Anderson Video - Ballistic Pendulum

Professor Anderson
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>> Hello, class. Good morning. Welcome to another edition of the learning glass lectures on physics. We're going to talk this morning about the ballistic pendulum. Which is one of the classic physics examples and it's a fun little homework problem, because it combines a bunch of different conservation principles. We have to use both conservation of momentum and conservation of energy. So, what is a ballistic pendulum? Well, ballistic, of course, refers to fire arms and so the ballistic pendulum is a device that is used to measure the speed of a bullet. Okay, so it looks like this. That's our ballistic part. There's our pendulum. And the pendulum is just a big block of wood, okay? You hang it from a string of length L, and now you fire the bullet into the block. Now, when the bullet goes into the block the block starts to move. So, it swings up to a height H, it's got the bullet in the block. And based on this information now you can figure out something about the speed of the bullet. Okay? We need to know a few more things here. We need to know the mass of the bullet. We need to know the mass of the block and we need to measure the height, H. And if we do all that then we can calculate V1. So, this is the ballistic pendulum problem. And the idea is that we need to combine a few conservation principles. So, let's think about the initial part where the bullet goes into the block. Okay? That is the first part. And when it enters the block it has to move a bunch of wood fibers out of the way, right? It's essentially got to bore a hole into that block. And when it does that it has to expend a lot of energy to bend all those wood fibers and annihilate them out of the way. So, when it enters we say this is an inelastic collision. And in the inelastic collision we only conserve momentum. We don't conserve mechanical energy. And it's because energy is lost in that process of going into the block, right. It has to bend stuff out of the way. Okay. The second part is during the swing. Okay. The bullet is now in the block and it's going to swing up to some height. And when it swings up to some height we are no longer conserving momentum in that system, okay? Why are we no longer conserving momentum? Brent? Grab the mic. Brent, why are we no longer conserving momentum during that swing? >> (student speaking) Changing vector. >> Okay. There's a change -- >> (student speaking) The velocity is changing because of that changing vector. >> Okay, good. So, momentum is a vector, absolutely right. When this block first starts to swing it, of course, has momentum, it has mass, it has velocity. But when it comes up to its maximum height it stops momentarily. So momentum goes to zero. So, there is no conservation of momentum and the reason is there's an external force acting on the system. That external force is, of course, gravity, right? Gravity is what brings that block to a stop. Gravity is what's eventually going to pull it back down, all right? So, we don't conserve momentum during the swing, but we do conserve energy. Energy is, of course, not a vector, so we don't put a vector sign on top of it. Energy is conserved during the swing. All right, those are the two principles that are involved here. Now let's see if we can apply a little math to figure out V1 in terms of all these other parameters. And we're going to say that this speed immediately after the collision is V2 final. And just to be clear, we're going to say this is V1 initial. All right. Let's see if we can solve this. Actually, I change my mind. Let's not do that. Let's just keep it as V1 and V2, just because I don't want to get confused with this as the final position. All right, let's see if we can apply some math now. So, the first part is conservation momentum. We have the initial momentum, has to equal the final momentum. This is immediately before and after the collision between the bullet and the block. And initially all we have is the bullet moving. The block is just hanging there at rest. So, on the left side we just have M1, V1. On the right side the bullet is now in the block, so we have two masses that are moving together, both at speed V2. All right? Let's solve this equation for V2, because we're going to need that later on. V2 is M1 over M1 plus M2. All of that times V1. And now the second part is when it swings up to a height H. And for that we are going to conserve energy. EI is going to equal EF. When it first starts to swing we have kinetic energy. The masses are together. So, it is one half, M1 plus M2, it is moving at we call it speed V2. This is right after the bullet enters the block. On the right side everything comes to rest, there's no kinetic energy anymore. But there is, of course, gravitational potential energy. So, we have M1 plus M2 times GH. And now we can take our solution for V2 and we can plug it into this equation and we can solve this for V1. Let's do that. First off the bat, let's simplify this. We can cross this out. And we can multiply by 2 on both sides. And so we get V2 squared equals 2GH and now let's plug in this solution for V2. So, we have M1 divided by M1 plus M2. Quantity squared times V1 squared equals 2GH. Let's take the square root of both sides, M1, divided by M1 plus M2, times V1 equals the square root of 2GH. And finally I want to multiply by the inverse of that and we'll move it over here. We get V1 equals M1 plus M2 all over M1 times the square root, 2GH. Now, when you get these solutions I encourage you, first off, to get the solutions in terms of variables, right? Make sure you keep the variables all the way through. It's really going to help you analyze this solution to see if it makes sense. The next thing that you should do is make sure that it looks right in terms of limits. So, let's say that block doesn't swing up at all. Let's say the wood block doesn't move at all. What would H be in that case? Zero. Right? If H is zero, this says that V1 is going to be zero. That makes sense. If the bullet doesn't even hit the block the block's not going to swing. Okay. It also makes sense that if the bullet swings -- I'm sorry, if the block swings up really high, then the bullet must have been going faster. So, as H goes up V1 goes up, that seems to make sense as well. You can also look at the limits of M1 and M2, the case of a really light bullet versus a heavy block or a heavy bullet versus a light block. And all of those limits should make sense as well. And finally you can double check the units. The Ms cancel out. We get square root of 2GH, G is meters per second squared. H is meters, so I'm going to get meters squared over second square, but I take the square root of that I get meters per second, which is, of course, velocity. Okay? So, this looks like the right answer in terms of limits and units and we're going to say, that's it. All right, questions about this problem? Everybody okay with this? All right. Good. Let's move onto the next.
>> Hello, class. Good morning. Welcome to another edition of the learning glass lectures on physics. We're going to talk this morning about the ballistic pendulum. Which is one of the classic physics examples and it's a fun little homework problem, because it combines a bunch of different conservation principles. We have to use both conservation of momentum and conservation of energy. So, what is a ballistic pendulum? Well, ballistic, of course, refers to fire arms and so the ballistic pendulum is a device that is used to measure the speed of a bullet. Okay, so it looks like this. That's our ballistic part. There's our pendulum. And the pendulum is just a big block of wood, okay? You hang it from a string of length L, and now you fire the bullet into the block. Now, when the bullet goes into the block the block starts to move. So, it swings up to a height H, it's got the bullet in the block. And based on this information now you can figure out something about the speed of the bullet. Okay? We need to know a few more things here. We need to know the mass of the bullet. We need to know the mass of the block and we need to measure the height, H. And if we do all that then we can calculate V1. So, this is the ballistic pendulum problem. And the idea is that we need to combine a few conservation principles. So, let's think about the initial part where the bullet goes into the block. Okay? That is the first part. And when it enters the block it has to move a bunch of wood fibers out of the way, right? It's essentially got to bore a hole into that block. And when it does that it has to expend a lot of energy to bend all those wood fibers and annihilate them out of the way. So, when it enters we say this is an inelastic collision. And in the inelastic collision we only conserve momentum. We don't conserve mechanical energy. And it's because energy is lost in that process of going into the block, right. It has to bend stuff out of the way. Okay. The second part is during the swing. Okay. The bullet is now in the block and it's going to swing up to some height. And when it swings up to some height we are no longer conserving momentum in that system, okay? Why are we no longer conserving momentum? Brent? Grab the mic. Brent, why are we no longer conserving momentum during that swing? >> (student speaking) Changing vector. >> Okay. There's a change -- >> (student speaking) The velocity is changing because of that changing vector. >> Okay, good. So, momentum is a vector, absolutely right. When this block first starts to swing it, of course, has momentum, it has mass, it has velocity. But when it comes up to its maximum height it stops momentarily. So momentum goes to zero. So, there is no conservation of momentum and the reason is there's an external force acting on the system. That external force is, of course, gravity, right? Gravity is what brings that block to a stop. Gravity is what's eventually going to pull it back down, all right? So, we don't conserve momentum during the swing, but we do conserve energy. Energy is, of course, not a vector, so we don't put a vector sign on top of it. Energy is conserved during the swing. All right, those are the two principles that are involved here. Now let's see if we can apply a little math to figure out V1 in terms of all these other parameters. And we're going to say that this speed immediately after the collision is V2 final. And just to be clear, we're going to say this is V1 initial. All right. Let's see if we can solve this. Actually, I change my mind. Let's not do that. Let's just keep it as V1 and V2, just because I don't want to get confused with this as the final position. All right, let's see if we can apply some math now. So, the first part is conservation momentum. We have the initial momentum, has to equal the final momentum. This is immediately before and after the collision between the bullet and the block. And initially all we have is the bullet moving. The block is just hanging there at rest. So, on the left side we just have M1, V1. On the right side the bullet is now in the block, so we have two masses that are moving together, both at speed V2. All right? Let's solve this equation for V2, because we're going to need that later on. V2 is M1 over M1 plus M2. All of that times V1. And now the second part is when it swings up to a height H. And for that we are going to conserve energy. EI is going to equal EF. When it first starts to swing we have kinetic energy. The masses are together. So, it is one half, M1 plus M2, it is moving at we call it speed V2. This is right after the bullet enters the block. On the right side everything comes to rest, there's no kinetic energy anymore. But there is, of course, gravitational potential energy. So, we have M1 plus M2 times GH. And now we can take our solution for V2 and we can plug it into this equation and we can solve this for V1. Let's do that. First off the bat, let's simplify this. We can cross this out. And we can multiply by 2 on both sides. And so we get V2 squared equals 2GH and now let's plug in this solution for V2. So, we have M1 divided by M1 plus M2. Quantity squared times V1 squared equals 2GH. Let's take the square root of both sides, M1, divided by M1 plus M2, times V1 equals the square root of 2GH. And finally I want to multiply by the inverse of that and we'll move it over here. We get V1 equals M1 plus M2 all over M1 times the square root, 2GH. Now, when you get these solutions I encourage you, first off, to get the solutions in terms of variables, right? Make sure you keep the variables all the way through. It's really going to help you analyze this solution to see if it makes sense. The next thing that you should do is make sure that it looks right in terms of limits. So, let's say that block doesn't swing up at all. Let's say the wood block doesn't move at all. What would H be in that case? Zero. Right? If H is zero, this says that V1 is going to be zero. That makes sense. If the bullet doesn't even hit the block the block's not going to swing. Okay. It also makes sense that if the bullet swings -- I'm sorry, if the block swings up really high, then the bullet must have been going faster. So, as H goes up V1 goes up, that seems to make sense as well. You can also look at the limits of M1 and M2, the case of a really light bullet versus a heavy block or a heavy bullet versus a light block. And all of those limits should make sense as well. And finally you can double check the units. The Ms cancel out. We get square root of 2GH, G is meters per second squared. H is meters, so I'm going to get meters squared over second square, but I take the square root of that I get meters per second, which is, of course, velocity. Okay? So, this looks like the right answer in terms of limits and units and we're going to say, that's it. All right, questions about this problem? Everybody okay with this? All right. Good. Let's move onto the next.