11. Momentum & Impulse

Push-Away Problems

# Sliding Football Player

Patrick Ford

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Hey guys, let's take a look this problem here. Hopefully get a chance to work it out on your own. Not here's a little bit help. So the idea here, we have a football player on frictionless ice who has a football. So they're both moving along together. He's got this football like this and they're both moving at two m per second afterwards. Them the player is actually going to throw the football ahead of about 15 m per second. So this is gonna be a push away problem. Both objects are together and now they're moving separately afterwards. Let's go ahead and draw our before and after. So this is our before, what is our after look like? Well, basically, now what happens is you have the football player like this and now he's throwing the football ahead of him And we actually can figure out what the speed is. So what I'm gonna do is I'm going to call this M1 which is 70 and I'm going to call this M2, which is 0.45 And we're told that the player is going to throw the ball with 15 m per second relative to the player. What does that mean? Well, this is gonna be my V. Two final here. What this means is that they are both initially were already moving at two m per second. Then the football player throws it an additional 15. So what actually happens is this is gonna be two plus 15 and that's going to equal 17. That's the new final velocity. We want to figure out what's the players velocity after that happens after this push away. All right. So let's go ahead and set up our momentum conservation equation. This is going to be M one V one initial plus M two V two initial equals M one V one final plus M two V two final. Okay, so basically I've got this 70 times something plus 0.45 times something equal, 70 times something plus 0.45 times something. All right, so, what what goes inside these parentheses? Remember initially they're actually both moving together. So these speeds are actually the same and they're both moving with initial two m per second. So that's what goes inside here. So that's two and two afterwards, we're looking for the final velocity of this V once. This is actually what goes inside here. That's my target variable. What about the football? Remember the football we just calculated is moving at 17 m per second forwards. That's what goes inside here. Now, we just go ahead and simplify the left and right sides. So that this becomes here is this becomes um let's see on the left side to get 140. And on the right side, I'm gonna get 70 times v one final plus. And then this is going to be 7.65. So all they do is just move and subtract the 7.65. I end up with 133.2 25 equals 70 times V one final and I can just go ahead and divide by the 70 once you get is you get V one final is equal to 1. m per second. So let's see basically what happens is that the football player is still moving to the rights just slightly slower than he was initially. Initially. Both objects were moving at two m per second. And then because the football is a lot lighter, the fact that it's going at 17 m per second now means there hasn't been a lot of recoil and the much heavier football player. So the football player has a final velocity of 1.9, so he's only slowed down a little bit, but he's given a lot of momentum to the football. All right, so that's the idea here, guys, let me know if you have any questions and I'll see the next one.

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