1 00:00:03,290 --> 00:00:06,310 We would like to now introduce a new methodological tool 2 00:00:06,310 --> 00:00:09,840 for analyzing problems that involve momentum transfers. 3 00:00:09,840 --> 00:00:13,800 And we call that tool momentum diagrams. 4 00:00:13,800 --> 00:00:17,310 Now, what we'd like to do is look 5 00:00:17,310 --> 00:00:22,620 at our fundamental idea, which was, for discrete objects, 6 00:00:22,620 --> 00:00:26,420 we have that-- involved in a collision-- we have 7 00:00:26,420 --> 00:00:30,010 that the external force integrated with respect 8 00:00:30,010 --> 00:00:34,890 to time-- in poles-- is equal to the change in momentum 9 00:00:34,890 --> 00:00:37,440 between two different states. 10 00:00:37,440 --> 00:00:41,570 So this is the momentum of the final state, 11 00:00:41,570 --> 00:00:46,560 and this is the momentum of the initial state. 12 00:00:46,560 --> 00:00:49,400 So when we want to analyze problems, 13 00:00:49,400 --> 00:00:52,780 how can we methodologically introduce a picture 14 00:00:52,780 --> 00:00:55,060 representation of our problems? 15 00:00:55,060 --> 00:00:58,800 So let's look at what we need to do first. 16 00:00:58,800 --> 00:01:03,650 We always need to choose a system that we're referring to. 17 00:01:03,650 --> 00:01:06,320 And with respect to the choice of that system, 18 00:01:06,320 --> 00:01:11,700 we need to also choose a reference frame. 19 00:01:11,700 --> 00:01:13,530 Now, once we've done that, we can now 20 00:01:13,530 --> 00:01:17,430 represent a collision with respect to this system. 21 00:01:17,430 --> 00:01:20,960 For instance, let's consider two objects-- as our system 22 00:01:20,960 --> 00:01:23,180 Object 1 and Object 2. 23 00:01:23,180 --> 00:01:26,000 They're moving on a frictionless, horizontal 24 00:01:26,000 --> 00:01:27,320 surface. 25 00:01:27,320 --> 00:01:30,330 And we're choosing as a reference frame, the ground 26 00:01:30,330 --> 00:01:32,600 frame. 27 00:01:32,600 --> 00:01:36,050 Now, what we'd like to do is identify two states. 28 00:01:36,050 --> 00:01:41,479 So for our initial state, we'll have-- 29 00:01:41,479 --> 00:01:43,670 if we're given some initial conditions-- what 30 00:01:43,670 --> 00:01:47,310 we'd like to do is represent each object with a velocity 31 00:01:47,310 --> 00:01:48,330 vector. 32 00:01:48,330 --> 00:01:51,070 So here, we'll write the 1 initial, 33 00:01:51,070 --> 00:01:54,800 and let's suppose this object is coming at it with the 2 34 00:01:54,800 --> 00:01:57,030 initial as an example. 35 00:01:57,030 --> 00:01:59,780 And once we've represented the velocities of the objects, 36 00:01:59,780 --> 00:02:02,980 we can write down their momentum in the initial state. 37 00:02:02,980 --> 00:02:08,220 Similarly, if these two objects collide and they're moving, 38 00:02:08,220 --> 00:02:11,920 we actually don't know which way those objects will end up 39 00:02:11,920 --> 00:02:13,110 moving. 40 00:02:13,110 --> 00:02:16,020 And so what we'd again like to do in our final state, 41 00:02:16,020 --> 00:02:20,600 after this collision, is to represent the velocities by, 42 00:02:20,600 --> 00:02:21,770 again, vectors. 43 00:02:21,770 --> 00:02:27,740 So this is V1 final and this is V2 final. 44 00:02:27,740 --> 00:02:31,220 And then, we can represent the change in momentum. 45 00:02:31,220 --> 00:02:35,400 So our momentum principle now becomes-- in this case, 46 00:02:35,400 --> 00:02:40,480 let's just assume that the external force here 47 00:02:40,480 --> 00:02:43,870 sums to zero-- we're assuming no friction. 48 00:02:43,870 --> 00:02:47,944 And then our momentum principle says that 0 49 00:02:47,944 --> 00:02:51,770 equals P final minus P initial. 50 00:02:51,770 --> 00:02:53,670 And now, we can read off those momentums 51 00:02:53,670 --> 00:02:57,350 as vectors on the diagram, and so what we have 52 00:02:57,350 --> 00:02:59,120 is that the final momentum will be 53 00:02:59,120 --> 00:03:00,870 equal to the initial momentum. 54 00:03:00,870 --> 00:03:05,330 So we can write down M1 final plus M2 55 00:03:05,330 --> 00:03:15,380 V2 final is equal to M1 V1 initial plus M2 V2 initial. 56 00:03:15,380 --> 00:03:19,910 And that's how we can represent a collision where there 57 00:03:19,910 --> 00:03:23,410 is no external forces and use our momentum principle 58 00:03:23,410 --> 00:03:26,710 to get an vector equation. 59 00:03:26,710 --> 00:03:28,940 Now, in many problems, you're given information. 60 00:03:28,940 --> 00:03:32,700 You might be given information about the speeds and magnitudes 61 00:03:32,700 --> 00:03:34,230 of the objects. 62 00:03:34,230 --> 00:03:36,740 And in order to then take this equation 63 00:03:36,740 --> 00:03:42,430 and represent it in speeds and directions or even components, 64 00:03:42,430 --> 00:03:46,230 we need to choose some coordinate system. 65 00:03:46,230 --> 00:03:50,220 So if we choose a coordinate system-- and that's 66 00:03:50,220 --> 00:03:56,290 the third step-- so suppose we choose a coordinate system, 67 00:03:56,290 --> 00:04:00,120 then we can start to look at two different representations 68 00:04:00,120 --> 00:04:01,890 for our problem. 69 00:04:01,890 --> 00:04:07,320 For instance, let's just choose this to be the i hat direction. 70 00:04:07,320 --> 00:04:13,680 Now, given that choice, we could describe the velocities 71 00:04:13,680 --> 00:04:15,570 in terms of components. 72 00:04:15,570 --> 00:04:22,210 Now this gets awkward-- V1i x component i hat. 73 00:04:22,210 --> 00:04:28,020 And similarly, we can write down all the velocities-- V2i-- 74 00:04:28,020 --> 00:04:35,680 as V2ix i hat, et cetera, for all the velocities. 75 00:04:35,680 --> 00:04:38,330 And our momentum equation in components 76 00:04:38,330 --> 00:04:44,000 then becomes, in the i hat direction, M1 V1ix 77 00:04:44,000 --> 00:04:46,310 plus-- well, here we have the final state, 78 00:04:46,310 --> 00:04:55,659 so let's make this consistent-- the final plus M2 V2 final x 79 00:04:55,659 --> 00:05:05,060 equals M1 V1 initial x plus M2 V2 initial x. 80 00:05:05,060 --> 00:05:08,290 And that's the same equation that we have as vectors, now 81 00:05:08,290 --> 00:05:10,370 expressed in terms of components. 82 00:05:10,370 --> 00:05:14,800 And recall that components can be positive or negative.