Two 500 g blocks of wood are 2.0 m apart on a frictionless table....

Question

Two 500 g blocks of wood are 2.0 m apart on a frictionless table. A 10 g bullet is fired at 400 m/s toward the blocks. It passes all the way through the first block, then embeds itself in the second block. The speed of the first block immediately afterward is 6.0 m/s . What is the speed of the second block after the bullet stops in it?

Accepted Answer

Hey, everyone in this problem, we have a 25 g pellet shot horizontally from a gun with a velocity of 325 m per second. It's going to cross a 0.75 kg wooden container and then continue its motion until it collides with a bulletproof plate of 1. kg. Ok? And it's going to stick to that bulletproof plate. The container moves with the velocity of 4.5 m per second immediately after the collision. And we're asked to find the velocity of the plate just after the pellet gets stuck in it. We're given four answer choices. Option A negative 3.7 m per second. I hat. Option B positive 3.7 m per second. I hat. Option C 190 m per second. I hat and option D negative 190 m per second. J hat. We're gonna start by writing out everything that we were given. So we have the mass of the pellet which is 25 g. We wanna convert that to kilograms. So we're gonna multiply it by one kg divided by 1000 g. OK? Because we know that there are 1000 g in every kilogram, the unit of gram divides out. And what we're essentially doing is dividing by 1000 forget that this is equal to 0.25 kg. Now let's look at the mass of the container and that we're, that's what we're told next and 0.75 kg. And then we have the mass of the bulletproof plate, which we're gonna call M B and that's 1. kg. Yeah. So we have the three masses we were given. Now let's write out the velocities and we have three time points to kind of consider. So we have the initial time point where we have the velocity of the pellet V, not P is equal to 325 m per second. We know that the container is initially not moving, ok? And so V not C it's just gonna be equal to zero m per second. Then we have this second time point after the first collision, ok. The pellet is gonna collide with that container after that collision. We don't know the velocity of the pellet. We know that the velocity of the container is equal to 4. m per second. And then we have this bulletproof plate, it is also not moving, ok. So it has an a velocity of zero m per second. So we have our initial time point. We have this second, intermediate time point after the first collision and then we have the final collision. Now the pellet is going to stick into that bulletproof plate. And so we're only gonna have one final velocity, that final velocity of the pellet is gonna be equal to the final velocity of the bulletproof plate. And we're just gonna call that V F, ok. But like I said, between the first time point and the second time point, we have this collision one in between the second two, we have collision too. Now we have collision problems. We want to think about the conservation of momentum. So let's look at the first collision. First conservation of momentum tells us that the initial velocity or sorry, the initial momentum P is gonna be equal to the momentum of that second time point P two. Our momentum is made up of two components because we have the momentum of the pellet and the momentum of the container. So we have P N P, the initial momentum of the pellet plus P N C, the initial momentum of the container, which is equal to P two P L S P two C. Now recall momentum is given by mass multiplied by velocity. So for each of the, these terms, we get the corresponding mass multiplied by the corresponding velocity. We have M PV naught P plus M CV, not C is equal to M PV two P plus M CV two C, the mass of the pellet substituting in our information, 0.25 kg multiplied by its initial velocity of 325 m per second. Ok. The container is initially at rest. And so that is going to go to zero the momentum there is gonna go to zero. So that's it. For the left hand side, on the right hand side, we have the mass of the pellet again, 0.25 kg multiplied by its velocity after the first collision, which we don't know what the mass of the container is. 0.75 kg multiplied by its velocity after the first collision, which we're told is 4.5 m per second. OK. Now let's move this B two P term to one side and everything else, all of those constants to the other side, we get 0.25 kg multiplied by V two P is equal to 8.125 kilogram meter per second minus 3. kg meters per second. OK. Simplifying on the right hand side and then dividing by 0. kg. We get that the velocity of the pellet after the first collision is 190 m per second. All right. So we figured out the velocity after the first collision. But remember that the problem is asking us to find the velocity of the plate after the pellet gets stuck in it. Ok. That's the second collision. So we now have V two P 190 m per second. And we wanna switch gears and look at collision two. And we're gonna use our conservation of momentum again. And in this time, we have the momentum at time 0.2, that's gonna be equal to that final momentum. After the second collision. The momentum again made up of two things. We have P two P plus two P two B, the momentum at the second time point of the pellet plus the momentum at the second time point of the bulletproof plate. Ok? And this is equal to the final momentum. Now, we're ignoring the container here because it's no, no longer involved involved in the collision. Ok? The pellet has already passed through it. And now we're looking at the isolated collision between that pellet and the bulletproof plate. Now we have this final momentum P F. We're not breaking that down into two components because we know that the bullet sticks into or the pellet sticks into that bulletproof plate. And so that's gonna, they're gonna move together. It's gonna be one total mass that's moving at one speed breaking this down further. We have MP multiplied by V two P plus M B multiplied by V two B is equal to, again, the sum of the masses that pellets sticking in the bulletproof plate. So we have to consider the mass of the pellet plus the mass of the bulletproof plate, all of that multiplied by that final velocity V F that they're gonna be traveling at together. Now that bulletproof plate is not moving before the collision. So the second term on the left hand side goes to zero, substituting in our values we're left with 0.2, five kg multiplied by 190 m per second. On the left hand side, that's equal to 0.25 kg plus 1.25 kg, all multiplied by V F. On the right hand side, we wanna isolate for V F. So we're gonna simplify and then we're gonna divide by that total mass that's out in front of our frontal velocity. We get that 4.75 kg meters per second divided by 1.275 kg. And the unit of kilogram is gonna divide out and we're left with a final velocity of 3.7255 m per second. OK? And that final velocity is the velocity of that pellet and bulletproof plate combined after the collision. OK. Which is exactly what we want. We want the velocity of that plate and that is what we found. Now, this is a positive velocity. We're looking in the horizontal direction, we were told that this was shot horizontally. So we know we're in the I ha direction. OK. That corresponds to the horizontal direction. So when we compare this to our answer choices, we can see that this corresponds to answer. Choice B 3.7 m per second. I hat thanks everyone for watching. I hope this video helped see you in the next one.

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