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California State Polytechnic University, Pomona

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Problem 76 Easy Difficulty

$0.100-\mathrm{kg}$ stone rests on a friction-less, horizontal surface. A bullet of mass 6.00 $\mathrm{g}$ , traveling horizontally at 350 $\mathrm{m} / \mathrm{s}$ , strikes the stone and rebounds horizontally at right angles to its original direction with a speed of 250 $\mathrm{m} / \mathrm{s}$ . (a) Compute the magnitude and direction of the velocity of the stone after it is struck. (b) Is the collision perfectly elastic?

Answer

(a) $v_{A 2}=25.8 \mathrm{m} / \mathrm{s}, 35.5$ o from the original direction of the bullet and
opposite to the rebouns direction.
(b) Not perfectly elastic

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Video Transcript

{'transcript': "once again welcome to a new problem. This time we're given this's the information will given given a bullet that's moving towards the right. Uh, this is the bullet. And it's travelling towards the right thie mass off the bullet, which I'm going to call MP happens to be, um, six grams and I'll change that right away to kilograms. So one kilogram over a thousand grams. And that gives us zero point uh, zero zero six kilograms. That's the mass of the bullet that's moving. It strikes a stone. So, you know, there's a, uh, this is stone right here. The bullet strikes the stone. Um, and something happens to the bullet. It tons perpendicular at an angle off nineteen. Okay, So this this bullet now, um, is moving at ninety. A shit of them that like that. So it strikes the stone and it's moving at an angle of ninety. Ah, the initial velocity of the bullet. The initial velocity of the bullet happens to be three hundred and fifty meters per second. And and you know, the positive ex direction is that to the right And then, you know, we also have the positive wide direction. That's outwards way. Do have the final velocity of the bullet, which happens to be two hundred fifty meters per second. Now, Khun, talk about the stone. When the bullet strikes the stone, the stone moves at a specific angle. Data Ah, we need to find what data is because so want. I want to find data and then the other thing that happens is that the stone is now moving with a certain velocity. It's moving with a certain final velocity s o final velocity off this tone which we're going to call V S is what is something else we're going to be looking for. So find Sita and also find the final velocity of the stone. Um, we don't know what Dada is. We do know that the mask off the stone happens to be zero point one zero zero kilograms. That's the muscle, the stone, the surfaces friction less. So you want to remember that not being affected by friction the initial velocity off the stone, zero meters per second. So that's the information we've given the initial velocity of the stone and soil in part A. You know, we want to find the magnitude and direction uh, I didn't off the stone and then in part B. We want to find the if the collision is perfectly elastic. Okay. Want to find if the collision is perfectly elastic? So, you know, this is all the information we were given in the next page. You're going to use the law of conservation of Momentum. Okay? And that's going to help us find the final velocity of the stone. So love conservation off momentum say, is that initial momentum is exactly the same as the final Momenta, Mme. So I'll say p I equals to a P final. The initial momentum. That's the mass of the stone. So the initial momentum off the stone off the bullet, plus the initial momentum off the stone equals to the final Momenta Mme Ah, off the bullet, Plus the final moment. Final momentum off the stone. That's the information we have. Remember the initial velocity of the stone zero right here. You know, this initial velocity zero and therefore we don't have any initial momentum for the stone. We just have the final momentum. Um, in terms ofthe variables, now we can move stuff. If we move this one to the right. We end up having the mass off the stone and the final velocity of the stone being equal to the initial momentum off the bullet minus the final momentum off the bullet and saw the final momentum of the final velocity off the stone is the mass of the bullet. The initial velocity minus the final velocity off the stone all over the mass of the stone. So this is this is the information that we have right now going to the next page. We get to see that plugging in numbers we have final. The last E off stone equals two mass ofthe bullet of a mass off stone mass of bullet. I was zero point zero zero six kilograms over the mass of the stone, which is zero point one zero zero per kilogram. So we've taken care of that on DH then Also, we do have the initial velocity of the bullet which was given us three hundred and, uh, and fifty meters per second. Uh, I had plus plus AA zero j had because it's not moving in the ex or the wind direction and then was subtracting thie, the the final, The final velocity off the, um, off off this. Sorry, the at the final velocity off the stone. So? So we have We have to change this a little bit. Let's let's just go back. So we have three hundred and fifty meters per second. I heard, um and then we're going to subtract two hundred and fifty. Yes, Yes, yes, Yes. Way have to subtract. Uh, two hundred and fifty. So this one this one has to go back. We have to subtract minus two hundred and fifty meters per second. Jae ha. Remember, I had and jihad is since I had its in the X and then jihad eyes in the white direction. That's how it's bouncing off. So you know when when we simplify these numbers, you know, three. Fifty, we'LL get some when zero zero six, divided by point one. So this is zero point zero six. It's a factor three fifty meters per second. I hade minus two fifty meters. Second jihad. We used the distributive property for this problem. And so we end up having twenty one meters per second. I had minus fifteen meters per second. Jae ha. You know, that's the final velocity in terms ofthe I had and they had But remember, they wanted the magnitude and direction. And so we have to change, you know, finding the magnitude of the final velocity. It's just the squirrel off the ex component. And do I component? So this is a squirrel off, uh, to have some space right there. It's some space. So this is this is the squirt off twenty one. Twenty one squared twenty one meters per second. I had squared plus fifteen meters a second. Uh, J hod squared. Okay, so those are the numbers we have, and then we end up getting a magnitude final velocity. The magnitude of the final velocity being equal to twenty five point, um, twenty five point eight meters per second. That's just the magnitude. Now we need to find the direction to find the direction. Remember, it's going that way, so we have on X component and a wide component. And this is the stone. So data eyes a ton in Vegas off. Um, a fifteen, which is the x fifteen meters per second. Oh, Jihad. Over twenty one meters per second. I had. Okay, that's the information that we have at this point and we're using that to find the, um, be angle. So data Ada happens to be thirty five point five degrees. So data angle is this one right here. And it happens to be a thirty five point five degrees. Um, in the next page, we want to check. So this is but be, you know, is the collision Is the collision perfectly elastic? So that's the question we're asking ourselves Is a collusion perfectly elastic? If it's perfectly elastic, then the initial kinetic energy will be equal to the final kinetic energy. So we want to get the initial kinetic energy of the system, and that's one half em be be initial plus one half this one squared m s, the S initial squared. Remember the initial velocity zero. So this whole thing will be zero and one plug in the numbers. We get one half mass ofthe bullet, zero point zero zero six kilograms and then the initial velocity bullet, five hundred fifty meters per second squared and and that's going to give us three sixty seven point five jewels. That's the initial kinetic energy. On the next page, you want to find the final kinetic energy, which is one half m b v b Final squared plus one half M s V s final squared The final The mass of the bullet. Zero point zero six kilogram. The final velocity of the bullet is two hundred and fifty meters per second. If you can recall, I want to square that and then plus one half muscles toned zero point zero zero one kilograms and then times a final velocity off the stone which is twenty five point eight meters per second. When second, we want to square that if you plug in the numbers you get one hundred eighty seven point five jewels. Therefore, we're going to say the initial kinetic energy, which happens to be three hundred and sixty seven point five, is not equal to the final kinetic energy which happens to be one hundred and eighty seven point five junes. And so we said the collision is not it is not perfectly elastic. Hope you enjoy the problem. We had a case where a bullet was fired towards stone bounced off a ninety and the stone bounced off at an angle, wanted to find the angle of bounce off. And we also wanted to find the the magnitude of the velocity of the stone. When it bounces off, we use the law of the conservation of momentum to isolate the final velocity of the stone. We have to remember that the last is have to be treated in terms of unit components. So the X and the Y And so that's what we did. We ended up using square root and the I ta gra ce photog ra sze tier um, comparing the sides of a triangle. And then we got the magnitude which is twenty five point eight meters per second. We went again and and got the angle using ton in of us, which is thirty five point five and then checking. If the collision is perfectly elastic, we checked the initial the last e off the system, or rather, the initial Connecticut image of the system, which is three. Sixty five point five, compared that to the final Connecticut and Jeff the system, they're not the same saw that the collision is not perfectly elastic. Hope you enjoyed the problem. Feel free to send any questions and have a wonderful day. Okay, thanks. Bye."}

California State Polytechnic University, Pomona
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