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Anderson Video - Collision of Box Sliding Down Hill

Professor Anderson
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>> Hello class. Professor Anderson here. Let's take a look at a problem that is going to combine both energy conservation and momentum conservation. And that idea is this. We're going to have a box that slides down this ramp. We'll pretend the ramp is frictionless, so we don't have to worry about any heat. It's going to collide with another box. And then one of two things is going to happen. Either they're going to stick together and move together, or they're going to bounce off each other, okay. And let's look at the case where this is an inelastic collision first. That's where they stick together. So, box slides down. We're going to have a picture right before they collide that looks like this. M1 is moving to the right at V1. M2 is stationary. And then after that collision, they're going to stick together. M1, M2, and they're both moving to the right with we'll call it V final. So this is the before that collision. This is after the collision. In this case we're dealing with the inelastic collision. So, there are sort of three separate pictures that we need here, right. Where it starts, how fast it's going right before the collision, how fast it's going after the collision. And let's start here at height H, and let's see if we can go from this initial to right before the collision. Alright. For that part, we need to conserve energy. So, when we're up here at height H, we have, of course, potential energy in that block M1G times H. If we set our altitude to zero at ground level, then there's no potential energy in M2. Right before the collision, the only block that's moving is M1. And we said it's moving with speed V1. Okay. Let's put a box around that. That's one equation that's going to help us quite a bit. Now we're going to deal with the collision. And the collision is going to conserve momentum but not energy. Because it's inelastic, alright. What's moving? M1. So the initial momentum is just M1 times V1. M2 is just sitting there at rest. On the right side, they are moving together. And so you get M1 plus M2, all times VF. And now we can put these two equations together, and we can solve for VF. Alright. Let's do that. VF is going to be the following. M1 divided by M1 plus M2, all of that times V1. And I can get V1 from this equation right here. If I look at this for just a second, I can probably see what V1 is. I can cross out the M1s. I can multiply by 2. I can take the square root. And so we get the following. M1 plus M2 times square root 2GH. Alright. So that's the final speed of this two block system after the collision. And all we need to know is the relative masses and the height H that block one started from. Now, let's do the harder case where they actually bounce off each other, okay. Okay, so we're looking at the problem where the box is going to slide down the ramp, the frictionless ramp. It's going to hit this other block. And now they're going to bounce off each other, okay. So immediately before the collision, box M1 is sliding at V1. And now afterwards they bounce off each other. We don't really know if they're going to separate like this, or they're both going to end up going to the right. But it's okay. We can just draw this arbitrarily. And if we get a minus sign in our answer, that means that V1 final is not going to the left. It's actually going to the right, okay. So as long as you are consistent with your math, you can draw these arrows in whichever direction you like. Okay. This first part, again, we need to conserve energy. And to conserve energy we have MGH initially. We have 1/2 M1V1 squared, right before the collision. Alright, and as we saw before, we can solve this for V1. V1 is just the square root 2GH. Now that in itself is kind of an interesting answer, right. The speed of this block is only dependent on the height H, and that's because it's a frictionless ramp. In fact, if you dropped this block straight down, it would hit the ground at square root of 2GH. If you made the ramp very shallow, it would be going square root of 2GH at the bottom of the ramp. Okay. And that's because it's frictionless. It doesn't matter how you get down there, it's going to be going the same speed at the bottom. What was your name again? >> Aiden? Can you pass the mic to Aiden? Aiden, you got a question? >> You said the square root of 2GH is going to be the same speed no matter where you drop the block or the box or how you drop it [inaudible] the ramp. Does the 2GH, does that velocity, is that dependent on direction? >> Okay, good question. So Aiden's asking about this 2GH here, the square root of 2GH. Is there any direction associated with this? Well the answer is no because I don't have any vector signs here, right. What we've really calculated is the speed of the block, alright. So the speed of the block is going to be the same whether I drop it straight down or whether it goes down the ramp, or whether it does this sliding to the right. It's always going to be the same. What will change is the direction, okay. >> Thank you. >> Yeah, and for that, you need to go back to the kinematic equations and figure out what direction it's really going. Okay. Good. So, before the collision, block one is moving at that speed. And now during the collision, we are going to conserve momentum. Okay. And before the collision we only have block one moving to the right, so it has momentum M1V1. After the collision now, it gets a little more complicated. We have M1V1F, but that's moving to the left in our pictures. And so we need a minus sign right in front of it. M2 is moving to the right, and so that has a plus sign on it. Okay. And so we are going to conserve momentum. Let's put a box around this equation. We will box up our former equation. But during this collision, it was elastic. And elastic, think of like a super ball, okay. A super ball is very springy. It's very elastic. When I bounce a super ball, when it hits the ground at V, it comes off the ground at pretty much V. Okay. And that means that you also conserve energy during this collision. Elastic collisions conserve mechanical energy. So any energy that went into the system initially, which was 1/2 M1V1 squared. That's the only block that's moving. Has to go into the final energy of the system. And now we have two blocks. We have 1/2 M1V1F squared. It's a positive number, right. Kinetic energy is always a positive number. And so it doesn't matter that it's moving to the left. You're going to square it and get rid of the negative sign anyway. So that's 1/2 M1V1F squared. But then we have to add the kinetic energy of the other one. Okay. And so this is our third equation that we need. And now we can solve this for V2F in terms of the initial parameter each, the initial masses M1 and M2. Alright, shall we see how that's done? Yeah? We'll take a look and see how that's done. Okay, let's say we are now faced with these three equations. And we want to combine them somehow to solve for V2F. Okay. How do we do that? Well let's start with the last equation, and let's start working from there and work our way backwards through the other equations. Okay, if I look at his equation, I can simplify it a little bit, right. First off, let's get rid of all the halves. We don't need all those halves anywhere. So let's make this M1V1 squared equals M1V1F squared plus M2V2F squared. And now, if I look at the left side here, right, I need to somehow get V1. I can do that from our first equation. But I also need to get V1F from this equation. Okay. And so there's a little bit of a complicated series here that we need to go through. Let's take the second equation and plug in for V1F. Alright. How do we do that? Well we've got M1V1 squared equals M1 times V1F. What is V1F from this equation? Well let's rewrite it. M1V1F is going to equal M2V2F minus M1V1. I just moved that over to the other side and moved that over to this side. And now I can write this as V1F is equal to M2 over M1V2F minus this M1 is going to go away. And so I just get V1. Okay, so now I have everything in terms of V1F. So let's put that in right here. M2 over M1V2F minus V1 quantity squared. We have M2V2F quantity squared right there. Okay. That looks potentially like a nightmare, right. Because now we have V2F squared. We have a V2F squared, but then we're going to get a V2F to the 1 power, and we have a V1 there, but we've got another V1 there. So we got to do something here, right. Let's get rid of the V1s, and let's do that from this first equation. So that's actually not so bad, right. M1 times V1 squared, we just put a 2GH right there. And now what do we have in here? We have the following. V1 we said was square root, 2GH. That whole thing is squared. And then we have M2V2F squared. Okay, let's see if we can just attack this algebra now and solve this thing for V2F, alright. What can we do? Well, let's simplify it a little bit. We can get rid of that M1 there, that M1 there if I put an M1 down there. Okay. And now we need to square this thing out. So what do we get? We get 2GH on the left side. We're going to get M2 over M1 quantity squared times V2F squared. We've got to square this term, and so we're going to add a 2GH. And then we have twice the cross product, and that's going to have a minus sign. And so we get minus 2 M2 over M1V2F square root, 2GH. And then we have our last term here. M2 over M1 times V2F squared. Aha. It looks like we have some stuff that is going to simplify a little bit. For instance, 2GH right there and 2GH right there. That looks good. We also have two terms that have V2F squared. So, let's put those together. This one has a M2 over M1. And we have M2 over M1 in that one. We're going to add one of those. And then we have this stuff right here, 2, I'm going to move this around a little bit, 2GH times M2 over M1. And all of that equals zero. Okay. It looks a little tricky, but it's actually not too bad, right. Because I'm just going to move, oh we forgot this V2F right here, right. Good, thanks. I'm just going to move this stuff over to the other side and rewrite this equation. M2 over M1 times M2 over M1 plus 1 times V2F squared equals this stuff, 2 square root 2GH, M2 over M1 times V2F. Can you see that okay, or did it go off? >> It went off a little bit. >> Okay. Let me scooch it over a little bit. Equals 2 square root 2GH, M2 over M1V2F. And now we can simplify this a bunch, okay. M2 over M1 on both sides. That's gone. V2F squared over here, V2F over there. Divide one of those. And so our equation now becomes the following, V2F equals 2 square root 2GH divided by this stuff. M2 over M1 plus 1. Alright. So it started out simple. It got a little tricky in the middle. It came back to something that is reasonably simple. Let's take a look at this answer and see it makes sense, okay. First off, let's say that the masses are equal, okay. If M1 equals M2, and let's go back to our picture for a second, right. We had a block that was sliding along. It fell from a height H. And then it's going to collide with M2, which is at rest. If M1 is equal to M2, what do we get for our final speed of the block 2? Ty, can somebody hand the mic to Ty? >> You just get V1, V1F again, or V1. See the initial velocity from the beginning, right. >> Okay. We get V2F for this guy. So, Ty, the question was, what is the final speed of block 2 if M1 equals M2? What do you think it is based on this equation? >> The square root of 2GH. >> The square root of 2GH, which is exactly how fast block one is moving at the bottom, right, which is what you said earlier. Does that make sense? >> Yeah. >> Okay. >> Because the masses are both the same. >> Okay. What's the final speed of block M1 then? >> It'd be zero. >> Yep, it is moving at zero. Okay. It's not moving, so we should just take this arrow off of it. Where do you see this sort of interaction? Two objects of same mass, one stationary, one coming in and hitting it. The second object taking off at the same speed. The first object staying at rest. Where do you see that sort of interaction? >> Yeah, a pool table. >> A pool table. Exactly right, okay. If I hit the cue ball directly on to the eight ball, and ignore spin and English and stuff like that for a second. If I hit it directly, all the speed of the cue ball is transferred to the eight ball. The cue ball comes to a stop. Okay. And so that's what happens in this elastic collision. If the masses are the same, we get two in the bottom. It cancels with that two up there. The final speed is exactly 2GH square root, which is the speed of M1 on the way in. Okay. So this answer looks like it makes sense. The units work out, of course, because we end up with the unit square root of 2GH, which we know is a velocity. And all the masses down here cancel. So it looks like it makes sense. Marcella, let's take a look at a different limit here. Let's say that we're going to do the following. We're going to let M2 be much, much bigger than M1. Okay. And so it's like we've rolled a little tiny block down the hill, and we hit this giant truck. What do you think the final speed of that mass M2 is going to be? >> It's going to be smaller. >> It should be small, right. It should go to zero. Is that what we see? Yes. M2 over M1 here. This number becomes very big. It's in the denominator. And so V2F is small. It's getting close to zero. What about in the other case? Let's say that M1 is the big one. M2 is the small one. So if M1 is much, much bigger than M2, what does our picture look like? We've got a very big block, very small block. And now when this thing slides down and hits it, how fast is this small block moving? >> Faster. >> Okay. Well, let's look at the problem right here. If M1 is much bigger than M2, then V2F becomes what? It's 2 times the square root 2GH divided by M2 over M1, we're going to say that is essentially zero. Okay, we're going to ignore this term. And so this thing ends up being 2, square root of 2GH. Okay. And so it will take off faster than the other cases that we looked at. And it's sort of interesting that it in fact goes faster than if itself had fallen from that height H, right. If it falls from height H, it goes square root of 2GH. But now with this complicated collision, it's in fact going to go 2 times the square root of 2GH. Which is kind of interesting, right. Because that says that this thing could go back up to a height that's even higher than the original block started from. You've somehow transferred the energy and the momentum from this one to this one and sent it back to a bigger height. Which is kind of cool I think. Alright. Any other questions about this one? Brent? >> Would that mean the bigger just stop moving in its spot? >> No. We know that it can't stop moving, right. Ask your question again, Brent. >> Would that mean the bigger one is stopped right there because it's perfectly elastic so it transferred all of its energy to the smaller block? >> Well, what do you think, right. Let's say a big truck comes and hits a really small car. Is that big truck going to come to a stop? >> [Inaudible] if it's perfectly elastic would that mean that it would? >> No. >> Okay. >> The only time it means it comes to a stop is when the masses are equal, alright. That's the only condition. If the masses are not equal, then it either bounces back off, or it continues in the same direction. So, even though V2F is going a lot faster, block M1 is still moving. And we know what's going to happen. It's going to keep moving in the same direction. Okay, so we manipulated those equations to calculate V2F. You can go back to those equations to plug them in and get V1F. Alright. Any questions about this one? Everybody alright with that? Alright, good.