 So, I've been thinking about momentum principle, and you know, the kinds of things I like to do with momentum principle don't really completely match with the textbook type of problems, but I think that it's important enough to look at some of them and do some examples, and then I will do some examples that would be things like in the textbook just because it usually approaches it from an F-net equals MA approach, and I think I like to talk about the momentum principle. I think it's a bit better, more useful. So, let me just make a couple situations and see if we can work some problems here. So, suppose you have a little, we're in space just to make things easier. We don't have to deal with friction, and we're way far away from objects. We don't have to deal with gravity, and I want to do it that way just because it will be a little bit easier for right now. So, let's suppose we're out in space, not in orbit, but just, there's nothing around. There's no gravitational forces at all, and I have a little model rocket that's only just one kilogram, and it has a little thruster on it. It actually has thrusters on all the sides. So, you can fire, you can make it move around. Actually, they do have these kinds of things they want to use in the space station, which has some issues with it. It looks just like there's no gravity, even though there is, and so you can go, and it can move around. So, suppose I have this, say that the thrusters F are called FT has a magnitude of, I'm just going to say two Newtons, just going to say that, made it up. So, suppose it's sitting here at rest, so that V1 equals zero x hat plus zero y hat, and now this little robot fires its rocket, so it has a thrust, so force this way FT, let's call this the X, Y direction, and let's say it does this for two seconds. I'm just completely making up stuff. I didn't want to use all one, so I picked the two, but I used the two up there too, so, oh well. Okay, so let's find out how fast is going after two seconds, and then we'll find out where it is, and then we'll do, we'll do something else. So, the momentum principle says this, F net is the change momentum over the change in time. Okay, so I know F net, in this case, is going to be just the thrust, which is going to be, it's pushing that way, so two X hat Newtons, you really need units, and the change momentum, I know the initial momentum, and I want to find the final momentum, so I can write this as two Newtons X hat equals the final momentum, mv2 minus mv1, all over the change in time. Remember, change momentum is always final minus initial, that's how we find the change, and there's my net force, I should have left it FT, but I didn't, and momentum is mass and velocity. So, I want to solve for v2, so let's first multiply by delta t, let me just write FT, F thrust, that way I can do it first just as a, I'll put any numbers in. So, I get FT times delta t equals mv2 minus m0 vector, because, and I can do it in this case, because the initial velocity was 0, so that's the zero vector, if I take mv minus the zero vector, mv2 minus the zero vector, I still get mv2. So, now I can solve for v2, you have to be careful when you're doing, dealing with vector equations, you can multiply vectors by scalars, you can add vectors together, you can divide in the same, it's multiplying, okay, but I can't divide by a vector, you can't do that, it's not, doesn't follow the same rules. So, but if I divide both sides by m, I get v2 equals FT delta t over m. Now, I can change this to a scalar equation by just looking at the x components, the x component of this should be equal to the x component of that. So, I have v2x equals FTx delta t over m, and now I can, I can find out what those things are. I had two Newtons, two seconds over one kilogram, and so that's gonna be four Newton seconds per kilogram, which is four meters per second, because one Newton is a kilogram meter per second squared. So, multiply by seconds, I get meters per second times a kilogram, I divide by kilograms, I get meters per second. So, after I fire this thruster for once, for two seconds, the rocket robot thing is moving four meters per second. The problem would be exactly the same if it was already moving at two meters per second. I would have found the change in velocity is four meters per second, so it actually be moving six meters per second. Okay, now, if it started at x equals zero and started at the origin, can I find out where it is after that time? Now, here I can use the following idea. I can say the average in the x direction, we only have x direction here, is delta x over delta t. That's true if my velocity is changing at a constant rate, meaning I have constant forces, then that's true. Or if the if the velocity is constant. Or if I can approximate it if the time interval is really small. So, it is true here. I can use that. So, the average velocity in this case is just going to be four meters per second plus zero meters per second over two. Two meters per second. And then I have x final, x two minus x one over delta t. So, that's zero. So, x two, you probably can't see that. I'm not sure where my camera is. I'm going to erase this. So, I get x two is equal to two meters per second times delta t of two seconds. So, it's going to be four meters. So, it started at x equals zero and it ends up at four meters. Maybe this is a bad idea because it ends up at four meters and its final velocity is four meters per second. But those aren't necessarily related. So, oh, I got a phone call. Okay. So, let me stop there.