On a frictionless, horizontal air table, puck A (with mass 0.250 ...

Question

On a frictionless, horizontal air table, puck A (with mass 0.250 kg) is moving toward puck B (with mass 0.350 kg), which is initially at rest. After the collision, puck A has a velocity of 0.120 m/s to the left, and puck B has a velocity of 0.650 m/s to the right. (a) What was the speed of puck A before the collision? (b) Calculate the change in the total kinetic energy of the system that occurs during the collision.

Accepted Answer

Hey everyone welcome back in this video. We have a problem where we have blocks that are being fired towards each other. Um were given some information about the blocks, which will come back to the question is asking us to determine the speed of block B before the collision. Okay. So we have a collision problem like this. The first thing we want to do is to draw out a picture of the system both before and after the collision. Okay. We're gonna take right to be the positive extraction and we'll get started with our diagram. So we're told that we have blocked A. Okay, before the collision we block A. It is .85 kg. So 0.85 kg and we're told that it's at rest. Okay, so the velocity of block a initially is zero. We have a second block as well A block B which we will make blue And we're told that it's 0.65 kg and it's being fired towards block So to the left, but we don't know what speed it's being fired with were not told that and that's actually what the question's asking. The questions asking. The speed of block B before the collision. So V. B not is that quantity we're looking for. Alright, now, after the collision, we have both blocks. Again, we have block A. It has the same mass as before, 0.85 kg. It's going 1.76 m per second to the left to the left and now we've taken the right to be our positive extraction. So if we have something moving to the left, the velocity is going to be negative. So the velocity of A. After the collision is going to be minus 1.76 m per second. And Block B Again is going to have the same mass as before 0.65 kg. And we're told that it's moving to the right with a speed or velocity and that's the final velocity. Not the initial velocity Of 0.202 m/s. Alright, so we have our diagram set out now. What can we do to find that missing velocity? Well, we know that we have conservation of momentum. Okay, no external forces. So we can write our conservation momentum equation. Conservation of momentum tells us that the momentum in the system initially is equal to the momentum in the system after the collision. Well, what's in our system initially before the collision, we have our two blocks. So we have to consider the momentum that comes from each of those blocks. So we have the momentum from block A. Initially plus the momentum from block B. Initially similarly after the collision, we still have our two blocks. So we have the momentum of block A. After the collision and momentum of block B. After the collision. Now let's recall what momentum is, momentum is, mass times velocity. So for each of these terms we're gonna have a corresponding mass and velocity. So we have the mass of blockade times the velocity of block A initially mass of B. Velocity of be initially it's equal to the mass of a. The velocity of a after the collision plus the mass of the velocity of the after the collision. Alright, so this quantity here, this VB not this is the vector the velocity we're looking for speed. So that's going to be related to that quantity we want to find. So let's go ahead and fill in what we know. Well, we know that the velocity of block A initially is zero, it's at rest and so its momentum is going to be zero. This whole term is going to go to zero. Okay, so on the left hand side we're gonna be left with the massive block B, which is going to be 0.65 kg times VB. Not. And on the right hand side we have the massive a which is 0.85 kg times the velocity of a after the collision. Okay, so this is going to be -1. m/s. Now, when we're looking at momentum direction matters just like for velocity. And so we need to include that negative to indicate the direction We have the massive B 0.65 kg. And finally the mass or the velocity of block b after the collision. 0.202 m/s. Alright, so, working this out on the left hand side, we still have 0.65 kg times v. B not on the right hand side, we're gonna have minus 1.496 kg meters per second plus 0.1313 in the same unit, kilogram meter per second. Alright, so just simplifying on the right hand side here, We will have -1.3647. Same unit, kilogram meters per second. All right now to isolate V B. Not. What we're gonna do is divide by 0.65. When we do that, the unit of kilogram will cancel. We're gonna be left with V V not is equal to -2.099. Like, and we're left with units of meters per second, which is a unit for velocity. So, that's great. Our units checked out. Okay. And now we just have to remember the questions asking, the question is asking for the speed of block B. Okay, so the speed of block B. When we're looking at speed, we just care about magnitude not direction. So the speed is going to be the the not which is going to be the magnitude of the The velocity. Okay, so the absolute value of velocity. So in this case it's going to be positive 2.1 m/s. Okay, so that's our speed of that block before the collision. That's going to be answer B. Thanks for watching everyone. See you in the next video

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