Question

A tennis player swings her 1000 g racket with a speed of 11.0 m/s . She hits a 60 g tennis ball that was approaching her at a speed of 13.0 m/s . The ball rebounds at 45.0 m/s . 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. If the tennis ball and racket are in contact for 11.0 ms , what is the average force that the racket exerts on the ball?

Answer 1

We can use the conservation of linear momentum to find the final velocity of the racket.
The initial total momentum of the system
P_i = m_r v_ir + m_b v_ib
Where the subscript r ,ir represent the racket and the subscript b represent the tennis ball.
P_i = 1000 g x 11.0 m/s - 60 g x 13.0 m/s
P_i = 10220 g m/s
The final momentum
P_f = m_r  v_fr +  m_b  v_fb
P_f = 1000 g x v_fr + 60.0 g x 45.0 m/s
P_f =  2700 + 1000 g x v_fr
Equating the momenta
P_i = P_f
2700 + 1000 g x v_fr = 10220 g m/s
v_fr = 7.52 m/s
b) The impulse on the ball is
F t = m v
Where the RHS is the change on momenta of the ball
The average force
F = m v /  t
F = 0.06 kg x (45 m/s - 13 m/s) / 11.0 x 10^-3 s
F =  174.54 N