11. Momentum & Impulse

Ballistic Pendulum

# Final Speed of Ballistic Pendulum Projectile

Patrick Ford

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Hey everybody. So let's take a look at our problem here. So we've got a bullet that's being fired into a wooden block. And so what we're told is that the length of this pendulum here is to the mass of the bullet, I'm gonna call this M one is 0.5. The mass of the block. M two is just gonna be one and this is sort of a ballistic pendulum type problem. You have a bullet is fired into a block. What's going to happen is when it hits, it's going to sort of travel in a curved path until it rises some heights. Right? And we're told here, is that when this sort of this sort of pendulum ends its swing, it's gonna be a distance of 11 centimeters higher. The center of mass rises by a distance of 11 centimeters. So if I sort of draw this line here from where it ends, then this distance between where it started and ended is gonna be 11 centimeters. I'm gonna call that Y and that's going to be 0.11. We want to do is we want to calculate the speed of the bullet as it travels and it emerges from the block. So this is gonna be a little bit different from normal project. Sorry, normal ballistic pendulum problems because in this case the bullet actually keeps on going once it's passed through the block. So it keeps on going this way and this v here, right. That's the speed, that's what we want to calculate. So remember these types of problems are really just conservation of momentum with some energy conservation as well. So the way we solve these is after we draw our diagram we're going to label some points of interest. So in this problem really there's sort of three sort of events going on. We have part A. Which is where the bullet is still traveling, hasn't hit the block. Part B is where the collision happens, and this is where the motion starts and then part C is where the block sort of reaches the end of its sort of swing upwards. Right? So in other words, this is before the collision, this is after the collision, this is also where the block starts swinging and this is finally where it ends like this. Right? So that's sort of like our timeline. And remember we use conservation of momentum in the A to B interval and then we use conservation of energy from the B two C interval. Alright, so now that we have that, let's go ahead and actually write out those conservation equations. So from A to B, what I've got here is I've got M one V one A plus M two V two A equals M one V one B plus M two V two B. We can't go ahead and assume that these things are going to stick together because remember what happens is the bullet actually travels through the block and it keeps on going so we can take a shortcut and actually combine those two masses together because they don't have the same velocity. Alright, now let's look at the B two C. Interval. Really? That's just conservation of energy and we've seen that stuff before. That's just kinetic and initial potential work done by non conservative and kinetic and potential final. Alright, so that's gonna be K B U B equals K C. Plus, you see. Okay, so now we set up our equations. Let's just go ahead and plug in the values and solve really. What we're trying to look for here is we're trying to look for the velocity of the bullets at point B. So let's go ahead and take a look at our and the momentum conservation equation here because this is ultimately what my target variable is. Right? So this is the speed of the bullets before the collision and this V one B will be the speed of the bullet after the collision. And that's really what we're looking for here. So it's V one B. So let's go ahead and start replacing some values. Remember this is just 0.5. Now, What's V one a. The speed of the bullets? Remember the bullet is initially shot with 450 m/s. So this is your V one a. So V one A. Is 4 50. So I'm gonna write that in here for 50 plus M two. That's just gonna be the block plus the times the initial velocity of the block. Now the block is initially at rest. So this V two A. Is equal to zero. So, I just cancel out that term there. Remember that's usually what happens in other words, we have one time zero and you cancel it out. All right. And then on the right side we have 0.5 V one B. That's my target variable Plus, what's the speed of the block after the collision? Well, if you think about this actually, we don't know what this is. After the collision the bullet goes off with some speed and the block is also going to have some speed because it's going to travel up its swing and eventually stop. So how do we calculate that? Remember when we get stuck, when we get stuck for one of these variables? We just have to go and look at the other intervals. So we have to go into the energy conservation to figure out the speed of the block. Alright, so now let's go ahead and expand our terms over here on energy conservation. If you look through your variables, which you've got here is we have kinetic energy. Right after the collision, the block is actually going upwards, we have no potential energy because remember we can just assume the lowest point of the problem is where y is equal to zero here. So that no potential energy. There's no done work done by non conservative forces, no friction, nothing like that. And then what about here at point C where the block stops swinging. Is there any kinetic energy there? No, it's basically transformed all the way to potential energy. So there's no kinetic energy there. So what happens is when you sort of simplify these terms, What you're going to get here, is that one half M two V two B squared equals. And on the right side we're gonna have M two G Y. C. Where this Y. C. Here is basically just the height of this block after the swing. So in other words, this distance here that we sort of indicated the vertical center of mass, distance that the distance that the center mass has risen, that's going to be Y. C. Alright. If you look at this problem, what happens is that M2 is going to cancel out? And remember we came here because we're looking for this v to be Alright. So what happens here? And then when you move the one half over to the other side, what you're gonna get here is that V two B is equal to the square root of two G. Someone over, right? Two times 9.8 times Y. C. Which is the the height that it rose? Remember this is 0.11. Just be careful. You're converting Uh and what you're gonna get here, is that the initial or sorry that the velocity of the block is equal to 1.47 m/s. Alright. So that's not our final answer because remember now we just have to plug that back into this over here. So now what we have here is that we have M. Two times V. Two which is gonna be one times. This is going to be 1.47. The one makes the numbers a little bit easier. Alright. So what you get when you sort of work this all out is you're gonna get to 0.25 when you multiply those two numbers on the left, this is going to equal 0.5 V. One B. Plus 1.47. What you're gonna get here is when you subtract this over, you're going to get 0.78 equals zero 0.5 V. One B. And then last but not least when you sort of divide what you're gonna get is that V. One B. Is equal to 156 m per second. That's pretty reasonable. The bullet impacts the block and it loses a lot of its speed and a lot of its initial speed because it's transferring it over to um because it's transferring it over to the block. Alright. So that's that's this one. Let me know if you have any questions

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