https://youtu.be/rToMmcsY9Ko
Let’s start our new material with an example. This is a classical example of the context in which you often see a collision. Usually a problem, the scope will involve more than just the collision, and you have to break the problem up into parts, meaning for the collision part only we’re going to use the tool of conservation of momentum. But then there’s going to be other motion in the problem perhaps that follows the collision or maybe motion that occurred before this abruptive end.
For the longer time scale motion, it’s going to be a different tool that we’re going to want to use, maybe energy conservation, that’s chapter six, or maybe kinematics, maybe it’s chapter three. So you’ll see that in this problem.
This problem, it’s about a ballistic pendulum. This is what is still used today in some cases to measure the speed of a projectile like a bullet. So here’s the picture of what goes on. I have a block of wood, and I know the mass of it. I have a bullet, and I know the mass of the bullet. The bullet comes in with its initial speed, and it embeds in this block. No question, this is an abrupt event, this collision.
The bullet stays embedded in the block, and then the block has this initial speed, it has this initial kinetic energy. It’s going to start moving forward. It swings up because it’s attached to this rope, so I have kind of circular motion or an arc motion that starts to happen. But this bullet’s going to reach some maximum height. All that kinetic energy turns in potential energy, and the block and bullet’s going to come to a stop ever so briefly.
What they do is they measure the height that the block-bullet system reaches, and from that you’re going to derive, or we’re going to derive, and realize that we can back-calculate what the speed was of the incoming projectile.
We have a lot of information in the problem. We know the mass of the bullet in the block, we know the height that the combination swings up to, so this is an altitude, and our goal is to find the speed of the incoming bullet.
Okay, as I suggested, we’re going to break this problem up into two parts. The very first part of the problem happens quick. It’s a dramatic collision between the bullet and the block, a millisecond, very fast. Immediately at the end of that collision the bullet is now — it’s stopped moving, it’s embedded in the block, or it’s stopped moving inside the block, and now the block and the bullet together swing up.
This is part two of the problem. This is going to take longer. This is a longer time-scale part to the motion.
My first quiz question to you is, what tool is the smartest tool to use to solve the first part of the problem? I gave you a collection here of the main tools that we’ve covered so far in this class. Here is chapter seven, here is chapter two or three, here is chapter seven. Choice D is chapter four, choice E is chapter six.
What is the basic starting equation that you want to apply to the first part of this problem?
On the next exam, on the final exam, it may not be as obvious to you because it won’t be in the context of a particular chapter. You’ll just be presented with a problem like this.
So, breaking it up into two parts, what tool are we going to through at the first part of the problem? Question four, what tool are we going to throw at the second part of the problem? What’s the smartest tool for us to use to describe the motion as the bullet and the block swings up after the collision? So starting at the end of the collision, and then the final instant is when that block is at its highest point.
An important concept when you break a problem up into parts like this is to be sure you recognize the physical connection between the parts, and in particular you have to recognize what that translates to in terms of the variables.
In other words, we’re going to use to capture the essence describe the result of the collision. It involves certain variables. If I use this one for example, it involves an f, a t, an m, v-final and mv-initial, et cetera. This involves position changes. There’s physical meaning attached of course to all of the variables that will show up in these equations.
What are the variables that are going to connect, or what are the, what’s the connection between the tool we’re going to use for the first part of the problem and the tool we’re going to use for the second part of the problem? How do we connect these two general equations that we’re going to be working with?
So I give you some choices here in question five to help you think that through. The most important point to question five is to have you realize that you do have to carefully think that through. If you don’t, it’ll seem like you don’t have enough information to solve the problem when actually you do.
Same thing for question six. Here I just want you to think a little bit more carefully about what some of the values would be for variables that you would use in parts of the problem, and here I’m particularly asking you about the m. There’s going to be an m that would show up when we work the problem for part two. Starting with our basic equation, we’re going to see mgh terms. G is clear enough, h is given in the problem. I’m going to have a starting altitude of 0 and an h-final that’s the height given that the ball swings up to.
What’s the right m to use for that part of the problem? Okay, here I’ve copied the problem over, at least a picture of the problem, and I’m going to divide my page in half and clearly keep the two parts separate. I’m going to have to connect them at some point you’ll see, but let’s try to be clear what part of the problem we’re working on, what our variables represent.
The left side is going to be where I handle the first part of the problem, the collision. On the right side of the page is where I’m going to tackle the motion, which is the swinging up here after the collision.
Off to the right here let’s make a list of our knowns and the variable that we want. Well, we knew the, know the masses. The bullet is light, 10 grams, or 0.01 kilograms. The block is heavier. It’s 2 1/2 kilograms. The final height I was given as 0.65 meters. And what do I want? I want the bullet’s incoming speed. The variable that the figure shows for me to use represent that is v0-1.
Okay, for the collision. A collision, no question, it’s abrupt event. Can I apply momentum conservation? During this collision, I have a force that the bullet and the block each exert on each other. They’re huge. They’re equal and opposite. I’ve understated the size of these.
In this horizontal direction those are the only forces in fact that are on these two objects, so for sure I can apply momentum conservation in this horizontal direction to this collision. And that’s good. In fact that’s all I want because I really just need to know the velocity of this combination after the collision because that’s going to tell me my initial kinetic energy because I’m going to use energy conservation to get this last part. Why am I going to do that? Energy conservation is a great tool to use once you get the hang of it and if it’ll work for you. And will it work? I claim yes, no question.
This is classic case of having no work done except for gravity as this object swings up. The tension in the rope is always perpendicular to the motion so the tension can’t do any work. It’s not helping to speed up or slow down this block. Gravity’s the only thing. That means mechanical energy is conserved.
I start out with kinetic energy that all turns into potential energy as gravity slows this thing down and brings it to a stop to reach this highest point. So all I need is the horizontal velocity. That’s the only velocity I’ll have right at this instant, and that’s what the collision, considering momentum conservation horizontally, is going to get for me.
Vertically the mass of the block and the tension balance. The size of those forces is going to be a hundredth of the size of these forces during the collision. Horizontally I definitely can apply momentum conservation. I only have these two dominant horizontal forces during the collision and they’re from the force the two objects exert on each other.
This means I’m going to apply capital P0 equals capital P-final. I’ve already got them labeled. My bullet is object one and my block is object two. My total initial momentum is the bullet’s mass times its initial velocity plus the block times its initial velocity, and that has to be equal to the total momentum just after.
Well just after the collision, this now I’m going to think of as one block that has a mass of m1 plus m2, with the bullet being stuck into the block. I may have lost some shards of wood or some splinters during the collision, but that’s not going to change the mass very much.
So I’m going to label this as the mass. I only have one object after the collision. It has a mass of m1 plus m2 and it has a speed of v-f. I’m not going to label this v-f with a 1 or a 2 because both objects have that same final velocity.
Well, let’s see what values I know. I was given the mass of the objects. Let me think about these initial velocities and what I’m after. Well, I’ll highlight in yellow the variable I need to solve for, so if I have values for all the other variables, I’m set.
I do know that the value for v2-0 is 0. This v2-0 is the initial velocity of object two. This is the velocity before the collision. The block is at rest. It’s sitting there. That’s 0, so this whole term is 0.
On the left then I end up with m1v1-0 is equal to m1 plus m2 over v-f. In fact, let me divide both sides by m1 just so I’ve done that algebra step to isolate or solve for what I want.
Okay, here’s the variable I need ultimately to answer the problem. It’s on the left side of my equation. What do I have on the right? I’ve got the masses, which I know, but I don’t know the velocity of the block and bullet after the collision.
Now I need to bring more information into the problem to try to solve it. What do I do? My first reaction is to try to bring in in a logical way the information I know about the second part to the problem.
The second part of the problem where the block swings up I’ve argued that energy conservation is the way to go. So I’m going to now move over to the right-hand side of the page, and I’m going to apply E conservation.
In fact, let me be more specific with the terminology. It’s actually mechanical energy conservation that I’m applying, and this is because there’s no non-conservative work being done. This means the force of gravity is the only force that’s slowing the object down. It’s the only, only force doing work on the object, so where, what’s my starting equation for that?
Well, my overall equation was work non-conservative equals the change in kinetic energy plus the change in potential energy. Left side of my equation is 0. Let me break out these terms here. It’s 1/2 m v0 squared — oops sorry, v-f squared, minus 1/2 m v0 squared plus mgh-final minus mgh-0.
Okay, let’s think now carefully about what each of those variables represents. Let me go to my — grab another copy of my picture. Okay, I’ve got a little snapshot here off to the left, so I can see it in the same screens as I think through these values.
Remember, for energy conservation, I assign my lowest altitude being equal to 0, so that’s h-0. H-final then, this is the end scope of my problem, is the 0.65 meters. The final velocity here for this part of the problem is 0. This term then goes to 0. This v0 for my energy conservation scope represents the speed of the block and bullet right at this instant. That’s just after the collision.
So I’m going to emphasize that this is the same as the v-f from part one. That’s what I have to solve for now in order to go back to the left side of my page here and complete the problem. Let me highlight that in green. This is what I need that connects these two parts of the problem. This is what I need to plug in right here to do my math to get the answer that I want.
At the end of the collision, the v-final from the collision is the v-0 for the energy conservation part to the problem.
Okay, then there’s another 0 here I can get rid of, my initial height, or altitude, I’ve called 0, so that term goes away, and this equation does simplify to 0 equals mgh-final minus 1/2 mv0 squared. If I just move this to the other side, it says mathematically what I said in words. I start out with kinetic energy. This is after the collision. This is at the start of my swinging up. That kinetic energy is equivalent to the potential energy I have at the end of the problem.
The masses cancel out, as often happened in this problem. If I do a little algebra now to solve for v0 I see that it’s the square root of 2gh-final. When you plug in numbers for that, you end up with a v0 that is 3.569 meters per second.
Let me note this is vf from first part. So I have that value. Let’s take that now over here to the left. So I have that v-f for part one is v-f is equal to 3.569 meters per second.
Now important safety tip. Energy conservation you remember we just dealt with speed, so we solve for the speed of the block and the bullet just as it starts to swing up here after the collision.
The left side, momentum conservation as a tool now deals with velocity. So these are — this is velocity information. We do want to do a quick check when we come over and substitute in for this that if we put a positive number in here that would have to indicate the positive direction that they’re headed. That’s going to set the same sign, or direction, for this velocity, and that does make sense.
I didn’t explicitly write it and I probably should have, but we can go with the standard convention of positive to the right. So, all velocities in this problem would be positive. Okay, so I am going to do that. I’m going to leave this as positive.
Here’s my equation. I just scrolled up on the screen and copied it and pasted again. I’m ready to plug numbers in now. I have my v-f. I’m going to erase this now and put a little check mark there, ch, and I’m ready to put in numbers. I end up with m1 plus m2 is 2.51 kilograms. I have to divide that by 0.01 kilograms, and that’s times then 3.569 meters per second.
I end up then with 896 meters per second, a very high speed. That’s almost 1,600 miles an hour for the velocity of the incoming bullet. It’s positive, meaning to the right. The problem directly asked me for the speed of the incoming bullet, so I would also report then the speed is 896 meters per second. Okay, that’s the end of that example.