A tennis player swings her 1000 {\ m g} racket with a speed of 12 {\ m m/s}. She

Question

A tennis player swings her 1000 { m g} racket with a speed of 12 { m m/s}. She hits a 60 { m g} tennis ball that was approaching her at a speed of 17 { m m/s}. The ball rebounds at 36 { m m/s} .

Part A -
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.
Express your answer using two significant figures.In m/s.

If the tennis ball and racket are in contact for 12 { m ms}, what is the average force that the racket exerts on the ball?
Express your answer using two significant figures.In N.

Explanation / Answer

Momentum is conserved: The sum of the momenta of the ball and the racket is the same before and after the collision: 1000*12g*m/s - 60*17g*m/s = 1000g*V + 60*36g*m/s. 10980g*m/s=1000g*V+2160g*m/s 8820g*m/s=1000g*V. V=8.8m/s Note that I used positive velocities to denote movement in the direction of the racket's swing, and negative velocities to denote movement opposite of the racket's swing. The change in the ball's speed is 17+36=53m/s. We added because the ball changed directions... it lost 17m/s of speed, and then it gained an additional 36m/s of speed in the reverse direction. So, the average acceleration was (53m/s)/(0.012sec) = 4416.7m/s/s And, since F=ma and m= .060kg, F=.060*4416.7 = 265N. Rounding to two sig figs, we get F=2.7*10^2 N

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