1

concept

## Ballistic Pendulum Problems

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Hey guys, so for this video, I'm gonna introduce you to a type of problem or a type of collision problem called a ballistic pendulum. Let's check this out. The basic set of a ballistic pendulum is you have an object that collides with a block and then that block swings up on a pendulum like this to some height. So we have here in this problem is we have a collision in the first part and then once these things collide the block moves upwards and starts changing speeds and heights. So you have a motion or energy parts of this problem as well. So what I'm gonna show you this video is that when we solve this problem is just like any other problem where we have collisions and then changing speeds and heights or motion and energy parts were really just going to combine both of our conservation equations, momentum and energy. We've seen this thing before, It's nothing new, remember this is just one of the special variations that you might see when you have a collision, you have collision, and then you have this pendulum here. So, because we're gonna talk about pendulum will also need this pendulum equation as well, which just relates the different variables in pendulums. So, let's take a look at our problem here. First thing we wanna do is just draw diagrams and then label the points of interest. So let's take a look here. Remember there's three parts of this problem. The first part is before the collision. Before the collision, I'm gonna call that point A The bullet is still firing inside of the blocker, it's still moving towards the block afterwards, when it embeds itself, that's gonna be after the collision. That's point B. That's where the motion or the swinging pendulum part of our problem starts. And then finally, when it reaches the maximum height over here, that's going to be with the motion ends. I'm gonna call that point, see what we're looking for. This problem is the maximum heights that the pendulum will reach. So what ends up happening is that this pendulum rises some heights and I'm going to call that Y. C. So that's really my target variable. So how do we solve for it? Remember we're just gonna write out both of our equations are conservation of momentum and energy equations. So let's go ahead and do that. So for this part a here I'm going to calculate um I'm gonna start write the equations. Um I'm gonna use conservation momentum in the interval from A to B. So I have M one V one initial, which is A plus M two V two initial equals M one V one final plus M two V two final. And then for the interval from B to C, that's going to be a conservation of energy equation. This is going to be K B plus you. B plus work done by non conservative forces equals K C. Plus, you see that's initial energy equals final energy. So which equation do we start off with? Well, hopefully you guys realize that we're trying to solve for Y. C. That's what we're going to come from the potential energy at point seed. So we're gonna use the conservation of energy equation first. So let's take a look at each for a terms and plug in values and start solving. So do we have any kinetic energy at point B? That's right here. Well, after the bullet hits the block, the block is going to move upwards. That's why it swings. So there's definitely some kinetic energy here. What about the gravitational potential member? There's no springs here in this problem. What happens is we actually don't know what this height is. Even at the lowest point of the problem, we have the lowest point of the pendulum. But what happens in these situations that we don't know the heights, we're just gonna actually set the lowest point of the pendulum to be where y equals zero, because then our potential energy equals zero and it makes our problems simpler. There's also no work done by non conservative forces, knows no work done by you or friction. What about K final? What happens here? Is that the maximum heights when the block just stops swinging, it stops and its velocity is equal to zero. So there's no kinetic energy, but there is some gravitational potential because it's risen some height here. So that's really what our conservation of energy equation boils down to. So now we just expand the terms. So this is gonna be one half and this is going to be M V B squared. Except what happens is there a little M actually becomes big? M the bullets is going to embed itself inside of the block. So this is a completely an elastic collision. So you have this bullet here that's now embedded itself inside of the block like this and they're both going to travel together. So what happens is your big M. Is equal to M. One plus M. Two. So this is gonna be one half of big. M V B squared equals big M. G Y C. So we're looking for this Y. C. We're gonna cancel out the masses and then write an expression for Y. C. We're gonna divide the G. Or to the other side and then Y C equals V. B squared over two G. We've come up with the kind of expression before, so before I can solve for this, Y. C. I actually first need to figure out what this VB. Is, what's the velocity at the start of the motion. Now remember what happens is this is the start of the motion, but it's also the end of the collision. So if I get stuck here and I don't know what this VB is, I can always go to my conservation of momentum equation and solve for it there because I have VBS here, that's all we have to do here. So we're gonna go ahead and write uh this M. One. So this is going to be the bullets and one equals 0.2. And then my block M two is equal equal to 40. The initial speed of the block V two is equal to zero. So what basically happens is this is going to be 0.2 Times plus 40 times zero because that rest And this is equal to now the blocks combined. We know this is a completely an elastic collision. So these things are the same, this is going to be 0.2 plus 40 times VB. All we have to do is solve for that speed and we can go ahead and plug it back to the equation. So this is going to be um let's see. This is going to be 140 on the, on the left side, divided by 40.2 equals V. B. And you're gonna end up getting three .48 m/s. So that's the speed. So now we just plug that back into here and this is going to be 3.48 squared divided by two times 9.8. And you go and work this out what you're gonna get 0.62 m. And that's our answer. So that's our answer for part A The bullet embeds itself into the block and the block rises to a height of 0.62 m. Right? So we just use conservation of energy and momentum. We've seen this kind of thing before, nothing new. So let's go ahead and talk about part B. Now, now we want to do is we want to calculate the angle of the pendulum makes with the vertical. So what happens is once the block swings up to heights, it's going to make some angle with the vertical, and I'm gonna call this fatal. Why here? So the state of Y is what I'm interested in. How do we solve for that? Or remember, we're just gonna solve, we have a relationship between the length of the pendulum, the height that it rises and also the angle, that's the pendulum equation here. This is l minus delta Y. Equals L. Cosign theta. Why? So I'm just gonna start off with that here, L minus delta Y. Except just to be consistent here. I'm gonna call this Y. C. So l minus the change in height. This is why C. Is equal to L. A. Times the cosine of theta Y. So I want to solve for this, but it's a little tricky because it's kind of wrapped up inside of this coastline term. So what I'm gonna do is I'm gonna sort of isolate the right side. I'm gonna move everything over to the left. So I'm gonna divide by L. This is going to be l minus Y. C. Or divided by L equals the cosine of theta. Y. Now, all I have to do is I have to take the inverse or arc co signed to get rid of that co sign on the right side. So now I'm gonna start plugging in my numbers. I get co sign in verse and then my L. Here is too so I've got to minus Y. C. Is the variable that I saw towards 0.62 here. And now we're gonna divide by the original length, which is to, So if you go ahead and plug this in, what you're gonna get is the angle for this. Um for the pendulum, make sure your calculator is in degrees, is going to be 46.4°.. So that's the answer. All right. So we're just really using again a momentum problem and an energy problem. We know how to solve both of these. And in this video we're just putting these things together. That's it for this one. Guys, let me know if you have any questions

2

example

## Final Speed of Ballistic Pendulum Projectile

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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