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A 5.00-g bullet is fired horizontally into a 1.20-kg wooden block resting on a horizontal surface. The coefficient of kinetic friction between block and surface is 0.20. The bullet remains embedded in the block, which is observed to slide 0.310 m along the surface before stopping. What was the initial speed of the bullet?

$266 \mathrm{m} / \mathrm{s},$ which rounds to $270 \mathrm{m} / \mathrm{s}$

Physics 101 Mechanics

Chapter 8

Momentum, Impulse, and Collisions

Section 3

Momentum Conservation and Collisions

Moment, Impulse, and Collisions

Cornell University

Hope College

University of Winnipeg

McMaster University

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welcome to a new problem. This time we're given a block the town of sliding surface. That's so it's sitting at rest on our sliding saw office. That means that the initial velocity off the block zero meters per second so its not moving. That's what we're given on DH. Then also, we're told that the coefficient off kinetic friction for this office zero point two, um, another piece of information that given is the mass off the block, which happens to be one point two zero kilograms who are also given the bullet. So there is a bullet that's fired. So this is kind of like before, like, you know what happens before. So a bullet, uh, the mass off the bullet we're going to call it hell is given us five grams and we can change that five grams right away to kilograms by saying this is more played that one kilogram of a thousand thousand grams and that gives us five times ten to the negative three kilograms. That's the mass of the bullet. The bullet is moving with an initial velocity. Oh, that happens to be time. The So while the bull the bullet hasn't initial velocity. We still we still don't know what that is, and that's what we want to find out. You know, What's the initial speed off the bullet? That's our goal. Given all this information, the second aspect of the problem is once the bullet hits the block and sits inside of it, the block starts to move in in in the right direction. And so the outcome of this movement is a distance travel equivalent to So we're going to call that as equals two zero point three one zero meters. That's the distance traveled. It's going to come to a stop, so the block is going to stop at some point. Okay, The block is going to stop at some point, and that's the distance covered. These almost some would say some resistance from friction. So there's a fictional force pointing in the opposite direction off the motion. Therefore, Beth and we can always compute that friction. Okay, we can compute that frictional force. So two ideas we're gonna use we're gonna use the law ofthe conservation of momentum. And also we're gonna use thie computations for frictional force. So that's the fasting. Wanna compute? You know, what's the frictional force between the block and this office. Remember, the force of friction happens to be, you know, we need the frictional force because it's going to help us determine the work done by fiction. We need that. We have the frictional forces mu times the normal force. Remember, when this thing is moving, there's a normal force pointing that way. And then the's are, uh, wait, you know the weight weight off the block Now here it's going to be the weight off the block and the bullet, you know? So we have the mass ofthe block plus the mass off the bullet together. So, you know, the fasting wanted to is obviously compute that friction off. Also, this becomes f of Africa. Lt's too mu times the mass. You know, this is this is kind of like the combined mass off the block and the bullet times the acceleration due to gravity. So that's that's the information we were given. And then on top of that, we want to find the walk down by friction. And that's just the friction ofthe force times the displacement, this s displacement right here. And so the work done by friction is mu times em be. Bus him ill rg times s That's the work done by friction. The second part of the problem we have to Wei have. So our goal is to find the initial velocity of the bullet law off conservation of energy. So here we're using law off a conservation of energy. Andi was saying that the home So we're using to laws, you know, faster. Voles love conservation of momentum because we're using the law of conservation of momentum. And then also, you know, the walk energy theory. So let's let's kind of, like, clarify that, huh? We want to know what that's what's so they're So they're two components here. So we have We have the work done by friction done by friction. It's going to be equivalent to the no changing kinetic energy off the system, which is the walk done. Bye. The bullet on the block. So it's kind of like that already have walked down by friction mu Um um you so so here were right. What done by friction is equal to walk down by the bullet plus the block and walk down by fiction for this side is you, um mbm l O M B plus m. Also, massive bullet and muscle block G has Then we want to find the change in kinetic energy off system. So that's one half, uh, M V squared. Well, we're involving the mass off the bullet, plus the moss off the block. No home, you know, one half. And these every squared men times the velocity, which is the final velocity off the system. Remember, if you go back, you see that you know the system that's going to be moving this way with the final velocity. Okay, It's going to be moving towards the right with a final glass. So we have in the initial velocity of the bullet and then the final blast, You off the off the off the bullet and the block. But this is the final blast of the off the block and the bullet come back. This is the final blast of the block and the bullet that was square that because we square that problem s o. Then our goal is to find that final velocity because it's going to help us solve certain through the initial velocity of the bullet. So final velocity off the block and the bullet becomes a notice. We can divide both sides by that. This's always they had Vonage off using variables up until the end because something special can happen like that. And so you're left with a simplified problem. We can multiply both sides by two. And so you know, we have, um too. Mew G. Yes. Two mute G s. This is squared. That's what we have right there. And then that can help us get the final velocity off the block and the bullet, which is just this quote of too mu gs. That problem. Um, if you plug in the numbers, you Khun, get an actual value in that case, in simplifying that problem. So, um, you point two times nine point eight. So this is screwed point two times nine point eight times when, two times ninety eight times the displacement, which is point three one zero. That's what it is. Let's just check if we got steps right. Thor Displacement is point three one zero. Wait, Miss out on something? I don't think so. Put two o. Yeah. We had a massive bullet Marcel's Doc. So point to those two combined weight. There's the future force. Yes. So going to come back and you can see we're doing the connection. The work done by friction, which is mu uh, m b m L G s. We have that right here. You can be. And then we also have the, um, the kinetic energy of the system. That's the change in kinetic energy. No, minus the initial Connecticut and just the system. Remember thee Initial kinetic energy of the system on ly involves the on ly involves two with a bullet. So the initial the initial kinetic energy of the system zero meters zero because the bullet, the bullet and blocked off the the initial velocity off the block I zero. You know, the who the wooden, but the old and block is at rest. Okay, so they wouldn't lock is his address. So we can We can always ignore that nose, even all the initial velocity on DH then you know, plugging in this numbers we get to bring the two up. So we have two new g s here we have two mute g s. Um okay. And then the final final mass. No, the final muss. So one half him him final squared, You know, just just to check that our numbers of right this is too mu. Yes. So let's just come from that. This is okay. The masses, the masses, the masses disappeared. Marcel's disappeared. So we got square root. Poof. Oh, two times you times nine point eight times as which is point three zero on. That should give us very good. Yeah. One the right truck. So, you know, this is this Ah, final velocity. You know, you can compute it just to test final velocity for the bullet on the block on. That's fine. But the best way is just to keep the variables up until up until the end. So we're going on and now go to use the law of conservation of momentum. So using the law ofthe conservation off momentum, we just say the initial momentum off the system across the final momentum of the systems are want to think about the initial momentum off system. And that's the mass off the bullet times the initial velocity of the bullet, plus the mass off the block times the initial velocity off the block because to the mask off the block, plus the bullet. Oh, time's a common final glass city off the block and the bullet. Remember, we already have this from this part of the problem. So come back and plug in numbers. We do know that the initial velocity off the block zero this one zero because it's addressed. So this is gonna be zero. And our target is to find the initial velocity off the bullet. So was saying mass off the bullet they should lost if the bullet equals two Marcel block to us. Marshal. Bullet times the velocity of the block and the bullet. Uh, this is our target who divide both sides by the mass ofthe bullet. And now it Khun comfortably plug in numbers. What? We have mass off the bullet, which is a computer to be zero point. Or rather, you know, we could We could write it in terms ofthe scientific figures because of the way then Abaza. So we have five times ten to the negative three kilograms. Thats the Marcel's bullet plus the moss off the block which is one point two kilograms. Actually, we should switch positions because that left side is the block on the bright side is well that just to make sure you comfortable with the process. So one point two kilograms and then plus five times ten to the negative three kilograms like that? No. All of this has to be multiplied by this velocity. Which one would go back? We find it one point one zero meters per second and all that's divided by the mass of the bullet, which is five times ten to the negative three kilograms. So you plug in the numbers and we can comfortably say that the initial velocity off the bullet happens to be two hundred and sixty five point one meters per second. Okay, so that's how you run that problem. We're given a problem. Bullet and the block. We have initial scenario and then we have a final scenario way. Use the love, you know, love, conservation, off energy at some point and then way also use the laugh conservation of momentum at some point. But we have to get the friction on force because that's the one that's going to help us to deal with the law of conservation, of energy and the frictional forces. The coefficient ofthe kinetic friction, which you call me okay, times the normal force. But this is the normal force involving about the mass of the bullet and the block. How you compute that, the work done by friction. And then we compare the work done by friction to the work done by the bullet and the block. We simplify always a good habit to keep their variables up until the end. See how it simplifies to a simple problem. You know, a simple expression That's the final velocity of the bullet in the block, its radical tomb U G s. And then you go back and, uh, use the law of conservation of momentum because your goal is to find the final velocity of the bullet. So you re organize the problem. One side, you have the initial momentum. And then on the other side of the final momentum you saw for the final velocity of the block off the bullet. Sorry. Using audible hope, you enjoy the problem, fans for watching it and see you next time. And if you have any questions that feel free to send them my way, aka things by

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