1 00:00:03,430 --> 00:00:05,530 So we analyzed our one-dimensional collision 2 00:00:05,530 --> 00:00:09,670 where we had an object 1 moving with some initial velocity, 3 00:00:09,670 --> 00:00:13,740 object 2 also moving with some initial velocity. 4 00:00:13,740 --> 00:00:16,570 And after the collision, we just arbitrarily 5 00:00:16,570 --> 00:00:19,840 said object 1 is moving this way and object 2 6 00:00:19,840 --> 00:00:22,250 was moving that way. 7 00:00:22,250 --> 00:00:25,620 And we called that our i hat direction. 8 00:00:25,620 --> 00:00:27,280 In this collision, we assume that it 9 00:00:27,280 --> 00:00:30,460 was a frictionless surface. 10 00:00:30,460 --> 00:00:32,080 And there were no external forces. 11 00:00:32,080 --> 00:00:33,730 So momentum is constant. 12 00:00:33,730 --> 00:00:36,100 Now what about energy? 13 00:00:36,100 --> 00:00:42,310 Well, we know that if there is no external forces, then 14 00:00:42,310 --> 00:00:46,790 in principle there could be no external work. 15 00:00:46,790 --> 00:00:49,300 However, during the collision, we've 16 00:00:49,300 --> 00:00:51,640 established different types of collisions. 17 00:00:51,640 --> 00:00:56,980 We said elastic collision is when 18 00:00:56,980 --> 00:01:03,670 we are assuming that the kinetic energy of the system-- that's 19 00:01:03,670 --> 00:01:10,360 K system final minus K system initial-- 20 00:01:10,360 --> 00:01:14,620 is 0, the statement that the kinetic energy is constant. 21 00:01:14,620 --> 00:01:22,560 We have an inelastic collision in which 22 00:01:22,560 --> 00:01:26,460 delta K system has decreased. 23 00:01:26,460 --> 00:01:28,896 And that can come from some deformation of the objects 24 00:01:28,896 --> 00:01:29,770 during the collision. 25 00:01:29,770 --> 00:01:30,660 Heat is generated. 26 00:01:30,660 --> 00:01:33,560 Some sound, some other source of energy 27 00:01:33,560 --> 00:01:37,170 in which energy is always constant, 28 00:01:37,170 --> 00:01:41,170 but the kinetic energy of the system could diminish. 29 00:01:41,170 --> 00:01:46,289 And we also talked about a super elastic collision. 30 00:01:46,289 --> 00:01:49,080 And in this type of collision, the kinetic energy 31 00:01:49,080 --> 00:01:52,390 of the system has increased. 32 00:01:52,390 --> 00:01:54,720 Now that's a little bit tricky in terms of what 33 00:01:54,720 --> 00:01:56,160 we're calling our system. 34 00:01:56,160 --> 00:01:59,280 But imagine that when these two objects collided, 35 00:01:59,280 --> 00:02:02,250 there was some type of chemicals on it that exploded. 36 00:02:02,250 --> 00:02:04,380 And so that was some chemical energy 37 00:02:04,380 --> 00:02:08,710 that was converted into extra kinetic energy. 38 00:02:08,710 --> 00:02:12,360 So in order to make a model of our problem, 39 00:02:12,360 --> 00:02:15,780 we have to beforehand make some assumptions about the nature 40 00:02:15,780 --> 00:02:17,020 of the collision. 41 00:02:17,020 --> 00:02:18,960 Now for our particular collision, 42 00:02:18,960 --> 00:02:22,540 let's assume that the collision is elastic. 43 00:02:22,540 --> 00:02:24,630 So we have two conservation principles. 44 00:02:24,630 --> 00:02:27,780 We have that the kinetic energy of the system is constant. 45 00:02:27,780 --> 00:02:29,700 And we've already said that because there's 46 00:02:29,700 --> 00:02:33,270 no external forces, the momentum of the system is constant. 47 00:02:33,270 --> 00:02:37,650 So now we can write down our kinetic energy condition. 48 00:02:37,650 --> 00:02:43,462 We have m1 V1-- now here, this can be a little bit tricky 49 00:02:43,462 --> 00:02:45,420 how we're going to write this, because remember 50 00:02:45,420 --> 00:02:50,220 kinetic energy is a magnitude squared 51 00:02:50,220 --> 00:03:00,360 plus 1/2 m2 V2 initial squared equals 1/2 m1 V1 final squared 52 00:03:00,360 --> 00:03:06,040 plus 1/2 m2 V2 final squared. 53 00:03:06,040 --> 00:03:09,810 Now because a component, although it 54 00:03:09,810 --> 00:03:12,900 can be positive or negative, in one dimension, 55 00:03:12,900 --> 00:03:16,170 if you square the component, you get the magnitude. 56 00:03:16,170 --> 00:03:18,300 So I can also write this equation-- 57 00:03:18,300 --> 00:03:21,390 and I'm going to divide through by the halves, 58 00:03:21,390 --> 00:03:27,750 cancel all those-- as 1x initial squared equal to the magnitude 59 00:03:27,750 --> 00:03:36,400 squared plus m2 V2x initial squared equals M1 60 00:03:36,400 --> 00:03:45,235 V1x final squared plus 1/2 m2 V2x final squared. 61 00:03:51,750 --> 00:03:54,500 And we've already canceled all the halves, 62 00:03:54,500 --> 00:03:57,230 so we have no half there. 63 00:03:57,230 --> 00:04:02,210 Now given this equation-- we'll call this equation 64 00:04:02,210 --> 00:04:08,780 1-- we can also had our momentum equation, 65 00:04:08,780 --> 00:04:13,640 which I'm going to write down as 2, m1 Vx 66 00:04:13,640 --> 00:04:25,790 initial plus m2 V2x initial equals m1 V1x final plus m2 V2x 67 00:04:25,790 --> 00:04:28,460 final, reminding us of our condition 68 00:04:28,460 --> 00:04:30,350 of the constancy of momentum. 69 00:04:30,350 --> 00:04:35,570 So now, if you look at this, if we're given the masses, all 70 00:04:35,570 --> 00:04:40,880 the mi's and we're also given the initial state 71 00:04:40,880 --> 00:04:45,950 of the system, V2x initial, then we 72 00:04:45,950 --> 00:04:47,940 have two equations and two unknowns. 73 00:04:47,940 --> 00:04:57,740 And we can solve algebraically for V1x final and V2x final. 74 00:04:57,740 --> 00:05:01,320 And by determining the signs of these components, 75 00:05:01,320 --> 00:05:04,850 we can figure out the actual final state of the system. 76 00:05:04,850 --> 00:05:10,070 So this is going to end up involving a quadratic equation. 77 00:05:10,070 --> 00:05:13,760 And what we'd like to do now is show an alternative way 78 00:05:13,760 --> 00:05:18,800 that gives us another kind of conservation, another principle 79 00:05:18,800 --> 00:05:21,410 to analyze these one dimensional collisions. 80 00:05:21,410 --> 00:05:24,160 So we'll begin that.