A tennis player swings her 1200 g racket with a speed of 9.1 m/s. She hits a 70

Question

A tennis player swings her 1200 g racket with a speed of 9.1 m/s. She hits a 70 g tennis ball that was approaching her at a speed of 21.7 m/s. The ball rebounds at 48.5 m/s. 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.

If the tennis ball and racket are in contact for 9.07 ms, what is the average force that the racket exerts on the ball? How does this compare to the gravitational force on the ball?

Explanation / Answer

We can apply conservation of momentum here:

m1u1 +m2u2 = m1v1 + m2v2

1200(9.1) + 70(-21.7) = 1200(v1) + 70(48.5)

v1 = 6006/1200 = 5 m/s (approx)

Change in momentum of ball = m(v-u) = 70/1000 * (48.5 + 21.7) = 4.914

Force = change in momentum/time = 4.914/9.07 * 1000 = 541.78 N

Gravitational force on the ball, G = mg = 70/1000 * 9.8 = 0.686 N

Thus, F = 541.78/0.686 = 789.77 times gravitaional force

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