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Ch 08: Momentum, Impulse, and Collisions

Chapter 8, Problem 8

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.

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Hey everyone welcome back in this video. We have two blocks on a horizontal frictionless surface. Okay, so we don't need to deal with friction. Um one of the blocks is going to be fired towards the other block and they're both going to be moving after the collision. So the question is asking us to find the change in total kinetic energy of the system due to the collision. Okay, so when we have a problem like this, the first thing we want to do is to draw out our system both before and after the collision and fill in any of the information that we're given. Alright, so before the collision, we know that our system contains two blocks. We have Block A here in red and we have Block B over here in blue. Okay, now we're told that block a has a mass of 3.5 kg. Okay, so 3.5 kg. We're not told anything about its movement other than it's being fired towards Block B. So, it's going to be going to the right but we don't know with what speed or velocity. Okay, so the velocity of and let me just switch that around velocity of Block A. Initially we don't know. Let's take right to be the positive X direction. Okay, so that's the direction that the block is going to be fired. Now for block B were told that it has a massive 2.5 kg and we're told that it's at rest. So what that means is that its initial speed or its initial velocity will just be zero. Great. Now, after the collision again, our system contains two things Both Block A With the same mass as before 3.5 kg. And Block B again with the same masses before 2.5 kg. Now this time we're told that both blocks are moving to the right, Okay, so Block A is going to be moving to the right and we're told it has a speed of 1. m per second. So, because it's going to the right, that's in our positive X direction. So the velocity of Block A will be positive. So a positive 1.29 m per second. Okay, right there. And Block B we're told has a speed of 1.96 m per second to the right. And again the right is our positive X direction. So that velocity will be positive. So the velocity of the final will be positive 1.96 m per second. Alright, so now we have our system drawn out, let's go ahead and figure out how to find that total change in kinetic energy. Alright, so, the change in total kinetic energy delta K. It's just going to be the final kinetic energy minus the initial kinetic energy. Right? The total change is just The difference between those two values. Well, how do we calculate kinetic energy recall that kinetic energy is going to be one half mv squared. Okay, so if we're looking at K final. Okay, it's going to be one half M V squared. But we have to remember what is in our system in our system after the collision we have two things. We have block A and block B. And so we're gonna have a kinetic energy from both of those. So the final kinetic energy is going to be the sum of the kinetic energy from both blocks A and B. So we're gonna have one half the mass of a times the speed of a final squared plus one half the mass of B times the speed of B final squared. Alright, so let's fill in what we know? We get one half the mass of a going to be 3. kg. The speed Of a is going to be 1.29 m/s squared. Okay. And then first B 2.5 kg and 1.96 m per second squared. Okay. Alright, so working all of this out, what we're going to get is we're going to get um 7.714175. Okay. And what about the units while we have kilograms? Okay, And we're gonna be multiplying by meters per second all squared. So that's gonna give us meters squared per second squared. Okay, Now we can think of this as kilogram meter per second squared times meters now kilogram meters per second squared. Well that's the unit for a newton. And then on the right here we have meters. Okay, And that's actually just going to be equal to a duel, which is what we want when we're looking at energy. So the units are good here. Alright, so I'm gonna go ahead and just race that now. Alright, so now what about our initial kinetic energy? It's okay. Not again. We have kinetic energy from block A. And kinetic energy from block B. So we're gonna have one half M. A initial V. A initial squared plus one half M. B initial V. B initial squared. Okay. Now remember that Block B. And sorry, the initial then should be not not eyes. Um the initial velocity of B is zero. Okay, so this second term is just going to go to zero. We're just gonna be left with one half M. A. Which is going to be 3.5 kg times V. A. Not squared. Okay, well, we don't know this value of V. A. Not so in order to find that total the change in total kinetic energy that we're looking for delta K. We're gonna need to find this V. A. Not value. Okay, well, we have a collision. So why don't we use the conservation of momentum to determine that value? So, if we're looking at our conservation of momentum, we know that the momentum in the system initially is going to be equal to the momentum in the system at the end. Okay, recall what momentum is, momentum is going to be mass times velocity. Okay, now the momentum in the system initially is going to come from two components Block A and block B. So we're going to have the momentum of A. Initially plus the momentum of be initially is going to equal the momentum of A at the end plus the momentum of block B at the end. Okay, so writing down that equation that I've just said mass times velocity for each corresponding component. We have the following and same on the right side. M A V A F plus M B V B. Okay. Now remember from our diagram and from the problem we know that VB not is zero. Okay, so this term here, the second term is going to go to zero. Okay, filling in the other information we know and again this VA not that's in here. That's that green value that we're looking for in order to find the total or the change in total connect energy. Okay, so the mass of a we know is 3.5 kg. Then we have V A. Not that value we're looking for. That's going to equal the massive a again 3.5 kg times the velocity of a final 1.29 m/s. And again, this is a velocity the momentum has a direction velocity has directions. We want to be sure that we're including direction in this case both values are positive 2.5 kg for block b with the final velocity of 1.96 m/s. Okay, now working out all those details, we're gonna get 3.5 kg times V. A not equals 4.515 plus 4.9. Okay. Which will give 9.415. Alright, so just going to continue over here. Alright, so we have 3.5 kg V. A. Not it's going to be 9.415 K dividing both sides and on the right hand side. We have kilogram meters per second. Okay, that's the unit here, dividing by both sides. We're going to be a Not the unit of kilogram is going to cancel And we're going to be left with 2.69 m per meters per second. Okay. Alright, so this is our green value. We know via Not now. So now we can figure out what the initial kinetic energy was. Okay, so let's go back here. Now, we can say that k not is equal to 1/2 times 3.5 Kg Times 2.69 m/s. All squared. Okay, well, that value is going to be 0.663175. And similar to k F. The units here are joules okay, we can work it out the same way. You're gonna get kilograms meter squared per second squared. So that's kilogram meter per second squared, which is a Newton times meter. Newton meter is a jewel. Okay, Alright, So the last thing we want to do is get back to this Delta K. Okay, this is the value we're looking for here, and we know that it's K final minus K initial and now we have the value of K final and can initial. Okay, So we have 7.714175 minus 12.663175. Okay, And that is going to equal minus 4.95 jewels. Okay, so that is going to be the total change in kinetic energy. It's negative. So what that tells us is that the system is losing kinetic energy due to the collision. Okay. And that's going to correspond with answer C. Alright, thanks for watching Guys see you in the next video.
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