1 00:00:04,010 --> 00:00:07,220 Let's consider a ball that is dropped 2 00:00:07,220 --> 00:00:13,160 from a certain height, h i, above the ground and this ball 3 00:00:13,160 --> 00:00:14,500 is falling. 4 00:00:14,500 --> 00:00:18,250 It hits the ground and it bounces up 5 00:00:18,250 --> 00:00:24,310 until it reaches some final height, h final. 6 00:00:24,310 --> 00:00:26,390 Now when the ball is colliding with the ground, 7 00:00:26,390 --> 00:00:27,820 there are collision forces. 8 00:00:27,820 --> 00:00:29,660 And in this problem what we like to do 9 00:00:29,660 --> 00:00:31,840 is figure out what the average force of the ground 10 00:00:31,840 --> 00:00:33,315 is on the ball. 11 00:00:36,310 --> 00:00:37,790 And that will be the normal force, 12 00:00:37,790 --> 00:00:43,120 the average normal force, on the ball during the collision. 13 00:00:43,120 --> 00:00:46,450 Now if we look at this ball dropping, 14 00:00:46,450 --> 00:00:50,020 it's going to lose a little bit of energy, 15 00:00:50,020 --> 00:00:52,460 because it's getting compressed at the collision. 16 00:00:52,460 --> 00:00:56,180 Let's look at an example of the actual ball dropping. 17 00:00:56,180 --> 00:00:58,390 As you can see in this high speed video, 18 00:00:58,390 --> 00:01:01,830 as the ball falls down, it collides with the ground. 19 00:01:01,830 --> 00:01:05,860 When it collides with the ground, it's compressed. 20 00:01:05,860 --> 00:01:09,230 And then as it rebounds upwards, the ball 21 00:01:09,230 --> 00:01:11,490 expands back to its original shape, 22 00:01:11,490 --> 00:01:14,450 but it doesn't quite get to the same height-- that's 23 00:01:14,450 --> 00:01:16,010 because when the ball is compressed, 24 00:01:16,010 --> 00:01:19,380 there's some deformation in the rubber structure of the ball, 25 00:01:19,380 --> 00:01:22,670 and it's not a completely elastic deformation. 26 00:01:22,670 --> 00:01:26,610 And so some of the energy is transformed into, first, 27 00:01:26,610 --> 00:01:30,420 molecular motions, which turn into thermal energy that's 28 00:01:30,420 --> 00:01:32,740 radiative into the environment. 29 00:01:32,740 --> 00:01:35,960 Let's look in particular at the details of the collision. 30 00:01:35,960 --> 00:01:39,750 If we look at it in slow motion what we have here-- 31 00:01:39,750 --> 00:01:43,350 and I'll draw a picture-- as the ball is colliding 32 00:01:43,350 --> 00:01:46,070 with the ground, ball compresses, 33 00:01:46,070 --> 00:01:48,100 expands as it goes upwards. 34 00:01:48,100 --> 00:01:51,000 And so we can draw a free body diagram of the ball 35 00:01:51,000 --> 00:01:55,070 with a normal force and a gravitational force. 36 00:01:55,070 --> 00:01:59,430 Now let's choose our positive direction up. 37 00:01:59,430 --> 00:02:02,900 So now what we'd like to do is apply the momentum principle 38 00:02:02,900 --> 00:02:06,540 to analyze the average normal force. 39 00:02:06,540 --> 00:02:10,889 And our momentum principle, remember is impulse. 40 00:02:10,889 --> 00:02:14,760 The force integrated over some time during the collision 41 00:02:14,760 --> 00:02:17,620 is equal to the change in momentum. 42 00:02:17,620 --> 00:02:20,110 So what we'd like to do is identify 43 00:02:20,110 --> 00:02:22,120 the states that are relevant. 44 00:02:22,120 --> 00:02:25,250 So it will have a state before, so what we'll do 45 00:02:25,250 --> 00:02:28,370 is we'll call this the before state, 46 00:02:28,370 --> 00:02:34,240 and that's right before the ball is hitting the ground. 47 00:02:34,240 --> 00:02:42,160 And we have an after state, and in the after state, 48 00:02:42,160 --> 00:02:46,040 the ball has now finished colliding with the ground, 49 00:02:46,040 --> 00:02:48,480 and it's now moving up with speed up. 50 00:02:48,480 --> 00:02:52,490 Now again, we're going to choose a positive up. 51 00:02:52,490 --> 00:02:55,180 Here on representing things as it speeds. 52 00:02:58,100 --> 00:03:00,950 One of the things, we need some times here, 53 00:03:00,950 --> 00:03:07,050 so let's say that t initial is zero, this is our final time. 54 00:03:07,050 --> 00:03:09,550 We'll call this time the before time, 55 00:03:09,550 --> 00:03:15,050 we'll just call this t before, and this is t after. 56 00:03:15,050 --> 00:03:18,690 And then our integral is going from before to 57 00:03:18,690 --> 00:03:20,950 after the momentum. 58 00:03:20,950 --> 00:03:25,370 And we can now apply the momentum principle. 59 00:03:25,370 --> 00:03:28,930 Well, this is a vector equation and we've chosen unit vectors 60 00:03:28,930 --> 00:03:32,300 up, so what we have here is the integral 61 00:03:32,300 --> 00:03:40,250 of from t before to t after of N minus mg, integrated over dt, 62 00:03:40,250 --> 00:03:43,220 and that's equal to the momentum at the y 63 00:03:43,220 --> 00:03:46,040 component of the momentum at t after, 64 00:03:46,040 --> 00:03:49,061 minus the y component of the momentum. 65 00:03:49,061 --> 00:03:50,560 We don't have a vector here anymore. 66 00:03:50,560 --> 00:03:54,329 The y component of the vector, t before. 67 00:03:54,329 --> 00:03:57,130 And so this is our expression of the momentum principle. 68 00:03:57,130 --> 00:03:59,570 Impulse causes momentum to change. 69 00:03:59,570 --> 00:04:06,380 Now we're assuming that the normal force just averaging it 70 00:04:06,380 --> 00:04:13,460 and so this intregral simply becomes N average minus 71 00:04:13,460 --> 00:04:19,320 ng, times the time of collision, is equal to-- now in here 72 00:04:19,320 --> 00:04:22,820 we can put the mass of the ball, we have the velocity. 73 00:04:22,820 --> 00:04:24,650 Now here's where we have to be a little bit 74 00:04:24,650 --> 00:04:28,160 careful, because we're looking at the y component. 75 00:04:28,160 --> 00:04:31,660 We chose speed downwards, that's in the negative y direction, 76 00:04:31,660 --> 00:04:36,270 so we have minus-- sorry, we're looking at after. 77 00:04:36,270 --> 00:04:40,680 We have plus V after, because this is going in the positive j 78 00:04:40,680 --> 00:04:42,500 direction. 79 00:04:42,500 --> 00:04:47,920 And over here we have a negative mass, 80 00:04:47,920 --> 00:04:51,140 but it's going in the minus direction, 81 00:04:51,140 --> 00:04:55,540 so we have a minus mV before, and so we 82 00:04:55,540 --> 00:05:00,170 get mass times V after plus V before. 83 00:05:00,170 --> 00:05:06,610 So our first result is that the normal force average. 84 00:05:06,610 --> 00:05:09,720 Let's bring the divide through by delta t, 85 00:05:09,720 --> 00:05:15,600 and bring the Ng term over, so we have m Va plus Vb, 86 00:05:15,600 --> 00:05:22,230 divided by delta t, plus mg. 87 00:05:22,230 --> 00:05:26,130 So we see that if the collision time is very short, 88 00:05:26,130 --> 00:05:29,220 then this average force is a little bit bigger. 89 00:05:29,220 --> 00:05:31,910 A long collision time, the average force 90 00:05:31,910 --> 00:05:33,940 a little bit smaller. 91 00:05:33,940 --> 00:05:37,530 Now from kinematics, we already have worked out the problem 92 00:05:37,530 --> 00:05:39,960 that the speed for an object that 93 00:05:39,960 --> 00:05:44,220 rises to a height, h final, this is the velocity afterwards, 94 00:05:44,220 --> 00:05:47,790 is just square root of 2g h final. 95 00:05:47,790 --> 00:05:52,500 And in a similar way, if an object is falling height h i, 96 00:05:52,500 --> 00:05:57,140 the speed when it gets to the bottom is 2ghi. 97 00:05:57,140 --> 00:06:00,640 And so now we can conclude with these substitutions 98 00:06:00,640 --> 00:06:08,780 that the average force equals m times square root of 2g h 99 00:06:08,780 --> 00:06:13,460 final, plus the square root of 2g h initial, 100 00:06:13,460 --> 00:06:18,970 over the collision time, plus and mg. 101 00:06:18,970 --> 00:06:21,580 And of course the collision time, 102 00:06:21,580 --> 00:06:25,750 we're saying is t after minus t before. 103 00:06:25,750 --> 00:06:28,580 And so that's how we can use the momentum principle 104 00:06:28,580 --> 00:06:33,430 to get an average expression for the normal force.