🎉 Announcing Numerade's $26M Series A, led by IDG Capital!Read how Numerade will revolutionize STEM Learning Numerade Educator ### Problem 77 Easy Difficulty # A movie stuntman (mass 80.0$\mathrm{kg}$) stands on a window ledge 5.0$\mathrm{m}$above the floor (Fig. 8.44 ). Grabbing a rope attached to a chandelier, he swings down to grapple with the movie's villain (mass 70.0$\mathrm{kg}$), who is standing directly under the chandelier. (Assume that thestuntman's center of mass moves downward 5.0$\mathrm{m} .$He releases the rope just as be reaches the villain.$)$(a) With what speed do the entwined foes start to slide across the floor? (b) If the coefficient of kinetic friction of their bodies with the floor is$\mu_{k}=0.250$, how far do they slide? ### Answer ## (a)$v_{2}=5.28 \mathrm{m} / \mathrm{s}$(b)$d=5.7 \mathrm{m}\$

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Moment, Impulse, and Collisions

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##### Andy C.

University of Michigan - Ann Arbor

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##### Jared E.

University of Winnipeg

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### Video Transcript

So in this question, we have a movie stunt, man. He stands on a window ledge of 5 m above before he swings down to grapple with the villain, Um, who is standing directly underneath the chandelier, which he is using to swing down with. Um, so we're to assume that the stunt man's center of mass moves downwards exactly 5 m when he does this, and he releases the rope just as he gets to the villain. So we're asked to calculate ah, couple things. Um, with what? Speed to the A stuntman and villain start to slide across the floor and were given the coefficient of kinetic friction and asked to calculate how far they slide. Um, S O, this is a very interesting question because it has a lot of different components to it. Um, we have to use quite a few different, um, different types of physics to solve this. So it's kind of a good one. Um, so here I'm going to draw someone as the stunt man. He starts 5 m above m two, which is the villain, and he kind of swings down so that he collides with the villain now in order to get there collective speed with which they're sliding across the floor. Um, we need Thio, analyze the collision, right? And in order to do that, we're gonna have to use conservation of momentum. But we'll need to know with what Speed the stuntman comes in and hits the villain first. So I'm first going to use a conservation of energy to calculate the speed of the stunt man just before he hits the villain. So we know that the final energy has to equal the initial energy. The initial energy here is just composed of gravitational potential. This is going to be M G H h i for height initial. We don't need to add a kinetic energy term because he starts from rest. And then the final energy is just going to be composed of through the kinetic energy of the stunt man because, um, he has gone through the complete height. He's fallen through the complete 5 m. Um, so we don't have any gravitational potential, and we've just got kinetic energy left at that point. So the EMS cancel from this equation on, we can arrange for V. We'll get the square root of two G h i and then it just plugging in 9.81 and 5 m for H. We got a value of 9.9 m per second. So just before, um, the stuntman hits the villain, he's going 9.9 m per second. Um and then they have a collision. And in order to figure out how much they are going to be, how fast they're going to be moving after the collision, we need to use conservation of momentum. So we're going to dio the same thing, but for momentum. So we're going to set the initial momentum equal to the final momentum and hear the initial is just going to be the stunt man swinging in at 9.9 meters per second. And then the final will be the two people sliding together. So initially, we just have the stunt man m one, um times his velocity initial velocity coming in, and then we have for the final momentum the two masses together because now they're stuck together times their final velocity, which is what we're trying to find. So we can easily rearrange this and all we need to do is just plug in those masses as well as the speed of the stunt man initially. So we're going to get 80/9 point times 9.9 divided by 1 50. That's the two masses added together 70 and 80 and so we'll get a speed here of 5.28 m per second. That is how fast they initially start sliding together at and then in part B. We need to calculate how far they're going to slide. So actually, in part B, this final velocity, the velocity after the collision, it becomes the initial velocity that we're considering here because now we're considering the motion of them just sliding across the floor. So the initial part of the slide they are going 5.28 m per second. Um, we're trying to find displacement, and we want to find the point where they're going to stop sliding. So the final velocity is going to be zero meters per second. So in order, Thio, um, complete this kinda Matics, um, calculation. We're going to need to know either time or acceleration. Since we have some information about the forces, Let's go ahead and try to calculate acceleration. So what we're going to use for that is f Net equals m A. And if we know that the couple of people are sliding to the right, then we know that the only force that's going to be acting on them in the horizontal direction is the kinetic friction. So F net is equal. Thio minus F k. Considering this to be the positive direction and that's gonna be equal Thio sometimes a So what is f k f k is mu k f n. And in this case, because we only have f n and F G in the Y direction and there's no movement in the Y direction, then FN is going to be equal to F. G. So we can just write this as u K F G is equal to m A. And we know F g s m times G. So the masses canceled here on. We can easily calculate acceleration as just negative. M u k Times G. So, um, UK was given in the question. It's 2.5 0.25 I apologize, and that's going to give a value for acceleration of negative. Yeah, 2.45 m per second squared And now that we have the acceleration, we can use just our Kinnah Matics to go ahead and calculate the distance so they can dramatics. Formula that has these four variables is VF squared equals V. I squared plus to 80. So if we rearrange this, we're going to bring the V I square to the other side and we're going to divide everything by two A. The initial speed is what we calculated from the previous part. The final speed is just zero. And so we got, um, minus 5.28 squared over two times negative 2.45 And that's great because the negatives will cancel. And numerically, we get a distance of 5.7 m. So they are going to slide 5.7 m based on that kinetic friction force. So that's your answer for Part B. And the answer for part A is appear and that's it.

McMaster University

#### Topics

Moment, Impulse, and Collisions

##### Andy C.

University of Michigan - Ann Arbor

LB
##### Jared E.

University of Winnipeg

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