Question

A tennis player swings her 1000 g racket with a speed of 15.0 m/s. She hits a 60 g tennis ball that was approaching her at a speed of 18.0 m/s. The ball rebounds at 40.0 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.
_________m/s
If the tennis ball and racket are in contact for 7.00 , what is the average force that the racket exerts on the ball?

Answer 1

Answer:

$v_r=13.68\ m.s^{-1}$ is the final velocity of the racket.

$F=18.86\ N$

Explanation:

Given:

• mass of the racket, $m_r=1000\ g$

• mass of ball, $m_b=60\ g$

• initial speed of racket, $u_r=15\ m.s^{-1}$

• initial speed of ball, $u_b=18\ m.s^{-1}$

• final speed of ball, $v_b=40\ m.s^{-1}$

• time of contact of racket with the ball, $t=0.07\ s$

By the law of conservation of momentum:

$m_r.u_r+m_b.u_b=m_r.v_r+m_b.v_b$

where: $v_r=$ final velocity of the racket

$1000\times 15+60\times 18=1000\times v_r+60\times 40$

$v_r=13.68\ m.s^{-1}$ is the final velocity of the racket.

By the Newton's second law of motion:

$F=\frac{dp}{dt}$ ............................(1)

where:

dp = change in momentum

dt = change in time

Change in momentum of ball:

$\Delta p_b=m_b.v_b-m_b.u_b$

$\Delta p_b=60\times 10^{-3}\times (40-18)$

$\Delta p_b=1.32\ kg.m.s^{-1}$

Now, using eq.(1):

$F=\frac{1.32}{0.07}$

$F=18.86\ N$