 Okay, I'm going to work out a problem, I'm not going to put the numbers because the numbers aren't important. This is a really classic problem about a ballistic pendulum. So one, and they actually use these things. So one way to measure the speed of a bullet is to take a bullet and it comes out of the gun and then it collides with let's say a block on some strings. So this is before the collision, let's call this Mb and let's call this Big M and this is V of the bullet when it's shot out of the gun. After it collides then we have next step is the bullet's embedded in there and this we'll call V2, that's in the next direction and then now it's a pendulum and it's moving so it swings up. So it goes like this, it's usually on two strings to prevent it from rotating like that. And so the thing about that is that it's easier to measure, this is going to be going slower after it collides but still you can't measure the velocity but you can record how high it goes, you can just be sitting by the side or have an angle measurement or something to determine how high it goes. So you're going to start with the mass of the bullet, the mass of the block and you're going to record how high it goes or the angle and from that say well how fast was the bullet going when it was shot. So we want to find Vb in terms of the mass of the bullet, the mass of the block and the height without measuring the velocity. So let's break this into pieces, in fact let's work backwards because backwards is good. So if I know that it collides and it swings up, can I find out how fast it was going down there? Yes, I can. So the first thing is, is this work energy a momentum principle? In this case it's work energy because I don't care about the time I have the height so that says you should use work energy. So work is a change in energy. Okay the next step, well what's my system? I'm going to say my system is the bullet, I can't spell, block, all those together. I can't worry about the strings. So this means the next thing is what's the work? Well do I do gravity, gravity is acting on that, do I tend to worry about that? No you don't because that's an internal force in my system, it's an interaction between these and the earth and so it doesn't do any work on the system, it's part of the system. What about the strings? They're pulling it this way, they do work and the answer to there is no because as the block's moving the strings pull perpendicular to the direction it's moving. So right here there's delta r and there's t of the tension and so t dot delta r is going to be zero so they don't do any work so the work done is zero. So I have change in kinetic plus change in gravitational potential. So at part two that's my final what's the kinetic energy? It stops when it gets to the highest point it stops. So I'm going to say zero equals k2 minus k1 plus u2 minus u1. So let's just put in the things that we know. Let's call this y equals zero so then I have zero equals k2 is zero minus k1 one half m plus mbv2 squared, I didn't leave enough room for planning on my part, okay let's put it down next. U2 is going to be m, I'm sorry, mb plus mgh and then u1 is zero. So I can solve this, I get one half m plus mbv2 squared equals mb plus mgh, these cancel so I get v2 squared equals 2gh, I'll just leave it right there, I'll just leave it like that. So some important things to note here, first it doesn't depend on the mass, it doesn't depend on the length of the string or anything like that because this is the same thing as just simple throwing something up in the air, we don't have any other type of internal energy, it's pretty straightforward. Okay let me write that over here, so v2 equals 2gh v2 squared, okay now I'm going to erase what I did so I can do some panels. Okay now I'm going backwards again, so now I know that velocity, now I can find, I can look at the collision between the bullet and the block. So the question is, show you work energy or momentum principle, this is a little tougher, so do I know the distance that these two interact? No, I don't. Do I know the time they interact? No, so what do you do? Well I know that if I look at the bullet and the block, they exert the same force for the same amount of time, so in that case it's, it is time, the time is the same. So I can say, if this is my system, then I can say delta P total equals f net to the whole thing, gravity is pulling down and the string is pulling up, so it's going to be zero vector. So in the x direction I can say P1x, the total initial momentum, I'm sorry, P2x, the total initial momentum minus P1x is zero. So finally the final momentum is going to be mass of the bullet plus mass of the block times V2 that way minus the initial momentum mass of the bullet, velocity of the bullet equals zero. So I can solve for the velocity of the bullet, it's going to be mass of the bullet plus mass of block V2 over mass of the bullet and then V2, I know that value right there, so velocity of the bullet is going to be mass of the bullet plus mass of the block times the square root of 2gh, all of that over mass of the bullet. Does this have the correct units? Well I have kilograms divided by kilograms and this is meters per second square times meter, meter square per second square, take the square root, so it does commute some velocity, so that's good. And so you can see that the higher the block goes up indicates a faster moving block, bullet. And you can put instant numbers here but I'm not going to do that. Okay, one more question, what about the increase in thermal energy of the block? Does the block increase in thermal energy? Well let's just look. So look at this and this. So before, what's the energy of the system? Well this is moving, so it's kinetic energy and this doesn't. Afterwards they're both moving. So do they have the same kinetic energy before and after? So let's call this K1, one half m bullet V bullet squared and then K2, it's going to be one half m bullet plus m V2 squared. Are they the same? Well I can put in the velocity of the bullet in terms of V2, one half the mass of the bullet cancels so I get mB plus m divided by mB and then I get VB squared, oh wait, I'm sorry, I skipped the steps. Okay, let's just do it the long way, one half mB and then put in VB squared I'm going to get mB plus m squared 2gh over mB squared. This cancels, look at that. So oh well that's V, I could get this in terms of the same thing too. Let me just go ahead and put that so they're similar. So right here I get one half mB plus m and then this squared is 2gh. So are they the same? No, they're not. This one, the kinetic energy is going to be, we're divided by small numbers so it's going to be different than this see, oh they should be squared, oh no it's not. So this is mB plus m and this is mB plus m squared over mB so it is different. So they're not the same energy before and after and in fact you're going to have more kinetic energy before than after. Well where does the other energy go? It has to go somewhere, in this case it goes into what is called thermal energy. So delta E thermal is going to be the difference in this, it's going to be k1 minus k2 because really we're saying delta E equals delta k plus delta E thermal. So delta E thermal is going to be the opposite of this. So I can take k1 minus k2 and that will give me my increase in thermal energy. Hopefully that helps.