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Baseball. A regulation 145 g baseball can be hit at speeds of100 mph. If a line drive is hit essentially horizontally at thisspeed and is caught by a 65 $\mathrm{kg}$ player who has leapt directlyupward into the air, what horizontal speed (in $\mathrm{cm} / \mathrm{s} )$ does heacquire by catching the ball?

9.97 $\mathrm{cm} / \mathrm{s}$

Physics 101 Mechanics

Chapter 8

Momentum

Physics Basics

Kinetic Energy

Potential Energy

Energy Conservation

Moment, Impulse, and Collisions

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this question. We're told that a baseball with a mass of 0.145 kilograms, or 145 grand's is traveling horizontally with the speed of 100 miles per hour of the unit conversion that's 44.7 meters per second is traveling at the speed of horizontally. It's caught by a player with the mass of 65 kilograms, and the player is not moving horizontally when the player catches the ball players in the air but jumping directly, vertically and so the player in the horizontal direction is at rest. So using this information, you need to calculate the final philosophy of the player after they catch this ball. So this is a simple conservation of momentum problem. So we just need to remember that the initial momentum of the system will be equal to the final moment, Um, of the system. So initially the player is at rest, so the only thing only object with velocity Is Thiebaud all so all of the momentum will belong to the ball. So initially we have the massive double, which I'm simply calling M times the initial velocity of the ball, which is V I is equal to three the in the final case. After the player catches the ball, both of them will have velocity, so they're both going to contribute to the momentum. So we will have the mass of the ball times uh, the final velocity of the ball, Um, which I'm calling VfB plus the mass of the player times the final velocity of the player. No, a few things to note is there? One thing to note is that Thebe player, after catching the ball and the ball itself, are going to move in unison because the ball is caught in the players glove, so thes two are actually the same. So let's say that v f b the final velocity of the ball is equal to the final velocity of the player, and we're just going to call it VF, since there's only one of them. In that case, we have M V, I equals and VF, plus the mass of the player times via we can factor out this VF then so M v. I is equal to VF times the mass of the ball, plus the mess of the player now to isolate VF on its own on solve this problem. Um, we just need to divide both sides by this in brackets. So VF is equal to mass time massive the ball times the initial speed of the ball divided by the mass of the ball, plus the mass of the player. Now we just need to plug in the Constance that were given. So the mass of the ball is 0.145 kilograms and the initial speed of a bomb is 44.7 meters per second. And on the, um, in the denominator, we have, uh, 0.145 kilograms, plus the mass of the player is 65 kilograms. Now plug in this all into a calculator. We find that the final speed of the player, which is equal to the total final speed, is 9.97 times 10 to the negative two meters per second, which with a simple unit convergence, is simply 9.97 centimetres per second, which is not very fast, because the player has much more mass in the bowl.

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