Question

A tennis player swings her 1000 g racket with a speed of 12 m/s. She hits a 60 g tennis ball that was approaching her at a speed of 15 m/s. The ball rebounds at 40 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 7.00, what is the average force that the racket exerts on the ball?
_________N

Answer 1

The velocity and force are required.

The speed of the racket is 8.7 m/s

The required force is 471.43 N.

$m_1$ = Mass of racket = 1000 g

$m_2$ = Mass of ball = 60 g

$u_1$ = Initial velocity of racket = 12 m/s

$u_2$ = Initial velocity of ball = -15 m/s

$v_1$ = Final velocity of racket

$v_2$ = Final velocity of ball = 40 m/s

$\Delta t$ = Time = 7 ms

The equation of the momentum will be

$m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\Rightarrow v_1=\dfrac{m_1u_1+m_2u_2-m_2v_2}{m_1}\\\Rightarrow v_1=\dfrac{1\times 12+0.06\times (-15)-0.06\times 40}{1}\\\Rightarrow v_1=8.7\ \text{m/s}$

Force is given by

$F=m_2\dfrac{v_2-u_2}{\Delta t}\\\Rightarrow F=0.06\times \dfrac{40-(-15)}{7\times 10^{-3}}\\\Rightarrow F=471.43\ \text{N}$

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