1 00:00:03,450 --> 00:00:06,330 We now would like to apply our energy momentum rule 2 00:00:06,330 --> 00:00:09,804 and momentum to analyze a one-dimensional elastic 3 00:00:09,804 --> 00:00:11,220 collision with no external forces. 4 00:00:11,220 --> 00:00:13,170 Let's remind ourselves, we'll call it 5 00:00:13,170 --> 00:00:20,340 the energy momentum equation said that V relative initial 6 00:00:20,340 --> 00:00:22,470 was equal to V relative final. 7 00:00:22,470 --> 00:00:28,530 So we have V1x initial minus V2x initial-- 8 00:00:28,530 --> 00:00:32,520 that's the x component of initial relative velocity-- 9 00:00:32,520 --> 00:00:37,920 is equal to the final x component 10 00:00:37,920 --> 00:00:40,170 of the relative velocity. 11 00:00:40,170 --> 00:00:42,270 And that was our energy momentum law. 12 00:00:42,270 --> 00:00:46,080 Now the momentum condition that it's constant 13 00:00:46,080 --> 00:00:53,810 was our equation V1x initial plus m2 V2x initial equals m1 14 00:00:53,810 --> 00:01:00,300 V1x final plus m2 V2x final. 15 00:01:00,300 --> 00:01:03,810 Now let's see how this linear system is much, much easier 16 00:01:03,810 --> 00:01:04,569 to solve. 17 00:01:04,569 --> 00:01:06,630 Let's look at the same problem that we solved 18 00:01:06,630 --> 00:01:11,130 before where m2 was equal 2m1. 19 00:01:11,130 --> 00:01:13,900 And also, we were in the laboratory frame, 20 00:01:13,900 --> 00:01:17,250 so V2x initial is 0. 21 00:01:17,250 --> 00:01:20,190 And that tells us that the initial velocity, 22 00:01:20,190 --> 00:01:23,370 relative velocity, is simply the velocity of object 1. 23 00:01:23,370 --> 00:01:26,670 So let's just write our two equations down again 24 00:01:26,670 --> 00:01:29,970 and see how much simpler our system is. 25 00:01:29,970 --> 00:01:38,610 V1x x initial is minus V1x final plus V2x final. 26 00:01:38,610 --> 00:01:40,710 So we have minus plus. 27 00:01:40,710 --> 00:01:45,000 And our momentum condition, remember V2x initial is 0. 28 00:01:45,000 --> 00:01:49,080 The m1 and 1, m2 will be 2m1. 29 00:01:49,080 --> 00:01:51,120 So we can cancel our m1s. 30 00:01:51,120 --> 00:01:57,600 And we get V1x initial equals V1x final. 31 00:01:57,600 --> 00:02:06,400 And m2 is twice m1, so there's a factor plus 2V2x final. 32 00:02:06,400 --> 00:02:10,740 Now I want to solve for our target variable. 33 00:02:10,740 --> 00:02:12,300 I look at these two equations. 34 00:02:12,300 --> 00:02:15,990 I can see almost immediately that if I add these two 35 00:02:15,990 --> 00:02:20,590 equations, V1x initial will cancel. 36 00:02:20,590 --> 00:02:29,930 And I get very simply by adding, we get 2V1x initial. 37 00:02:29,930 --> 00:02:35,160 And this is 3V2x final. 38 00:02:35,160 --> 00:02:41,190 Or V2x final is 2/3 V1x initial. 39 00:02:41,190 --> 00:02:45,390 And let's just call this equation 1 and 2. 40 00:02:45,390 --> 00:02:47,430 So we added. 41 00:02:47,430 --> 00:02:57,390 And now to find V1x final, let's see what we'll do there. 42 00:02:57,390 --> 00:03:00,370 So we can do this a variety of different ways. 43 00:03:05,990 --> 00:03:09,370 I think the simplest thing here to do, 44 00:03:09,370 --> 00:03:13,630 we could eliminate V2x final by multiplying through by minus 2. 45 00:03:13,630 --> 00:03:19,900 Or we can simply substitute in V2x final right here. 46 00:03:19,900 --> 00:03:24,970 And we get-- so let's do that. 47 00:03:24,970 --> 00:03:28,660 Let's substitute that in right there. 48 00:03:28,660 --> 00:03:34,270 And we get V1x initial equals V1x final 49 00:03:34,270 --> 00:03:42,100 plus 2 times 2/3 V1x initial. 50 00:03:42,100 --> 00:03:46,040 When we bring that over to the other side, 1 minus 4/3 51 00:03:46,040 --> 00:03:51,590 is minus 1/3 V1x initial equals V1x final. 52 00:03:54,400 --> 00:03:59,550 And at the cost of introducing a new concept, 53 00:03:59,550 --> 00:04:03,010 we've found the algebra much, much simpler 54 00:04:03,010 --> 00:04:05,600 to solve in this problem. 55 00:04:05,600 --> 00:04:08,050 And we can just double check our result 56 00:04:08,050 --> 00:04:11,380 that the initial velocity, relative velocity, 57 00:04:11,380 --> 00:04:13,450 was simply Vx1. 58 00:04:13,450 --> 00:04:19,750 And the final relative velocity, V relative final, 59 00:04:19,750 --> 00:04:27,610 is minus V1x final minus-- let's see, 60 00:04:27,610 --> 00:04:34,990 the final relative velocity is V1x final V2x final i hat. 61 00:04:34,990 --> 00:04:42,600 And when we put that in, we have minus 1/3 V1x initial 62 00:04:42,600 --> 00:04:50,800 minus 2/3 V1x initial i hat. 63 00:04:50,800 --> 00:04:54,840 And we have minus V1x initial i hat, which 64 00:04:54,840 --> 00:04:59,070 is minus V relative initial. 65 00:04:59,070 --> 00:05:03,000 And so we see that the relative velocity simply 66 00:05:03,000 --> 00:05:04,710 changed direction. 67 00:05:04,710 --> 00:05:06,840 This approach is much, much easier. 68 00:05:06,840 --> 00:05:10,470 And keep in mind that the energy momentum and the momentum laws 69 00:05:10,470 --> 00:05:14,940 are just rewriting our two fundamental constants 70 00:05:14,940 --> 00:05:18,470 of motion, kinetic energy and momentum.