Question
Asked Oct 8, 2019
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A baseball player throws a ball at a nearby boulder. The ball hits the boulder with a speed of 33 m/s, and then bounces straight up to a height of 15.0 m. Assuming that the rotational motion of the ball is negligible, what fraction of the ball\'s initial mechanical energy was dissipated in the collision with the boulder? Use g = 9.8 m/s2 for the acceleration due to gravity.

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Step 1

Let the mass of the baseball is m.

The initial mechanical energy when the ball hits the boulder with a velocity is,

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1 K -ту 2 1 m(33m/s) = 544.5m J

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Step 2

Potential energy after bouncing to a height h is,

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U = mgh -m(9.8m/s2)(15m) 147m J

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Step 3

Then the energy dis...

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E= 544.5m J-147m J 397.5m J

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