a tennis player swings her 800 gram racket with a speed of 15m/s. she hits a 50

Question

a tennis player swings her 800 gram racket with a speed of 15m/s. she hits a 50 gram tennis ball that was approaching her at a speed of 30m/s. the ball rebounds at 50m/s.

A. how fast is her racket moving immediately after the impact? ( ignore the interaction of the racket with her hand for the breif duration of the collision)

b. if the tennis ball and racket are in contact for 5ms, what is the average force that the racket exerts on the ball?

please give detail answers (along with the formulas used)

Explanation / Answer

Momentum = mass*velocity

Initial momentum of racket = 0.8*15 = 12 kg-m/s

Initial momentum of ball = 0.05*(-30) = -1.5 kg-m/s (negative sign because ball is coming from opposite direction)

Total initial mometum = 12 + (-1.5) = 10.5 kg-m/s

Final momentum of ball = 0.05*50 = 2.5 kg-m/s

Final momentum of racket = 0.8*v (assuming v is the velocity with which racket is moving after collision)

Total final momentum = 2.5 + 0.8*v

(a) By momentum consrvation, total initial momentum = total final momentum

Thus, 10.5 = 2.5 + 0.8*v

So, v = 10 m/s

(b) Impulse on ball = change in momentum of ball = 2.5 - (-1.5) = 4 kg-m/s

Impulse = Force*time

So, 4 = force*0.005

force = 800 N

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