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 tennis ball that was approaching her at a speed of 22 m/s. The ball rebounds at 44 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?
N

(c) How does this compare to the ball's weight?
Favg / wball

Explanation / Answer

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 = -22 m/s (note negative since ball is moving opposite direction of racket.)
v1 = final vel of racket = ?
v2 = final vel of ball = 44 m/s

plug values into above equation and solve for v1
1000 g * 9 m/s + 60 g * (-22 m/s) = 1000 g * v1 + 60 g * 44 m/s
solve for v1
v1 = 5.04 m/s

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 * (44 m/s + 22 m/s)/0.01 s
F = 396 N

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