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A 15.0-kg block is attached to a very light horizontal spring of force constant 500.0 N>m and is resting on a frictionless horizontal table ($\textbf{Fig. E8.44}$). Suddenly it is struck by a 3.00-kg stone traveling horizontally at 8.00 m/s to the right, whereupon the stone rebounds at 2.00 m/s horizontally to the left. Find the maximum distance that the block will compress the spring after the collision.

$x=0.346 \mathrm{m}$

Physics 101 Mechanics

Chapter 8

Momentum, Impulse, and Collisions

Section 3

Momentum Conservation and Collisions

Moment, Impulse, and Collisions

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Rutgers, The State University of New Jersey

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so once again, we have a block so well given, uh, well given the saw office and the block want to think of it as a frictional surface and isa block sitting on right there? Um, the the mass off the block happens to be fifteen kilograms, so there's a spring, a touch to the block and the mass of the block happens to be fifteen kilograms like that. Oh, what you can say fifteen point zero kilograms on dead thing that's happening is that he's a bowl that's moving towards the block and the muss off the ball, um, happens to be the well. It's, you know, it's it's it's going to be struck by a stone. So this is a stone and the mass of the stone happens to be three kilograms. The stone is moving with an initial velocity towards the right, which is positive, and that initial blast e um, happens to be on. Well, this stone is going to go, you know it's going towards the right, and it's travelling at a horizontal velocity of eight meters per second. That's to the right, the initial velocity of the block. As you can see zero meters per second because it's not moving. Um, the stone is gonna bounce back. So at some point, at some point, this is kind of like what's happening before and then what's happening afterwards? Uh, you know this The block is going to collide with the not the block, but the stone will collide with the block. And obviously you see what's going to happen when there's a collision. Eyes going to be a push. So we still we still have this him scenario like that. But this time something new has happened. The ball, the stone has collided with the block, and, um, and the there is a movement. Okay, so if you think about this, there's a movement. There's a maximum distance off compression X, and that's what we want to find out. One to find, uh, ex. Remember, the ball is, is is bouncing off, so the ball is going to bounce off that way with the final velocity, not the ball, but the stone. The stone will bound off, bounce off with the final velocity. That happens to be two minutes for second. I mean, you could say negative to cause it's moving towards the left right now? No, On the block itself, as you can see, is moving towards the right with the final velocity off. We don't know. But you know we can. We can always determine that. So two types off laws that were using in this particular problem, the fast one is the law. Oh, off conservation, off energy. And then the second one is obviously the law ofthe momentum. Conservation of momentum. Okay, we're going to start with the Law of Conservation of Momentum, which say's that the initial momentum? Because to the final momentum, the initial momentum ofthe system is M s, the S I. That's the momentum off the stone, the initial plus the initial momentum off the block that's going to be equal to the final momentum off the stone. Plus the final. This is this is stone, plus the final momentum off the block. Okay, so that's that's what we have in this problem. We have the numbers, the numbers. We need eight meters per second. Remember? That's the initial velocity of the off the block off the off the stones are, and so of one assault for the final velocity off the block. You know, moving, moving things around. You get to see that the final velocity off the block, it becomes equal to the initial velocity of the block zero. So this whole thing will be zero. So then we have the the initial momentum off the stone, minus the final momentum off the stone. This one, when it moves to the left, becomes negative. Off are the the moss off the block. Okay, so that's the you know. That's kind of what we have in terms ofthe simplifying this problem. In the next page, we can show the, you know, the final certification steps. So the the final velocity off the block becomes equivalent to going back to the formula muss over stone. We can pull out that. Then we're left with the initial velocity off the stone, minus the final velocity. Oh, off the stone over the moss off the blocks must mass off. The stone is three kilograms. Our times must off the block. Rather, initial velocity of car off the stone is eight meters per second. Final of velocity. No, off of Stoney's negative two meters per second. And you can see that from the other side. This is negative too. Meters per second, and then you want to divide that by the mass of the block. And that's fifteen kilograms, fifteen kilograms. And so you know, the final velocity, which is a step in helping us get the distance. X for the block is two meters per second if you go back or goes to find that distance. So this time round in the next page, we're gonna use the laugh conservation off energy. Which means that Ah, this time it's focused on the spring itself. You know, the spring is, um the spring has elastic potential energy on the blockers. Connecticut remember the initial kinetic energy of the Glock zero, So we don't even have to think about that. So the final, um, this is the final Connecticut niche of the blocks of the final kinetic energy of the block us to be the same mers the elastic potential energy Come on off the spring. And that's what we're looking for. Um, and so, you know, one half massive block final velocity of block a squared equals to one health. Okay, X squared. Something else We also have to remember is that well, given the spring constant for the block and the spring constant happens to be five hundred. Oh, Newtons, the media. And so if we go back, uh, you know, we have that number we're trying to solve for X. We can divide both sides by one half. Okay, one half k, the house cancel and the case cancel on this side. And so we have of X squared because to Marcel Block and a final velocity of look and so off. Okay. And so the value of axes radical off. Massive block, kind of lost his block squared. All of okay. Ah, the the other thing that happens is now we can plug in the numbers. So fifteen kilograms. Time's final blast, You lock. We found two meters per second. Uh, we have to square that over five hundred Newton meter, so, you know, going back two. Using a calculator to see that this is very call fifteen times two squared. Close. That. Actually, I'm gonna go back and do divide by five hundred, then close that throwing, giving us that fifteen times two screwed right by five hundred. That gives us a zero point three four six. That's the distance. X zero point three four six uh, meters. So you know, once again, we use the law of conservation, of momentum and the law of conservation of energy. The first step was to find the final velocity off the block, and that happens to be two meters per second and then use that are together with the law of conservation, of energy to compare the kinetic energy of the block and elastic potential energy of spring helping us solve for the distance X hope you enjoy the video. Feel free to ask any questions and have a wonderful day. Okay, thanks. Bye.

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