1 00:00:03,480 --> 00:00:05,850 Let's compare kinetic energies in a two particle 2 00:00:05,850 --> 00:00:09,190 one-dimensional collision in different reference frames. 3 00:00:09,190 --> 00:00:11,040 So we could have one reference frame 4 00:00:11,040 --> 00:00:15,870 in which particle 1 is coming in and particle 2 is 5 00:00:15,870 --> 00:00:17,370 moving like that. 6 00:00:17,370 --> 00:00:21,750 And in particular-- so we can call this the ground frame. 7 00:00:21,750 --> 00:00:25,850 And now let's consider the center of mass frame. 8 00:00:25,850 --> 00:00:27,780 And in the center of mass frame, let's 9 00:00:27,780 --> 00:00:32,670 remind you that when we have two different reference frames, 10 00:00:32,670 --> 00:00:36,510 the velocity-- we'll call this the ground frame 11 00:00:36,510 --> 00:00:42,900 g-- in the ground frame is equal to the velocity of the object-- 12 00:00:42,900 --> 00:00:45,990 we'll actually write it this way, just unprime, 13 00:00:45,990 --> 00:00:50,890 V1-- the velocity in the center of mass frame-- 14 00:00:50,890 --> 00:00:55,560 so this is the velocity in the cm frame-- 15 00:00:55,560 --> 00:00:58,770 plus the relative velocities between the frames. 16 00:00:58,770 --> 00:01:01,290 And that's why that's the velocity of the center of mass. 17 00:01:01,290 --> 00:01:04,080 So this was our rule for describing 18 00:01:04,080 --> 00:01:07,077 how velocities change in different reference frames. 19 00:01:07,077 --> 00:01:08,910 And so we can draw the picture in the center 20 00:01:08,910 --> 00:01:17,400 of mass frame, V1 initial prime and V2 initial prime. 21 00:01:17,400 --> 00:01:19,650 Now let's compare kinetic energies 22 00:01:19,650 --> 00:01:21,460 in these different frames. 23 00:01:21,460 --> 00:01:24,600 So we know that the kinetic energy 24 00:01:24,600 --> 00:01:30,690 in the center of mass frame is just 1/2 m1 V1 initial squared 25 00:01:30,690 --> 00:01:37,650 prime-- put the prime there-- plus 1/2 26 00:01:37,650 --> 00:01:42,150 V2 initial prime squared, kinetic energy 27 00:01:42,150 --> 00:01:43,680 in the center of mass frame. 28 00:01:43,680 --> 00:01:47,789 How do we calculate the kinetic energy in the lab frame? 29 00:01:47,789 --> 00:01:50,190 Well, that's a little bit more complicated. 30 00:01:50,190 --> 00:01:52,710 And we'll need a little algebra to start that. 31 00:01:52,710 --> 00:01:56,640 So let's put kinetic energy in the ground frame. 32 00:01:56,640 --> 00:02:01,230 We know is 1/2 m1 V1 initial squared 33 00:02:01,230 --> 00:02:06,370 plus 1/2 m2 V2 initial squared. 34 00:02:06,370 --> 00:02:10,978 Now what I have to do is use the law, the velocity relationship. 35 00:02:10,978 --> 00:02:13,620 And this is going to take a little bit algebra. 36 00:02:13,620 --> 00:02:21,900 We have m1-- I'll write V1 prime plus Vcm. 37 00:02:21,900 --> 00:02:26,300 And remember that any quantity squared is the dot product, 38 00:02:26,300 --> 00:02:32,130 V dot V. So I'm going to dot this with itself. 39 00:02:32,130 --> 00:02:35,960 V1 initial prime plus the center of mass. 40 00:02:38,800 --> 00:02:40,579 And I have the second term which looks 41 00:02:40,579 --> 00:02:42,320 identical to this first term. 42 00:02:42,320 --> 00:02:43,820 I'll write it all the way down here. 43 00:02:43,820 --> 00:02:50,240 1/2 m2 V2 initial prime plus Vcm. 44 00:02:50,240 --> 00:02:59,810 Vector dot scalar dot product of V2 initial prime plus Vcm. 45 00:02:59,810 --> 00:03:02,300 Now when you take a dot product, remember 46 00:03:02,300 --> 00:03:03,860 there's four terms here. 47 00:03:03,860 --> 00:03:09,050 There's V1 prime dot V1 prime, which is just V1 prime squared. 48 00:03:09,050 --> 00:03:12,422 There's V center of mass dot V center of mass. 49 00:03:12,422 --> 00:03:13,880 So that's V center of mass squared. 50 00:03:13,880 --> 00:03:15,620 And then there's the cross term. 51 00:03:15,620 --> 00:03:18,290 And because they're identical, there's a factor of 2. 52 00:03:18,290 --> 00:03:20,690 And it will be repeated below. 53 00:03:20,690 --> 00:03:28,340 So the kinetic energy in the ground frame is 1/2 m1. 54 00:03:28,340 --> 00:03:31,760 So we'll take v1 i prime dotted with itself. 55 00:03:31,760 --> 00:03:34,940 That's V1 i prime squared. 56 00:03:34,940 --> 00:03:36,890 We have the cross term, which is a factor 57 00:03:36,890 --> 00:03:39,050 of 2, which will cancel this. 58 00:03:39,050 --> 00:03:42,810 So the cross term is . m1. 59 00:03:42,810 --> 00:03:44,990 That canceled the factor of 2. 60 00:03:44,990 --> 00:03:54,720 V1 i prime dot Vcm plus the Vcm with itself. 61 00:03:54,720 --> 00:03:58,980 So that's 1/2 m1 Vcm squared. 62 00:03:58,980 --> 00:04:01,769 Now I have exactly the same thing on the next one. 63 00:04:01,769 --> 00:04:02,810 So we'll write that down. 64 00:04:02,810 --> 00:04:11,240 1/2 V2 i prime squared plus m2 V2i prime dot Vcm. 65 00:04:11,240 --> 00:04:13,050 That's the same in both. 66 00:04:13,050 --> 00:04:18,420 Plus 1/2 m2 Vcm squared. 67 00:04:18,420 --> 00:04:21,260 Now let's look carefully at what we. 68 00:04:21,260 --> 00:04:25,260 Have 1/2 m1 V1 prime squared. 69 00:04:25,260 --> 00:04:27,050 This is an m2. 70 00:04:27,050 --> 00:04:29,960 1/2 m2 V2 i prime squared. 71 00:04:29,960 --> 00:04:36,010 So we have 1/2 m1 V1i prime squared 72 00:04:36,010 --> 00:04:39,722 plus 1/2 m2 V2i prime squared. 73 00:04:39,722 --> 00:04:41,930 And you're already noticing that's the kinetic energy 74 00:04:41,930 --> 00:04:44,090 in the center of mass frame. 75 00:04:44,090 --> 00:04:52,159 We have the total mass, 1/2 m1 plus m2 Vcm squared. 76 00:04:52,159 --> 00:04:54,530 I can just put a little check to show 77 00:04:54,530 --> 00:04:56,659 which terms I've done so far. 78 00:04:56,659 --> 00:04:58,850 And now here's the interesting one. 79 00:04:58,850 --> 00:05:12,710 We have m1 V1i prime plus m2 V2i prime dot Vcm. 80 00:05:12,710 --> 00:05:15,920 That represents this term and this term added together. 81 00:05:15,920 --> 00:05:20,360 But recall that the center of mass reference frame 82 00:05:20,360 --> 00:05:25,280 is defined by the condition that the total momentum 83 00:05:25,280 --> 00:05:28,190 in that frame is zero. 84 00:05:28,190 --> 00:05:33,800 So this term is zero. 85 00:05:33,800 --> 00:05:37,640 And thus, we get that the kinetic energy in the ground 86 00:05:37,640 --> 00:05:41,780 frame is equal to the kinetic energy 87 00:05:41,780 --> 00:05:49,280 in the center of mass frame plus 1/2 m1 plus m2 times Vcm 88 00:05:49,280 --> 00:05:51,050 squared. 89 00:05:51,050 --> 00:05:56,120 And that's how kinetic energy is in different reference frames. 90 00:05:56,120 --> 00:05:57,620 And the next thing we'll look at is 91 00:05:57,620 --> 00:06:01,810 how that changes when we have a collision.