A 0.200-kg billiard ball that is moving at 4.10 m/s strikes the bumper of a pool

Question

A 0.200-kg billiard ball that is moving at 4.10 m/s strikes the bumper of a pool table and bounces straight back at 3.28 m/s (80% of its original speed). The collision lasts 0.0250 s. (Assume that the ball moves in the positive direction initially.) (a) Calculate the average force exerted on the ball by the bumper. (Indicate the direction with the sign of your answer.) (b) How much kinetic energy in joules is lost during the collision? (Enter the magnitude.) (c) What percent of the original energy is left?

Explanation / Answer

a) Impulse = Change in momentum

=> Favgt = m(vf - vi)

=> Favg = m(vf - vi)/t = 0.200 * [(-3.28) - 4.10] / 0.0250 = -59 N

b) KE = KEi - KEf = m(vi2 - vf2)/2 = 0.200 * (4.102 - 3.282) / 2 = 0.6 J

c) KEf = mvf2/2 = m(0.8vi)2/2 = 0.64(mvi2/2) = 0.64KEi

So. 64% of original energy is left.

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