WEBVTT 00:00:04.000 --> 00:00:06.370 Let's look at these two blocks here. 00:00:06.370 --> 00:00:11.470 Block 1 has a mass of 3m and Block 2 has a mass of 1 m. 00:00:11.470 --> 00:00:13.870 And they're colliding and sticking together. 00:00:13.870 --> 00:00:16.239 And then they're going to move in this direction 00:00:16.239 --> 00:00:20.230 and eventually, slide onto this rough surface here. 00:00:20.230 --> 00:00:22.480 What we want to figure out is, what 00:00:22.480 --> 00:00:26.250 was that initial velocity v0 with which 00:00:26.250 --> 00:00:28.990 the two blocks collided? 00:00:28.990 --> 00:00:34.450 So up to this part here-- before this x equals 0-- 00:00:34.450 --> 00:00:37.750 there's no external force acting on the system. 00:00:37.750 --> 00:00:40.510 So conservation of momentum holds, 00:00:40.510 --> 00:00:47.260 which means that p final equals p initial. 00:00:47.260 --> 00:00:49.240 So let's put this together. 00:00:49.240 --> 00:00:54.260 We're going to have 3m v0. 00:00:54.260 --> 00:00:56.530 That block goes in the positive i hat direction. 00:00:56.530 --> 00:01:01.550 Block 2 goes the opposite way, so we have m v0. 00:01:01.550 --> 00:01:05.830 And then at the end, when they're stuck together 00:01:05.830 --> 00:01:09.280 before here, we have to consider only this portion here 00:01:09.280 --> 00:01:12.220 that pertains to a frictionless surface. 00:01:12.220 --> 00:01:16.690 We have 4m v final. 00:01:16.690 --> 00:01:19.990 And so the m's fall out. 00:01:19.990 --> 00:01:21.230 This is 2. 00:01:21.230 --> 00:01:22.630 This is 4. 00:01:22.630 --> 00:01:29.440 So we'll see that the final velocity is v0 half. 00:01:29.440 --> 00:01:33.460 Now, in order to get to this v0, we also 00:01:33.460 --> 00:01:35.650 have to consider energy. 00:01:35.650 --> 00:01:39.009 And in particular, we have to use the work energy principle 00:01:39.009 --> 00:01:43.840 here, because this joint block is going to slide 00:01:43.840 --> 00:01:45.490 onto this rough surface here. 00:01:45.490 --> 00:01:48.315 So we can make use of the work energy principle, 00:01:48.315 --> 00:01:54.100 as we know what the initial conditions are at this point 00:01:54.100 --> 00:01:58.900 here, using the conservation of momentum 00:01:58.900 --> 00:02:02.090 from here in the first place. 00:02:02.090 --> 00:02:06.100 So we want to use work equals delta K, the work energy 00:02:06.100 --> 00:02:07.150 principle. 00:02:07.150 --> 00:02:10.090 Let's look at the work first. 00:02:10.090 --> 00:02:16.150 Work is the integral of F dot dS. 00:02:16.150 --> 00:02:18.520 And we're dealing with a one-dimensional problem 00:02:18.520 --> 00:02:24.040 here, so we can just write Fx dx. 00:02:24.040 --> 00:02:29.910 And we want to have this run from 0 to d. 00:02:29.910 --> 00:02:32.380 I should say here that this block eventually 00:02:32.380 --> 00:02:36.460 comes to rest at x equals d. 00:02:36.460 --> 00:02:39.290 We're going to use that in a second. 00:02:39.290 --> 00:02:42.760 So let's look at what this F is. 00:02:42.760 --> 00:02:45.850 That's the frictional force here at work. 00:02:45.850 --> 00:02:49.870 And actually, which direction does it go? 00:02:49.870 --> 00:02:52.840 It opposes the motion fk. 00:02:56.290 --> 00:03:04.630 So we're going to get a minus sign here, integral fk dx, 00:03:04.630 --> 00:03:10.600 again going from x prime equals 0 to x prime equals d. 00:03:10.600 --> 00:03:20.090 Now what is this fk ?We know that fk equals mu k N. 00:03:20.090 --> 00:03:23.230 And from doing a quick free body diagram, 00:03:23.230 --> 00:03:27.280 we'll see that we have 4mg going down, 00:03:27.280 --> 00:03:30.670 and we have a normal force going up. 00:03:30.670 --> 00:03:37.060 So we'll see that N equals 4mg, which we can then plug in here. 00:03:37.060 --> 00:03:38.980 And then that goes over there. 00:03:38.980 --> 00:03:41.470 And the coefficient of kinetic friction 00:03:41.470 --> 00:03:42.680 is actually given to us. 00:03:42.680 --> 00:03:47.140 So that nicely works out. 00:03:47.140 --> 00:03:52.560 So we're going to have minus integral x prime 0, x prime 00:03:52.560 --> 00:04:05.980 equals d, the coefficient first, b x squared, and then N 4mg dx. 00:04:05.980 --> 00:04:12.340 If we integrate that, we're going to get minus 4bmg. 00:04:15.820 --> 00:04:22.970 And we'll integrate this one, so that's x cubed over 3. 00:04:22.970 --> 00:04:24.860 And we're going to plug in this value here, 00:04:24.860 --> 00:04:28.370 so we're going to get a d, and that second term goes away. 00:04:28.370 --> 00:04:31.700 So we are left with d cubed here. 00:04:31.700 --> 00:04:33.430 So this is our work. 00:04:33.430 --> 00:04:38.380 And now we need to look at the change in kinetic energy 00:04:38.380 --> 00:04:40.720 between here and there. 00:04:40.720 --> 00:04:45.580 We already know that this block is at rest at x equals d. 00:04:45.580 --> 00:04:50.050 So if the velocity is 0, our kinetic energy will be 0, 00:04:50.050 --> 00:04:54.720 and so our K final minus K initial 00:04:54.720 --> 00:04:58.240 is going to be just minus initial. 00:04:58.240 --> 00:05:01.240 And that's initial here, which is the final situation 00:05:01.240 --> 00:05:02.800 of our collision. 00:05:02.800 --> 00:05:06.170 So we're going to look at the combined block. 00:05:06.170 --> 00:05:14.450 So we'll have minus 1/2 and we have 4mv final squared. 00:05:14.450 --> 00:05:18.550 We determined here already that the final velocity is 00:05:18.550 --> 00:05:25.030 the initial 1/2, so have minus. 00:05:25.030 --> 00:05:32.740 Here we're going to have 2m, and now v0 squared over 4. 00:05:32.740 --> 00:05:43.846 And that actually gives us minus 1/2 m v0 squared. 00:05:48.310 --> 00:05:51.610 OK, so now we can put this all together. 00:05:51.610 --> 00:05:53.810 We apply this work energy principle. 00:05:53.810 --> 00:06:02.080 So we're going to get minus 4/3 bmg d cubed 00:06:02.080 --> 00:06:08.860 equals minus 1/2 m v0 squared. 00:06:08.860 --> 00:06:14.810 We'll see that this and this and this and this goes. 00:06:14.810 --> 00:06:23.230 And so we're going to get v0 equals 8/3, 00:06:23.230 --> 00:06:30.640 and then we have bgd cubed and the square root of that. 00:06:30.640 --> 00:06:34.600 So this is our initial velocity with which the two blocks 00:06:34.600 --> 00:06:38.110 collided initially, then eventually went 00:06:38.110 --> 00:06:42.750 onto this rough surface and came to rest here at x equals d.