A tennis player swings her 1000 g racket with a speed of 12 m/s. She hits a 60 g

Question

A tennis player swings her 1000 g racket with a speed of 12 m/s. She hits a 60 g tennis ball that was approaching her at a speed of 19 m/s. The ball rebounds at 38 m/s. (a) What is the magnitude of the change in momentum of the tennis ball? (Remember that momentum is a vector quantity.) 13.38 X kg m/s (b) How fast is her racket moving immediately after the impact? You can ignore the interaction of the racket with her hand for the brief duration of the collision. HINT: Think about Newton's Third Law for the ball and the racquet. 3.42 x m/s (c) If the tennis ball and racket are in contact for 10 ms, what is the average force that the racket exerts on the ball? HINT: Remember that we can approximate the Impulse as the product of some average force times the time duration of the collision. 488.6

Explanation / Answer

A) Change in momentum, delta P = 0.06*(19 -(-38)) = 0.06*57 = 3.42 kg.m/s

B) 1000*12 - 60*19 = 60*38 + 1000*v

(12000 - 1140 - 2280)/1000 = v

v = 8.58 m/s

C) change in momentum = 3.42 = F*10*10^-3

F = 342 N

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