2-dimensional momentum problem

Welcome back. We will now do a momentum problem in two dimensions. So let's see what we have here. So we have this ball A and we could maybe even think of it as maybe what's going on on the surface of a pool table. We have ball A and it's moving with its 10 kilograms. So these numbers are the mass of the balls. This is a 10 kilogram ball and it's moving to the right at 3 meters per second. And then it hits this ball, B, which is a 5 kilogram ball. And then we know that ball A, ball A kind of ricochets off of ball B and gets set onto this new trajectory. Now, instead of going due right, it's going at a 30 degree angle to, I guess we could say, horizontal. It's going in a 30 degree angle at 2 meters per second. And the question is, what is the velocity of ball B? So velocity is both magnitude and direction. So we need to figure out essentially, what is ball B doing? Ball B is going to be going-- We can just think about it. If you ever played pool, we could guess that ball B is going to go roughly in that direction. But we need to figure out exactly what the angle is and exactly what its velocity is. So let's do this problem. So at first you're saying, oh, Sal, this looks confusing. You know, I know momentum should be conserved and all that. But now we have these vectors and there's two dimensions and how do I do that? And the key here is that there's just really one more step when you're working on it in two dimensions or really three dimensions or an arbitrary number of dimensions. When we did one dimension, you made sure that momentum was conserved in that one dimension. So when you do two dimensions, what you do is you figure out the initial momentum in each of the dimensions. So you break it up into the x and y components. And then you say the final momentum of both objects are going to equal the initial x momentum and are going to equal the initial y momentum. So let's figure out the initial x momentum. So P for momentum. Because the m is for mass. So let's say the initial momentum in the x direction-- And we don't have to write initial or final because really, the total momentum in the x direction is always going to be the same. So let's say what the initial-- Actually, let me write initial just so it hits the point home that initial and final don't change. So the initial momentum in the x direction. So i for initial, x-- I should do something better than keep writing these subscripts --is equal to what? Well ball B has no initial velocity, so it has no momentum. Ball A is 10 kilograms. And what is its velocity in the x direction? Well all of its velocity is in the x direction. So it's 3. I mean, this is only moving in the x direction. So the momentum in the x direction is 30 kilogram meters per second. Mass times velocity, kilogram meters per second. And what's the initial momentum in the y direction? Well B isn't moving at all, so it has no momentum in any direction. And A, all of A's movement is in the x direction. It's not moving at an angle or up at all, so it has no momentum in the y direction. So we immediately know that after the collision, the combined momentum of both of these balls in the x direction has to be 30, and the combined momentum of both of these balls in the y direction has to be 0. So let's figure out what A's momentum in the x and y directions are. So to figure out what A's momentum is, we just have to figure out what A's velocity in the x and y directions are and then multiply that times the mass. Because mass doesn't have any direction. So let's figure out the x and y components of this velocity. Let's do the x component of the vector first. So the x is just this vector. Change colors to keep things interesting. The y is this vector. That is the y component. And so, what are those? And this hopefully, is going to be almost second nature to you if you've been watching all of the other videos on Newton's laws. This is just our trigonometry and we can write out our SOH-CAH-TOA again. And I reassure you, this is the hardest part of any of these multi-dimensional trig problems-- Multi-dimensional physics problems, which really are just trig problems. So if we want to figure out the x component, so the velocity of A in the x direction, what is it equal to? Well this is adjacent to the angle. We know the hypotenuse. So we know VA sub x or the velocity of A in the x direction over the hypotenuse, over 2 meters per second, is equal to what? Adjacent over hypotenuse. Cosine. Is equal to cosine of 30 degrees. Or the velocity of A in the x direction is equal to 2 cosine of 30 degrees. What's cosine of 30 degrees? It's square root of 3 over 2. This is square root of 3 over 2. And square root of 3 over 2 times 2 is equal to square root of 3. So this is equal to the square root of 3 meters per second. And what is the velocity of A in the y direction? Well hopefully, this second nature to you as well. But since opposite over hypotenuse is equal to the sine of 30. So VA in the y direction is equal to 2 times the sine of 30 degrees. The sine of 30 degrees is 1/2. So this is 1/2. 1/2 times 2 is equal to 1 meter per second. So after the collision, A is moving at 1 meter per second up. One meter per second in the upwards direction. And it's moving at square root of 3 meters per second in the rightwards direction. So what is going to be A's momentum in each of the directions? Well, we figured out its velocity, so we just multiply each of the velocities times the mass. So A has a mass of 10 kilograms. And this is going to be the final momentum. Momentum of A in the x direction is going to equal square root of 3 times 10. Square root of 3 is the velocity, 10 is the mass. So it's 10 square roots of 3 kilogram meters per second. And the momentum of A in the y direction is going to be-- and since it's going up, we'll say its positive --it's 1 meters per second is the velocity times the mass. So 10 times 1 is 10 kilogram meter per second. So now let's figure out B. Let's do the y direction first because they add up to 0. I'm going to switch colors. We know that the momentum of-- and this is after the collision. The momentum of A in the y direction plus momentum of B in the y direction have to equal what? What was the initial momentum in the y direction? Right, it was 0. There was no movement in the y direction initially. We know the momentum of A in the y direction. It's 10. 10 kilogram meters per second plus the momentum of B in the y direction is equal to 0. So solving for this, just subtract 10 from both sides. So the momentum of B in the y direction is equal to 10 kilogram meters per second. You know the units. So if its momentum is 10 in the y direction, what is its velocity in the y direction? Well, momentum is equal to mass times velocity. So we know that 5 times the velocity in the y direction-- that's its mass --is equal to 10. 10 is its momentum. So the velocity of the y direction of B must be 2 meters per second. So there we go. We figured out B's velocity. And so let's say this is B's velocity vector in the y direction is-- And this is a minus because this is equal to minus 10. So it's going down. It was a velocity of positive 1 going up and then the minuses carry through and this is a velocity of minus 2 meters per second for B in the y direction. So now let's figure out the velocity of B in the x direction. And I'm running out of space and it's getting messy. But we just have to remember that the momentum of B in the-- The momentum of A in the x direction, which is 10 square root of 3, plus momentum of B in the x direction has to equal what? It has to equal the initial momentum in the x direction, which is 30. So to figure out the momentum of B in the x direction, we just subtract 10 square root of 3 from 30. And let's do that. So let's figure out 3 square root times 10 equals. And then subtract that from 30. And we get let's just say 12.7. So we know that the momentum of B in the x direction is equal to 12.7, 12.7 kilogram meters per second. And we know the momentum, so we just divide by the mass and we get its velocity in the x direction. So 12.7 divided by 5. So velocity of B in the x direction is 12.7 divided by 5. 12.7 divided by 5 is equal to 2.54 meters per second. So its velocity in the x direction is 2.54 meters per second. So it's going faster in both directions. I'm not going to figure out the angle here because I've actually run out of time. But if you were to add these two vectors, you'd get an angle something like this. And you could figure out the angle by taking the arc tan. Well, I won't go into the-- that's a complexity right now. Actually, I'll do that in the next video just so I won't leave you hanging. But we know what the x and y components of B's velocity is. See you in the next video.