1 00:00:04,000 --> 00:00:06,370 Let's look at these two blocks here. 2 00:00:06,370 --> 00:00:11,470 Block 1 has a mass of 3m and Block 2 has a mass of 1 m. 3 00:00:11,470 --> 00:00:13,870 And they're colliding and sticking together. 4 00:00:13,870 --> 00:00:16,239 And then they're going to move in this direction 5 00:00:16,239 --> 00:00:20,230 and eventually, slide onto this rough surface here. 6 00:00:20,230 --> 00:00:22,480 What we want to figure out is, what 7 00:00:22,480 --> 00:00:26,250 was that initial velocity v0 with which 8 00:00:26,250 --> 00:00:28,990 the two blocks collided? 9 00:00:28,990 --> 00:00:34,450 So up to this part here-- before this x equals 0-- 10 00:00:34,450 --> 00:00:37,750 there's no external force acting on the system. 11 00:00:37,750 --> 00:00:40,510 So conservation of momentum holds, 12 00:00:40,510 --> 00:00:47,260 which means that p final equals p initial. 13 00:00:47,260 --> 00:00:49,240 So let's put this together. 14 00:00:49,240 --> 00:00:54,260 We're going to have 3m v0. 15 00:00:54,260 --> 00:00:56,530 That block goes in the positive i hat direction. 16 00:00:56,530 --> 00:01:01,550 Block 2 goes the opposite way, so we have m v0. 17 00:01:01,550 --> 00:01:05,830 And then at the end, when they're stuck together 18 00:01:05,830 --> 00:01:09,280 before here, we have to consider only this portion here 19 00:01:09,280 --> 00:01:12,220 that pertains to a frictionless surface. 20 00:01:12,220 --> 00:01:16,690 We have 4m v final. 21 00:01:16,690 --> 00:01:19,990 And so the m's fall out. 22 00:01:19,990 --> 00:01:21,230 This is 2. 23 00:01:21,230 --> 00:01:22,630 This is 4. 24 00:01:22,630 --> 00:01:29,440 So we'll see that the final velocity is v0 half. 25 00:01:29,440 --> 00:01:33,460 Now, in order to get to this v0, we also 26 00:01:33,460 --> 00:01:35,650 have to consider energy. 27 00:01:35,650 --> 00:01:39,009 And in particular, we have to use the work energy principle 28 00:01:39,009 --> 00:01:43,840 here, because this joint block is going to slide 29 00:01:43,840 --> 00:01:45,490 onto this rough surface here. 30 00:01:45,490 --> 00:01:48,315 So we can make use of the work energy principle, 31 00:01:48,315 --> 00:01:54,100 as we know what the initial conditions are at this point 32 00:01:54,100 --> 00:01:58,900 here, using the conservation of momentum 33 00:01:58,900 --> 00:02:02,090 from here in the first place. 34 00:02:02,090 --> 00:02:06,100 So we want to use work equals delta K, the work energy 35 00:02:06,100 --> 00:02:07,150 principle. 36 00:02:07,150 --> 00:02:10,090 Let's look at the work first. 37 00:02:10,090 --> 00:02:16,150 Work is the integral of F dot dS. 38 00:02:16,150 --> 00:02:18,520 And we're dealing with a one-dimensional problem 39 00:02:18,520 --> 00:02:24,040 here, so we can just write Fx dx. 40 00:02:24,040 --> 00:02:29,910 And we want to have this run from 0 to d. 41 00:02:29,910 --> 00:02:32,380 I should say here that this block eventually 42 00:02:32,380 --> 00:02:36,460 comes to rest at x equals d. 43 00:02:36,460 --> 00:02:39,290 We're going to use that in a second. 44 00:02:39,290 --> 00:02:42,760 So let's look at what this F is. 45 00:02:42,760 --> 00:02:45,850 That's the frictional force here at work. 46 00:02:45,850 --> 00:02:49,870 And actually, which direction does it go? 47 00:02:49,870 --> 00:02:52,840 It opposes the motion fk. 48 00:02:56,290 --> 00:03:04,630 So we're going to get a minus sign here, integral fk dx, 49 00:03:04,630 --> 00:03:10,600 again going from x prime equals 0 to x prime equals d. 50 00:03:10,600 --> 00:03:20,090 Now what is this fk ?We know that fk equals mu k N. 51 00:03:20,090 --> 00:03:23,230 And from doing a quick free body diagram, 52 00:03:23,230 --> 00:03:27,280 we'll see that we have 4mg going down, 53 00:03:27,280 --> 00:03:30,670 and we have a normal force going up. 54 00:03:30,670 --> 00:03:37,060 So we'll see that N equals 4mg, which we can then plug in here. 55 00:03:37,060 --> 00:03:38,980 And then that goes over there. 56 00:03:38,980 --> 00:03:41,470 And the coefficient of kinetic friction 57 00:03:41,470 --> 00:03:42,680 is actually given to us. 58 00:03:42,680 --> 00:03:47,140 So that nicely works out. 59 00:03:47,140 --> 00:03:52,560 So we're going to have minus integral x prime 0, x prime 60 00:03:52,560 --> 00:04:05,980 equals d, the coefficient first, b x squared, and then N 4mg dx. 61 00:04:05,980 --> 00:04:12,340 If we integrate that, we're going to get minus 4bmg. 62 00:04:15,820 --> 00:04:22,970 And we'll integrate this one, so that's x cubed over 3. 63 00:04:22,970 --> 00:04:24,860 And we're going to plug in this value here, 64 00:04:24,860 --> 00:04:28,370 so we're going to get a d, and that second term goes away. 65 00:04:28,370 --> 00:04:31,700 So we are left with d cubed here. 66 00:04:31,700 --> 00:04:33,430 So this is our work. 67 00:04:33,430 --> 00:04:38,380 And now we need to look at the change in kinetic energy 68 00:04:38,380 --> 00:04:40,720 between here and there. 69 00:04:40,720 --> 00:04:45,580 We already know that this block is at rest at x equals d. 70 00:04:45,580 --> 00:04:50,050 So if the velocity is 0, our kinetic energy will be 0, 71 00:04:50,050 --> 00:04:54,720 and so our K final minus K initial 72 00:04:54,720 --> 00:04:58,240 is going to be just minus initial. 73 00:04:58,240 --> 00:05:01,240 And that's initial here, which is the final situation 74 00:05:01,240 --> 00:05:02,800 of our collision. 75 00:05:02,800 --> 00:05:06,170 So we're going to look at the combined block. 76 00:05:06,170 --> 00:05:14,450 So we'll have minus 1/2 and we have 4mv final squared. 77 00:05:14,450 --> 00:05:18,550 We determined here already that the final velocity is 78 00:05:18,550 --> 00:05:25,030 the initial 1/2, so have minus. 79 00:05:25,030 --> 00:05:32,740 Here we're going to have 2m, and now v0 squared over 4. 80 00:05:32,740 --> 00:05:43,846 And that actually gives us minus 1/2 m v0 squared. 81 00:05:48,310 --> 00:05:51,610 OK, so now we can put this all together. 82 00:05:51,610 --> 00:05:53,810 We apply this work energy principle. 83 00:05:53,810 --> 00:06:02,080 So we're going to get minus 4/3 bmg d cubed 84 00:06:02,080 --> 00:06:08,860 equals minus 1/2 m v0 squared. 85 00:06:08,860 --> 00:06:14,810 We'll see that this and this and this and this goes. 86 00:06:14,810 --> 00:06:23,230 And so we're going to get v0 equals 8/3, 87 00:06:23,230 --> 00:06:30,640 and then we have bgd cubed and the square root of that. 88 00:06:30,640 --> 00:06:34,600 So this is our initial velocity with which the two blocks 89 00:06:34,600 --> 00:06:38,110 collided initially, then eventually went 90 00:06:38,110 --> 00:06:42,750 onto this rough surface and came to rest here at x equals d.