"Start":"00:43.540 ","End":"00:49.655","Text":"In this situation, each ball is at a different angle with respect to the x-axis."},{"Start":"00:49.655 ","End":"00:51.170","Text":"If like I said,"},{"Start":"00:51.170 ","End":"00:55.970","Text":"Ball 1 and Ball 2 are of equal mass and Ball 2 starts at rest,"},{"Start":"00:55.970 ","End":"01:00.335","Text":"then we know that the angle between the 2 will be 90 degrees."},{"Start":"01:00.335 ","End":"01:05.930","Text":"That means that if we know the angle between Ball 2 and the x-axis to be say 15"},{"Start":"01:05.930 ","End":"01:09.455","Text":"degrees then we can also determine"},{"Start":"01:09.455 ","End":"01:14.290","Text":"that the angle between Ball 1 and the x-axis will complete that,"},{"Start":"01:14.290 ","End":"01:15.560","Text":"will make 90 degrees,"},{"Start":"01:15.560 ","End":"01:18.960","Text":"so that\u0027s 75 degrees in this case."},{"Start":"01:19.270 ","End":"01:22.790","Text":"From here, I want to provide a mathematical proof."},{"Start":"01:22.790 ","End":"01:24.860","Text":"Again, like in the last example,"},{"Start":"01:24.860 ","End":"01:27.185","Text":"if you feel like you\u0027re pressed for time,"},{"Start":"01:27.185 ","End":"01:29.330","Text":"this is not the most essential thing,"},{"Start":"01:29.330 ","End":"01:31.190","Text":"but if you\u0027re interested in why this works,"},{"Start":"01:31.190 ","End":"01:33.260","Text":"this is still critical."},{"Start":"01:33.260 ","End":"01:35.435","Text":"Please bear with me."},{"Start":"01:35.435 ","End":"01:37.565","Text":"If you really, really don\u0027t want to watch,"},{"Start":"01:37.565 ","End":"01:39.770","Text":"of course, I can\u0027t stop you from moving on."},{"Start":"01:39.770 ","End":"01:42.170","Text":"Remember we\u0027re talking about a system where"},{"Start":"01:42.170 ","End":"01:48.635","Text":"m_1=m_2 and we also have a system where v_2=0."},{"Start":"01:48.635 ","End":"01:52.160","Text":"That means that this goes to 0 and this goes to 0,"},{"Start":"01:52.160 ","End":"01:53.690","Text":"and if our masses are equal,"},{"Start":"01:53.690 ","End":"01:58.365","Text":"we can take out our m_1\u0027s and m_2\u0027s on both equations."},{"Start":"01:58.365 ","End":"02:02.615","Text":"We can also, in our energy equation, forget the halves."},{"Start":"02:02.615 ","End":"02:04.100","Text":"We don\u0027t have to deal with them."},{"Start":"02:04.100 ","End":"02:08.640","Text":"What we\u0027re left with is in conservation of momentum,"},{"Start":"02:09.820 ","End":"02:14.910","Text":"v_1=u_1 plus u_2,"},{"Start":"02:15.560 ","End":"02:18.300","Text":"and in terms of energy,"},{"Start":"02:18.300 ","End":"02:23.910","Text":"v_1^2=u_1^2"},{"Start":"02:23.910 ","End":"02:28.500","Text":"plus u_2^2."},{"Start":"02:28.500 ","End":"02:31.055","Text":"At this point, I\u0027m going to take my equation on top,"},{"Start":"02:31.055 ","End":"02:32.960","Text":"my conservation of momentum equation,"},{"Start":"02:32.960 ","End":"02:40.700","Text":"and square both sides so I get v_1^2 on the left and on the right, u_1 plus u_2^2."},{"Start":"02:40.700 ","End":"02:43.815","Text":"My result is v,"},{"Start":"02:43.815 ","End":"02:44.940","Text":"here going down here,"},{"Start":"02:44.940 ","End":"02:52.920","Text":"v_1^2=u_1^2"},{"Start":"02:53.390 ","End":"02:56.860","Text":"plus 2u_1.u_2,"},{"Start":"02:56.860 ","End":"03:03.335","Text":"this is a scalar multiplication because these are vectors, plus u_2^2."},{"Start":"03:03.335 ","End":"03:09.815","Text":"Now if I subtract the energy equation from my new momentum equation squared,"},{"Start":"03:09.815 ","End":"03:15.530","Text":"the result is that my V_1^2 is dropout so 0 is my result on the left side,"},{"Start":"03:15.530 ","End":"03:21.750","Text":"and on the right side, u_1^2 and u_2^2 both zero out."},{"Start":"03:21.750 ","End":"03:25.750","Text":"What I\u0027m left with is 0=2u_1,"},{"Start":"03:26.150 ","End":"03:30.450","Text":"scalar multiplication by u_2,"},{"Start":"03:30.450 ","End":"03:34.625","Text":"and I know that if the result of a scalar multiplication is 0,"},{"Start":"03:34.625 ","End":"03:39.630","Text":"then the angle between these 2 vectors has to be 90 degrees."},{"Start":"03:39.880 ","End":"03:44.930","Text":"This solution could also work for 180-degree collision"},{"Start":"03:44.930 ","End":"03:47.180","Text":"but we know that that\u0027s not the case here"},{"Start":"03:47.180 ","End":"03:49.700","Text":"because we\u0027re talking about a non-head-on collision,"},{"Start":"03:49.700 ","End":"03:53.215","Text":"and 180-degree collision would be a head-on collision."},{"Start":"03:53.215 ","End":"03:59.475","Text":"This proves that the angle between the 2 balls after the collision is 90 degrees."},{"Start":"03:59.475 ","End":"04:01.670","Text":"Just to repeat that 1 more time,"},{"Start":"04:01.670 ","End":"04:06.500","Text":"the angle between 2 objects after a non-head-on elastic collision,"},{"Start":"04:06.500 ","End":"04:10.475","Text":"if both objects have equal mass and 1 object starts at rest,"},{"Start":"04:10.475 ","End":"04:12.930","Text":"will be 90 degrees."}],"ID":9347},{"Watched":false,"Name":"Why Recoil and Plastic Collisions Don\u0026#39;t Conserve Energy","Duration":"5m 38s","ChapterTopicVideoID":9075,"CourseChapterTopicPlaylistID":5394,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.815","Text":"Hello, in this lecture,"},{"Start":"00:01.815 ","End":"00:06.405","Text":"I want to show you why recoil and plastic collisions don\u0027t conserve energy."},{"Start":"00:06.405 ","End":"00:09.510","Text":"Essentially, I\u0027ll use a mathematical proof to show that"},{"Start":"00:09.510 ","End":"00:13.095","Text":"these 2 equations can\u0027t coexist in terms of a collision."},{"Start":"00:13.095 ","End":"00:15.600","Text":"For these 2 equations to work together,"},{"Start":"00:15.600 ","End":"00:20.670","Text":"you need V_1 equal V_2 and if"},{"Start":"00:20.670 ","End":"00:23.430","Text":"V_1 equals V_2 that means the velocity"},{"Start":"00:23.430 ","End":"00:26.340","Text":"of our 2 objects are the same before the collision."},{"Start":"00:26.340 ","End":"00:29.835","Text":"If that\u0027s the case, then physically they should never collide."},{"Start":"00:29.835 ","End":"00:33.630","Text":"Ultimately it means that they don\u0027t make sense and these 2 equations can"},{"Start":"00:33.630 ","End":"00:37.140","Text":"both be true if we have a collision."},{"Start":"00:37.140 ","End":"00:41.030","Text":"We will do this is by finding the value of"},{"Start":"00:41.030 ","End":"00:45.980","Text":"u and then plugging it into our conservation of energy equation."},{"Start":"00:45.980 ","End":"00:50.390","Text":"Just to note, these equations are specific to plastic collisions."},{"Start":"00:50.390 ","End":"00:53.225","Text":"You can see we only have 1 u on the right side."},{"Start":"00:53.225 ","End":"00:55.384","Text":"We can prove the same thing with recoil,"},{"Start":"00:55.384 ","End":"00:57.520","Text":"but we\u0027ll do it with plastic in this case."},{"Start":"00:57.520 ","End":"01:00.965","Text":"The first step is we\u0027ll find the value of u."},{"Start":"01:00.965 ","End":"01:08.959","Text":"We do that by dividing both sides by m_1 plus m_2 and our answer is that u equals"},{"Start":"01:08.959 ","End":"01:15.410","Text":"m_1 v_1 plus m_2 v_2"},{"Start":"01:15.410 ","End":"01:20.225","Text":"over m_1 plus m_2."},{"Start":"01:20.225 ","End":"01:24.325","Text":"Now we can take this and plug it into our conservation of energy."},{"Start":"01:24.325 ","End":"01:26.380","Text":"We can get rid of these one-halves."},{"Start":"01:26.380 ","End":"01:30.870","Text":"Our answer is m_1 v_1"},{"Start":"01:30.870 ","End":"01:37.800","Text":"squared plus m_2 v_2 squared equals,"},{"Start":"01:37.800 ","End":"01:41.935","Text":"and you\u0027ll notice m_1 plus m_2 is also the denominator for u."},{"Start":"01:41.935 ","End":"01:43.780","Text":"This will reduce a little bit,"},{"Start":"01:43.780 ","End":"01:45.670","Text":"but because we\u0027re squaring u,"},{"Start":"01:45.670 ","End":"01:54.875","Text":"our answer is 1 over m_1 plus m_2 times in parentheses or brackets,"},{"Start":"01:54.875 ","End":"02:05.235","Text":"m_1 v_1 plus m_2 v_2 squared."},{"Start":"02:05.235 ","End":"02:12.560","Text":"The next step is to multiply both sides by m_1 plus m_2 to bring this over to the left."},{"Start":"02:12.560 ","End":"02:15.460","Text":"Also, open up our brackets on the right side."},{"Start":"02:15.460 ","End":"02:17.700","Text":"Our solution is as follows."},{"Start":"02:17.700 ","End":"02:27.590","Text":"m_1 v_1 squared times m_1 plus m_2"},{"Start":"02:27.590 ","End":"02:32.075","Text":"plus m_2 v_2 squared"},{"Start":"02:32.075 ","End":"02:37.950","Text":"times m_1 plus m_2 equals."},{"Start":"02:37.950 ","End":"02:39.455","Text":"When we open up these brackets,"},{"Start":"02:39.455 ","End":"02:46.880","Text":"we get m_1 squared v_1 squared minus"},{"Start":"02:46.880 ","End":"02:51.870","Text":"2 m_1 v_1 scalar"},{"Start":"02:51.870 ","End":"02:58.840","Text":"multiplication dot m_2 v_2"},{"Start":"03:01.280 ","End":"03:07.165","Text":"plus m_2 squared v_2 squared."},{"Start":"03:07.165 ","End":"03:10.280","Text":"This equation is clearly long and unwieldy,"},{"Start":"03:10.280 ","End":"03:11.974","Text":"so we need to start simplifying."},{"Start":"03:11.974 ","End":"03:14.855","Text":"The first thing we can do is see that m_1 times"},{"Start":"03:14.855 ","End":"03:18.960","Text":"m_1 v_1 squared equals m_1 squared v_1 squared."},{"Start":"03:18.960 ","End":"03:21.635","Text":"This will drop out and our m_1 will drop out."},{"Start":"03:21.635 ","End":"03:24.755","Text":"Secondly, the same is true for m_2,"},{"Start":"03:24.755 ","End":"03:31.115","Text":"m_2 times m_2 v_2 squared equals m_2 squared v_2 squared. This will drop out."},{"Start":"03:31.115 ","End":"03:36.605","Text":"Now you\u0027ll notice that every term we have left is in terms of m_1 times m_2."},{"Start":"03:36.605 ","End":"03:38.030","Text":"We can take those both out."},{"Start":"03:38.030 ","End":"03:40.760","Text":"Here we have m_1 times m_2,"},{"Start":"03:40.760 ","End":"03:47.100","Text":"here we have m_1 times m_2 and here we have m_1 times m_2."},{"Start":"03:47.100 ","End":"03:52.114","Text":"Essentially all of our masses have fallen out and we\u0027re left with only velocity."},{"Start":"03:52.114 ","End":"03:55.865","Text":"If we take all of our velocity and bring it to the left side,"},{"Start":"03:55.865 ","End":"04:04.590","Text":"we find v_1 squared from here minus 2v_1.v_2,"},{"Start":"04:04.590 ","End":"04:07.505","Text":"this is a vector and that\u0027s from over here,"},{"Start":"04:07.505 ","End":"04:13.865","Text":"minus 2v_1.v_2 plus v_2 squared."},{"Start":"04:13.865 ","End":"04:16.550","Text":"That\u0027s from here, equals 0."},{"Start":"04:16.550 ","End":"04:22.035","Text":"Now you may already see that this equation can be rewritten as"},{"Start":"04:22.035 ","End":"04:29.955","Text":"v_1 minus v_2 squared equals 0."},{"Start":"04:29.955 ","End":"04:33.655","Text":"If v_1 minus v_2 squared is equal 0,"},{"Start":"04:33.655 ","End":"04:37.505","Text":"that can only be true when v_1 minus v_2 equals 0."},{"Start":"04:37.505 ","End":"04:43.650","Text":"We can say therefore v_1 minus v_2 equals 0,"},{"Start":"04:43.650 ","End":"04:49.615","Text":"and therefore v_1 equals v_2."},{"Start":"04:49.615 ","End":"04:53.255","Text":"This is our term from above here, v_1 equals v_2."},{"Start":"04:53.255 ","End":"04:56.230","Text":"Just to reiterate, that means that the velocity before"},{"Start":"04:56.230 ","End":"04:59.210","Text":"the collision of our 2 objects has to be the same."},{"Start":"04:59.210 ","End":"05:01.315","Text":"If that\u0027s true, they should never collide,"},{"Start":"05:01.315 ","End":"05:04.150","Text":"meaning that this can\u0027t be true in the case of a collision."},{"Start":"05:04.150 ","End":"05:06.745","Text":"For these 2 equations to work together,"},{"Start":"05:06.745 ","End":"05:10.520","Text":"for there to be both a conservation of momentum and a conservation of energy."},{"Start":"05:10.520 ","End":"05:12.035","Text":"We can\u0027t have a collision."},{"Start":"05:12.035 ","End":"05:16.070","Text":"Therefore, recoil and plastic collisions don\u0027t conserve energy."},{"Start":"05:16.070 ","End":"05:17.690","Text":"One last note is that,"},{"Start":"05:17.690 ","End":"05:18.845","Text":"well, this is straightforward."},{"Start":"05:18.845 ","End":"05:20.630","Text":"It can get a little complicated."},{"Start":"05:20.630 ","End":"05:24.620","Text":"I would encourage you to return to the beginning of the video and now"},{"Start":"05:24.620 ","End":"05:29.090","Text":"look at any of the work I\u0027ve done here and try to solve the equations yourself."},{"Start":"05:29.090 ","End":"05:32.240","Text":"It\u0027s a really good way to learn the material quickly."},{"Start":"05:32.240 ","End":"05:35.310","Text":"With that, I invite you to try your hand at solving this,"},{"Start":"05:35.310 ","End":"05:38.610","Text":"and when you finished, move on to the next lecture."}],"ID":9348},{"Watched":false,"Name":"Summary of Types of Collisions","Duration":"4m 35s","ChapterTopicVideoID":9076,"CourseChapterTopicPlaylistID":5394,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.050 ","End":"00:05.055","Text":"I want to do a quick summary of the types of collisions and the things we\u0027ve discussed."},{"Start":"00:05.055 ","End":"00:09.990","Text":"First and foremost is the difference between inelastic and elastic collisions."},{"Start":"00:09.990 ","End":"00:12.786","Text":"If you can do this, you can really solve any problem."},{"Start":"00:12.786 ","End":"00:18.278","Text":"Remember, inelastic collisions are those where we only have conservation of momentum."},{"Start":"00:18.278 ","End":"00:21.750","Text":"In that case, we can only use the conservation of momentum formula."},{"Start":"00:21.750 ","End":"00:23.595","Text":"With elastic collisions,"},{"Start":"00:23.595 ","End":"00:25.740","Text":"there\u0027s also a conservation of energy,"},{"Start":"00:25.740 ","End":"00:28.230","Text":"so we can use the conservation of momentum formula"},{"Start":"00:28.230 ","End":"00:30.990","Text":"and the formula for the conservation of energy."},{"Start":"00:30.990 ","End":"00:33.855","Text":"If we\u0027re not told anything about a problem,"},{"Start":"00:33.855 ","End":"00:38.835","Text":"we can\u0027t assume it to be elastic and therefore we have to treat it as inelastic."},{"Start":"00:38.835 ","End":"00:44.405","Text":"Again, if you know the difference between these 2 types of collisions and can use them,"},{"Start":"00:44.405 ","End":"00:46.625","Text":"then you can solve any of these problems."},{"Start":"00:46.625 ","End":"00:49.940","Text":"The rest of what we talked about are specific cases that give us"},{"Start":"00:49.940 ","End":"00:54.630","Text":"shortcuts and deeper understanding of what\u0027s going on in a collision."},{"Start":"00:55.000 ","End":"00:58.460","Text":"Let\u0027s delve into these specific cases."},{"Start":"00:58.460 ","End":"01:00.350","Text":"First, we add plastic collisions."},{"Start":"01:00.350 ","End":"01:03.230","Text":"That\u0027s when the objects move together after the collision."},{"Start":"01:03.230 ","End":"01:05.930","Text":"The special thing we can do is change"},{"Start":"01:05.930 ","End":"01:09.200","Text":"our conservation of momentum equations that it\u0027s m_1,"},{"Start":"01:09.200 ","End":"01:12.300","Text":"v_1 plus m_2,"},{"Start":"01:12.300 ","End":"01:20.820","Text":"v_2 equals m_1 plus m_2 times 1 velocity u."},{"Start":"01:20.820 ","End":"01:22.925","Text":"Because the objects are traveling together,"},{"Start":"01:22.925 ","End":"01:25.355","Text":"they have 1 velocity after the collision."},{"Start":"01:25.355 ","End":"01:29.240","Text":"Some examples of this were a bullet or arrow hitting a tree trunk and getting"},{"Start":"01:29.240 ","End":"01:33.845","Text":"stuck or a 2 balls that hit each other and stick together afterwards."},{"Start":"01:33.845 ","End":"01:37.440","Text":"Similar to plastic collisions are recoil collisions."},{"Start":"01:37.440 ","End":"01:38.900","Text":"In some ways, they\u0027re the opposite."},{"Start":"01:38.900 ","End":"01:41.510","Text":"Recoil collisions are when 2 objects move together"},{"Start":"01:41.510 ","End":"01:44.495","Text":"before a collision and separate after the collision."},{"Start":"01:44.495 ","End":"01:49.100","Text":"Just like we had u instead of u_1 and u_2 for plastic collisions,"},{"Start":"01:49.100 ","End":"01:53.270","Text":"for recoil collisions, we have v instead of v_1 and v_2."},{"Start":"01:53.270 ","End":"01:59.090","Text":"We can change our conservation of momentum equation to be m_1"},{"Start":"01:59.090 ","End":"02:05.730","Text":"plus m_2 times v equals m_1,"},{"Start":"02:05.730 ","End":"02:10.740","Text":"u_1 plus m_2, u_2."},{"Start":"02:10.740 ","End":"02:16.925","Text":"Again, examples of recoil are a bullet being shot from a gun or explosions."},{"Start":"02:16.925 ","End":"02:21.650","Text":"Now we can move on to some special cases dealing with elastic collisions."},{"Start":"02:21.650 ","End":"02:23.550","Text":"First, we add head-on collisions,"},{"Start":"02:23.550 ","End":"02:26.675","Text":"and what we said here was that in the case of a head-on collision,"},{"Start":"02:26.675 ","End":"02:29.315","Text":"regardless of the masses of the 2 objects,"},{"Start":"02:29.315 ","End":"02:32.390","Text":"we can simplify our conservation of energy formula,"},{"Start":"02:32.390 ","End":"02:39.915","Text":"and we can simplify it to be v_1 plus u_1 equals v_2 plus u_2."},{"Start":"02:39.915 ","End":"02:42.920","Text":"Once again, this is regardless of mass."},{"Start":"02:42.920 ","End":"02:45.790","Text":"Now we had a second special case with head-on collisions,"},{"Start":"02:45.790 ","End":"02:49.220","Text":"and this was more specific when we have a head-on collision between"},{"Start":"02:49.220 ","End":"02:53.285","Text":"2 objects with equal masses when 1 object starts at rest."},{"Start":"02:53.285 ","End":"02:57.795","Text":"Ball 1 starts going towards ball 2,"},{"Start":"02:57.795 ","End":"02:59.325","Text":"ball 2 is at rest,"},{"Start":"02:59.325 ","End":"03:01.020","Text":"and after the collision,"},{"Start":"03:01.020 ","End":"03:05.150","Text":"ball 1 is at rest and ball 2 is"},{"Start":"03:05.150 ","End":"03:10.006","Text":"moving with the same velocity that ball 1 had before in the exact same direction."},{"Start":"03:10.006 ","End":"03:14.570","Text":"Remember the takeaway here is if we have 2 objects with equal mass,"},{"Start":"03:14.570 ","End":"03:17.855","Text":"1 object is at rest and they have a head-on collision,"},{"Start":"03:17.855 ","End":"03:20.660","Text":"then the second object will continue with"},{"Start":"03:20.660 ","End":"03:24.575","Text":"the same velocity that the first object had on the same axis,"},{"Start":"03:24.575 ","End":"03:27.910","Text":"and the first object will come to a complete stop."},{"Start":"03:27.910 ","End":"03:32.615","Text":"Lastly, we had a special case where we had collisions that were not head-on,"},{"Start":"03:32.615 ","End":"03:37.580","Text":"but were between 2 objects of equal mass when 1 object starts at rest."},{"Start":"03:37.580 ","End":"03:41.675","Text":"ball 1 hits ball 2, not head-on,"},{"Start":"03:41.675 ","End":"03:43.700","Text":"a little above or below its center,"},{"Start":"03:43.700 ","End":"03:46.565","Text":"and the 2 balls go in opposite directions,"},{"Start":"03:46.565 ","End":"03:47.630","Text":"ball 2 to up to here,"},{"Start":"03:47.630 ","End":"03:49.340","Text":"ball 1 down to here."},{"Start":"03:49.340 ","End":"03:56.040","Text":"The takeaway here was that the angle between the 2 objects had to equal 90 degrees."},{"Start":"03:56.040 ","End":"03:58.640","Text":"Remember, if we have collisions that are not"},{"Start":"03:58.640 ","End":"04:03.140","Text":"head-on between 2 objects of equal mass when 1 object starts at rest,"},{"Start":"04:03.140 ","End":"04:08.905","Text":"then the angle between the 2 objects after the collision is 90 degrees."},{"Start":"04:08.905 ","End":"04:11.870","Text":"Once again, these are some special cases,"},{"Start":"04:11.870 ","End":"04:15.110","Text":"and knowing the special cases can save us some time and work."},{"Start":"04:15.110 ","End":"04:18.320","Text":"However, the most important thing here is to differentiate between"},{"Start":"04:18.320 ","End":"04:21.470","Text":"elastic and inelastic collisions and know how to"},{"Start":"04:21.470 ","End":"04:23.990","Text":"use your conservation of momentum equation in"},{"Start":"04:23.990 ","End":"04:28.415","Text":"both cases and your conservation of energy equation in elastic cases."},{"Start":"04:28.415 ","End":"04:31.190","Text":"Usually, they\u0027ll tell you if you\u0027re dealing with an elastic case,"},{"Start":"04:31.190 ","End":"04:35.700","Text":"and with that, we\u0027ve come to the end of our lecture on types of collisions."}],"ID":9349}],"Thumbnail":null,"ID":5394},{"Name":"Conservation of Momentum During Short Timed Collisions","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Conservation of Momentum During Short Collisions","Duration":"8m 41s","ChapterTopicVideoID":9077,"CourseChapterTopicPlaylistID":5395,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.505","Text":"Hello. In this lecture,"},{"Start":"00:02.505 ","End":"00:05.685","Text":"I want to show you a few specific situations"},{"Start":"00:05.685 ","End":"00:08.220","Text":"where we still have conservation of momentum,"},{"Start":"00:08.220 ","End":"00:10.755","Text":"even though we have external forces."},{"Start":"00:10.755 ","End":"00:12.705","Text":"If you recall, up until this point,"},{"Start":"00:12.705 ","End":"00:20.970","Text":"we\u0027ve said that if the sum of external forces equals 0,"},{"Start":"00:20.970 ","End":"00:24.405","Text":"that means that P_T,"},{"Start":"00:24.405 ","End":"00:30.135","Text":"or the total momentum is constant and that means that we have conservation of momentum."},{"Start":"00:30.135 ","End":"00:33.690","Text":"In this lecture, I want to show you a couple of cases where we do have"},{"Start":"00:33.690 ","End":"00:37.755","Text":"external forces and yet we still have conservation of momentum."},{"Start":"00:37.755 ","End":"00:41.525","Text":"It\u0027s important to know these situations or these scenarios,"},{"Start":"00:41.525 ","End":"00:46.160","Text":"because you might get a question like this on a test in the future and it can"},{"Start":"00:46.160 ","End":"00:47.870","Text":"save you a lot of time and help you solve"},{"Start":"00:47.870 ","End":"00:50.705","Text":"the problem if you know that you can still consider the case,"},{"Start":"00:50.705 ","End":"00:53.260","Text":"one with conservation of momentum."},{"Start":"00:53.260 ","End":"00:59.675","Text":"The classic example of this is that you have 2 balls and they collide in the air."},{"Start":"00:59.675 ","End":"01:01.880","Text":"This is ball 1, this is ball 2."},{"Start":"01:01.880 ","End":"01:04.622","Text":"The first thing I want to choose my x and y-axis."},{"Start":"01:04.622 ","End":"01:06.064","Text":"Let\u0027s go like this,"},{"Start":"01:06.064 ","End":"01:11.890","Text":"x is towards the right and y is upwards."},{"Start":"01:12.920 ","End":"01:19.160","Text":"Now I need to check on each axis whether I have conservation of momentum."},{"Start":"01:19.160 ","End":"01:22.540","Text":"The way I check that is if I have external force."},{"Start":"01:22.540 ","End":"01:25.984","Text":"On the x-axis there are no external forces."},{"Start":"01:25.984 ","End":"01:31.910","Text":"Sum of F external on the x-axis equals 0."},{"Start":"01:31.910 ","End":"01:39.090","Text":"Therefore, P_x is constant."},{"Start":"01:39.090 ","End":"01:42.515","Text":"I have conservation of momentum along the x-axis."},{"Start":"01:42.515 ","End":"01:45.125","Text":"However, along the y-axis,"},{"Start":"01:45.125 ","End":"01:48.230","Text":"I have a downward force that\u0027s gravity mg."},{"Start":"01:48.230 ","End":"01:52.055","Text":"I can say that the sum of"},{"Start":"01:52.055 ","End":"02:00.110","Text":"external forces along my y-axis does not equal 0."},{"Start":"02:00.110 ","End":"02:02.075","Text":"In fact, it equals the total mass,"},{"Start":"02:02.075 ","End":"02:08.330","Text":"m_1 plus m_2 times g and"},{"Start":"02:08.330 ","End":"02:12.270","Text":"therefore P_y should not be"},{"Start":"02:12.270 ","End":"02:18.800","Text":"constant and I should not have conservation of momentum along my y-axis."},{"Start":"02:18.840 ","End":"02:23.260","Text":"What I want to demonstrate for you here is that even"},{"Start":"02:23.260 ","End":"02:27.040","Text":"though we have an external force on the y-axis,"},{"Start":"02:27.040 ","End":"02:29.560","Text":"we can still relate to this problem and"},{"Start":"02:29.560 ","End":"02:33.220","Text":"the external forces on the y-axis as though they are equal to 0."},{"Start":"02:33.220 ","End":"02:36.310","Text":"Here\u0027s how it works. We know that we have Delta P,"},{"Start":"02:36.310 ","End":"02:38.050","Text":"that is a change in momentum."},{"Start":"02:38.050 ","End":"02:45.285","Text":"That means that P_f minus P_i does not equal 0 and of course,"},{"Start":"02:45.285 ","End":"02:47.775","Text":"this is all along the y-axis."},{"Start":"02:47.775 ","End":"02:53.045","Text":"This means that the final momentum is different than the initial momentum,"},{"Start":"02:53.045 ","End":"02:56.480","Text":"meaning P_f minus P_i does not equal 0."},{"Start":"02:56.480 ","End":"03:03.320","Text":"What this also equals is the impulse J of the force mg, the force of gravity."},{"Start":"03:03.320 ","End":"03:06.900","Text":"Of course, along the y-axis."},{"Start":"03:07.210 ","End":"03:10.235","Text":"This option makes sense up until now."},{"Start":"03:10.235 ","End":"03:12.140","Text":"If you recall that J,"},{"Start":"03:12.140 ","End":"03:14.745","Text":"where our impulse equals Delta P,"},{"Start":"03:14.745 ","End":"03:16.365","Text":"the change in momentum."},{"Start":"03:16.365 ","End":"03:18.950","Text":"In a case with conservation of momentum,"},{"Start":"03:18.950 ","End":"03:22.835","Text":"there is no difference between P_Fy and P_iy."},{"Start":"03:22.835 ","End":"03:26.450","Text":"There are no external forces acting so J equals 0."},{"Start":"03:26.450 ","End":"03:30.290","Text":"In this case, J equals something Delta p and the way we\u0027re going to find"},{"Start":"03:30.290 ","End":"03:36.270","Text":"our impulse is to take an integral of the external forces."},{"Start":"03:36.270 ","End":"03:40.880","Text":"Forces external along the y-axis over d_t."},{"Start":"03:40.880 ","End":"03:44.300","Text":"We\u0027re going to take that over the time 0"},{"Start":"03:44.300 ","End":"03:49.345","Text":"to Delta t. This is the period of time of the collision."},{"Start":"03:49.345 ","End":"03:54.155","Text":"If we plug in the coefficients from this problem,"},{"Start":"03:54.155 ","End":"03:57.725","Text":"we get an integral of mg,"},{"Start":"03:57.725 ","End":"04:06.795","Text":"which is m_1 plus m_2 times g d t from time 0 to"},{"Start":"04:06.795 ","End":"04:11.425","Text":"time Delta t. The answer for this is"},{"Start":"04:11.425 ","End":"04:16.705","Text":"because m_1 and m_2 are constant and g is the gravitational constant, it\u0027s constant."},{"Start":"04:16.705 ","End":"04:22.745","Text":"They can go to the outside of our integral and what we get is m_1 plus"},{"Start":"04:22.745 ","End":"04:32.035","Text":"m_2 times g Delta t. This is my answer here."},{"Start":"04:32.035 ","End":"04:34.420","Text":"But let\u0027s suppose for a second that Delta t,"},{"Start":"04:34.420 ","End":"04:35.920","Text":"this is the time of the collision,"},{"Start":"04:35.920 ","End":"04:39.070","Text":"the amount of time the 2 balls are touching each other."},{"Start":"04:39.070 ","End":"04:42.925","Text":"Let\u0027s say that that\u0027s very close to 0 or even equal to 0."},{"Start":"04:42.925 ","End":"04:45.130","Text":"If you think about a collision we\u0027re talking about a"},{"Start":"04:45.130 ","End":"04:48.220","Text":"100th of a second at most, in many situations."},{"Start":"04:48.220 ","End":"04:52.595","Text":"If Delta t is close to 0 or essentially 0,"},{"Start":"04:52.595 ","End":"04:56.000","Text":"that means that we have 0 times a constant and so"},{"Start":"04:56.000 ","End":"05:00.650","Text":"this whole answer goes to 0 or functionally to 0."},{"Start":"05:00.650 ","End":"05:05.165","Text":"And that means that P_Fy minus P_ iy equals 0,"},{"Start":"05:05.165 ","End":"05:09.575","Text":"or that we have no value for Delta p. Delta p goes to 0."},{"Start":"05:09.575 ","End":"05:13.115","Text":"Therefore functionally, we can assume that we actually do have"},{"Start":"05:13.115 ","End":"05:18.390","Text":"a constant momentum and we do have conservation of momentum."},{"Start":"05:18.580 ","End":"05:22.310","Text":"Functionally, what this means for us is that when we talk"},{"Start":"05:22.310 ","End":"05:25.565","Text":"about short collisions and many of our collisions are short,"},{"Start":"05:25.565 ","End":"05:30.570","Text":"we can consider mg and other forces sometimes as well, to be negligible."},{"Start":"05:30.570 ","End":"05:35.135","Text":"The value here of mg is so close to 0 that it really has no effect,"},{"Start":"05:35.135 ","End":"05:38.000","Text":"no influence on the momentum of the collision."},{"Start":"05:38.000 ","End":"05:39.830","Text":"This is the upshot here."},{"Start":"05:39.830 ","End":"05:42.815","Text":"Mg can be considered negligible,"},{"Start":"05:42.815 ","End":"05:48.214","Text":"meaning we don\u0027t have to worry about its effect on momentum during short collisions."},{"Start":"05:48.214 ","End":"05:50.300","Text":"This isn\u0027t the only case."},{"Start":"05:50.300 ","End":"05:52.385","Text":"We can also look at another example."},{"Start":"05:52.385 ","End":"05:58.205","Text":"Let\u0027s assume these 2 balls are on some sort of surface and they collide on the surface."},{"Start":"05:58.205 ","End":"06:03.365","Text":"Well, there should be some amount of friction f k with the surface."},{"Start":"06:03.365 ","End":"06:05.920","Text":"It doesn\u0027t really matter which direction the friction is going."},{"Start":"06:05.920 ","End":"06:10.810","Text":"But the point here is that to find the influence of the friction,"},{"Start":"06:10.810 ","End":"06:12.334","Text":"to find its impact,"},{"Start":"06:12.334 ","End":"06:15.320","Text":"we have to take J_FK,"},{"Start":"06:15.320 ","End":"06:17.195","Text":"the impulse of the friction,"},{"Start":"06:17.195 ","End":"06:20.704","Text":"equals an integral of the friction,"},{"Start":"06:20.704 ","End":"06:27.410","Text":"which if you recall, is Mu_kndt."},{"Start":"06:27.410 ","End":"06:36.800","Text":"I\u0027m going to do for this from 0 to Delta t and so our answer is Mu kn Delta t,"},{"Start":"06:36.800 ","End":"06:38.990","Text":"and if Delta t is close to 0,"},{"Start":"06:38.990 ","End":"06:42.140","Text":"then this whole influence is close to 0."},{"Start":"06:42.140 ","End":"06:44.495","Text":"Meaning again, just like with mg,"},{"Start":"06:44.495 ","End":"06:45.770","Text":"the force of gravity,"},{"Start":"06:45.770 ","End":"06:48.005","Text":"if we have friction in this case,"},{"Start":"06:48.005 ","End":"06:50.255","Text":"when our time is close to 0,"},{"Start":"06:50.255 ","End":"06:52.730","Text":"the influence is negligible."},{"Start":"06:52.730 ","End":"06:55.505","Text":"Now you do have to be careful when talking about friction,"},{"Start":"06:55.505 ","End":"06:59.030","Text":"because sometimes your friction will not be a constant coefficient,"},{"Start":"06:59.030 ","End":"07:02.810","Text":"rather some other formula and sometimes it will have a large value,"},{"Start":"07:02.810 ","End":"07:04.880","Text":"meaning that even when time is small,"},{"Start":"07:04.880 ","End":"07:07.970","Text":"it still does have a discernible influence,"},{"Start":"07:07.970 ","End":"07:12.065","Text":"a discernible effect on your conservation of momentum."},{"Start":"07:12.065 ","End":"07:14.450","Text":"For the sake of illustration,"},{"Start":"07:14.450 ","End":"07:16.700","Text":"I want to show you a counterexample where"},{"Start":"07:16.700 ","End":"07:20.965","Text":"the normal force is extremely large and is not negligible."},{"Start":"07:20.965 ","End":"07:26.150","Text":"Let\u0027s assume we have a ball and the ball is on a collision course with the floor"},{"Start":"07:26.150 ","End":"07:31.910","Text":"and it hits the floor at an angle and it goes back up in this direction after bouncing."},{"Start":"07:31.910 ","End":"07:33.230","Text":"Now in this case,"},{"Start":"07:33.230 ","End":"07:35.975","Text":"the normal force is acting upwards."},{"Start":"07:35.975 ","End":"07:39.800","Text":"However, if we\u0027re to find the impulse of the normal force,"},{"Start":"07:39.800 ","End":"07:43.160","Text":"which is an integral of Ndt."},{"Start":"07:43.160 ","End":"07:47.644","Text":"N, the normal force will be much greater than mg."},{"Start":"07:47.644 ","End":"07:49.940","Text":"In fact, this is what\u0027s changing your momentum."},{"Start":"07:49.940 ","End":"07:53.435","Text":"In this case, the normal force is not negligible."},{"Start":"07:53.435 ","End":"07:55.700","Text":"It has a very serious effect in fact."},{"Start":"07:55.700 ","End":"07:58.475","Text":"If I were to graph the normal force N,"},{"Start":"07:58.475 ","End":"08:01.355","Text":"over time, it would look something like this."},{"Start":"08:01.355 ","End":"08:04.160","Text":"It would go up towards infinity and back down to 0."},{"Start":"08:04.160 ","End":"08:05.570","Text":"It\u0027s a peak of sorts."},{"Start":"08:05.570 ","End":"08:07.385","Text":"In that case N,"},{"Start":"08:07.385 ","End":"08:10.240","Text":"the normal force is not negligible."},{"Start":"08:10.240 ","End":"08:14.150","Text":"What we get from this is that in certain cases when"},{"Start":"08:14.150 ","End":"08:17.975","Text":"our normal force or another force is large and has a large impact,"},{"Start":"08:17.975 ","End":"08:20.465","Text":"it is not negligible and we have to account for it."},{"Start":"08:20.465 ","End":"08:22.985","Text":"We can\u0027t say we have conservation of momentum."},{"Start":"08:22.985 ","End":"08:29.300","Text":"However, if we have a weak force that is constant or any constant force,"},{"Start":"08:29.300 ","End":"08:33.500","Text":"and our time is close to 0 our Delta t is close to 0,"},{"Start":"08:33.500 ","End":"08:35.930","Text":"we can consider that forces effect to be"},{"Start":"08:35.930 ","End":"08:39.035","Text":"negligible and assume that for all intents and purposes,"},{"Start":"08:39.035 ","End":"08:42.000","Text":"we have conservation of momentum."}],"ID":9350}],"Thumbnail":null,"ID":5395},{"Name":"Momentum Conclusion","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Summary of Momentum","Duration":"8m 24s","ChapterTopicVideoID":9078,"CourseChapterTopicPlaylistID":5396,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.565","Text":"Hello. In this video,"},{"Start":"00:02.565 ","End":"00:07.365","Text":"I want to give a little summary of all the topics we discussed regarding momentum."},{"Start":"00:07.365 ","End":"00:10.680","Text":"Let\u0027s start where we began with Newton\u0027s Second Law,"},{"Start":"00:10.680 ","End":"00:12.690","Text":"and we wrote it in a new way,"},{"Start":"00:12.690 ","End":"00:17.190","Text":"saying that the sum of forces equals d P,"},{"Start":"00:17.190 ","End":"00:22.065","Text":"d t. The change in momentum over time."},{"Start":"00:22.065 ","End":"00:31.325","Text":"We define momentum P as equal to m v. This is the equation for a one momentum,"},{"Start":"00:31.325 ","End":"00:32.900","Text":"the momentum of one object."},{"Start":"00:32.900 ","End":"00:36.910","Text":"If we have multiple objects with multiple momenta, then P_T,"},{"Start":"00:36.910 ","End":"00:40.235","Text":"the total momentum equals P_1,"},{"Start":"00:40.235 ","End":"00:41.825","Text":"the momentum of object 1,"},{"Start":"00:41.825 ","End":"00:45.110","Text":"plus P2, the momentum of object 2,"},{"Start":"00:45.110 ","End":"00:47.345","Text":"plus each momentum of each object,"},{"Start":"00:47.345 ","End":"00:52.265","Text":"so on and so forth until you\u0027ve added all the momenta of all of your objects."},{"Start":"00:52.265 ","End":"00:56.810","Text":"Later on, we said that if the sum of external forces,"},{"Start":"00:56.810 ","End":"01:00.530","Text":"F external equals 0,"},{"Start":"01:00.530 ","End":"01:06.960","Text":"this is if, that means that P_T,"},{"Start":"01:06.960 ","End":"01:09.640","Text":"a total momentum is constant,"},{"Start":"01:09.640 ","End":"01:12.235","Text":"or we have conservation of momentum."},{"Start":"01:12.235 ","End":"01:14.560","Text":"When we have conservation of momentum,"},{"Start":"01:14.560 ","End":"01:17.665","Text":"it\u0027s important because we can use certain formula."},{"Start":"01:17.665 ","End":"01:21.715","Text":"Now we usually see this in the collision between two objects."},{"Start":"01:21.715 ","End":"01:24.100","Text":"We were talking about balls of equal mass,"},{"Start":"01:24.100 ","End":"01:26.424","Text":"perfect spheres of equal mass oftentimes,"},{"Start":"01:26.424 ","End":"01:28.270","Text":"or of different masses really."},{"Start":"01:28.270 ","End":"01:31.660","Text":"The reason is when you have a collision between two objects,"},{"Start":"01:31.660 ","End":"01:33.475","Text":"you have a lot of internal forces."},{"Start":"01:33.475 ","End":"01:37.780","Text":"One object, one sphere is exerting some force on the other sphere,"},{"Start":"01:37.780 ","End":"01:42.025","Text":"and the other sphere is exerting an equal and opposite force on the first sphere."},{"Start":"01:42.025 ","End":"01:48.815","Text":"Here we have a lot of internal forces but oftentimes no external force."},{"Start":"01:48.815 ","End":"01:50.405","Text":"If this is the case,"},{"Start":"01:50.405 ","End":"01:55.420","Text":"then we can say that P_i equals P_f,"},{"Start":"01:55.420 ","End":"01:57.890","Text":"meaning our initial momentum,"},{"Start":"01:57.890 ","End":"01:59.780","Text":"our momentum before the collision,"},{"Start":"01:59.780 ","End":"02:02.375","Text":"equals the momentum after the collision,"},{"Start":"02:02.375 ","End":"02:05.165","Text":"meaning we have a conservation of momentum."},{"Start":"02:05.165 ","End":"02:08.315","Text":"We should also remember that momentum is a vector,"},{"Start":"02:08.315 ","End":"02:10.970","Text":"therefore, it has elements along each axis."},{"Start":"02:10.970 ","End":"02:13.645","Text":"You could have MV_x,"},{"Start":"02:13.645 ","End":"02:15.785","Text":"you could have MV_y,"},{"Start":"02:15.785 ","End":"02:18.020","Text":"which is along the y-axis, of course,"},{"Start":"02:18.020 ","End":"02:23.555","Text":"and you can have MV_z along the z-axis."},{"Start":"02:23.555 ","End":"02:28.805","Text":"It could be that you have conservation of momentum along the x-axis,"},{"Start":"02:28.805 ","End":"02:32.810","Text":"meaning the sum of external forces along the x-axis equals 0,"},{"Start":"02:32.810 ","End":"02:35.740","Text":"the sum of external forces along the y-axis could"},{"Start":"02:35.740 ","End":"02:38.590","Text":"equal 0 and you could have conservation along the y-axis,"},{"Start":"02:38.590 ","End":"02:40.710","Text":"and the same is true for the z-axis,"},{"Start":"02:40.710 ","End":"02:45.960","Text":"or it could be true long two or even all three of the axis at the same time."},{"Start":"02:46.340 ","End":"02:49.505","Text":"When I have conservation of momentum,"},{"Start":"02:49.505 ","End":"02:52.355","Text":"I can use my equation for the conservation of momentum,"},{"Start":"02:52.355 ","End":"02:55.735","Text":"and that is m1_ v_1,"},{"Start":"02:55.735 ","End":"03:00.105","Text":"plus m_2 v_2 equals"},{"Start":"03:00.105 ","End":"03:07.140","Text":"m_1 u_1, plus m_2 u_2."},{"Start":"03:07.140 ","End":"03:11.375","Text":"In this case, v is the velocity before the collision,"},{"Start":"03:11.375 ","End":"03:13.820","Text":"so v_1 is the velocity for object 1,"},{"Start":"03:13.820 ","End":"03:15.395","Text":"v_2 for object 2,"},{"Start":"03:15.395 ","End":"03:18.595","Text":"and u is velocity after the collision,"},{"Start":"03:18.595 ","End":"03:21.260","Text":"u_1 is the post collision velocity of object 1,"},{"Start":"03:21.260 ","End":"03:23.465","Text":"u_2, the same for object 2."},{"Start":"03:23.465 ","End":"03:27.530","Text":"We also talked about a specific case where you have an elastic collision,"},{"Start":"03:27.530 ","End":"03:31.115","Text":"if you recall the difference between elastic and inelastic collisions,"},{"Start":"03:31.115 ","End":"03:33.515","Text":"and in the case of elastic collisions,"},{"Start":"03:33.515 ","End":"03:36.485","Text":"I can add in my conservation of energy formula,"},{"Start":"03:36.485 ","End":"03:45.180","Text":"and that is 1/2 of m_1 v_1 squared plus 1/2 of m_2 v_2 squared,"},{"Start":"03:45.180 ","End":"03:55.335","Text":"equals 1/2 of m_1 u_1 squared plus 1/2 of m_2 u_2 squared."},{"Start":"03:55.335 ","End":"04:00.280","Text":"Of course here v and u represent the same velocities they represent above."},{"Start":"04:00.280 ","End":"04:01.910","Text":"In both of these situations,"},{"Start":"04:01.910 ","End":"04:04.370","Text":"we\u0027re talking about a collision between two objects."},{"Start":"04:04.370 ","End":"04:05.780","Text":"If there are three or more objects,"},{"Start":"04:05.780 ","End":"04:07.295","Text":"you have to add in more elements,"},{"Start":"04:07.295 ","End":"04:08.945","Text":"one for each object."},{"Start":"04:08.945 ","End":"04:12.050","Text":"One more note is that it\u0027s important to keep in"},{"Start":"04:12.050 ","End":"04:15.380","Text":"mind that when we\u0027re talking about the conservation of energy,"},{"Start":"04:15.380 ","End":"04:18.160","Text":"we only mean kinetic energy."},{"Start":"04:18.160 ","End":"04:23.165","Text":"Two important differences between this formula is that our first formula,"},{"Start":"04:23.165 ","End":"04:26.645","Text":"the formula for the conservation of momentum, is a vector,"},{"Start":"04:26.645 ","End":"04:27.740","Text":"meaning it has directions,"},{"Start":"04:27.740 ","End":"04:31.250","Text":"so we can talk about conservation of momentum along the x-axis,"},{"Start":"04:31.250 ","End":"04:33.950","Text":"along the y-axis, along the z-axis,"},{"Start":"04:33.950 ","End":"04:35.465","Text":"or in any direction."},{"Start":"04:35.465 ","End":"04:38.300","Text":"However, in the case of conservation of energy,"},{"Start":"04:38.300 ","End":"04:39.845","Text":"energy has no directions."},{"Start":"04:39.845 ","End":"04:44.750","Text":"This is just one formula and can\u0027t be broken down into axes are elements."},{"Start":"04:44.750 ","End":"04:49.459","Text":"The second difference is that as long as you have no external forces,"},{"Start":"04:49.459 ","End":"04:53.225","Text":"the conservation of momentum is present in all collisions,"},{"Start":"04:53.225 ","End":"04:59.110","Text":"whereas the conservation of energy is only present or applicable in elastic collisions."},{"Start":"04:59.110 ","End":"05:04.691","Text":"We also talked about an even more special case of a head-on elastic collision,"},{"Start":"05:04.691 ","End":"05:07.340","Text":"meaning that two objects are along"},{"Start":"05:07.340 ","End":"05:11.450","Text":"the same axis and collide and stay along that same axis."},{"Start":"05:11.450 ","End":"05:17.720","Text":"In this case, we can switch out our conservation of energy formula with a new version,"},{"Start":"05:17.720 ","End":"05:22.985","Text":"v_1 plus u_1 equals v_2 plus u_2."},{"Start":"05:22.985 ","End":"05:25.885","Text":"This is only in the case of a head-on collision."},{"Start":"05:25.885 ","End":"05:27.890","Text":"We like to use this because it\u0027s a little easier,"},{"Start":"05:27.890 ","End":"05:29.480","Text":"we don\u0027t have to square elements,"},{"Start":"05:29.480 ","End":"05:30.740","Text":"we don\u0027t have to do with 1/2,"},{"Start":"05:30.740 ","End":"05:31.970","Text":"we don\u0027t have to deal with mass."},{"Start":"05:31.970 ","End":"05:37.140","Text":"So if you can, it\u0027s best to use v_1 plus u_1 equals v_2 plus u_2,"},{"Start":"05:37.140 ","End":"05:41.095","Text":"of course, only in the case of head-on collisions."},{"Start":"05:41.095 ","End":"05:44.840","Text":"We also discussed a couple of other special cases."},{"Start":"05:44.840 ","End":"05:46.565","Text":"However, those are more marginal."},{"Start":"05:46.565 ","End":"05:48.920","Text":"Now let\u0027s move on to cases where you do have"},{"Start":"05:48.920 ","End":"05:53.735","Text":"external forces and we may not have conservation of momentum."},{"Start":"05:53.735 ","End":"05:56.465","Text":"The first thing to remember is impulse."},{"Start":"05:56.465 ","End":"06:01.864","Text":"Impulse or J equals an integral of the force exerted"},{"Start":"06:01.864 ","End":"06:04.655","Text":"times d t. What you end up with is an"},{"Start":"06:04.655 ","End":"06:07.820","Text":"integral in the same direction as the force exerted."},{"Start":"06:07.820 ","End":"06:11.150","Text":"We can also find the impulse of the total force,"},{"Start":"06:11.150 ","End":"06:12.985","Text":"or the sum of forces,"},{"Start":"06:12.985 ","End":"06:18.715","Text":"and that equals an integral of the sum of forces d t,"},{"Start":"06:18.715 ","End":"06:21.680","Text":"and can also be calculated by taking"},{"Start":"06:21.680 ","End":"06:24.350","Text":"the impulse of each force and adding them all together."},{"Start":"06:24.350 ","End":"06:30.825","Text":"So you take the sum of all impulses and this equals Delta P_T,"},{"Start":"06:30.825 ","End":"06:34.080","Text":"or the change in total momentum."},{"Start":"06:34.080 ","End":"06:38.460","Text":"You can talk about that in terms of one object or entire system,"},{"Start":"06:38.460 ","End":"06:40.685","Text":"but when you take J,"},{"Start":"06:40.685 ","End":"06:42.925","Text":"the impulse of the sum of forces,"},{"Start":"06:42.925 ","End":"06:45.130","Text":"it equals Delta P_T."},{"Start":"06:45.130 ","End":"06:47.180","Text":"The next question is,"},{"Start":"06:47.180 ","End":"06:49.695","Text":"what is Delta P?"},{"Start":"06:49.695 ","End":"06:51.185","Text":"Really what Delta P is,"},{"Start":"06:51.185 ","End":"06:53.930","Text":"is the difference between P final and P initial."},{"Start":"06:53.930 ","End":"06:55.520","Text":"On the left side here,"},{"Start":"06:55.520 ","End":"06:57.635","Text":"P final and P initial are equal,"},{"Start":"06:57.635 ","End":"06:59.660","Text":"but if we have external forces,"},{"Start":"06:59.660 ","End":"07:06.620","Text":"Delta P_T equals P final minus P initial,"},{"Start":"07:06.620 ","End":"07:12.295","Text":"that is the difference between the final momentum and the initial momentum."},{"Start":"07:12.295 ","End":"07:16.985","Text":"These are all the basic things we discussed on the topic of momentum."},{"Start":"07:16.985 ","End":"07:19.790","Text":"Just to add in the other special cases we discussed,"},{"Start":"07:19.790 ","End":"07:24.860","Text":"we talked about plastic collisions which are inelastic, and in that case,"},{"Start":"07:24.860 ","End":"07:26.990","Text":"u_1 equals u_2,"},{"Start":"07:26.990 ","End":"07:32.015","Text":"meaning that the two objects travel together at the same velocity after the collision,"},{"Start":"07:32.015 ","End":"07:36.450","Text":"and in this case, energy is never conserved."},{"Start":"07:36.650 ","End":"07:40.385","Text":"We also talked about a special head-on case"},{"Start":"07:40.385 ","End":"07:43.400","Text":"when you have a head-on collision between equal masses,"},{"Start":"07:43.400 ","End":"07:48.085","Text":"m_1 equals m_2 and m_2 starts at rest."},{"Start":"07:48.085 ","End":"07:53.000","Text":"What happens is all of the velocity from m_1 is transferred to m_2."},{"Start":"07:53.000 ","End":"07:56.390","Text":"M_1 is that at rest and m_2 travels at"},{"Start":"07:56.390 ","End":"08:02.300","Text":"the exact same velocity that m_1 had in the same direction that m_1 was traveling."},{"Start":"08:02.300 ","End":"08:08.120","Text":"Lastly, we talked about a non head-on collision between spheres of equal mass,"},{"Start":"08:08.120 ","End":"08:11.870","Text":"and what happens is these two spheres go in"},{"Start":"08:11.870 ","End":"08:16.535","Text":"opposite directions and the angle between them has to equal 90 degrees."},{"Start":"08:16.535 ","End":"08:21.455","Text":"This is a pretty complete summary of the subjects we discussed regarding momentum,"},{"Start":"08:21.455 ","End":"08:24.330","Text":"and that ends our lecture."}],"ID":9351}],"Thumbnail":null,"ID":5396},{"Name":"Advanced Exercises","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"An Object on a Trolley and Loss of Energy","Duration":"26m 57s","ChapterTopicVideoID":9079,"CourseChapterTopicPlaylistID":5397,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.530 ","End":"00:03.195","Text":"Let\u0027s look at this exercise."},{"Start":"00:03.195 ","End":"00:05.655","Text":"In the exercise, we\u0027re given force F,"},{"Start":"00:05.655 ","End":"00:10.455","Text":"which is a constant force pulling a trolley with a mass of m_1 without friction."},{"Start":"00:10.455 ","End":"00:11.700","Text":"The trolley is on wheels,"},{"Start":"00:11.700 ","End":"00:13.890","Text":"and there\u0027s no friction with the ground."},{"Start":"00:13.890 ","End":"00:17.340","Text":"On top of the trolley is the mass m_2."},{"Start":"00:17.340 ","End":"00:20.145","Text":"There is friction between the 2 objects."},{"Start":"00:20.145 ","End":"00:22.170","Text":"Between m_2 and m_1,"},{"Start":"00:22.170 ","End":"00:24.060","Text":"we do have some friction."},{"Start":"00:24.060 ","End":"00:27.495","Text":"The force F is pulling the trolley to the right."},{"Start":"00:27.495 ","End":"00:31.395","Text":"We\u0027re given the coefficients for static and kinetic friction,"},{"Start":"00:31.395 ","End":"00:35.095","Text":"the force F, mass 1 and mass 2."},{"Start":"00:35.095 ","End":"00:37.505","Text":"In part a, it asks,"},{"Start":"00:37.505 ","End":"00:42.965","Text":"what is the maximum force that F can reach without m_2 slipping relative to m_1?"},{"Start":"00:42.965 ","End":"00:49.770","Text":"What is the maximum force F that we can reach before m_2 begins slipping?"},{"Start":"00:49.770 ","End":"00:53.870","Text":"We\u0027ll start with part a, and then we\u0027ll move on to the rest of the question."},{"Start":"00:53.870 ","End":"00:56.809","Text":"Part a is asking me about a maximum,"},{"Start":"00:56.809 ","End":"00:59.180","Text":"which involves force and friction."},{"Start":"00:59.180 ","End":"01:01.880","Text":"The first thing that jumps into my head is that we\u0027re"},{"Start":"01:01.880 ","End":"01:05.360","Text":"probably talking about static friction. Why is that?"},{"Start":"01:05.360 ","End":"01:08.615","Text":"Because f_s, the force of static friction,"},{"Start":"01:08.615 ","End":"01:12.540","Text":"is equal to or less than mu_sN,"},{"Start":"01:12.560 ","End":"01:16.480","Text":"as opposed to kinetic friction,"},{"Start":"01:16.480 ","End":"01:19.810","Text":"which is always equal to mu_kN."},{"Start":"01:19.810 ","End":"01:23.005","Text":"Static friction is not constant."},{"Start":"01:23.005 ","End":"01:25.405","Text":"It can be up to a certain amount,"},{"Start":"01:25.405 ","End":"01:27.190","Text":"and therefore it has a maximum."},{"Start":"01:27.190 ","End":"01:30.880","Text":"Whereas f_k is always equal to the same value."},{"Start":"01:30.880 ","End":"01:35.945","Text":"The first thing that jumps into my head is we\u0027re probably using static friction."},{"Start":"01:35.945 ","End":"01:38.035","Text":"Looking at my system,"},{"Start":"01:38.035 ","End":"01:39.655","Text":"I have 1 force, F,"},{"Start":"01:39.655 ","End":"01:42.370","Text":"and it\u0027s only exerting force on mass 1,"},{"Start":"01:42.370 ","End":"01:45.265","Text":"the trolley itself not on mass 2."},{"Start":"01:45.265 ","End":"01:48.940","Text":"I know that only friction is going to move mass 2."},{"Start":"01:48.940 ","End":"01:50.875","Text":"The question is basically asking me,"},{"Start":"01:50.875 ","End":"01:53.740","Text":"what is the maximum force we can exert F,"},{"Start":"01:53.740 ","End":"01:58.180","Text":"where the friction upon m_2 is still static friction."},{"Start":"01:58.850 ","End":"02:01.170","Text":"Enough beating around the bush,"},{"Start":"02:01.170 ","End":"02:03.830","Text":"it\u0027s time to address the question directly."},{"Start":"02:03.830 ","End":"02:06.394","Text":"The question is asking me about forces,"},{"Start":"02:06.394 ","End":"02:07.910","Text":"so I don\u0027t have to hesitate."},{"Start":"02:07.910 ","End":"02:12.155","Text":"I know the first thing I need to do is start making some free body diagrams,"},{"Start":"02:12.155 ","End":"02:14.690","Text":"writing out where my forces are coming from, where they\u0027re going."},{"Start":"02:14.690 ","End":"02:16.295","Text":"There\u0027s 2 ways we can do this."},{"Start":"02:16.295 ","End":"02:19.235","Text":"The first way is to do this as a whole system,"},{"Start":"02:19.235 ","End":"02:20.735","Text":"m_1 and m_2 together."},{"Start":"02:20.735 ","End":"02:22.160","Text":"We know there\u0027s some static friction,"},{"Start":"02:22.160 ","End":"02:25.400","Text":"we might be able to treat them as the same object."},{"Start":"02:25.400 ","End":"02:28.010","Text":"We can also treat each object separately,"},{"Start":"02:28.010 ","End":"02:32.510","Text":"and look at the forces operating or exerting themselves on each object separately."},{"Start":"02:32.510 ","End":"02:35.660","Text":"Because we\u0027re dealing with static friction between the 2 objects,"},{"Start":"02:35.660 ","End":"02:38.735","Text":"I\u0027m going to choose to address each object separately."},{"Start":"02:38.735 ","End":"02:40.620","Text":"Let\u0027s start with m_1."},{"Start":"02:40.620 ","End":"02:43.400","Text":"M_1, we have the force F that we know about."},{"Start":"02:43.400 ","End":"02:48.980","Text":"We also have f_s or static friction that is a result of m_2,"},{"Start":"02:48.980 ","End":"02:53.345","Text":"and it\u0027s pushing to the left backwards for the sake of m_1, our trolley."},{"Start":"02:53.345 ","End":"02:57.425","Text":"Think of m_2 as some sort of ceiling and the trolley is going past it."},{"Start":"02:57.425 ","End":"02:59.345","Text":"F_s is going to the left."},{"Start":"02:59.345 ","End":"03:01.160","Text":"We also have the normal force,"},{"Start":"03:01.160 ","End":"03:04.570","Text":"N_1 going upwards, and m_1g,"},{"Start":"03:04.570 ","End":"03:07.580","Text":"the force of gravity pushing downwards,"},{"Start":"03:07.580 ","End":"03:12.230","Text":"and we also have another force that\u0027s coming from m_2 and pushing downwards,"},{"Start":"03:12.230 ","End":"03:14.245","Text":"we\u0027ll call that N_2."},{"Start":"03:14.245 ","End":"03:18.605","Text":"Now in blue, we\u0027ll do a free body diagram for m_2."},{"Start":"03:18.605 ","End":"03:20.225","Text":"First of all, we have f_s."},{"Start":"03:20.225 ","End":"03:23.960","Text":"F_s is the force of static friction,"},{"Start":"03:23.960 ","End":"03:26.780","Text":"and it\u0027s going in the opposite direction of this f_s."},{"Start":"03:26.780 ","End":"03:28.925","Text":"They\u0027re the same force in opposite directions."},{"Start":"03:28.925 ","End":"03:30.650","Text":"We also have N_2,"},{"Start":"03:30.650 ","End":"03:32.914","Text":"the normal force which is going upwards."},{"Start":"03:32.914 ","End":"03:35.990","Text":"It\u0027s the equal and opposite force of this N_2."},{"Start":"03:35.990 ","End":"03:38.030","Text":"We have m_2g,"},{"Start":"03:38.030 ","End":"03:42.180","Text":"the force of gravity forcing downwards."},{"Start":"03:42.950 ","End":"03:46.250","Text":"Now I\u0027ve listed my forces for each object,"},{"Start":"03:46.250 ","End":"03:49.220","Text":"and I can apply Newton\u0027s second law to each object."},{"Start":"03:49.220 ","End":"03:51.350","Text":"By the way, you should notice here you don\u0027t need"},{"Start":"03:51.350 ","End":"03:53.990","Text":"that much intuitive understanding of the system,"},{"Start":"03:53.990 ","End":"03:56.225","Text":"rather if you just know where your forces are going,"},{"Start":"03:56.225 ","End":"03:57.725","Text":"and you do your list of forces,"},{"Start":"03:57.725 ","End":"03:59.560","Text":"everything seems to work itself out."},{"Start":"03:59.560 ","End":"04:02.345","Text":"Let\u0027s start doing force equations."},{"Start":"04:02.345 ","End":"04:05.270","Text":"Let\u0027s start with our first object, m_1."},{"Start":"04:05.270 ","End":"04:08.075","Text":"We\u0027ll do a sum of forces along each axis."},{"Start":"04:08.075 ","End":"04:16.550","Text":"The sum of forces along the x-axis equals F minus f_s,"},{"Start":"04:16.550 ","End":"04:21.770","Text":"and that equals m_1a_1."},{"Start":"04:21.770 ","End":"04:25.880","Text":"Along the y-axis, we don\u0027t really care so much in terms of m_1,"},{"Start":"04:25.880 ","End":"04:27.650","Text":"this object, it\u0027s not so important."},{"Start":"04:27.650 ","End":"04:31.395","Text":"We\u0027ll move on to the second object, object 2 m_2."},{"Start":"04:31.395 ","End":"04:33.320","Text":"It\u0027s along the x-axis."},{"Start":"04:33.320 ","End":"04:38.245","Text":"Sum of forces along the x-axis equals f_s,"},{"Start":"04:38.245 ","End":"04:40.795","Text":"that\u0027s your only force along the x-axis,"},{"Start":"04:40.795 ","End":"04:43.880","Text":"and that equals m_2a_2."},{"Start":"04:44.310 ","End":"04:55.755","Text":"Along the y-axis, you have 2 forces, N_2 minus m_2g."},{"Start":"04:55.755 ","End":"05:01.760","Text":"This has to equal 0 because our object is not moving along the y-axis."},{"Start":"05:02.620 ","End":"05:05.090","Text":"These are my equations."},{"Start":"05:05.090 ","End":"05:08.555","Text":"One more thing is if there is static friction,"},{"Start":"05:08.555 ","End":"05:10.730","Text":"and these 2 objects are moving together,"},{"Start":"05:10.730 ","End":"05:12.845","Text":"we can assume they have the same acceleration."},{"Start":"05:12.845 ","End":"05:15.980","Text":"Instead of a_1 and a_2, we just have a."},{"Start":"05:15.980 ","End":"05:19.590","Text":"Now at this point we have 2 unknown variables,"},{"Start":"05:19.590 ","End":"05:21.079","Text":"and we have 2 equations."},{"Start":"05:21.079 ","End":"05:24.570","Text":"My unknown variables are a and f_s."},{"Start":"05:24.570 ","End":"05:28.520","Text":"What I can do is solve for one and plug it in, and solve for the other."},{"Start":"05:28.520 ","End":"05:32.240","Text":"We already have a solution for f_s, f_s equals m_2a."},{"Start":"05:32.240 ","End":"05:36.710","Text":"We can plug that in here and solve for a. F"},{"Start":"05:36.710 ","End":"05:43.260","Text":"minus m_2a equals m_1a."},{"Start":"05:43.260 ","End":"05:45.605","Text":"In that case, if we isolate a,"},{"Start":"05:45.605 ","End":"05:53.370","Text":"then we get a equals F over m_1 plus m_2."},{"Start":"05:53.370 ","End":"05:58.515","Text":"If we want to describe f_s in terms of m_1, m_2 and F,"},{"Start":"05:58.515 ","End":"06:01.770","Text":"we can rewrite f_s as equal"},{"Start":"06:01.770 ","End":"06:10.420","Text":"to m_2F over m_1 plus m_2."},{"Start":"06:11.540 ","End":"06:16.600","Text":"At this point, I have solutions for a and for f_s."},{"Start":"06:16.600 ","End":"06:20.875","Text":"Now I need some condition that allows this to be true."},{"Start":"06:20.875 ","End":"06:27.085","Text":"I know that this is only true when f_s is less than or equal to mu_sN."},{"Start":"06:27.085 ","End":"06:32.675","Text":"I can say if this is less than or equal to mu_sN,"},{"Start":"06:32.675 ","End":"06:34.630","Text":"then this is true."},{"Start":"06:34.630 ","End":"06:36.455","Text":"Now we\u0027re talking about N_2 remember,"},{"Start":"06:36.455 ","End":"06:40.320","Text":"because this is the static friction between m_2 and m_1."},{"Start":"06:40.320 ","End":"06:42.120","Text":"We can say N_2,"},{"Start":"06:42.120 ","End":"06:46.165","Text":"and we know from these equations that N_2 equals m_2g."},{"Start":"06:46.165 ","End":"06:51.140","Text":"I can replace this N_2 with m_2g."},{"Start":"06:51.140 ","End":"06:57.000","Text":"My solution is that F is less than or equal to"},{"Start":"06:57.000 ","End":"07:04.390","Text":"mu_sg times m_1 plus m_2."},{"Start":"07:05.580 ","End":"07:07.915","Text":"If this is my condition,"},{"Start":"07:07.915 ","End":"07:14.815","Text":"that means that the maximum is when F equals Musg times m1 plus m2."},{"Start":"07:14.815 ","End":"07:17.200","Text":"That\u0027s the solution to part a."},{"Start":"07:17.200 ","End":"07:19.990","Text":"Now let\u0027s move on to part b."},{"Start":"07:19.990 ","End":"07:22.540","Text":"Before we get to part b,"},{"Start":"07:22.540 ","End":"07:27.355","Text":"we\u0027re told to assume that force F is larger than the maximum from part a."},{"Start":"07:27.355 ","End":"07:30.670","Text":"This means that m2 is now moving backwards."},{"Start":"07:30.670 ","End":"07:33.505","Text":"There\u0027s kinetic friction instead of static friction,"},{"Start":"07:33.505 ","End":"07:37.105","Text":"and m2 is moving backwards relative to m1."},{"Start":"07:37.105 ","End":"07:42.145","Text":"We\u0027re also told to assume that F is acting for a given time big T,"},{"Start":"07:42.145 ","End":"07:45.400","Text":"and that m2 doesn\u0027t fall off of m1."},{"Start":"07:45.400 ","End":"07:48.535","Text":"M2 is moving backwards for a time T,"},{"Start":"07:48.535 ","End":"07:52.645","Text":"but it hasn\u0027t yet fallen off of m1.The specific section,"},{"Start":"07:52.645 ","End":"07:54.415","Text":"part b, asks us,"},{"Start":"07:54.415 ","End":"07:56.619","Text":"what is the acceleration, velocity,"},{"Start":"07:56.619 ","End":"08:00.370","Text":"and position of m1 and m2 as a function of"},{"Start":"08:00.370 ","End":"08:04.690","Text":"time until the time T. We need to find the acceleration,"},{"Start":"08:04.690 ","End":"08:07.750","Text":"position, and velocity as functions in"},{"Start":"08:07.750 ","End":"08:11.770","Text":"terms of time before the block falls off the back of the Trolley."},{"Start":"08:11.770 ","End":"08:15.505","Text":"The first thing I\u0027m going to do is adjust my free body diagrams."},{"Start":"08:15.505 ","End":"08:16.825","Text":"I\u0027m going to write out my forces,"},{"Start":"08:16.825 ","End":"08:21.430","Text":"and the only change here is that my static friction is now kinetic friction."},{"Start":"08:21.430 ","End":"08:23.305","Text":"This changes to f_k,"},{"Start":"08:23.305 ","End":"08:27.280","Text":"and this is also going to change to f_k,"},{"Start":"08:27.280 ","End":"08:31.000","Text":"and other than that, my force diagrams are the same."},{"Start":"08:31.000 ","End":"08:36.115","Text":"From here, I can now list out my forces just like I did in part a,"},{"Start":"08:36.115 ","End":"08:37.330","Text":"and it\u0027ll be rather similar,"},{"Start":"08:37.330 ","End":"08:38.950","Text":"just we\u0027re going to replace our forces,"},{"Start":"08:38.950 ","End":"08:41.890","Text":"and accelerations are no longer equal."},{"Start":"08:41.890 ","End":"08:44.305","Text":"For object 1,"},{"Start":"08:44.305 ","End":"08:48.910","Text":"the sum of forces along the x-axis equals F"},{"Start":"08:48.910 ","End":"08:54.655","Text":"minus f_k equals m1, a1."},{"Start":"08:54.655 ","End":"08:58.150","Text":"Remember, a1 does not equal a2 anymore,"},{"Start":"08:58.150 ","End":"09:00.070","Text":"and for m2,"},{"Start":"09:00.070 ","End":"09:09.370","Text":"the sum of forces along the x-axis equals f_k equals m2, a2,"},{"Start":"09:09.370 ","End":"09:14.590","Text":"and the sum of forces along the y-axis equals"},{"Start":"09:14.590 ","End":"09:21.880","Text":"m2 minus m2g equals 0."},{"Start":"09:21.880 ","End":"09:25.825","Text":"At first we have a problem because we have 3 unknown variables,"},{"Start":"09:25.825 ","End":"09:28.780","Text":"f_k, a2, and a1."},{"Start":"09:28.780 ","End":"09:35.960","Text":"However, we know that f_k is a constant and f_k always equals muk_N,"},{"Start":"09:36.380 ","End":"09:38.680","Text":"and that\u0027s N2 to of course,"},{"Start":"09:38.680 ","End":"09:40.900","Text":"and we know that N2 equals m2g,"},{"Start":"09:40.900 ","End":"09:44.960","Text":"so we can rewrite this as muk_m2g."},{"Start":"09:45.630 ","End":"09:50.860","Text":"We can plug this straight into our equations and then solve for a1 and a2,"},{"Start":"09:50.860 ","End":"09:52.180","Text":"and I\u0027ll give you the solution."},{"Start":"09:52.180 ","End":"09:53.740","Text":"We\u0027re going to skip some algebra here,"},{"Start":"09:53.740 ","End":"09:55.825","Text":"but again, we\u0027re here to learn the physics."},{"Start":"09:55.825 ","End":"10:06.910","Text":"A1 equals F over m1 minus m2 over m1 times muk_g,"},{"Start":"10:06.910 ","End":"10:16.180","Text":"and a2 equals muk_g."},{"Start":"10:16.180 ","End":"10:22.720","Text":"One thing to note is that both a1 and a2 are positive because we have kinetic friction,"},{"Start":"10:22.720 ","End":"10:24.400","Text":"F should be greater than this,"},{"Start":"10:24.400 ","End":"10:29.320","Text":"and even though our object m2 is moving backwards relative to m1,"},{"Start":"10:29.320 ","End":"10:31.780","Text":"it\u0027s still moving forwards overall."},{"Start":"10:31.780 ","End":"10:34.090","Text":"Now, both of these are constants."},{"Start":"10:34.090 ","End":"10:39.520","Text":"We can immediately plug it into our equation for V(t ) when a is constant,"},{"Start":"10:39.520 ","End":"10:43.585","Text":"and that is V0 plus a_t."},{"Start":"10:43.585 ","End":"10:45.730","Text":"Each of these starts at rest,"},{"Start":"10:45.730 ","End":"10:47.725","Text":"so V0 will be 0,"},{"Start":"10:47.725 ","End":"10:55.570","Text":"so V1 equals a1t and V2 equals a2t."},{"Start":"10:55.570 ","End":"10:59.110","Text":"I\u0027m not going to write out a1 and a2 because they\u0027re just constants,"},{"Start":"10:59.110 ","End":"11:00.280","Text":"and we have them above,"},{"Start":"11:00.280 ","End":"11:03.370","Text":"but we now found V1 and V2."},{"Start":"11:03.370 ","End":"11:08.155","Text":"The same process that we just went through to go from acceleration to velocity."},{"Start":"11:08.155 ","End":"11:12.970","Text":"We can now do from velocity to position because we have constant acceleration."},{"Start":"11:12.970 ","End":"11:17.770","Text":"The function to go from constant acceleration to position is as follows,"},{"Start":"11:17.770 ","End":"11:19.090","Text":"and you may want to remember it."},{"Start":"11:19.090 ","End":"11:24.310","Text":"It\u0027s x(t) equals X0 plus"},{"Start":"11:24.310 ","End":"11:31.022","Text":"V0t plus 0.5at^2,"},{"Start":"11:31.022 ","End":"11:33.655","Text":"and we can assume that X0,"},{"Start":"11:33.655 ","End":"11:36.175","Text":"that they both start at position 0,"},{"Start":"11:36.175 ","End":"11:38.620","Text":"and we know that V0 equals 0,"},{"Start":"11:38.620 ","End":"11:44.740","Text":"so X1 equals 0.5 of a1t ^2"},{"Start":"11:44.740 ","End":"11:52.410","Text":"and X2 equals 0.5 a2t^2."},{"Start":"11:52.410 ","End":"11:55.140","Text":"Now this only work because we had constant acceleration,"},{"Start":"11:55.140 ","End":"11:57.450","Text":"but now we\u0027ve found a1 and a2,"},{"Start":"11:57.450 ","End":"11:59.985","Text":"V1 and V2, and X1 and X2,"},{"Start":"11:59.985 ","End":"12:01.740","Text":"so we\u0027ve solved part b."},{"Start":"12:01.740 ","End":"12:06.580","Text":"Now let\u0027s move on to part c. In part c,"},{"Start":"12:06.580 ","End":"12:09.940","Text":"I\u0027m asked how much energy is lost during this time."},{"Start":"12:09.940 ","End":"12:12.145","Text":"There are two ways to calculate this."},{"Start":"12:12.145 ","End":"12:16.690","Text":"The first is to say that E, the energy lost,"},{"Start":"12:16.690 ","End":"12:22.615","Text":"equals Q. Q is generally how we symbolize energy lost to heat,"},{"Start":"12:22.615 ","End":"12:28.100","Text":"and that equals the work of kinetic friction f_k."},{"Start":"12:28.160 ","End":"12:35.115","Text":"The second way is to take the same E. E Again in this case is energy lost."},{"Start":"12:35.115 ","End":"12:40.090","Text":"We\u0027re going to set that equal to the work of force F. The work of"},{"Start":"12:40.090 ","End":"12:45.595","Text":"force F is essentially how much energy force F has put into our system."},{"Start":"12:45.595 ","End":"12:49.465","Text":"Some of that energy goes into the system and is used efficiently,"},{"Start":"12:49.465 ","End":"12:53.725","Text":"and other parts of that are used as heat and they escaped the system."},{"Start":"12:53.725 ","End":"12:56.635","Text":"We\u0027re going to subtract from that delta E,"},{"Start":"12:56.635 ","End":"12:58.495","Text":"which is the change in energy."},{"Start":"12:58.495 ","End":"13:00.760","Text":"This is to say how much energy was used"},{"Start":"13:00.760 ","End":"13:05.300","Text":"efficiently and what\u0027s leftover will be our energy lost."},{"Start":"13:05.550 ","End":"13:09.010","Text":"First, I\u0027ll use our first method and show you how we"},{"Start":"13:09.010 ","End":"13:13.520","Text":"calculate this using the work of kinetic friction."},{"Start":"13:14.070 ","End":"13:19.000","Text":"When we\u0027re calculating the work of the kinetic friction,"},{"Start":"13:19.000 ","End":"13:21.250","Text":"we need to look at both objects,"},{"Start":"13:21.250 ","End":"13:23.770","Text":"because if you notice here,"},{"Start":"13:23.770 ","End":"13:28.075","Text":"friction is acting both on m2 and on m1,"},{"Start":"13:28.075 ","End":"13:29.980","Text":"so we need to account for both."},{"Start":"13:29.980 ","End":"13:32.845","Text":"Calculating using the work of kinetic friction,"},{"Start":"13:32.845 ","End":"13:35.815","Text":"we\u0027re going to need to account for the kinetic friction both on m2,"},{"Start":"13:35.815 ","End":"13:40.615","Text":"that\u0027s this blue arrow and the kinetic friction on m1, this red arrow."},{"Start":"13:40.615 ","End":"13:43.570","Text":"The way that we\u0027re going to do that is using equation."},{"Start":"13:43.570 ","End":"13:47.680","Text":"This is the equation to calculate the work of a given force."},{"Start":"13:47.680 ","End":"13:56.050","Text":"The work of kinetic friction equals an integral of the force F dot d_r."},{"Start":"13:56.050 ","End":"14:02.650","Text":"In our case, the force is f_k kinetic friction will take that dot d_r,"},{"Start":"14:02.650 ","End":"14:07.360","Text":"and we can actually switch out d_r with d_x because all of my motion,"},{"Start":"14:07.360 ","End":"14:11.925","Text":"all of my work, and all of my force is along the x-axis."},{"Start":"14:11.925 ","End":"14:14.210","Text":"For the sake of simplicity from here,"},{"Start":"14:14.210 ","End":"14:19.270","Text":"I\u0027ll break this down into my 2 objects and calculate the work of each object separately."},{"Start":"14:19.270 ","End":"14:23.880","Text":"Let me start with the work of kinetic friction along object 2,"},{"Start":"14:23.880 ","End":"14:28.245","Text":"and that equals f_k and the dot product with d_x."},{"Start":"14:28.245 ","End":"14:31.505","Text":"Now, to do a dot product or scalar multiplication,"},{"Start":"14:31.505 ","End":"14:34.190","Text":"I need to find the angle between these 2."},{"Start":"14:34.190 ","End":"14:38.000","Text":"I know f_k moves to the right along the x-axis and"},{"Start":"14:38.000 ","End":"14:41.975","Text":"the direction of the trajectory is also to the right along the x-axis."},{"Start":"14:41.975 ","End":"14:44.525","Text":"Therefore the angle between these two is 0,"},{"Start":"14:44.525 ","End":"14:49.075","Text":"and my answer is simply f_k times d_x."},{"Start":"14:49.075 ","End":"14:51.760","Text":"Now I know that f_k is a constant,"},{"Start":"14:51.760 ","End":"14:54.415","Text":"so I can move f_k to the outside,"},{"Start":"14:54.415 ","End":"14:58.660","Text":"and I\u0027m left with f_k delta X."},{"Start":"14:58.660 ","End":"15:01.090","Text":"Because we started at X equals 0,"},{"Start":"15:01.090 ","End":"15:05.680","Text":"that means that delta X is the same as X over time T. I can"},{"Start":"15:05.680 ","End":"15:10.645","Text":"say that it equals f_k X (t),"},{"Start":"15:10.645 ","End":"15:13.480","Text":"so this was all for the second mass,"},{"Start":"15:13.480 ","End":"15:15.445","Text":"so we can make these all 2."},{"Start":"15:15.445 ","End":"15:20.680","Text":"When we calculate the work of the kinetic friction of object 1,"},{"Start":"15:20.680 ","End":"15:22.585","Text":"we see it equals almost the same thing."},{"Start":"15:22.585 ","End":"15:25.600","Text":"The only difference is the angle between d_x and f_k."},{"Start":"15:25.600 ","End":"15:30.070","Text":"In the case of object 1 f_k is acting towards the left,"},{"Start":"15:30.070 ","End":"15:34.285","Text":"meaning there\u0027s a 180 degree angle between f_k and d_x."},{"Start":"15:34.285 ","End":"15:43.765","Text":"What we get is an integral of f_k times d_x times the cosine"},{"Start":"15:43.765 ","End":"15:45.880","Text":"of 180 degrees."},{"Start":"15:45.880 ","End":"15:49.430","Text":"The cosine of 180 degrees is negative 1,"},{"Start":"15:49.430 ","End":"15:51.950","Text":"so result instead of positive f_k,"},{"Start":"15:51.950 ","End":"15:57.115","Text":"X (t) is the work of kinetic friction,"},{"Start":"15:57.115 ","End":"16:04.810","Text":"on object 1 is negative f_k X1T."},{"Start":"16:06.240 ","End":"16:11.300","Text":"1 small correction here is we\u0027re not doing lowercase t time in general."},{"Start":"16:11.300 ","End":"16:14.655","Text":"We\u0027re using our uppercase T Specific time."},{"Start":"16:14.655 ","End":"16:17.420","Text":"Now, if I want to get the total work of kinetic friction,"},{"Start":"16:17.420 ","End":"16:20.930","Text":"that\u0027s f_k1 and f_k2 combined."},{"Start":"16:20.930 ","End":"16:24.035","Text":"I can add these together and that will be as follows."},{"Start":"16:24.035 ","End":"16:28.085","Text":"Wf_k equals"},{"Start":"16:28.085 ","End":"16:35.810","Text":"f_k times X2 (T) large T"},{"Start":"16:35.810 ","End":"16:39.220","Text":"minus X1(T)."},{"Start":"16:39.220 ","End":"16:44.095","Text":"If I really want, we know that X1 and X2 equal the following,"},{"Start":"16:44.095 ","End":"16:46.610","Text":"so I can also put those in and plug it in,"},{"Start":"16:46.610 ","End":"16:47.810","Text":"and the answer is that,"},{"Start":"16:47.810 ","End":"16:52.925","Text":"that equals f_k times"},{"Start":"16:52.925 ","End":"17:01.510","Text":"0.5 of T^2 multiplied by a2 minus a1."},{"Start":"17:01.510 ","End":"17:07.550","Text":"1 final note here is that if you do your calculations and plug in all of your values,"},{"Start":"17:07.550 ","End":"17:10.550","Text":"you will find that this is a negative number."},{"Start":"17:10.550 ","End":"17:13.925","Text":"That\u0027s because we\u0027re talking here about the work of f_k."},{"Start":"17:13.925 ","End":"17:19.350","Text":"This is the kind of work that prevents our objects from being as efficient as possible."},{"Start":"17:19.350 ","End":"17:22.370","Text":"It is a negative force in some senses."},{"Start":"17:22.370 ","End":"17:25.700","Text":"If we\u0027re talking about the amount of energy that\u0027s lost in the system,"},{"Start":"17:25.700 ","End":"17:27.560","Text":"Q or E, in that case,"},{"Start":"17:27.560 ","End":"17:31.650","Text":"we need to just take the magnitude of the absolute value."},{"Start":"17:31.650 ","End":"17:39.680","Text":"In fact, sometimes you write that the energy lost equals the negative work of f_k,"},{"Start":"17:39.680 ","End":"17:42.630","Text":"the kinetic friction on the system."},{"Start":"17:42.630 ","End":"17:46.580","Text":"1 more thing you may notice is that X2T minus"},{"Start":"17:46.580 ","End":"17:51.515","Text":"X1T is essentially the relative position of our 2 objects."},{"Start":"17:51.515 ","End":"17:55.205","Text":"It is how m2 has moved relative to m1."},{"Start":"17:55.205 ","End":"17:57.470","Text":"That means that if they\u0027re in the same place,"},{"Start":"17:57.470 ","End":"17:59.460","Text":"if neither of them has moved,"},{"Start":"17:59.460 ","End":"18:01.305","Text":"this equals 0,"},{"Start":"18:01.305 ","End":"18:03.870","Text":"and therefore the work of the kinetic friction is 0,"},{"Start":"18:03.870 ","End":"18:08.720","Text":"and that makes intuitive sense because we\u0027d still be using static friction."},{"Start":"18:08.720 ","End":"18:12.425","Text":"Now, I want to show you how to do the same calculation"},{"Start":"18:12.425 ","End":"18:16.505","Text":"using the work of the force F minus the change in energy."},{"Start":"18:16.505 ","End":"18:20.610","Text":"We\u0027ll do is take E and that\u0027s our energy lost"},{"Start":"18:21.790 ","End":"18:30.780","Text":"equals the work of the force F minus delta E. The change in energy."},{"Start":"18:31.860 ","End":"18:35.470","Text":"Once again, the work of the force F tells us"},{"Start":"18:35.470 ","End":"18:39.220","Text":"how much energy force F is putting into the system."},{"Start":"18:39.220 ","End":"18:43.555","Text":"Some of that energy will not be used efficiently and will come off as heat."},{"Start":"18:43.555 ","End":"18:49.659","Text":"Delta E tells us how much of the energy was used efficiently."},{"Start":"18:49.659 ","End":"18:53.890","Text":"This works because F is an external force."},{"Start":"18:53.890 ","End":"18:57.505","Text":"Let\u0027s start by calculating the work of the force F,"},{"Start":"18:57.505 ","End":"19:02.930","Text":"and that equals the integral of the scalar multiplication of the force F.dr."},{"Start":"19:03.210 ","End":"19:08.050","Text":"Again, the force F is only along the x-axis and"},{"Start":"19:08.050 ","End":"19:12.190","Text":"the trajectory of dr is also along the x-axis to the right,"},{"Start":"19:12.190 ","End":"19:16.840","Text":"so our answer is F times Delta x."},{"Start":"19:16.840 ","End":"19:20.845","Text":"Because the force is only acting on m_1,"},{"Start":"19:20.845 ","End":"19:23.275","Text":"we can talk about this as x_1."},{"Start":"19:23.275 ","End":"19:28.630","Text":"We know that Delta x is the change between x final and x initial,"},{"Start":"19:28.630 ","End":"19:30.430","Text":"and we know that x initial is 0,"},{"Start":"19:30.430 ","End":"19:34.360","Text":"so we can rewrite this as F times x final,"},{"Start":"19:34.360 ","End":"19:36.640","Text":"which is x_1 at the time T,"},{"Start":"19:36.640 ","End":"19:41.710","Text":"because we\u0027re only looking until the given time T. In that case,"},{"Start":"19:41.710 ","End":"19:43.750","Text":"because we know this to be x_1,"},{"Start":"19:43.750 ","End":"19:45.460","Text":"we can plug it in from over here."},{"Start":"19:45.460 ","End":"19:55.300","Text":"We can rewrite this one more time as F times 1/2 of a_1 T^2."},{"Start":"19:55.460 ","End":"19:58.675","Text":"I won\u0027t plug in a_1 here,"},{"Start":"19:58.675 ","End":"20:02.350","Text":"but you can see that this is the work of the force"},{"Start":"20:02.350 ","End":"20:06.130","Text":"F. Now we can move on to Delta E. Delta E,"},{"Start":"20:06.130 ","End":"20:07.300","Text":"the change in energy,"},{"Start":"20:07.300 ","End":"20:10.735","Text":"equals E final minus E initial."},{"Start":"20:10.735 ","End":"20:14.695","Text":"Now we know that E initial is 0 because our system started at rest,"},{"Start":"20:14.695 ","End":"20:17.275","Text":"so we only need to calculate E final."},{"Start":"20:17.275 ","End":"20:20.650","Text":"E final is the kinetic energy we have in the system,"},{"Start":"20:20.650 ","End":"20:25.555","Text":"so that\u0027s 1/2 of m_2v_2^2,"},{"Start":"20:25.555 ","End":"20:27.190","Text":"and we\u0027re talking at time T,"},{"Start":"20:27.190 ","End":"20:33.070","Text":"plus 1/2 m_1v_1^2,"},{"Start":"20:33.070 ","End":"20:37.225","Text":"also at time T. For now I\u0027ll leave the answer like this."},{"Start":"20:37.225 ","End":"20:42.145","Text":"Of course you can plug in your larger solution for v_1 and v_2,"},{"Start":"20:42.145 ","End":"20:43.840","Text":"and you can add them together,"},{"Start":"20:43.840 ","End":"20:46.840","Text":"and you\u0027ll see that the solution is the same as we found before."},{"Start":"20:46.840 ","End":"20:48.399","Text":"But for now I\u0027ll leave it like this,"},{"Start":"20:48.399 ","End":"20:56.410","Text":"and we can move on to part d. Part d asks us to find"},{"Start":"20:56.410 ","End":"21:00.430","Text":"the terminal velocity of the objects m_1 and m_2 at"},{"Start":"21:00.430 ","End":"21:05.830","Text":"the time t\u003eT before the mass m_2 falls."},{"Start":"21:05.830 ","End":"21:10.150","Text":"The idea is that at some point after the time T,"},{"Start":"21:10.150 ","End":"21:12.040","Text":"our force F stops,"},{"Start":"21:12.040 ","End":"21:14.260","Text":"and these 2 objects are moving together,"},{"Start":"21:14.260 ","End":"21:18.550","Text":"and we have to assume that m_2 has not fallen off the back of the trolley."},{"Start":"21:18.550 ","End":"21:21.370","Text":"First we need to understand how this works."},{"Start":"21:21.370 ","End":"21:26.680","Text":"At first, the object start moving and the force F pulls m_1 forwards,"},{"Start":"21:26.680 ","End":"21:28.510","Text":"and m_1 is moving forwards."},{"Start":"21:28.510 ","End":"21:31.795","Text":"m_2 is still moving forwards relative to our lab,"},{"Start":"21:31.795 ","End":"21:34.615","Text":"but it\u0027s moving backwards relative to m_1."},{"Start":"21:34.615 ","End":"21:38.050","Text":"At some point along the way at time T,"},{"Start":"21:38.050 ","End":"21:42.205","Text":"the force F stops exerting any force on m_1,"},{"Start":"21:42.205 ","End":"21:48.565","Text":"and m_1 starts to slow down while m_2 continues to move forward and eventually,"},{"Start":"21:48.565 ","End":"21:51.550","Text":"v_1 will equal v_2."},{"Start":"21:51.550 ","End":"21:54.790","Text":"When v_1 equals v_2,"},{"Start":"21:54.790 ","End":"21:57.145","Text":"the 2 objects will be moving together."},{"Start":"21:57.145 ","End":"21:59.635","Text":"There\u0027ll no longer be any force F,"},{"Start":"21:59.635 ","End":"22:02.455","Text":"there\u0027ll no longer be any kinetic friction,"},{"Start":"22:02.455 ","End":"22:05.050","Text":"and the static friction will equal 0."},{"Start":"22:05.050 ","End":"22:09.380","Text":"They\u0027ll just move together on and on without stopping."},{"Start":"22:09.390 ","End":"22:13.925","Text":"Really what we\u0027re asked to do is find that velocity."},{"Start":"22:13.925 ","End":"22:15.610","Text":"There are 2 ways to do this."},{"Start":"22:15.610 ","End":"22:17.980","Text":"The first is to look after time T,"},{"Start":"22:17.980 ","End":"22:20.635","Text":"how the dynamics of the problem workout,"},{"Start":"22:20.635 ","End":"22:22.210","Text":"how m_1 slows down,"},{"Start":"22:22.210 ","End":"22:24.100","Text":"how m_2\u0027s velocity changes,"},{"Start":"22:24.100 ","End":"22:26.440","Text":"and at what time v_1 equals v_2."},{"Start":"22:26.440 ","End":"22:30.925","Text":"This is a very work and labor-intensive process and you can find your answer,"},{"Start":"22:30.925 ","End":"22:33.790","Text":"but there is a shortcut that is useful to know."},{"Start":"22:33.790 ","End":"22:36.595","Text":"This shortcut will save you a lot of time."},{"Start":"22:36.595 ","End":"22:38.575","Text":"Let\u0027s work using the shortcut."},{"Start":"22:38.575 ","End":"22:40.240","Text":"Let\u0027s start with the following."},{"Start":"22:40.240 ","End":"22:44.260","Text":"The initial momentum in the problem P_i is 0."},{"Start":"22:44.260 ","End":"22:47.875","Text":"We know that because the whole system starts at rest."},{"Start":"22:47.875 ","End":"22:50.800","Text":"However, P final has some value."},{"Start":"22:50.800 ","End":"22:52.660","Text":"We know that there is a final momentum,"},{"Start":"22:52.660 ","End":"22:57.385","Text":"and that is mass 1 plus mass 2 times u,"},{"Start":"22:57.385 ","End":"23:04.010","Text":"and u is some velocity that the 2 objects share because they\u0027re moving together."},{"Start":"23:05.490 ","End":"23:10.465","Text":"Now, we know that if P_f does not equal P_i,"},{"Start":"23:10.465 ","End":"23:13.285","Text":"we have a change in momentum, and Delta p,"},{"Start":"23:13.285 ","End":"23:16.375","Text":"or the changed momentum equals J, our impulse."},{"Start":"23:16.375 ","End":"23:24.410","Text":"We calculate our impulse by taking an integral of the sum of external forces, dt."},{"Start":"23:25.440 ","End":"23:28.210","Text":"I can rewrite this in a different way."},{"Start":"23:28.210 ","End":"23:29.425","Text":"I know that in our case,"},{"Start":"23:29.425 ","End":"23:33.507","Text":"Delta P equals P_f because the initial momentum equals 0,"},{"Start":"23:33.507 ","End":"23:41.245","Text":"so Delta P equals m_1 plus m_2 times u, the final velocity."},{"Start":"23:41.245 ","End":"23:45.760","Text":"That has to equal the result of this integral."},{"Start":"23:45.760 ","End":"23:47.140","Text":"We know that in our case,"},{"Start":"23:47.140 ","End":"23:50.685","Text":"the only external force is the force F. Remember"},{"Start":"23:50.685 ","End":"23:55.590","Text":"that friction is an internal force because it\u0027s between Object 1 and Object 2,"},{"Start":"23:55.590 ","End":"23:58.635","Text":"so F is our only external force and F is constant,"},{"Start":"23:58.635 ","End":"24:00.480","Text":"so it can go on the outside of the integral,"},{"Start":"24:00.480 ","End":"24:02.845","Text":"and we\u0027re left with F and an integral of dt,"},{"Start":"24:02.845 ","End":"24:05.840","Text":"which is the time T."},{"Start":"24:09.120 ","End":"24:14.245","Text":"We can actually consider this u to be u_f, u final."},{"Start":"24:14.245 ","End":"24:16.195","Text":"If we want to solve for u_f,"},{"Start":"24:16.195 ","End":"24:25.510","Text":"we find that u_f equals FT divided by m_1 plus m_2."},{"Start":"24:25.510 ","End":"24:29.455","Text":"With this shortcut, we can find our answer rather quickly."},{"Start":"24:29.455 ","End":"24:33.715","Text":"With that, we\u0027ve solved part d and ended the exercise."},{"Start":"24:33.715 ","End":"24:37.345","Text":"If the direction of static friction was confusing to you,"},{"Start":"24:37.345 ","End":"24:40.216","Text":"I want to take an opportunity here to explain it."},{"Start":"24:40.216 ","End":"24:41.649","Text":"If you\u0027re interested in explanation,"},{"Start":"24:41.649 ","End":"24:44.200","Text":"please watch the following portion."},{"Start":"24:44.200 ","End":"24:48.445","Text":"I know this can be confusing because friction tends to act"},{"Start":"24:48.445 ","End":"24:52.240","Text":"against the direction of our force or the direction of our general movement,"},{"Start":"24:52.240 ","End":"24:55.885","Text":"but this is a great example to show that that\u0027s not always the case."},{"Start":"24:55.885 ","End":"24:59.125","Text":"The first thing to understand is Newton\u0027s third law;"},{"Start":"24:59.125 ","End":"25:03.055","Text":"any action has an equal and opposite reaction,"},{"Start":"25:03.055 ","End":"25:05.500","Text":"so if our force is acting to the right for 1 object,"},{"Start":"25:05.500 ","End":"25:07.750","Text":"it must be acting to the left for another,"},{"Start":"25:07.750 ","End":"25:10.645","Text":"and vice versa, if it\u0027s acting to the left for 1 object,"},{"Start":"25:10.645 ","End":"25:13.240","Text":"it must be acting to the right for another."},{"Start":"25:13.240 ","End":"25:14.890","Text":"Now the question is,"},{"Start":"25:14.890 ","End":"25:17.110","Text":"on which object is the force acting to"},{"Start":"25:17.110 ","End":"25:20.710","Text":"the right and on which object is the force acting to the left?"},{"Start":"25:20.710 ","End":"25:22.615","Text":"There\u0027s 2 ways to explain this."},{"Start":"25:22.615 ","End":"25:24.805","Text":"The first one is rather mathematical."},{"Start":"25:24.805 ","End":"25:26.965","Text":"Let\u0027s look at m_2 for a second."},{"Start":"25:26.965 ","End":"25:28.435","Text":"We know that at some point,"},{"Start":"25:28.435 ","End":"25:31.203","Text":"force F starts acting on m_1,"},{"Start":"25:31.203 ","End":"25:34.450","Text":"and takes the whole system to the right,"},{"Start":"25:34.450 ","End":"25:36.490","Text":"so m_2 is also moving towards the right,"},{"Start":"25:36.490 ","End":"25:38.290","Text":"and we have to figure out how."},{"Start":"25:38.290 ","End":"25:40.570","Text":"Now if we take Newton\u0027s second law,"},{"Start":"25:40.570 ","End":"25:44.960","Text":"the sum of forces equals m_2a_2."},{"Start":"25:45.210 ","End":"25:48.295","Text":"Now if we\u0027re looking along the x-axis,"},{"Start":"25:48.295 ","End":"25:51.055","Text":"we know that along the x-axis,"},{"Start":"25:51.055 ","End":"25:54.850","Text":"the only force is the force of static friction,"},{"Start":"25:54.850 ","End":"25:59.500","Text":"f_s, and we also know that that equals m_2a_2."},{"Start":"25:59.500 ","End":"26:03.265","Text":"Now Newton\u0027s second law also accounts for direction."},{"Start":"26:03.265 ","End":"26:06.970","Text":"We have to have our force going in the same direction as our acceleration."},{"Start":"26:06.970 ","End":"26:08.335","Text":"Mass doesn\u0027t have direction,"},{"Start":"26:08.335 ","End":"26:11.380","Text":"so in the same direction as acceleration."},{"Start":"26:11.380 ","End":"26:16.000","Text":"In this case, we know that m_2 is accelerating to the right, so therefore,"},{"Start":"26:16.000 ","End":"26:18.805","Text":"f_s must be pushing m_2 to the right,"},{"Start":"26:18.805 ","End":"26:22.150","Text":"and also pushing m_1 to the left."},{"Start":"26:22.150 ","End":"26:26.110","Text":"The second way to think about this is much more intuitive."},{"Start":"26:26.110 ","End":"26:29.650","Text":"Let\u0027s imagine that m_1 is going to the right as it is,"},{"Start":"26:29.650 ","End":"26:32.200","Text":"and even if m_2 were stuck in place,"},{"Start":"26:32.200 ","End":"26:33.385","Text":"say it was a ceiling,"},{"Start":"26:33.385 ","End":"26:36.340","Text":"it would be causing kinetic friction against m_1,"},{"Start":"26:36.340 ","End":"26:40.900","Text":"pushing it backwards, pushing it to the left or preventing it from going to the right."},{"Start":"26:40.900 ","End":"26:44.199","Text":"That only makes sense intuitively, and therefore,"},{"Start":"26:44.199 ","End":"26:46.690","Text":"f_s is moving to the left on m_1,"},{"Start":"26:46.690 ","End":"26:50.050","Text":"and it must be moving to the right for m_2."},{"Start":"26:50.050 ","End":"26:54.895","Text":"I hope that that made everything clear and that you understand where this is coming from."},{"Start":"26:54.895 ","End":"26:57.980","Text":"From here, I\u0027ll move on to the next topic."}],"ID":9352},{"Watched":false,"Name":"Ball in a Trolley-Solution Part A","Duration":"6m 28s","ChapterTopicVideoID":9080,"CourseChapterTopicPlaylistID":5397,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.649","Text":"Hello, I wanted to take the opportunity to walk through this exercise."},{"Start":"00:04.649 ","End":"00:07.335","Text":"In the exercise, we\u0027re given a trolley."},{"Start":"00:07.335 ","End":"00:12.045","Text":"The trolley has a mass M and a length L and starts at rest."},{"Start":"00:12.045 ","End":"00:17.265","Text":"The trolley is on wheels and has no friction and is free to move."},{"Start":"00:17.265 ","End":"00:19.545","Text":"Inside the trolley is a ball,"},{"Start":"00:19.545 ","End":"00:22.320","Text":"and the ball has the mass m and"},{"Start":"00:22.320 ","End":"00:26.070","Text":"an initial velocity of v_0 towards the front of the trolley."},{"Start":"00:26.070 ","End":"00:29.990","Text":"Now we\u0027re told that the relationship between the mass M of"},{"Start":"00:29.990 ","End":"00:33.995","Text":"the trolley and the mass m of the ball is large M"},{"Start":"00:33.995 ","End":"00:37.760","Text":"equals Alpha m. This ball will"},{"Start":"00:37.760 ","End":"00:41.845","Text":"eventually reach the front of the trolley and there will be an elastic collision."},{"Start":"00:41.845 ","End":"00:46.280","Text":"In Part a, I\u0027m asked to solve for what is the velocity of"},{"Start":"00:46.280 ","End":"00:52.040","Text":"the object after the collision and I\u0027m asked to check the solution at Alpha=0,"},{"Start":"00:52.040 ","End":"00:55.400","Text":"Alpha=1, and Alpha equals infinity."},{"Start":"00:55.400 ","End":"00:58.310","Text":"The way that I\u0027ll find the velocity of the objects after"},{"Start":"00:58.310 ","End":"01:01.594","Text":"the collision is by using my 2 equations."},{"Start":"01:01.594 ","End":"01:05.795","Text":"One for the conservation of momentum and the second for the conservation of energy,"},{"Start":"01:05.795 ","End":"01:08.170","Text":"because I know this is an elastic equation."},{"Start":"01:08.170 ","End":"01:09.915","Text":"Let\u0027s start with momentum."},{"Start":"01:09.915 ","End":"01:13.760","Text":"I\u0027ll only be measuring along the x-axis because the ball is moving along"},{"Start":"01:13.760 ","End":"01:18.195","Text":"the x-axis and I don\u0027t have to worry about external forces."},{"Start":"01:18.195 ","End":"01:21.690","Text":"P_x, the momentum along the x-axis,"},{"Start":"01:21.690 ","End":"01:27.360","Text":"equals m(V_0) plus M times 0 is"},{"Start":"01:27.360 ","End":"01:36.405","Text":"0 equals m(U_1) plus M(U_2)."},{"Start":"01:36.405 ","End":"01:45.720","Text":"For energy, 1/2(m)V_0^2 is my initial energy plus"},{"Start":"01:45.720 ","End":"01:48.090","Text":"0 equals"},{"Start":"01:48.090 ","End":"01:51.560","Text":"1/2(mU_1)^2"},{"Start":"01:51.560 ","End":"01:58.500","Text":"plus 1/2(MU_2)^2."},{"Start":"01:58.500 ","End":"01:59.780","Text":"This is a head-on collision."},{"Start":"01:59.780 ","End":"02:02.990","Text":"Again, my ball is traveling across the x-axis with"},{"Start":"02:02.990 ","End":"02:07.655","Text":"no external forces and it\u0027s hitting head-on against the side of the trolley."},{"Start":"02:07.655 ","End":"02:10.265","Text":"Because this is a head-on collision,"},{"Start":"02:10.265 ","End":"02:20.815","Text":"I can switch out my energy formula to this version, V_1 plus U_1=V_2 plus U_2."},{"Start":"02:20.815 ","End":"02:29.985","Text":"In our case, this becomes V_0 plus U_1=0 plus U_2."},{"Start":"02:29.985 ","End":"02:31.895","Text":"This is my first equation,"},{"Start":"02:31.895 ","End":"02:34.115","Text":"and this is my second equation."},{"Start":"02:34.115 ","End":"02:40.260","Text":"Now I have 2 equations and 2 unknown variables and I can solve."},{"Start":"02:40.260 ","End":"02:42.200","Text":"In my second equation,"},{"Start":"02:42.200 ","End":"02:44.670","Text":"I can see that U_2=V_0 plus U_1."},{"Start":"02:45.340 ","End":"02:50.590","Text":"I\u0027ll plug that into the first equation and the result is as follows,"},{"Start":"02:50.590 ","End":"02:53.715","Text":"mV_0 from the left side"},{"Start":"02:53.715 ","End":"03:01.380","Text":"equals mU_1 plus M and instead of U_2,"},{"Start":"03:01.380 ","End":"03:05.205","Text":"I\u0027ll put in V_0, U_1."},{"Start":"03:05.205 ","End":"03:09.480","Text":"Now what I want do is isolate U_1 and the result"},{"Start":"03:09.480 ","End":"03:15.060","Text":"is U_1=m minus M"},{"Start":"03:15.060 ","End":"03:22.200","Text":"divided by m plus M times V_0."},{"Start":"03:22.200 ","End":"03:24.885","Text":"If we now, instead of using M,"},{"Start":"03:24.885 ","End":"03:28.620","Text":"use Alpha m, my result,"},{"Start":"03:28.620 ","End":"03:33.490","Text":"is m times 1 minus Alpha"},{"Start":"03:33.590 ","End":"03:41.685","Text":"over m times 1 plus Alpha times V_0."},{"Start":"03:41.685 ","End":"03:44.190","Text":"Here my m\u0027s dropout."},{"Start":"03:44.190 ","End":"03:49.650","Text":"This is my general answer for U_1 and my general answer for U_2 will be"},{"Start":"03:49.650 ","End":"03:58.150","Text":"that U_2=2(V_0) over 1 plus Alpha."},{"Start":"03:59.120 ","End":"04:03.600","Text":"Now let\u0027s solve for Alpha=0."},{"Start":"04:03.600 ","End":"04:09.780","Text":"At Alpha=0, we get 1 minus 0, 1 p plus 0."},{"Start":"04:09.780 ","End":"04:19.030","Text":"Basically U_1=V_0 and U_2=2V_0."},{"Start":"04:22.870 ","End":"04:25.865","Text":"This makes intuitive sense actually,"},{"Start":"04:25.865 ","End":"04:29.405","Text":"because if M=0m,"},{"Start":"04:29.405 ","End":"04:32.915","Text":"that means that the trolley has no mass, and therefore,"},{"Start":"04:32.915 ","End":"04:37.265","Text":"our ball should continue at the same velocity after the collision."},{"Start":"04:37.265 ","End":"04:40.530","Text":"Now let\u0027s check for Alpha=1."},{"Start":"04:42.500 ","End":"04:52.040","Text":"If Alpha=1, that means that U_1=1 minus 1 over 1 plus 1, that\u0027s 0."},{"Start":"04:52.040 ","End":"05:00.700","Text":"U_2=V_0, 2V over 2=V_0."},{"Start":"05:03.380 ","End":"05:08.760","Text":"This actually makes a lot of sense because you have 2 balls of equal mass."},{"Start":"05:10.210 ","End":"05:14.530","Text":"M=m times 1. You have 2 objects of equal mass."},{"Start":"05:14.530 ","End":"05:22.220","Text":"The first starts in motion and the second at rest along one axis and after the collision,"},{"Start":"05:22.220 ","End":"05:24.740","Text":"the first ball will be at rest and the second ball will"},{"Start":"05:24.740 ","End":"05:27.365","Text":"move in the same direction with the same velocity."},{"Start":"05:27.365 ","End":"05:32.225","Text":"Now let\u0027s move on to our last scenario where Alpha equals infinity."},{"Start":"05:32.225 ","End":"05:38.680","Text":"In this case, we\u0027re saying that the trolley is enormously heavy or that M is"},{"Start":"05:38.680 ","End":"05:42.080","Text":"much greater than m. We can"},{"Start":"05:42.080 ","End":"05:45.860","Text":"think of it almost like throwing a tennis ball against the wall."},{"Start":"05:45.860 ","End":"05:48.380","Text":"What we\u0027d expect is that the wall wouldn\u0027t move and that"},{"Start":"05:48.380 ","End":"05:51.190","Text":"the ball would bounce back with equal velocity."},{"Start":"05:51.190 ","End":"05:54.540","Text":"Let\u0027s see what happens, when Alpha equals infinity,"},{"Start":"05:54.540 ","End":"05:57.245","Text":"1 minus infinity is negative infinity,"},{"Start":"05:57.245 ","End":"05:59.285","Text":"1 plus infinity is infinity,"},{"Start":"05:59.285 ","End":"06:03.275","Text":"so its negative infinity over infinity is negative 1."},{"Start":"06:03.275 ","End":"06:07.700","Text":"What you get is U_1 equals negative V_0."},{"Start":"06:07.700 ","End":"06:08.750","Text":"This makes sense again,"},{"Start":"06:08.750 ","End":"06:10.490","Text":"our ball is bouncing back."},{"Start":"06:10.490 ","End":"06:16.320","Text":"For U_2, 2V_0 over infinity equals 0,"},{"Start":"06:16.320 ","End":"06:18.950","Text":"so U_2=0. Again, this makes sense."},{"Start":"06:18.950 ","End":"06:24.185","Text":"Our wall is not moving and our ball is bouncing back with equal velocity."},{"Start":"06:24.185 ","End":"06:26.150","Text":"That\u0027s it for Part 1,"},{"Start":"06:26.150 ","End":"06:28.680","Text":"now let\u0027s move on to Part 2."}],"ID":9353},{"Watched":false,"Name":"Solution Part B-From the Trolley\u0027s Perspective","Duration":"2m 42s","ChapterTopicVideoID":9081,"CourseChapterTopicPlaylistID":5397,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.140 ","End":"00:07.435","Text":"Let\u0027s assume that U_ 1 and U_ 2 from part a are not given and remember U_ 1 equals this,"},{"Start":"00:07.435 ","End":"00:08.785","Text":"U_ 2 equals this,"},{"Start":"00:08.785 ","End":"00:11.920","Text":"and Alpha can equal any given value."},{"Start":"00:11.920 ","End":"00:14.840","Text":"Now we need to move on to part b."},{"Start":"00:14.840 ","End":"00:18.580","Text":"We\u0027re asked how much time will pass between the 1st and 2nd collisions."},{"Start":"00:18.580 ","End":"00:21.610","Text":"What that means is that our ball has gone from starting"},{"Start":"00:21.610 ","End":"00:25.090","Text":"point and hit the far wall of the trolley,"},{"Start":"00:25.090 ","End":"00:28.000","Text":"and now it\u0027s returning to the 1st wall of the trolley."},{"Start":"00:28.000 ","End":"00:31.945","Text":"If the moment when it hits the firewall is t is equals 0,"},{"Start":"00:31.945 ","End":"00:37.530","Text":"and the time between the 1st wall and the 2nd wall collisions is Delta t. Then we\u0027re"},{"Start":"00:37.530 ","End":"00:43.855","Text":"asked to find Delta t. Remember that both the ball and the trolley are now moving."},{"Start":"00:43.855 ","End":"00:46.805","Text":"There are 2 ways to solve this problem."},{"Start":"00:46.805 ","End":"00:52.070","Text":"The first is to find the velocity of the trolley to the right and the velocity"},{"Start":"00:52.070 ","End":"00:54.590","Text":"of the ball to the left and figure out at what point"},{"Start":"00:54.590 ","End":"00:57.679","Text":"they meet because both are moving in opposite directions."},{"Start":"00:57.679 ","End":"01:01.190","Text":"This is of course from the perspective of our mythical lab."},{"Start":"01:01.190 ","End":"01:04.865","Text":"In my opinion, it\u0027s a slightly more difficult way of doing this."},{"Start":"01:04.865 ","End":"01:10.120","Text":"The second way is to start from the perspective of an observer within the trolley."},{"Start":"01:10.120 ","End":"01:11.840","Text":"From this person\u0027s perspective,"},{"Start":"01:11.840 ","End":"01:13.610","Text":"the trolley is not moving."},{"Start":"01:13.610 ","End":"01:17.090","Text":"All that the observer sees is the ball moving towards the back of"},{"Start":"01:17.090 ","End":"01:20.855","Text":"the trolley for a distance of L with a velocity U_ 1."},{"Start":"01:20.855 ","End":"01:22.640","Text":"If we can find U_ 1,"},{"Start":"01:22.640 ","End":"01:25.435","Text":"we can solve from this perspective."},{"Start":"01:25.435 ","End":"01:27.935","Text":"Now of course, from our perspective,"},{"Start":"01:27.935 ","End":"01:30.860","Text":"this velocity U is a relative velocity."},{"Start":"01:30.860 ","End":"01:33.510","Text":"It\u0027s actually U tag."},{"Start":"01:34.310 ","End":"01:40.715","Text":"Let\u0027s begin. The first thing I need to find is U_ 1, 2."},{"Start":"01:40.715 ","End":"01:46.120","Text":"That is the relative velocity of the ball relative to the trolley."},{"Start":"01:46.120 ","End":"01:51.435","Text":"What that equals is U_ 1 minus U_ 2."},{"Start":"01:51.435 ","End":"01:56.660","Text":"I can also symbolize this with U tag and I\u0027m going to do that from now on."},{"Start":"01:56.660 ","End":"01:59.165","Text":"This will be U_ 1 tag. Either way works."},{"Start":"01:59.165 ","End":"02:02.045","Text":"U_ 1, 2 or U_ 1 tag."},{"Start":"02:02.045 ","End":"02:06.145","Text":"From here on out, I\u0027ll symbolize this with U_ 1 tag."},{"Start":"02:06.145 ","End":"02:10.655","Text":"This is my relative velocity and I can use it to solve for t time,"},{"Start":"02:10.655 ","End":"02:12.430","Text":"which is what we\u0027re trying to find here."},{"Start":"02:12.430 ","End":"02:15.215","Text":"We know that x, the distance traveled,"},{"Start":"02:15.215 ","End":"02:18.305","Text":"equals V, the velocity times t time."},{"Start":"02:18.305 ","End":"02:22.490","Text":"Well, in our case, L is the distance traveled V,"},{"Start":"02:22.490 ","End":"02:28.085","Text":"the velocity is U_ 1 tag and t time is what we\u0027re looking for."},{"Start":"02:28.085 ","End":"02:36.755","Text":"We could also say in other terms that t equals L divided by U_ 1 tag."},{"Start":"02:36.755 ","End":"02:40.730","Text":"Now if I plug in my values for U_ 1 tag and L,"},{"Start":"02:40.730 ","End":"02:42.900","Text":"I can find t."}],"ID":9354},{"Watched":false,"Name":"Solution Part C-From the Lab\u0027s Perspective","Duration":"1m 37s","ChapterTopicVideoID":9082,"CourseChapterTopicPlaylistID":5397,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.290 ","End":"00:04.950","Text":"Now let\u0027s see how we would do this from the perspective of the laboratory."},{"Start":"00:04.950 ","End":"00:10.290","Text":"The first thing we want to do is move our observer outside of the trolley."},{"Start":"00:10.290 ","End":"00:13.305","Text":"Now the observer is on the outside."},{"Start":"00:13.305 ","End":"00:18.945","Text":"Now we need to describe from this perspective the position of the ball and the trolley."},{"Start":"00:18.945 ","End":"00:21.600","Text":"Let\u0027s do that starting with the trolley."},{"Start":"00:21.600 ","End":"00:24.461","Text":"We\u0027re going to describe the position of the far left wall,"},{"Start":"00:24.461 ","End":"00:27.997","Text":"the wall that\u0027s not on the screen here because that\u0027s what the ball is meeting."},{"Start":"00:27.997 ","End":"00:33.660","Text":"x_2=x_0, which is 0,"},{"Start":"00:33.660 ","End":"00:38.100","Text":"the initial position of the wall, plus u_2t,"},{"Start":"00:38.100 ","End":"00:42.195","Text":"and u_2 is the velocity with which it\u0027s traveling to the right,"},{"Start":"00:42.195 ","End":"00:44.075","Text":"times t the time."},{"Start":"00:44.075 ","End":"00:47.735","Text":"Now we can also describe x_1, our ball."},{"Start":"00:47.735 ","End":"00:55.560","Text":"It starts the position L. It is L distance away from the left wall plus u_1t."},{"Start":"00:55.560 ","End":"00:58.520","Text":"We don\u0027t know if u_1 is going to be positive or negative."},{"Start":"00:58.520 ","End":"01:02.030","Text":"It all depends on the solution from the first equation."},{"Start":"01:02.030 ","End":"01:05.190","Text":"Now we need to find when these two objects meet each other."},{"Start":"01:05.190 ","End":"01:13.125","Text":"We need to set the equations equal and we get u_2t=L plus u_1t."},{"Start":"01:13.125 ","End":"01:17.570","Text":"Now we need to solve for t to find out what time these two objects mean."},{"Start":"01:17.570 ","End":"01:24.015","Text":"t equals L divided by U_2 minus U_1."},{"Start":"01:24.015 ","End":"01:28.070","Text":"If you recall from our other relative method,"},{"Start":"01:28.070 ","End":"01:30.770","Text":"this was the exact same solution we got."},{"Start":"01:30.770 ","End":"01:35.705","Text":"We\u0027ve solved for t now using both methods and this ends the exercise."},{"Start":"01:35.705 ","End":"01:37.860","Text":"I hope everything was clear."}],"ID":9355},{"Watched":false,"Name":"Two Masses on a Pulley with a Loose Rope","Duration":"42m 54s","ChapterTopicVideoID":9083,"CourseChapterTopicPlaylistID":5397,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.590 ","End":"00:04.650","Text":"Let\u0027s look at this exercise, 2 masses,"},{"Start":"00:04.650 ","End":"00:08.969","Text":"m_1 and m_2 are hanging from an ideal frictionless pulley."},{"Start":"00:08.969 ","End":"00:13.439","Text":"M_1 is resting on the ground while m_2 is hanging in the air."},{"Start":"00:13.439 ","End":"00:18.060","Text":"We have this pulley here and m_1 is resting on the ground"},{"Start":"00:18.060 ","End":"00:22.964","Text":"when the rope is fully tense and m_2 is hanging in the air."},{"Start":"00:22.964 ","End":"00:27.494","Text":"From this point, m_2 is raised an additional distance h from the ground,"},{"Start":"00:27.494 ","End":"00:30.645","Text":"such that the rope holding the 2 masses hangs loose."},{"Start":"00:30.645 ","End":"00:32.339","Text":"We take m_2 from here,"},{"Start":"00:32.339 ","End":"00:35.459","Text":"this dotted box, and raise it another height h,"},{"Start":"00:35.459 ","End":"00:41.359","Text":"so that it\u0027s hanging in the air and the rope holding onto the mass is loose."},{"Start":"00:41.359 ","End":"00:45.679","Text":"The system is at rest as described until m_2 is dropped."},{"Start":"00:45.679 ","End":"00:48.394","Text":"The setup for our problem is as drawn here."},{"Start":"00:48.394 ","End":"00:51.979","Text":"M_2 is raised an additional distance h"},{"Start":"00:51.979 ","End":"00:56.269","Text":"from its initial resting spot in the air and m_1 is on the ground."},{"Start":"00:56.269 ","End":"00:57.770","Text":"From this point of rest,"},{"Start":"00:57.770 ","End":"00:58.895","Text":"m_2 is dropped,"},{"Start":"00:58.895 ","End":"01:00.729","Text":"and then our system starts to move."},{"Start":"01:00.729 ","End":"01:06.305","Text":"In part a, we\u0027re asked to find the velocity of m_2 before the rope is fully taught."},{"Start":"01:06.305 ","End":"01:09.559","Text":"That is to say, what is the velocity of m_2 right"},{"Start":"01:09.559 ","End":"01:13.830","Text":"before it reaches the point where our rope has full tension?"},{"Start":"01:14.210 ","End":"01:16.649","Text":"Essentially in part a,"},{"Start":"01:16.649 ","End":"01:20.400","Text":"what we\u0027re looking at is the entire fall of mass 2,"},{"Start":"01:20.400 ","End":"01:22.125","Text":"where the rope is still loose."},{"Start":"01:22.125 ","End":"01:23.314","Text":"If the rope is loose,"},{"Start":"01:23.314 ","End":"01:26.000","Text":"we can assume that it\u0027s not influencing the fall."},{"Start":"01:26.000 ","End":"01:28.490","Text":"We can look at this as free fall,"},{"Start":"01:28.490 ","End":"01:33.149","Text":"and therefore we have conservation of energy. Let\u0027s start part a."},{"Start":"01:33.310 ","End":"01:39.385","Text":"Let\u0027s assume that m_2 starts at a height of h and that equals 0."},{"Start":"01:39.385 ","End":"01:42.920","Text":"If that\u0027s the case, then the height at the beginning is H."},{"Start":"01:42.920 ","End":"01:46.385","Text":"We can write its energy at the beginning as m_2,"},{"Start":"01:46.385 ","End":"01:49.400","Text":"the mass times gH."},{"Start":"01:49.400 ","End":"01:51.880","Text":"This is the gravitational constant."},{"Start":"01:51.880 ","End":"01:54.770","Text":"Its energy at the end is the kinetic energy,"},{"Start":"01:54.770 ","End":"01:57.499","Text":"1.5 of m_2,"},{"Start":"01:57.499 ","End":"02:02.015","Text":"and we\u0027ll call the velocity v_2^2."},{"Start":"02:02.015 ","End":"02:05.930","Text":"From here, we can find the value of v_2."},{"Start":"02:05.930 ","End":"02:07.384","Text":"Our mass drops out,"},{"Start":"02:07.384 ","End":"02:09.200","Text":"m_2 drops out from both sides,"},{"Start":"02:09.200 ","End":"02:13.950","Text":"and v_2 equals the square root of 2gH."},{"Start":"02:18.260 ","End":"02:21.705","Text":"This is the answer to part a."},{"Start":"02:21.705 ","End":"02:24.940","Text":"Now we can move on to part b."},{"Start":"02:25.910 ","End":"02:29.205","Text":"Now let\u0027s move on to part b."},{"Start":"02:29.205 ","End":"02:32.840","Text":"In part b, we\u0027re told to assume that the rope has reached"},{"Start":"02:32.840 ","End":"02:36.889","Text":"full tension and that it is not elastic. What does this mean?"},{"Start":"02:36.889 ","End":"02:42.050","Text":"First of all, our object has now reached its initial spot before we raised it."},{"Start":"02:42.050 ","End":"02:44.254","Text":"M_2 is now down here."},{"Start":"02:44.254 ","End":"02:46.460","Text":"Second of all, our rope is tense now."},{"Start":"02:46.460 ","End":"02:48.319","Text":"Our rope isn\u0027t loose."},{"Start":"02:48.319 ","End":"02:51.349","Text":"When it says that the rope is not elastic,"},{"Start":"02:51.349 ","End":"02:55.924","Text":"what it means is that when m_2 reaches its position down here,"},{"Start":"02:55.924 ","End":"02:59.194","Text":"we\u0027re going to forget about that moment where the rope may seem to jump."},{"Start":"02:59.194 ","End":"03:02.629","Text":"We\u0027re going to assume for the sake of our problem that m_2"},{"Start":"03:02.629 ","End":"03:06.349","Text":"reaches its final position and then it starts its new motion."},{"Start":"03:06.349 ","End":"03:08.780","Text":"The second thing we\u0027re going to assume is that the rope"},{"Start":"03:08.780 ","End":"03:11.194","Text":"does not get longer as it gets tenser,"},{"Start":"03:11.194 ","End":"03:12.409","Text":"whereas it gets tighter."},{"Start":"03:12.409 ","End":"03:15.154","Text":"Once m_2 reaches its destination here,"},{"Start":"03:15.154 ","End":"03:18.200","Text":"that\u0027s how long the rope will be for the rest of our problem."},{"Start":"03:18.200 ","End":"03:21.440","Text":"The tension may change and the direction or movement may change,"},{"Start":"03:21.440 ","End":"03:25.010","Text":"but the rope will stay the exact same length no matter what."},{"Start":"03:25.010 ","End":"03:27.770","Text":"This of course assumes that our rope stays tense."},{"Start":"03:27.770 ","End":"03:30.184","Text":"If there is no tension then our rope can get shorter."},{"Start":"03:30.184 ","End":"03:34.609","Text":"But as soon as we have full tension our rope will always remain the same length."},{"Start":"03:34.609 ","End":"03:39.994","Text":"Assuming that our rope has reached full tension and that the rope is not elastic,"},{"Start":"03:39.994 ","End":"03:42.320","Text":"we now need to find the total change in"},{"Start":"03:42.320 ","End":"03:46.009","Text":"momentum of both objects from the moment just before"},{"Start":"03:46.009 ","End":"03:48.650","Text":"the rope reaches full tension to the moment just"},{"Start":"03:48.650 ","End":"03:52.509","Text":"after the rope reaches full tension and both masses move."},{"Start":"03:52.509 ","End":"03:55.909","Text":"Essentially, we\u0027re looking at the moment when m_2 reaches"},{"Start":"03:55.909 ","End":"04:00.984","Text":"this little box and tension starts exerting its force upon our system."},{"Start":"04:00.984 ","End":"04:05.119","Text":"In some sense, we\u0027re looking for the effect of the force of tension."},{"Start":"04:05.119 ","End":"04:08.014","Text":"We know that just before the rope is tense,"},{"Start":"04:08.014 ","End":"04:11.374","Text":"the velocity of m_2 is this v_2 here."},{"Start":"04:11.374 ","End":"04:16.164","Text":"What we\u0027re looking for is the effect of tension on that velocity."},{"Start":"04:16.164 ","End":"04:19.189","Text":"The first thing we can do to think about this is take"},{"Start":"04:19.189 ","End":"04:22.854","Text":"our block m_2 and bring it down to where it should be."},{"Start":"04:22.854 ","End":"04:26.705","Text":"We know that a moment before our block reached this point,"},{"Start":"04:26.705 ","End":"04:29.030","Text":"the velocity was v_2."},{"Start":"04:29.030 ","End":"04:35.579","Text":"Now, we have some force T or tension that\u0027s exerting force on the whole system really."},{"Start":"04:35.579 ","End":"04:38.344","Text":"It starts very small and it should get larger."},{"Start":"04:38.344 ","End":"04:41.130","Text":"But the magnitude isn\u0027t what concerns us at the moment."},{"Start":"04:41.130 ","End":"04:43.909","Text":"For now, all that we need to know is that we\u0027re using"},{"Start":"04:43.909 ","End":"04:48.004","Text":"the force T. A moment before m_2 reaches this point,"},{"Start":"04:48.004 ","End":"04:51.859","Text":"we know its velocity is v_2 and the rope is completely loose."},{"Start":"04:51.859 ","End":"04:53.000","Text":"At some point,"},{"Start":"04:53.000 ","End":"04:55.909","Text":"the rope starts to gain tension and this force T"},{"Start":"04:55.909 ","End":"04:59.375","Text":"starts getting larger and exerting force upwards on m_2."},{"Start":"04:59.375 ","End":"05:01.429","Text":"Let\u0027s actually draw out this force."},{"Start":"05:01.429 ","End":"05:05.300","Text":"T is acting upwards on m_2."},{"Start":"05:05.300 ","End":"05:09.065","Text":"It\u0027s also acting in an upwards direction on m_1."},{"Start":"05:09.065 ","End":"05:13.854","Text":"T, the force of tension is still the same throughout this rope."},{"Start":"05:13.854 ","End":"05:17.974","Text":"What I\u0027m looking at is from the first moment when our velocity"},{"Start":"05:17.974 ","End":"05:21.335","Text":"is still v_2 until we\u0027ve reached full tension."},{"Start":"05:21.335 ","End":"05:24.070","Text":"At that point, both of our masses should have moved"},{"Start":"05:24.070 ","End":"05:27.095","Text":"and they\u0027ll have moved at the same velocity,"},{"Start":"05:27.095 ","End":"05:29.119","Text":"because the length of the rope is not"},{"Start":"05:29.119 ","End":"05:32.919","Text":"changing and tension is unified throughout this rope."},{"Start":"05:32.919 ","End":"05:38.675","Text":"One understanding to take away is that whatever velocity m_1 moves at,"},{"Start":"05:38.675 ","End":"05:41.539","Text":"is the same velocity with which m_2 will move."},{"Start":"05:41.539 ","End":"05:43.280","Text":"We can think of it as velocity,"},{"Start":"05:43.280 ","End":"05:47.515","Text":"u, shared velocity between the 2 objects."},{"Start":"05:47.515 ","End":"05:50.315","Text":"It doesn\u0027t matter how T changes,"},{"Start":"05:50.315 ","End":"05:53.975","Text":"the velocity of m_1 and m_2 will remain the same."},{"Start":"05:53.975 ","End":"05:58.805","Text":"This fact doesn\u0027t have anything to do with the force T or the masses m,"},{"Start":"05:58.805 ","End":"06:01.174","Text":"rather it has to do with the fact that the length"},{"Start":"06:01.174 ","End":"06:03.755","Text":"of the rope is constant, it stays the same."},{"Start":"06:03.755 ","End":"06:07.039","Text":"If the mass m_2 has to move 1 meter,"},{"Start":"06:07.039 ","End":"06:11.230","Text":"then the mass m_1 also has to move 1 meter in the same amount of time."},{"Start":"06:11.230 ","End":"06:14.014","Text":"Their velocity will always be the same."},{"Start":"06:14.014 ","End":"06:17.555","Text":"Let\u0027s actually write down these 2 important facts."},{"Start":"06:17.555 ","End":"06:20.419","Text":"The first is that when the rope is tense,"},{"Start":"06:20.419 ","End":"06:25.510","Text":"the tension T is equal throughout the entire length of rope."},{"Start":"06:25.510 ","End":"06:28.069","Text":"As I wrote here, this is also true while"},{"Start":"06:28.069 ","End":"06:31.219","Text":"the tension is growing while the rope is getting tighter."},{"Start":"06:31.219 ","End":"06:34.324","Text":"Our second assumption is that when the rope is tense,"},{"Start":"06:34.324 ","End":"06:38.045","Text":"the velocity of both objects is also equal."},{"Start":"06:38.045 ","End":"06:40.859","Text":"These are 2 assumptions that are always true whenever"},{"Start":"06:40.859 ","End":"06:43.685","Text":"we have a rope with 2 objects and full tension."},{"Start":"06:43.685 ","End":"06:46.819","Text":"The reason I wanted to write them down is because we start with"},{"Start":"06:46.819 ","End":"06:50.645","Text":"a non-tense rope and we\u0027re talking about the process of tightening."},{"Start":"06:50.645 ","End":"06:52.850","Text":"It can sometimes be confusing to students."},{"Start":"06:52.850 ","End":"06:56.254","Text":"I want you to know that when we\u0027re going through the process of the rope tightening,"},{"Start":"06:56.254 ","End":"06:58.900","Text":"these assumptions are still true."},{"Start":"06:58.900 ","End":"07:02.929","Text":"Essentially what I\u0027m saying is we\u0027ll be using the same assumptions we always"},{"Start":"07:02.929 ","End":"07:07.325","Text":"use when dealing with a pulley and a rope that is tense with 2 masses."},{"Start":"07:07.325 ","End":"07:09.380","Text":"I\u0027m going to make this a little smaller to give us"},{"Start":"07:09.380 ","End":"07:12.899","Text":"more room to work and then we can address the problem."},{"Start":"07:14.390 ","End":"07:17.865","Text":"Let\u0027s start by looking at mass 2."},{"Start":"07:17.865 ","End":"07:22.490","Text":"During this short time when we go from totally loose to totally tense,"},{"Start":"07:22.490 ","End":"07:24.905","Text":"the velocity of mass 2 changes,"},{"Start":"07:24.905 ","End":"07:28.405","Text":"and therefore, so does the momentum."},{"Start":"07:28.405 ","End":"07:33.020","Text":"The change momentum, or Delta P of object 2,"},{"Start":"07:33.020 ","End":"07:39.330","Text":"we\u0027ll call that Delta P_2 equals the impulse applied to object 2."},{"Start":"07:39.330 ","End":"07:41.175","Text":"We\u0027ll call that J_2."},{"Start":"07:41.175 ","End":"07:43.645","Text":"What is the impulse J_2?"},{"Start":"07:43.645 ","End":"07:48.365","Text":"Essentially, it\u0027s the impulse applied by the force T tension."},{"Start":"07:48.365 ","End":"07:52.130","Text":"In theory, gravity is also having some effect here,"},{"Start":"07:52.130 ","End":"07:55.460","Text":"but because we\u0027re talking about a very short period of time,"},{"Start":"07:55.460 ","End":"07:58.520","Text":"we can consider gravity to be a negligible force,"},{"Start":"07:58.520 ","End":"08:01.590","Text":"just like we do with regular collisions."},{"Start":"08:01.640 ","End":"08:06.690","Text":"We can rewrite delta P_2=J_2 in a different form."},{"Start":"08:06.690 ","End":"08:08.104","Text":"It is as follows."},{"Start":"08:08.104 ","End":"08:11.899","Text":"If you recall, Delta P_2 or change momentum can be broken"},{"Start":"08:11.899 ","End":"08:16.085","Text":"down into the final momentum minus the initial momentum."},{"Start":"08:16.085 ","End":"08:19.609","Text":"That\u0027s m_2, v_2 final,"},{"Start":"08:19.609 ","End":"08:24.965","Text":"that\u0027s the velocity of m_2 after this whole process takes place,"},{"Start":"08:24.965 ","End":"08:30.290","Text":"minus m_2, v_2 initial,"},{"Start":"08:30.290 ","End":"08:33.560","Text":"that is the velocity before this collision,"},{"Start":"08:33.560 ","End":"08:35.614","Text":"so to speak, takes place."},{"Start":"08:35.614 ","End":"08:38.465","Text":"That is v_2 from above here."},{"Start":"08:38.465 ","End":"08:46.099","Text":"This has to equal an integral of Tdt our force over time."},{"Start":"08:46.099 ","End":"08:47.840","Text":"Before we do this integral,"},{"Start":"08:47.840 ","End":"08:49.975","Text":"we need to choose our direction."},{"Start":"08:49.975 ","End":"08:55.265","Text":"We could say that upwards is the y-axis that\u0027s positive,"},{"Start":"08:55.265 ","End":"08:57.815","Text":"and downwards is negative along the y-axis,"},{"Start":"08:57.815 ","End":"08:59.750","Text":"but for now, I want to do something different."},{"Start":"08:59.750 ","End":"09:03.894","Text":"Let\u0027s go with the direction of movement and say that on the left side,"},{"Start":"09:03.894 ","End":"09:07.060","Text":"y is going upwards,"},{"Start":"09:07.130 ","End":"09:09.510","Text":"and on the right side,"},{"Start":"09:09.510 ","End":"09:13.299","Text":"y is going downward so that at least at first,"},{"Start":"09:13.299 ","End":"09:16.369","Text":"our axis goes with the movement."},{"Start":"09:16.369 ","End":"09:19.730","Text":"Imagine that we\u0027re taking this rope and stretching it out without"},{"Start":"09:19.730 ","End":"09:23.499","Text":"the pulley along an axis up and down or side to side."},{"Start":"09:23.499 ","End":"09:25.594","Text":"This can be a little confusing."},{"Start":"09:25.594 ","End":"09:26.929","Text":"We have to be a little careful,"},{"Start":"09:26.929 ","End":"09:28.310","Text":"but for our purposes here,"},{"Start":"09:28.310 ","End":"09:30.904","Text":"this can be rather useful too. Let\u0027s try this."},{"Start":"09:30.904 ","End":"09:33.005","Text":"If these are the directions we\u0027ve chosen,"},{"Start":"09:33.005 ","End":"09:34.295","Text":"that means that T,"},{"Start":"09:34.295 ","End":"09:35.855","Text":"in this case, is negative."},{"Start":"09:35.855 ","End":"09:39.260","Text":"It\u0027s going upwards, whereas the y-axis goes downwards."},{"Start":"09:39.260 ","End":"09:42.170","Text":"We can say this is an integral of negative T,"},{"Start":"09:42.170 ","End":"09:45.169","Text":"or we can take the negative 1 and bring it to the outside,"},{"Start":"09:45.169 ","End":"09:48.130","Text":"say it\u0027s a negative integral of Tdt."},{"Start":"09:48.130 ","End":"09:51.754","Text":"For now, this is written about as simply as I can write it."},{"Start":"09:51.754 ","End":"09:56.300","Text":"Let\u0027s move on to the change in momentum of Object 1."},{"Start":"09:56.300 ","End":"10:03.240","Text":"Delta P1=J1, the impulse applied to mass 1."},{"Start":"10:03.240 ","End":"10:08.950","Text":"This can be rewritten as m_1v_1f, again,"},{"Start":"10:08.950 ","End":"10:15.079","Text":"the final velocity times mass 1 minus the initial velocity m_1 times 0."},{"Start":"10:15.079 ","End":"10:16.969","Text":"This can really be erased."},{"Start":"10:16.969 ","End":"10:19.580","Text":"We don\u0027t need to deal with it, equals 0."},{"Start":"10:19.580 ","End":"10:22.055","Text":"This equals, once again,"},{"Start":"10:22.055 ","End":"10:24.475","Text":"the integral of Tdt."},{"Start":"10:24.475 ","End":"10:28.819","Text":"In this case, T is going upwards,"},{"Start":"10:28.819 ","End":"10:30.349","Text":"the y-axis goes upwards,"},{"Start":"10:30.349 ","End":"10:32.314","Text":"so this is positive."},{"Start":"10:32.314 ","End":"10:34.714","Text":"Now that we\u0027ve reached this point,"},{"Start":"10:34.714 ","End":"10:38.270","Text":"I want to integrate these 2 assumptions from before. They\u0027re important for me."},{"Start":"10:38.270 ","End":"10:41.915","Text":"The first assumption is when the rope is tense or tightening,"},{"Start":"10:41.915 ","End":"10:44.420","Text":"the tension T is equal throughout the rope."},{"Start":"10:44.420 ","End":"10:47.659","Text":"That means that this T and this T are equal."},{"Start":"10:47.659 ","End":"10:50.074","Text":"This integral equals negative,"},{"Start":"10:50.074 ","End":"10:51.935","Text":"this integral, and vice versa."},{"Start":"10:51.935 ","End":"10:55.639","Text":"The second assumption is that when the rope is tense,"},{"Start":"10:55.639 ","End":"10:57.995","Text":"both objects have the same velocity."},{"Start":"10:57.995 ","End":"11:01.895","Text":"That means that our velocity at the end of this collision, so to speak,"},{"Start":"11:01.895 ","End":"11:06.590","Text":"is going to be the same. That means v_1f=v_2f."},{"Start":"11:06.590 ","End":"11:12.405","Text":"We could also say, v_1f=v_2f."},{"Start":"11:12.405 ","End":"11:14.650","Text":"We could even just call this u,"},{"Start":"11:14.650 ","End":"11:17.000","Text":"for the sake of simplicity."},{"Start":"11:17.000 ","End":"11:21.739","Text":"At this point, I can replace my v_2f and my v_1f with"},{"Start":"11:21.739 ","End":"11:26.544","Text":"u and say that these 2 things are equal to the negative of one another."},{"Start":"11:26.544 ","End":"11:28.460","Text":"If I try to plug this in,"},{"Start":"11:28.460 ","End":"11:35.400","Text":"I get that m_2, u_2 minus m_2."},{"Start":"11:35.400 ","End":"11:37.139","Text":"Instead of v_2i, for now,"},{"Start":"11:37.139 ","End":"11:39.680","Text":"I\u0027m going to write our velocity from above that we found."},{"Start":"11:39.680 ","End":"11:44.400","Text":"Remember this equals this root 2gH."},{"Start":"11:44.800 ","End":"11:50.405","Text":"This equals, instead of saying the negative integral of Tdt,"},{"Start":"11:50.405 ","End":"11:54.065","Text":"I can say that\u0027s equal to negative integral Tdt,"},{"Start":"11:54.065 ","End":"11:59.636","Text":"which equals negative m_1u."},{"Start":"11:59.636 ","End":"12:04.329","Text":"I can rearrange this equation by adding m_1 u to both sides and"},{"Start":"12:04.329 ","End":"12:09.804","Text":"then adding m_2 root 2gH to both sides and the result is as follows."},{"Start":"12:09.804 ","End":"12:18.640","Text":"M_2 plus m_1 times u equals m_2 and instead of writing"},{"Start":"12:18.640 ","End":"12:21.280","Text":"the root I\u0027ll rewrite this as it was before"},{"Start":"12:21.280 ","End":"12:27.370","Text":"v_2i and what we have here is something that looks a lot like a plastic equation."},{"Start":"12:27.370 ","End":"12:29.799","Text":"We have the final velocity,"},{"Start":"12:29.799 ","End":"12:32.310","Text":"u times the 2 masses,"},{"Start":"12:32.310 ","End":"12:34.342","Text":"this is our final momentum,"},{"Start":"12:34.342 ","End":"12:36.114","Text":"equals the initial momentum,"},{"Start":"12:36.114 ","End":"12:38.319","Text":"the initial velocity times the initial mass and"},{"Start":"12:38.319 ","End":"12:41.065","Text":"the other initial velocity of course is 0."},{"Start":"12:41.065 ","End":"12:44.350","Text":"I can describe this in the sense of a plastic equation."},{"Start":"12:44.350 ","End":"12:46.697","Text":"Imagine that I have 1 mass here,"},{"Start":"12:46.697 ","End":"12:48.189","Text":"we\u0027ll call this m_1,"},{"Start":"12:48.189 ","End":"12:51.190","Text":"on the end of a loose rope is m_2."},{"Start":"12:51.190 ","End":"12:56.680","Text":"M_2 starts with a velocity of v_2i and eventually"},{"Start":"12:56.680 ","End":"13:04.730","Text":"m_2 reaches tension over here and both objects will move together with a velocity of u."},{"Start":"13:04.830 ","End":"13:08.815","Text":"In this case it looks a lot like a plastic equation,"},{"Start":"13:08.815 ","End":"13:10.524","Text":"we could even consider it functionally"},{"Start":"13:10.524 ","End":"13:16.040","Text":"a plastic equation and we also know that we have conservation of momentum."},{"Start":"13:16.040 ","End":"13:20.520","Text":"Now we can write out the value of u and if the question"},{"Start":"13:20.520 ","End":"13:24.705","Text":"doesn\u0027t really ask us for this but it can be useful for later stages here,"},{"Start":"13:24.705 ","End":"13:30.650","Text":"so u equals m_2 root 2gH"},{"Start":"13:30.830 ","End":"13:38.110","Text":"over m_1 plus m_2."},{"Start":"13:38.110 ","End":"13:42.865","Text":"What we\u0027ve found is u and u is our shared velocity for both of our objects,"},{"Start":"13:42.865 ","End":"13:47.005","Text":"m_1 and m_2 after the rope has reached full tension."},{"Start":"13:47.005 ","End":"13:50.110","Text":"Now what we can do is use u to find what"},{"Start":"13:50.110 ","End":"13:53.379","Text":"part b is asking us for which is the change in momentum,"},{"Start":"13:53.379 ","End":"14:01.810","Text":"Delta p. The change in momentum we\u0027re trying to find is the total change in momentum,"},{"Start":"14:01.810 ","End":"14:05.305","Text":"p total or P_T, so that\u0027s for both objects."},{"Start":"14:05.305 ","End":"14:09.070","Text":"Well, you may think from our last part of the problem that you could just say that"},{"Start":"14:09.070 ","End":"14:13.299","Text":"because we have a conservation of momentum Delta p_ T is 0,"},{"Start":"14:13.299 ","End":"14:15.580","Text":"that\u0027s not really the case and the problem"},{"Start":"14:15.580 ","End":"14:18.070","Text":"here is that when we\u0027re calculating the change in"},{"Start":"14:18.070 ","End":"14:23.094","Text":"total momentum we have to have 1 axis for both of our objects."},{"Start":"14:23.094 ","End":"14:27.719","Text":"We can\u0027t have y going up on the left and y going down on the right,"},{"Start":"14:27.719 ","End":"14:32.260","Text":"we have to choose 1 direction for both of them otherwise it\u0027s very easy to get confused."},{"Start":"14:32.260 ","End":"14:34.164","Text":"For the sake of this problem,"},{"Start":"14:34.164 ","End":"14:38.785","Text":"let\u0027s assume that from now on our y-axis is always going upwards,"},{"Start":"14:38.785 ","End":"14:44.480","Text":"so it\u0027s going upwards on the left and it\u0027s going upwards on the right."},{"Start":"14:47.460 ","End":"14:52.753","Text":"First things first, let\u0027s rewrite this y, it\u0027s rather ugly."},{"Start":"14:52.753 ","End":"14:55.374","Text":"That\u0027s a little better I think."},{"Start":"14:55.374 ","End":"14:59.740","Text":"Now what we need to do is calculate Delta P total."},{"Start":"14:59.740 ","End":"15:03.339","Text":"The way we do that is take the final total momentum minus"},{"Start":"15:03.339 ","End":"15:06.010","Text":"the initial total momentum or you can break it"},{"Start":"15:06.010 ","End":"15:08.919","Text":"down and find the change in momentum of m_1"},{"Start":"15:08.919 ","End":"15:11.680","Text":"plus the change in momentum of m_2 and we have to"},{"Start":"15:11.680 ","End":"15:14.529","Text":"keep in mind that now the signs for m_2 are"},{"Start":"15:14.529 ","End":"15:17.379","Text":"reversed because the axis has changed from going"},{"Start":"15:17.379 ","End":"15:20.739","Text":"in a downward direction to going in an upwards direction."},{"Start":"15:20.739 ","End":"15:24.440","Text":"Let\u0027s make a little bit of room and begin."},{"Start":"15:25.560 ","End":"15:32.095","Text":"Delta p_ T equals m_1 final minus m_1 initial,"},{"Start":"15:32.095 ","End":"15:40.899","Text":"so p (m_1) final is m_1 times u and p( m_1) initial is 0 and"},{"Start":"15:40.899 ","End":"15:49.510","Text":"now we can do the same for m_2 and we get that it\u0027s m_2 times negative u for p final."},{"Start":"15:49.510 ","End":"15:52.449","Text":"Remember it\u0027s negative u because our action is going"},{"Start":"15:52.449 ","End":"15:56.320","Text":"downwards whereas our y-axis is now facing upwards"},{"Start":"15:56.320 ","End":"16:03.775","Text":"minus the initial momentum of m_2 which is m_2 times v_2 initial."},{"Start":"16:03.775 ","End":"16:06.219","Text":"That\u0027s also going to be negative because again,"},{"Start":"16:06.219 ","End":"16:08.950","Text":"v_2 initial was going downwards or in"},{"Start":"16:08.950 ","End":"16:14.149","Text":"the opposite direction of the current orientation of the y-axis."},{"Start":"16:15.210 ","End":"16:19.584","Text":"Now we can solve this but there\u0027s an easier way to find the solution."},{"Start":"16:19.584 ","End":"16:24.835","Text":"We can also think of Delta p_T as equaling J or the impulse."},{"Start":"16:24.835 ","End":"16:28.479","Text":"In our case, we\u0027ll call it J collision because again,"},{"Start":"16:28.479 ","End":"16:29.860","Text":"for all intents and purposes,"},{"Start":"16:29.860 ","End":"16:34.359","Text":"we can assume that we\u0027re talking about the impulse between when"},{"Start":"16:34.359 ","End":"16:37.769","Text":"the m_2 mass starts giving"},{"Start":"16:37.769 ","End":"16:41.459","Text":"the rope tension to the point when the rope is reaching full tension."},{"Start":"16:41.459 ","End":"16:43.485","Text":"We can think of this as a collision."},{"Start":"16:43.485 ","End":"16:47.515","Text":"We know that J or the impulse equals the"},{"Start":"16:47.515 ","End":"16:51.610","Text":"integral of the force that is applying the impulse and in this case,"},{"Start":"16:51.610 ","End":"16:55.929","Text":"for both m_1 and m_2 reason the force T and in"},{"Start":"16:55.929 ","End":"17:01.160","Text":"both cases the force T is acting in the same direction as the y-axis."},{"Start":"17:01.160 ","End":"17:05.820","Text":"Instead of canceling out we actually have 2 times the integral of T,"},{"Start":"17:05.820 ","End":"17:10.690","Text":"so the impulse equals 2 times the integral of Tdt."},{"Start":"17:11.070 ","End":"17:14.874","Text":"Now that we have our impulse setup we can solve."},{"Start":"17:14.874 ","End":"17:17.844","Text":"First of all, we know that the integral of Tdt"},{"Start":"17:17.844 ","End":"17:22.989","Text":"equals m_1 v_1F and v_1F is a final velocity,"},{"Start":"17:22.989 ","End":"17:26.085","Text":"meaning at the end of this collision so to speak."},{"Start":"17:26.085 ","End":"17:29.115","Text":"We know that the velocity at the end of the collision"},{"Start":"17:29.115 ","End":"17:32.130","Text":"we can also write as u v_1F equals u,"},{"Start":"17:32.130 ","End":"17:39.350","Text":"so we an rewrite this as 2m_1 times u."},{"Start":"17:39.350 ","End":"17:48.579","Text":"If we plug in our value for u from here we find that our final answer is"},{"Start":"17:48.579 ","End":"17:54.429","Text":"2m_1 times m_2 over"},{"Start":"17:54.429 ","End":"18:01.820","Text":"m_1 plus m_2 times the square root of 2gH."},{"Start":"18:04.350 ","End":"18:08.829","Text":"This is the answer to part b and we can"},{"Start":"18:08.829 ","End":"18:12.955","Text":"now move on to part c and I\u0027ll clean up a little bit so we have some more room to work."},{"Start":"18:12.955 ","End":"18:16.345","Text":"Here I have the answers to part a and part b."},{"Start":"18:16.345 ","End":"18:20.200","Text":"We have v_2 which is our initial velocity"},{"Start":"18:20.200 ","End":"18:24.924","Text":"for m_2 that\u0027s the velocity of our block before the rope is tense."},{"Start":"18:24.924 ","End":"18:30.640","Text":"We also have u, which is the final shared velocity of both m_1 and m_2."},{"Start":"18:30.640 ","End":"18:33.850","Text":"That\u0027s the velocity with which the 2 blocks move once"},{"Start":"18:33.850 ","End":"18:37.975","Text":"the rope is tense and we also have this,"},{"Start":"18:37.975 ","End":"18:44.785","Text":"the change in total momentum of m_1 and m_2 of our blocks on the pulleys."},{"Start":"18:44.785 ","End":"18:49.270","Text":"Now what we want to do is in part c we need to find"},{"Start":"18:49.270 ","End":"18:53.740","Text":"the impulse applied to the pulley by the ceiling during part a."},{"Start":"18:53.740 ","End":"18:56.079","Text":"We can really think of part a and part b as"},{"Start":"18:56.079 ","End":"18:58.809","Text":"the same time frame in the sense that we want to"},{"Start":"18:58.809 ","End":"19:04.015","Text":"know what impulse was applied before our rope reaches full tension,"},{"Start":"19:04.015 ","End":"19:06.890","Text":"that\u0027s the final moment in part b."},{"Start":"19:07.020 ","End":"19:14.049","Text":"What we need to do here is find another impulse we\u0027ll call this J ceiling because in"},{"Start":"19:14.049 ","End":"19:16.210","Text":"this case it\u0027s the ceiling that\u0027s applying"},{"Start":"19:16.210 ","End":"19:21.069","Text":"the impulse and the way we\u0027re going to do that is we need to do a sum of forces,"},{"Start":"19:21.069 ","End":"19:23.605","Text":"a free body diagram for our pulley."},{"Start":"19:23.605 ","End":"19:30.429","Text":"First we know that T is acting downwards against our y-axis on"},{"Start":"19:30.429 ","End":"19:36.894","Text":"this side and we also have T on the left side acting downwards against the y-axis."},{"Start":"19:36.894 ","End":"19:38.754","Text":"This T equals this T, this T equals this T,"},{"Start":"19:38.754 ","End":"19:41.045","Text":"they\u0027re all the same force,"},{"Start":"19:41.045 ","End":"19:47.710","Text":"1 just acts in reverse of the other and we also have T_2 will call it acting upwards."},{"Start":"19:47.710 ","End":"19:50.335","Text":"This is the force applied by the ceiling."},{"Start":"19:50.335 ","End":"19:53.469","Text":"We\u0027ll call it T_2, we could call it T ceiling as well."},{"Start":"19:53.469 ","End":"19:55.254","Text":"This is acting upwards."},{"Start":"19:55.254 ","End":"19:58.735","Text":"Now, we can do a sum of the forces."},{"Start":"19:58.735 ","End":"20:03.100","Text":"Sum of the forces acting on the pulley equals"},{"Start":"20:03.100 ","End":"20:10.555","Text":"T_2 minus 2T and we know that the pulley doesn\u0027t move."},{"Start":"20:10.555 ","End":"20:12.879","Text":"That means that there is no change in momentum,"},{"Start":"20:12.879 ","End":"20:16.250","Text":"that the sum of forces has to equal 0."},{"Start":"20:16.620 ","End":"20:23.875","Text":"If that\u0027s the case, then we can also understand that T_2 equals 2T."},{"Start":"20:23.875 ","End":"20:28.119","Text":"These 2 have to be equal for the sum of forces to be 0."},{"Start":"20:28.119 ","End":"20:29.710","Text":"Now if this is the case,"},{"Start":"20:29.710 ","End":"20:33.250","Text":"this actually helps us solve J ceiling,"},{"Start":"20:33.250 ","End":"20:37.390","Text":"so we know that J ceiling is the impulse of"},{"Start":"20:37.390 ","End":"20:41.725","Text":"the forces from the ceiling and the force and the ceiling is T_2,"},{"Start":"20:41.725 ","End":"20:45.850","Text":"so we\u0027re going to take a derivative of T_2dt,"},{"Start":"20:45.850 ","End":"20:49.180","Text":"but we know that T_2 equals 2T,"},{"Start":"20:49.180 ","End":"20:52.075","Text":"so we\u0027ll take instead a derivative of"},{"Start":"20:52.075 ","End":"21:00.565","Text":"2Tdt and it turns out that 2T is the exact same force that we have for J collision,"},{"Start":"21:00.565 ","End":"21:04.030","Text":"the impulse of the collision as we called it before."},{"Start":"21:04.030 ","End":"21:09.250","Text":"That means that the answer for c is the same as the answer for part b."},{"Start":"21:09.250 ","End":"21:17.860","Text":"The answer for part c is 2m_1 times m_2 over m_1 plus m_2 times square root of 2gH."},{"Start":"21:17.860 ","End":"21:23.560","Text":"This also makes sense intuitively because for the pulley not to move,"},{"Start":"21:23.560 ","End":"21:26.829","Text":"that means that the force that the ceiling exerts has to be equal"},{"Start":"21:26.829 ","End":"21:30.129","Text":"to the force that the masses exert on the pulley and the force"},{"Start":"21:30.129 ","End":"21:32.784","Text":"that the masses exert on the pulley have to be equal and"},{"Start":"21:32.784 ","End":"21:36.985","Text":"opposite to the forces that the pulley exerts on the masses."},{"Start":"21:36.985 ","End":"21:40.030","Text":"In that sense, the force exerted by the pulley on"},{"Start":"21:40.030 ","End":"21:42.850","Text":"the masses has to equal the force exerted by"},{"Start":"21:42.850 ","End":"21:50.390","Text":"the ceiling on the pulley and therefore we can say that J ceiling equals J collision."},{"Start":"21:50.940 ","End":"21:54.850","Text":"Of course the 1 thing we haven\u0027t yet mentioned is"},{"Start":"21:54.850 ","End":"21:57.880","Text":"the direction that our impulse or our forces are"},{"Start":"21:57.880 ","End":"22:04.695","Text":"acting and in the case of the total momentum change or the impulse of the collision,"},{"Start":"22:04.695 ","End":"22:10.269","Text":"we know that is going to be in the direction of the y-axis as we defined it and the same"},{"Start":"22:10.269 ","End":"22:12.610","Text":"is going to be true for J ceiling because"},{"Start":"22:12.610 ","End":"22:15.699","Text":"the force of the ceiling is acting upwards therefore,"},{"Start":"22:15.699 ","End":"22:17.515","Text":"the impulse is applied upwards."},{"Start":"22:17.515 ","End":"22:21.770","Text":"This is also in the direction of the y-axis."},{"Start":"22:23.670 ","End":"22:29.320","Text":"In part d, I\u0027m asked how high will m_1 rise assuming"},{"Start":"22:29.320 ","End":"22:34.480","Text":"that m_1 is greater than m_2 and that m_2 doesn\u0027t reach the ground."},{"Start":"22:34.480 ","End":"22:36.789","Text":"What I have is a pretty basic problem in"},{"Start":"22:36.789 ","End":"22:40.479","Text":"kinematics and the way I\u0027m going to approach this is the way that I"},{"Start":"22:40.479 ","End":"22:43.495","Text":"approached part b. I\u0027ll use my blue axis"},{"Start":"22:43.495 ","End":"22:47.960","Text":"and I can erase everything I used in part c that has to do with the collision."},{"Start":"22:49.680 ","End":"22:53.289","Text":"Now let\u0027s approach the problem in the following way."},{"Start":"22:53.289 ","End":"22:55.765","Text":"Let\u0027s assume that part d starts"},{"Start":"22:55.765 ","End":"23:03.935","Text":"the moment after our collision."},{"Start":"23:03.935 ","End":"23:05.440","Text":"By that I mean,"},{"Start":"23:05.440 ","End":"23:07.869","Text":"we\u0027re starting the moment that our rope has reached"},{"Start":"23:07.869 ","End":"23:11.380","Text":"full tension and the masses have reached the velocity u."},{"Start":"23:11.380 ","End":"23:14.410","Text":"Now we\u0027re dealing with the velocity u and we can say that"},{"Start":"23:14.410 ","End":"23:18.400","Text":"m_2 is going downwards with a velocity of u,"},{"Start":"23:18.400 ","End":"23:22.555","Text":"and that m_1 is going upwards with a velocity of u."},{"Start":"23:22.555 ","End":"23:25.959","Text":"Again, each has going with the y-axis because on the right side,"},{"Start":"23:25.959 ","End":"23:29.289","Text":"our y-axis points towards the ground and on the left side,"},{"Start":"23:29.289 ","End":"23:32.065","Text":"the y-axis points upwards."},{"Start":"23:32.065 ","End":"23:34.720","Text":"What I expect to happen here,"},{"Start":"23:34.720 ","End":"23:36.880","Text":"because m_1 is heavier than m_2,"},{"Start":"23:36.880 ","End":"23:38.289","Text":"is a greater mass,"},{"Start":"23:38.289 ","End":"23:42.129","Text":"is that m_1 will rise and slow down, and at the same time,"},{"Start":"23:42.129 ","End":"23:47.139","Text":"m_2 will fall and slow down and at some point before m_2 reaches the ground,"},{"Start":"23:47.139 ","End":"23:51.369","Text":"m_1 will reach its maximum height and m_1 will start falling again,"},{"Start":"23:51.369 ","End":"23:54.260","Text":"and m_2 will start rising again."},{"Start":"23:54.900 ","End":"23:59.545","Text":"From this point, I want to try to find my maximum for m_1,"},{"Start":"23:59.545 ","End":"24:03.655","Text":"the maximum height, and the way to do this is to first find acceleration."},{"Start":"24:03.655 ","End":"24:06.459","Text":"We need to do a free body diagram and account for all of"},{"Start":"24:06.459 ","End":"24:11.200","Text":"our forces because now we don\u0027t only have T going upwards for both,"},{"Start":"24:11.200 ","End":"24:14.185","Text":"we also have on the mass m_2,"},{"Start":"24:14.185 ","End":"24:18.389","Text":"the force of gravity m2g going"},{"Start":"24:18.389 ","End":"24:23.654","Text":"downwards and because our problem starts the moment after the two masses reach tension,"},{"Start":"24:23.654 ","End":"24:25.560","Text":"m_1 is no longer on the ground,"},{"Start":"24:25.560 ","End":"24:27.779","Text":"so that two will have to deal with gravity."},{"Start":"24:27.779 ","End":"24:33.640","Text":"We\u0027ll draw it like this downwards, m_1g."},{"Start":"24:33.640 ","End":"24:36.295","Text":"Now that we have our free body diagrams,"},{"Start":"24:36.295 ","End":"24:43.255","Text":"we can write out our sum of forces and the sum of forces remember,"},{"Start":"24:43.255 ","End":"24:46.225","Text":"for mass 2 our y-axis is pointing downwards,"},{"Start":"24:46.225 ","End":"24:52.779","Text":"so we have m2g going with the y-axis minus t,"},{"Start":"24:52.779 ","End":"24:54.820","Text":"which is going against the y-axis,"},{"Start":"24:54.820 ","End":"24:56.154","Text":"the opposite direction,"},{"Start":"24:56.154 ","End":"24:58.879","Text":"and that equals m2a."},{"Start":"24:59.250 ","End":"25:04.135","Text":"And the sum of forces for object 1, mass 1."},{"Start":"25:04.135 ","End":"25:05.679","Text":"Remember for mass 1,"},{"Start":"25:05.679 ","End":"25:08.095","Text":"the y-axis is pointing upwards."},{"Start":"25:08.095 ","End":"25:15.190","Text":"We have T going with the y-axis minus m_1g going against the y-axis,"},{"Start":"25:15.190 ","End":"25:19.730","Text":"going downwards, and this equals m_1a."},{"Start":"25:20.820 ","End":"25:25.930","Text":"Now we have two equations and we have two unknown variables,"},{"Start":"25:25.930 ","End":"25:29.470","Text":"a and t. First we can solve for a by"},{"Start":"25:29.470 ","End":"25:33.145","Text":"putting the two equations together and when we add the two equations,"},{"Start":"25:33.145 ","End":"25:35.380","Text":"the result is m2g."},{"Start":"25:35.380 ","End":"25:37.330","Text":"We have minus T plus T,"},{"Start":"25:37.330 ","End":"25:42.085","Text":"so those two will zero out minus m_1g"},{"Start":"25:42.085 ","End":"25:49.509","Text":"equals m_1 plus m_2 in parentheses times a."},{"Start":"25:49.509 ","End":"25:51.849","Text":"When we solve for a,"},{"Start":"25:51.849 ","End":"26:00.339","Text":"our result is that a equals m_2 minus m_1 over m_2"},{"Start":"26:00.339 ","End":"26:10.209","Text":"plus m_1 times g. This is our answer for a and keep in mind that according to part d,"},{"Start":"26:10.209 ","End":"26:12.460","Text":"m_1 is greater than m_2."},{"Start":"26:12.460 ","End":"26:16.870","Text":"That means that our acceleration will be negative and this makes intuitive sense."},{"Start":"26:16.870 ","End":"26:18.459","Text":"So we know we\u0027re on the right track,"},{"Start":"26:18.459 ","End":"26:20.589","Text":"because if the acceleration is negative,"},{"Start":"26:20.589 ","End":"26:21.699","Text":"it means, for example,"},{"Start":"26:21.699 ","End":"26:25.075","Text":"with m_2, if Y is going downwards,"},{"Start":"26:25.075 ","End":"26:30.280","Text":"that the deceleration is going upwards or you\u0027re decelerating so that m_2 is"},{"Start":"26:30.280 ","End":"26:33.039","Text":"slowing down as it goes towards the ground and then eventually"},{"Start":"26:33.039 ","End":"26:35.800","Text":"it will start rising again and the same with m_1,"},{"Start":"26:35.800 ","End":"26:38.485","Text":"but in reverse, the Y-axis faces"},{"Start":"26:38.485 ","End":"26:42.744","Text":"upwards and our acceleration is negative so as m_1 gets higher,"},{"Start":"26:42.744 ","End":"26:45.549","Text":"it starts decelerating and starts going slower until it"},{"Start":"26:45.549 ","End":"26:48.699","Text":"reaches its maximum and turns around and starts falling again."},{"Start":"26:48.699 ","End":"26:52.370","Text":"This result is in line with what we expected."},{"Start":"26:52.620 ","End":"26:55.945","Text":"Our next step to find our maximum,"},{"Start":"26:55.945 ","End":"26:57.984","Text":"let me put a little more room here."},{"Start":"26:57.984 ","End":"27:01.495","Text":"Our next step is to use a kinematic equation."},{"Start":"27:01.495 ","End":"27:05.649","Text":"This is a kinematic equation we can use when we have constant acceleration."},{"Start":"27:05.649 ","End":"27:10.854","Text":"That is v_f squared equals v_i"},{"Start":"27:10.854 ","End":"27:18.564","Text":"squared plus two a times the displacement delta x."},{"Start":"27:18.564 ","End":"27:21.370","Text":"In our case, we need to find v_f,"},{"Start":"27:21.370 ","End":"27:24.850","Text":"which is the final velocity and for both of our objects,"},{"Start":"27:24.850 ","End":"27:26.695","Text":"because we\u0027re looking at a maximum point,"},{"Start":"27:26.695 ","End":"27:29.199","Text":"the final velocity is 0."},{"Start":"27:29.199 ","End":"27:32.725","Text":"Because we know v initial, that\u0027s u,"},{"Start":"27:32.725 ","End":"27:37.800","Text":"we can write that in plus 2a and delta x,"},{"Start":"27:37.800 ","End":"27:40.650","Text":"our displacement, is the maximum height we\u0027re going to reach."},{"Start":"27:40.650 ","End":"27:41.955","Text":"We\u0027ll write that is little h,"},{"Start":"27:41.955 ","End":"27:44.355","Text":"remember this is not the same as large H,"},{"Start":"27:44.355 ","End":"27:46.305","Text":"the distance that m_2 drops."},{"Start":"27:46.305 ","End":"27:52.139","Text":"This is a different h, little h. From here we need to solve for little h and what we"},{"Start":"27:52.139 ","End":"27:59.440","Text":"get is that little h equals negative u squared over 2a."},{"Start":"27:59.940 ","End":"28:03.789","Text":"Again, the negative makes sense because remember a,"},{"Start":"28:03.789 ","End":"28:07.840","Text":"our acceleration is negative so ultimately our height will be positive."},{"Start":"28:07.840 ","End":"28:12.759","Text":"What we can do now is plug in our values for u and a and find"},{"Start":"28:12.759 ","End":"28:18.805","Text":"our final solution for h. If we plug in our value for u and for a,"},{"Start":"28:18.805 ","End":"28:20.440","Text":"we get that u,"},{"Start":"28:20.440 ","End":"28:21.610","Text":"if you\u0027re a call from above,"},{"Start":"28:21.610 ","End":"28:24.380","Text":"I can scroll up so you can see,"},{"Start":"28:25.500 ","End":"28:33.385","Text":"u equals m_2 square root"},{"Start":"28:33.385 ","End":"28:40.300","Text":"of 2gH over m_1 plus m_2,"},{"Start":"28:40.300 ","End":"28:43.570","Text":"I\u0027m a scroll back down so we have some more room,"},{"Start":"28:43.570 ","End":"28:46.554","Text":"and a, as we solved above,"},{"Start":"28:46.554 ","End":"28:50.619","Text":"is m_2 minus m_1 over m_2 plus"},{"Start":"28:50.619 ","End":"28:55.360","Text":"m_1 times g. Because we\u0027re dividing by a we\u0027ll take the inverse and remember,"},{"Start":"28:55.360 ","End":"28:58.464","Text":"we\u0027re going to incorporate this negative sign, I almost forgot."},{"Start":"28:58.464 ","End":"29:06.895","Text":"What we get is m_2 plus m_1 over and our whole denominator is"},{"Start":"29:06.895 ","End":"29:16.300","Text":"m_2 minus m_1 times g. We\u0027re going to take that times 2 because we\u0027re dividing by 2a."},{"Start":"29:16.300 ","End":"29:19.359","Text":"When we do our whole problem here,"},{"Start":"29:19.359 ","End":"29:26.454","Text":"we find that our solution for h equals m_2"},{"Start":"29:26.454 ","End":"29:30.490","Text":"over m_1 minus m_2 and the reason it\u0027s m_1"},{"Start":"29:30.490 ","End":"29:35.410","Text":"minus m_2 is I just incorporated our negative sign from here into our denominator."},{"Start":"29:35.410 ","End":"29:39.354","Text":"That is multiplied by h,"},{"Start":"29:39.354 ","End":"29:42.550","Text":"the square root of h over 2g."},{"Start":"29:42.550 ","End":"29:48.100","Text":"This is our solution for the maximum height reached by m_1 given the conditions"},{"Start":"29:48.100 ","End":"29:53.440","Text":"of Part d. Now we can move on to part e. In part e,"},{"Start":"29:53.440 ","End":"29:56.950","Text":"I\u0027m asked how much impulse does the ceiling apply to"},{"Start":"29:56.950 ","End":"30:01.914","Text":"the pulley from the time T equals 0 until m_1 reaches its maximum height."},{"Start":"30:01.914 ","End":"30:04.839","Text":"Basically, I want to know the impulse applied by"},{"Start":"30:04.839 ","End":"30:09.025","Text":"the ceiling over the course of the whole problem that we\u0027ve talked about so far."},{"Start":"30:09.025 ","End":"30:12.294","Text":"What I\u0027ve done is kept some relevant information up here."},{"Start":"30:12.294 ","End":"30:14.290","Text":"I didn\u0027t just do this by luck,"},{"Start":"30:14.290 ","End":"30:15.895","Text":"these are things that we\u0027re going to use."},{"Start":"30:15.895 ","End":"30:17.845","Text":"First of all, from part c,"},{"Start":"30:17.845 ","End":"30:21.129","Text":"we calculated the impulse applied by the ceiling during"},{"Start":"30:21.129 ","End":"30:24.940","Text":"the time that we call the collision and said that that equals j collision,"},{"Start":"30:24.940 ","End":"30:27.219","Text":"the impulse of the forces"},{"Start":"30:27.219 ","End":"30:34.300","Text":"t. I also left some pieces of information from part d because during part d,"},{"Start":"30:34.300 ","End":"30:39.055","Text":"when we were trying to find how high m_1 would rise or its maximum height,"},{"Start":"30:39.055 ","End":"30:43.135","Text":"we did that by calculating the acceleration and the forces"},{"Start":"30:43.135 ","End":"30:47.530","Text":"after the collision until the time when m_1 reaches its maximum height."},{"Start":"30:47.530 ","End":"30:51.830","Text":"This will also be relevant and we\u0027ll use all of these pieces of information."},{"Start":"30:52.020 ","End":"30:55.659","Text":"What I want to find here is j total,"},{"Start":"30:55.659 ","End":"31:00.279","Text":"and that is my total impulse applied by the ceiling."},{"Start":"31:00.279 ","End":"31:05.035","Text":"And the way I can do that is breakdown J total into three parts."},{"Start":"31:05.035 ","End":"31:08.305","Text":"If you recall, we can find J total by taking the sum of"},{"Start":"31:08.305 ","End":"31:13.060","Text":"all the different impulses of different periods of time or different objects."},{"Start":"31:13.060 ","End":"31:15.864","Text":"First we\u0027ll break it down into J_1."},{"Start":"31:15.864 ","End":"31:21.805","Text":"J_1 is going to be the impulse from the time that the rope is loose."},{"Start":"31:21.805 ","End":"31:23.440","Text":"This is from the time that the rope is"},{"Start":"31:23.440 ","End":"31:26.500","Text":"loose until the time that the rope bridge is full tension."},{"Start":"31:26.500 ","End":"31:29.259","Text":"This is what we talked about in part a."},{"Start":"31:29.259 ","End":"31:32.710","Text":"Then I\u0027m going to add to that J_2 and this"},{"Start":"31:32.710 ","End":"31:36.250","Text":"is going to be during the time of the collision."},{"Start":"31:36.250 ","End":"31:41.710","Text":"We can actually rename this J collision because it equals J collision from above."},{"Start":"31:41.710 ","End":"31:46.599","Text":"This is the impulse applied by the ceiling during the point when the rope reaches"},{"Start":"31:46.599 ","End":"31:53.155","Text":"full tension and the velocity of the masses changes and reaches u,"},{"Start":"31:53.155 ","End":"31:57.610","Text":"so we\u0027ll call this J collision."},{"Start":"31:57.610 ","End":"32:01.105","Text":"We need to add to that J_3,"},{"Start":"32:01.105 ","End":"32:05.755","Text":"and we\u0027ll call this after the collision or when we\u0027re reaching our maximum."},{"Start":"32:05.755 ","End":"32:08.125","Text":"This is after the collision,"},{"Start":"32:08.125 ","End":"32:10.614","Text":"until m_1 reaches its maximum height,"},{"Start":"32:10.614 ","End":"32:14.680","Text":"when m_2 is still falling and m_1 is still rising."},{"Start":"32:14.680 ","End":"32:18.740","Text":"So after collision until max."},{"Start":"32:18.750 ","End":"32:21.534","Text":"Let\u0027s start with J_1."},{"Start":"32:21.534 ","End":"32:26.440","Text":"With J_1 and when calculating the impulse for any period of time,"},{"Start":"32:26.440 ","End":"32:29.530","Text":"we know that what we\u0027re looking at is the impulse of"},{"Start":"32:29.530 ","End":"32:32.784","Text":"the force applied by the ceiling and that force is always,"},{"Start":"32:32.784 ","End":"32:36.145","Text":"as we identified earlier T_2 we\u0027ll call it."},{"Start":"32:36.145 ","End":"32:37.930","Text":"As we also said earlier,"},{"Start":"32:37.930 ","End":"32:44.294","Text":"T_2 equals 2T and T of course,"},{"Start":"32:44.294 ","End":"32:46.319","Text":"is the force of tension."},{"Start":"32:46.319 ","End":"32:51.435","Text":"Now in J_1 we\u0027re talking about the time before the rope reaches tension,"},{"Start":"32:51.435 ","End":"32:54.120","Text":"when the rope is loose."},{"Start":"32:54.120 ","End":"32:57.270","Text":"When the rope is loose, we know that tension equals 0,"},{"Start":"32:57.270 ","End":"32:58.334","Text":"there is no tension,"},{"Start":"32:58.334 ","End":"33:01.964","Text":"therefore 2T equals 0 and T_2,"},{"Start":"33:01.964 ","End":"33:04.805","Text":"or the force that we\u0027re calculating equals 0."},{"Start":"33:04.805 ","End":"33:07.389","Text":"In that case, J_1, the impulse of the force"},{"Start":"33:07.389 ","End":"33:11.890","Text":"T_2 until the rope reaches tension equals 0."},{"Start":"33:11.890 ","End":"33:19.314","Text":"This handles J_1 and we already know that J_2 = J collision from above."},{"Start":"33:19.314 ","End":"33:22.375","Text":"We have that here. Let\u0027s move on to J_3."},{"Start":"33:22.375 ","End":"33:25.284","Text":"Now J_3 is a little tricky. Why is that?"},{"Start":"33:25.284 ","End":"33:27.280","Text":"Because in the case of J-3,"},{"Start":"33:27.280 ","End":"33:33.025","Text":"J_3 does not equal the change in momentum m_1 + m_2."},{"Start":"33:33.025 ","End":"33:35.425","Text":"Remember for J collision,"},{"Start":"33:35.425 ","End":"33:38.485","Text":"we just used the change in momentum for m_1 and m_2,"},{"Start":"33:38.485 ","End":"33:40.810","Text":"but with J_3, that\u0027s not the case."},{"Start":"33:40.810 ","End":"33:43.750","Text":"That\u0027s because m_2 and m_1 are"},{"Start":"33:43.750 ","End":"33:46.749","Text":"both off the ground and they both have to deal with gravity."},{"Start":"33:46.749 ","End":"33:49.780","Text":"We have m_2g and m_1g to deal with."},{"Start":"33:49.780 ","End":"33:55.749","Text":"The change in total momentum for m_1 and m_2 = J_3"},{"Start":"33:55.749 ","End":"34:03.925","Text":"plus J of the force m_1g plus the force of m_2g."},{"Start":"34:03.925 ","End":"34:08.800","Text":"We could do is find the total change in momentum for masses 1 and 2."},{"Start":"34:08.800 ","End":"34:13.644","Text":"But then we\u0027d also need to find the impulse of gravity on each object and subtract that."},{"Start":"34:13.644 ","End":"34:15.310","Text":"There\u0027s a little bit complicated."},{"Start":"34:15.310 ","End":"34:19.059","Text":"Now the reason we have to account for the force of gravity here and"},{"Start":"34:19.059 ","End":"34:22.675","Text":"not forget about it is because of the time period we\u0027re discussing,"},{"Start":"34:22.675 ","End":"34:24.414","Text":"if you recall, for the collision,"},{"Start":"34:24.414 ","End":"34:27.745","Text":"we could disregard the force of gravity in terms of impulse,"},{"Start":"34:27.745 ","End":"34:30.850","Text":"because the time period was infinitesimally short or"},{"Start":"34:30.850 ","End":"34:34.180","Text":"so short that the result was very small,"},{"Start":"34:34.180 ","End":"34:36.895","Text":"approaching 0, and it was negligible."},{"Start":"34:36.895 ","End":"34:39.190","Text":"In this case, that\u0027s not true."},{"Start":"34:39.190 ","End":"34:41.290","Text":"In general, just to remind you,"},{"Start":"34:41.290 ","End":"34:51.190","Text":"J equals an integral of m_1 plus m_2,"},{"Start":"34:51.190 ","End":"34:55.660","Text":"so on so forth, times g dt."},{"Start":"34:55.660 ","End":"35:01.377","Text":"The result of that is m_1 plus"},{"Start":"35:01.377 ","End":"35:10.449","Text":"m_2 times g times the change in time Delta t. When the change in time is close to 0,"},{"Start":"35:10.449 ","End":"35:12.010","Text":"then because this is all a constant,"},{"Start":"35:12.010 ","End":"35:15.625","Text":"the whole thing is so close to 0 that it doesn\u0027t affect our problem."},{"Start":"35:15.625 ","End":"35:19.674","Text":"That was true with the collision when we were talking about a very quick turnaround time,"},{"Start":"35:19.674 ","End":"35:21.685","Text":"100th of a second at most."},{"Start":"35:21.685 ","End":"35:24.520","Text":"However, when we\u0027re talking about J_3,"},{"Start":"35:24.520 ","End":"35:28.674","Text":"the time period after the collision until m_1 reaches its maximum height."},{"Start":"35:28.674 ","End":"35:30.160","Text":"It could be a number of seconds."},{"Start":"35:30.160 ","End":"35:31.524","Text":"It could be longer than that."},{"Start":"35:31.524 ","End":"35:33.910","Text":"The point is that the time is long enough,"},{"Start":"35:33.910 ","End":"35:40.660","Text":"Delta t that is large enough that we can\u0027t neglect m_1g and m_2g."},{"Start":"35:40.660 ","End":"35:42.445","Text":"We have to account for these forces."},{"Start":"35:42.445 ","End":"35:45.970","Text":"These are not approaching 0 and therefore they are irrelevant."},{"Start":"35:45.970 ","End":"35:48.594","Text":"To bring us back to our problem,"},{"Start":"35:48.594 ","End":"35:50.604","Text":"in the case of J_3,"},{"Start":"35:50.604 ","End":"35:55.150","Text":"our impulse after the collision until we reach the maximum height we"},{"Start":"35:55.150 ","End":"36:01.040","Text":"do have to account for Jm_1g, and Jm_2g."},{"Start":"36:01.140 ","End":"36:05.859","Text":"Moving forward, what we can do is instead of calculating"},{"Start":"36:05.859 ","End":"36:11.023","Text":"the change in total momentum of m_1 and m_2 and calculating m_1g,"},{"Start":"36:11.023 ","End":"36:14.425","Text":"m_2g and the impulse of those and subtracting them."},{"Start":"36:14.425 ","End":"36:18.520","Text":"We\u0027re going to go directly and calculate the impulse of J_3 and the way we do"},{"Start":"36:18.520 ","End":"36:23.560","Text":"that is say that J_3 equals an integral of T_2. That\u0027s our force."},{"Start":"36:23.560 ","End":"36:27.579","Text":"Remember it\u0027s always the same force in this case, times dt."},{"Start":"36:27.579 ","End":"36:32.710","Text":"We can also say, because we know that T_2=2T"},{"Start":"36:32.710 ","End":"36:35.169","Text":"and in this case it does not equal 0 because"},{"Start":"36:35.169 ","End":"36:37.975","Text":"we\u0027re now onto the third portion of our problem."},{"Start":"36:37.975 ","End":"36:46.659","Text":"We can say that an integral of T_2 dt = an integral of 2T dt."},{"Start":"36:46.659 ","End":"36:51.490","Text":"The way that I\u0027ll calculate this is take the 2 equations above that we had,"},{"Start":"36:51.490 ","End":"36:52.810","Text":"if you recall in part d,"},{"Start":"36:52.810 ","End":"36:56.050","Text":"we said we had 2 unknown variables,"},{"Start":"36:56.050 ","End":"36:58.434","Text":"a and t and 2 equations."},{"Start":"36:58.434 ","End":"37:01.674","Text":"We can solve for t and then take an integral of that."},{"Start":"37:01.674 ","End":"37:08.380","Text":"Let\u0027s do that below. We know that T minus m_1g = m_1a."},{"Start":"37:08.380 ","End":"37:10.225","Text":"If we want to solve for T,"},{"Start":"37:10.225 ","End":"37:18.039","Text":"we can say that T=m_1 times g plus a."},{"Start":"37:18.039 ","End":"37:26.935","Text":"In that case, T=m_1 times g and we can incorporate our a value from above,"},{"Start":"37:26.935 ","End":"37:31.539","Text":"plus m_2 minus m_1 over"},{"Start":"37:31.539 ","End":"37:38.430","Text":"m_2 plus m_1 times g. We can take our g here to the outside."},{"Start":"37:38.430 ","End":"37:44.385","Text":"Our answer is that T=m_1g"},{"Start":"37:44.385 ","End":"37:53.499","Text":"times 2m_2 over m_1 plus m_2."},{"Start":"37:54.180 ","End":"37:58.089","Text":"Now that I know T, I can go back to this integral and"},{"Start":"37:58.089 ","End":"38:01.524","Text":"say that the integral equals 2 goes to the outside"},{"Start":"38:01.524 ","End":"38:10.825","Text":"T times Delta t. Delta t is my time period that we discussed earlier."},{"Start":"38:10.825 ","End":"38:16.540","Text":"In this case, our time period is from the moment"},{"Start":"38:16.540 ","End":"38:21.729","Text":"when the rope reaches full tension to the point when m_1 reaches its maximum height,"},{"Start":"38:21.729 ","End":"38:23.964","Text":"let\u0027s say it\u0027s here, t his is our maximum height,"},{"Start":"38:23.964 ","End":"38:28.690","Text":"small h. The whole time period that it takes to go from m_1 just leaving"},{"Start":"38:28.690 ","End":"38:34.795","Text":"the ground to m_1 reaching the height small h. Knowing all the information we know,"},{"Start":"38:34.795 ","End":"38:39.234","Text":"we can use kinematic equations to find Delta t or a range of time."},{"Start":"38:39.234 ","End":"38:46.129","Text":"Remember Delta t equals the time between the end of collision."},{"Start":"38:47.550 ","End":"38:57.460","Text":"I\u0027m going to do that coll and the max height."},{"Start":"38:57.460 ","End":"39:00.115","Text":"I have constant acceleration,"},{"Start":"39:00.115 ","End":"39:01.405","Text":"I have the masses,"},{"Start":"39:01.405 ","End":"39:03.999","Text":"I have u my initial velocity,"},{"Start":"39:03.999 ","End":"39:05.845","Text":"I have a final velocity that I know."},{"Start":"39:05.845 ","End":"39:09.040","Text":"I have enough information that I can use a kinematic equation."},{"Start":"39:09.040 ","End":"39:12.040","Text":"The kinematic equation I\u0027ll use is to find"},{"Start":"39:12.040 ","End":"39:16.090","Text":"m_1 maximum height and the time at which that occurs."},{"Start":"39:16.090 ","End":"39:22.960","Text":"V_1(t) as a function of time equals"},{"Start":"39:22.960 ","End":"39:32.739","Text":"V_1i initial plus a acceleration times Delta t,"},{"Start":"39:32.739 ","End":"39:36.144","Text":"our change in time and that is what we\u0027re trying to find."},{"Start":"39:36.144 ","End":"39:39.069","Text":"We know that V_1 as a function of time and we\u0027re"},{"Start":"39:39.069 ","End":"39:42.714","Text":"looking for the time at the maximum equals 0."},{"Start":"39:42.714 ","End":"39:47.965","Text":"We know that V_1i our initial velocity equals u."},{"Start":"39:47.965 ","End":"39:49.824","Text":"We can add to that a,"},{"Start":"39:49.824 ","End":"39:50.904","Text":"which we have over here."},{"Start":"39:50.904 ","End":"39:57.399","Text":"We can plug in when we need times Delta t. From here,"},{"Start":"39:57.399 ","End":"39:59.229","Text":"I can solve for Delta t."},{"Start":"39:59.229 ","End":"40:08.125","Text":"Delta t = u divided by a and its negative u divide it by a."},{"Start":"40:08.125 ","End":"40:10.929","Text":"Once again, this negative sign makes sense because"},{"Start":"40:10.929 ","End":"40:14.515","Text":"acceleration a is negative and our change in time should be positive."},{"Start":"40:14.515 ","End":"40:16.960","Text":"Negative times Negative equals a positive."},{"Start":"40:16.960 ","End":"40:21.280","Text":"Now we can solve for Delta t in terms of familiar terms,"},{"Start":"40:21.280 ","End":"40:23.350","Text":"in terms of given information."},{"Start":"40:23.350 ","End":"40:29.680","Text":"Because we know that u=m_1 plus m_2 over"},{"Start":"40:29.680 ","End":"40:38.335","Text":"m_2 times the square root of 2g large H,"},{"Start":"40:38.335 ","End":"40:40.300","Text":"they\u0027ll be over negative a."},{"Start":"40:40.300 ","End":"40:41.710","Text":"Because I\u0027ve put in the negative sign,"},{"Start":"40:41.710 ","End":"40:43.270","Text":"I\u0027ll switch around my signs here."},{"Start":"40:43.270 ","End":"40:47.710","Text":"It\u0027s now m_1 minus m_2 over"},{"Start":"40:47.710 ","End":"40:55.089","Text":"m_1 plus m_2 times g. When I do the math here,"},{"Start":"40:55.089 ","End":"41:03.029","Text":"what I\u0027ll find is that Delta t=m_1 plus m_2"},{"Start":"41:03.029 ","End":"41:13.759","Text":"squared over m_2 times m_1 minus m_2."},{"Start":"41:13.830 ","End":"41:17.740","Text":"All of this multiplied by the square root of"},{"Start":"41:17.740 ","End":"41:26.305","Text":"2H over g. Now that we have both Delta t and large T,"},{"Start":"41:26.305 ","End":"41:31.430","Text":"we can solve for J_3 and find our impulse during our third period of time."},{"Start":"41:33.150 ","End":"41:41.004","Text":"J_3=2T Delta t and I\u0027ll just write this all out at once for you."},{"Start":"41:41.004 ","End":"41:43.150","Text":"What you\u0027ll find is certain terms,"},{"Start":"41:43.150 ","End":"41:45.550","Text":"dropout m_1 plus m_2."},{"Start":"41:45.550 ","End":"41:48.760","Text":"We get rid of the squared here r_2 will dropout,"},{"Start":"41:48.760 ","End":"41:50.785","Text":"rg will simplify with the root."},{"Start":"41:50.785 ","End":"41:55.840","Text":"m_2 drops out here and here."},{"Start":"41:55.840 ","End":"41:58.495","Text":"What we\u0027re left with is the following."},{"Start":"41:58.495 ","End":"42:06.892","Text":"J_3=m_1 times m_1 plus m_2 over"},{"Start":"42:06.892 ","End":"42:13.225","Text":"m_1 minus m_2 times"},{"Start":"42:13.225 ","End":"42:17.515","Text":"root 32gH."},{"Start":"42:17.515 ","End":"42:20.845","Text":"At this point we have answers for J1 = 0,"},{"Start":"42:20.845 ","End":"42:23.484","Text":"J2 = J collision,"},{"Start":"42:23.484 ","End":"42:25.675","Text":"and J_3 we have written here."},{"Start":"42:25.675 ","End":"42:27.490","Text":"If we add all these together,"},{"Start":"42:27.490 ","End":"42:34.015","Text":"0 plus our n term here plus J collision will get J total for the ceiling."},{"Start":"42:34.015 ","End":"42:35.739","Text":"This will be your answer."},{"Start":"42:35.739 ","End":"42:38.709","Text":"I\u0027m going to skip this last little bit of algebra."},{"Start":"42:38.709 ","End":"42:41.094","Text":"I encourage you to do it on your own."},{"Start":"42:41.094 ","End":"42:42.819","Text":"Now that we\u0027ve done this,"},{"Start":"42:42.819 ","End":"42:46.285","Text":"we\u0027ve solved the problem and the whole exercise is done."},{"Start":"42:46.285 ","End":"42:50.155","Text":"With that, we\u0027ve finished Part E and we finished the exercise."},{"Start":"42:50.155 ","End":"42:52.224","Text":"This was not a simple exercise,"},{"Start":"42:52.224 ","End":"42:55.249","Text":"but in the end, I hope that it was all clear."}],"ID":9356}],"Thumbnail":null,"ID":5397}]

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