A tennis player swings her 1000 g racket with a speed of 9 m/s. She hits a 60 g

Question

A tennis player swings her 1000 g racket with a speed of 9 m/s. She hits a 60 g tenis ball that was approaching her at a speed of 23 m/s. The ball rebounds at 36 m/s. (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. m/s (b) If the tennis ball and racket are in contact for 10 ms, what is the average force that the racket exerts on the ball? (c) How does this compare to the ball's weight? Favo/ wball =

Explanation / Answer

a)

Use conservation of linear momentum

m1u1 + m2u2 = m1v1 + m2v2

m1 = mass of racket = 1000g

u1 = initial vel of racket = 9 m/s

m2 = mass of ball = 60 g

u2 = initial velo of ball = -23 m/s (note negative since ball is moving opposite direction of racket.)

v1 = final vel of racket = ?

v2 = final vel of ball = 36 m/s

plug values into above equation and solve for v1

1000 g * 9 m/s + 60 g * (-23 m/s) = 1000 g * v1 + 60 g * 36 m/s  

v1 = 5.46 m/s

b)

To calculate average force use impulse-momentum theorem

I = F* (delta t)

I = m (delta v)

equate the above and solve for F

F * (delta t) = m * (delta v)

F = m * (delta v)/(delta t)

F = 0.060 kg * (36 m/s + 23 m/s)/0.010 s

F = 354 N

c)

Favg / Wball = 354*1000/(60*9.8) = 602 times

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