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2-dimensional momentum problem (part 2)

We finish the 2-dimensional momentum problem. Created by Sal Khan.

Video transcript

Welcome back. When I left off I was rushing at the end of this problem because I tend to rush at the end of problems when I am getting close to the YouTube 10 minute limit. But I just wanted to review the end of it because I feel like I rushed it. And then, actually continue with it and actually solve for the angle and then, introduce a little bit of-- a little more trigonometry. So just to review what we did, we said momentum is conserved and in two dimensions that means momentum is conserved in each of the dimensions. So we figured out what the initial momentum of the entire system was and we said, well, in the x direction, the initial momentum-- and all the momentum was coming from the ball A right. Because ball B wasn't moving, so its velocity was 0. So its momentum was 0. So ball A in the x direction and it was only moving in the x direction. So it's momentum in the x direction was 3 meters per second times 10 kilogram meters per second. And we got 30 kilogram meters per second. And then there was no momentum in the y direction. And then we knew that well after they hit each other, ball A kind of ricochets off at a 30 degree angle at 2 meters per second. We used that information to figure out the x and y components of A's velocity. So A's velocity in the y direction was 1 meter per second and A's velocity in the x direction was square root of 3. And we used that information to figure out A's momentum in each direction. We said well, the momentum in the y direction must be 1 meter per second times A's mass, which is 10 kilogram meters per second. Which I wrote-- what I wrote here. And then we figured out A's momentum in the B direction and we said well, that's just going to be square root of 3 times 10. And that's 10 square root of 3. And then we used that information to solve for B's momentum. Because we said well, B's momentum plus A's momentum in the x direction has to add up to 30. This was the x direction before. And we knew that B's momentum plus A's momentum in the y direction had to add up to 0, right? And so, since y's momentum going upwards was 10 kilogram meters per second, we knew that B's momentum going downwards would also have to be 10 kilogram meters per second. Or you could even say it's negative 10. And we figure that out based on the fact that B had half the mass. That its velocity going down was 2 meters per second. And similarly, we knew that A's momentum in the x direction, which was 10 square root of 3 kilogram meters per second, plus B's momentum in the x direction is equal to 30. And then we just subtracted out and we got B's momentum in the x direction. And then we divided by B's mass to get its velocity. Which we got as 2.54. So that's where I left off and we were rushing. And already, this gives you a sense of what B is doing. Although it's broken up into the x and y direction. Now if we wanted to simplify this, if we wanted to kind of write B's new velocity the same way that the problem gave us A's velocity, right? They told us A's velocity was 2 meters per second at an angle of 30 degrees. We now have to use this information to figure out B's velocity and the angle of it. And how do we do that? Well this is just straight up trigonometry at this point, or really just straight up geometry. Let me clear all of this. And let's remember these two numbers, 2.54 and minus 2. So B, we learned that in the x direction its velocity-- this is all for B-- is equal to 2.54 meters per second and then y direction, it was moving down. We could write this as minus 2. But I'll just write this as 2 meters per second downwards. Right? Same thing. Minus 2 up is the same thing as 2 meters per second down. So the resulting vector's going to look something like this. When you add two vectors you just put them-- put the one's end at the beginning of the other-- put them front to end, like we did here. And then you add them together and this is the resulting vector. And I think you're used to that at this point. And now we have to figure out this angle and this side. Well this side is easy because this is a right angle, so we use Pythagorean theorem. So this is going to be the square root of 2.54 squared plus 2 squared. And what's 2.54 squared? 2.54 times-- whoops. 2.54 times 2.54 is equal to 6.45. So that's the square root of 6.45 plus 4, which equals the square root of 10.45. And take the square root of that. So that's 3.2, roughly. So the resulting velocity in this direction, whatever angle this is, is 3.2 meters per second. And I just used Pythagorean theorem. So now all we have to do is figure out the angle. We could use really any of the trig ratios because we know all of the sides. So I don't know, let's use one that you feel comfortable with. Well let's use sine. So sine of theta is equal to what? SOH CAH TOA. Sine is opposite over hypotenuse. So the opposite side is the y direction, so that's 2, over the hypotenuse, 3.2. So 2 divided by 2 divided by 3.2 is equal to 0.625, which equals 0.625. So sine of theta equals 0.625. And maybe you're not familiar with arcsine yet because I don't think I actually have covered yet in the trig modules, although I will eventually. So we know it's just the inverse function of sine. So sine of theta is equal to 0.625. Then we know that theta is equal to the arcsine of 0.625. This is essentially saying, when you say arcsine, this says, tell me the angle whose sine is this number? That's what arcsine is. And we can take out Google because it actually happens that Google has a-- let's see. Google actually-- it's an automatic calculator. So you could type in arcsine on Google of 0.625. Although I think the answer they give you will be in radians. So I'll take that answer that will be in radians and I want to convert to degrees, so I multiply it times 180 over pi. That's just how I convert from radians to degrees. And let's see what I get. So Google, you see, Google says 38.68 degrees. They multiplied the whole thing times 180 and then divided by pi, but that should be the same thing. So roughly 38.7 degrees is theta. Hope you understand that. You could pause it here if you don't, but let me just write that down. So it's 38 degrees. So theta is equal to 38.7 degrees. So then we're done. We figured out that ball B gets hit. This is ball B and it got hit by ball A. Ball A went off in that direction at a 30 degree angle, at a 30 degree angle at 2 meters per second. And now ball B goes at 38.-- or we could say roughly 39 degrees below the horizontal at a velocity of 3.2 meters per second. And does this intuitively make sense to you? Well if you remember the problem from before-- and I know I erased everything. Ball A had a mass of 10 kilograms while ball B had a mass of 5 kilograms. So it makes sense. So let's think about just the y direction. Ball A, we figured out, the y component of its velocity was 1 meter per second. And ball B's y component is 2 meters per second downwards. And does that makes sense? Well sure. Because their momentums have to add up to 0. There was no y component of the momentum before they hit each other. And in order for B to have the same momentum going downwards in the y direction as A going upwards, its velocity has to be essentially double, because its mass is half. And a similar logic, although the cosine-- it doesn't work out exactly like that. But a similar logic would mean that its overall velocity is going to be faster than the- than A's velocity. And so what was I just-- oh yeah. My phone was ringing and I got caught up. My brain starts to malfunction. But anyway, as I was saying, so just intuitively it makes sense. B has a smaller mass than A, so it makes sense that-- one, B will be going faster and that it gets deflected a little bit more as well. The reason why it seems like it gets deflected more is because its y component is more. But anyway, that last piece is just to kind of hopefully give you a sense of what's happening and I will see you in the next video.