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

`v_r=13.68\ m.s^(-1)` is the final velocity of the racket.

`F=18.86\ N`

Step-by-step 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* 15+60* 18=1000* v_r+60* 40`

`v_r=13.68\ m.s^(-1)` is the final velocity of the racket.

By the Newton's second law of motion:

`F=(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* 10^(-3)* (40-18)`

`\Delta p_b=1.32\ kg.m.s^(-1)`

Now, using eq.(1):

`F=(1.32)/(0.07)`

`F=18.86\ N`