1 00:00:03,860 --> 00:00:06,580 We would now like to use Newton's second law to relate 2 00:00:06,580 --> 00:00:08,630 impulse to change in momentum. 3 00:00:08,630 --> 00:00:10,900 So again, let's look at our set up. 4 00:00:10,900 --> 00:00:14,560 We have an object m, a velocity V, 5 00:00:14,560 --> 00:00:19,150 and, let's say, here the picture as t initial. 6 00:00:19,150 --> 00:00:21,230 And we have some initial velocity. 7 00:00:21,230 --> 00:00:27,700 And then a little bit later in time, we have time t final. 8 00:00:27,700 --> 00:00:30,070 The moment the velocity has changed. 9 00:00:30,070 --> 00:00:32,729 And that's because throughout this time interval, 10 00:00:32,729 --> 00:00:34,860 we're applying an impulse. 11 00:00:34,860 --> 00:00:38,500 We can call this the i hat direction. 12 00:00:38,500 --> 00:00:41,460 Now recall that our definition of impulse, 13 00:00:41,460 --> 00:00:43,440 it's a vector quantity. 14 00:00:43,440 --> 00:00:47,110 It's equal to the integral of the force. 15 00:00:47,110 --> 00:00:49,300 Now when I write force of t, I mean force 16 00:00:49,300 --> 00:00:51,280 as a function of time. 17 00:00:51,280 --> 00:00:53,410 And that's our dummy variable t prime. 18 00:00:53,410 --> 00:00:56,920 It's not force times time, but force is a function of time. 19 00:00:56,920 --> 00:00:59,860 And we're integrating from the initial 20 00:00:59,860 --> 00:01:02,800 to the final time period. 21 00:01:02,800 --> 00:01:06,400 Now here's where we use the second-- 22 00:01:06,400 --> 00:01:09,880 the version of Newton's second law, 23 00:01:09,880 --> 00:01:13,690 which is that force causes the momentum of an object 24 00:01:13,690 --> 00:01:15,000 to change. 25 00:01:15,000 --> 00:01:20,500 So we can write this integral t prime t initial, t prime equals 26 00:01:20,500 --> 00:01:24,050 t final of dp dt. 27 00:01:24,050 --> 00:01:26,350 I'll just make a note that we've now 28 00:01:26,350 --> 00:01:30,960 applied Newton's second law, and because we're using 29 00:01:30,960 --> 00:01:33,490 our dummy variable dt prime 30 00:01:33,490 --> 00:01:37,259 And you can see that the two dt primes cancel. 31 00:01:37,259 --> 00:01:39,470 And this just becomes the integral 32 00:01:39,470 --> 00:01:46,780 from t initial t prime to t final of dp. 33 00:01:46,780 --> 00:01:48,720 And this is a pure differential. 34 00:01:48,720 --> 00:01:55,870 And so we end up with impulse is the momentum at t final minus 35 00:01:55,870 --> 00:01:59,280 the momentum at t initial. 36 00:01:59,280 --> 00:02:03,960 Now recall, this is a vector integral. 37 00:02:03,960 --> 00:02:06,420 But a vector integral is just three separate intervals 38 00:02:06,420 --> 00:02:07,700 for each component. 39 00:02:07,700 --> 00:02:12,290 So each component of impulse satisfies this equation. 40 00:02:12,290 --> 00:02:15,450 For instance, the x component of impulse 41 00:02:15,450 --> 00:02:21,710 is how the x component of momentum is changing in time. 42 00:02:21,710 --> 00:02:25,170 Now generally, when we write a final momentum 43 00:02:25,170 --> 00:02:29,430 in our final state minus the momentum during initial state, 44 00:02:29,430 --> 00:02:33,760 we can call this the change in momentum. 45 00:02:33,760 --> 00:02:38,800 So down here, we would have change in the x direction. 46 00:02:38,800 --> 00:02:44,079 And so in conclusion, we now have an integral way 47 00:02:44,079 --> 00:02:48,410 of casting Newton's second law. 48 00:02:48,410 --> 00:02:52,260 We have that impulse causes momentum to change. 49 00:02:52,260 --> 00:02:57,840 And so we can see that the si units of impulse 50 00:02:57,840 --> 00:03:05,020 are the same as the si units of momentum, 51 00:03:05,020 --> 00:03:10,140 which we saw before was kilogram meter inverse seconds.