A tennis player swings her $1000$g racket with a speed of 10 m/s. She hits a $60$g tennis ball that was approaching her at a speed of 20 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.

b. If the tennis ball and racket are in contact for 10 m/s, what is the average force that the racket exerts on the ball? How does this compare to the gravitational force on the ball?

Mass of racket, $m=1000g$

Speed of racket, $10m/s$

Mass of tennis ball$60g$$60g$

Speed of tennis ball, $20m/s$

Final speed of tennis ball,$40m/s$

The collision in this issue is an elastic collision in which the laws of momentum conservation and energy conservation are applied. Mechanical energy is preserved in a perfectly elastic collision. The conservation of momentum law, which is expressed as an equation in the form

${\overrightarrow{p}}_i={\overrightarrow{p}}_f$

The momentum is given by, $p=mv$

Considering the parameters,

$$\begin{gathered} m_1{v}_{1i}+m_2{v}_{2i}=m_1{v}_{1f}+m_2{v}_{2f} \\ {m}_1{v}_{1f}=m_1{v}_{1i}+m_2{v}_{2i}-m_2{v}_{2f} \\ {v}_{1f}=\frac{m_1{v}_{1i}+m_2\left(v_{2i}-v_{2f}\right)}{m_1} \end{gathered}$$

On substitution of values,

$$\begin{gathered} v_{1,f}=\frac{1g\times 10m/s+0.06g\times (-20m/s-40m/s)}{1g} \\ {v}_{1,f}=6.4m/s \end{gathered}$$

When force act for a small interval of time, impulse acts on the area,

so the equation becomes,

$p_{f,x}=p_{i,x}+J_x$

So evaluating the value of impulse,

$J_x=0.06\times 40-0.06\times (-20)=3.6kg.m/s$

As we know,

$$\begin{gathered} J_x=\frac{F}{△t} \\ F=\frac{3.6kg\times m/s}{{10}^{-2}s}=360N \end{gathered}$$

Now the gravitational force of the ball,

$F_g=mg=60g\times 9.8=0.59N$