A tennis player swings her 1090 g racket with a speedof 10.2 m/s. She hits a 53

Question

A tennis player swings her 1090 g racket with a speedof 10.2 m/s. She hits a 53 g tennis ball that was approaching herat a speed of 18.7 m/s. The ball rebounds at 38.3 m/s. How fast isher racket moving immediately after the impact? You can ignore theinteraction of the racket with her hand for the brief duration ofthe collision.





If the tennis ball and racket are in contact for 9.3ms, 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

The question implies that the "center of mass" of the racketis moving at 10.2 m/s so we are basically considering the collision of the center of mass ofthe racket with that of the ball. m1 v1 =m2 v2    Thechange of momentum of the racket equals that of the ball v1 = m2 v2 / m1 = .053 * 56 / 1.09 = 2.72m/s   (the magnitude of the change in velocity of theball is 56 m/s) So the racket after impact is moving at 10.2 - 2.72 = 7.48m/s p = m v / t = F    wherethe applied force produces the change in momentum p F = .053 * 56 / .0093 = 319 N The gravitational force on the ball m g = .053 * 9.8 =.519 N So the force applied by the racket is 319 / .519 = 614 timesthat of the force of gravity

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