A tennis ball has a mass of 0.057kg.A professional tennis player hits the ball hard enough to give it a speed of 50 m/s (about 120 mi/h). The ball hits a wall and bounces back with almost the same speed (50m/s). As indicated in Figure 2.55 , high-speed photography shows that the ball is crushed 2 cm (0.02 m) at the instant when its speed is moment0arily zero, before rebounding.

Making the very rough approximation that the large force that the wall exerts on the ball is approximately constant during contact, determine the approximate magnitude of this force. Hint: Think about the approximate amount of time it takes for the ball to come momentarily to rest. (For comparison note that the gravitational force on the ball is quite small, only about (0.057 kg) (9.8 N/kg) $\mathbf{\approx }\mathbf{0}\mathbf{.}\mathbf{6}$ N. A force of5N is approximately the same as a force of one pound.)

Given that the mass $m$ of the tennis ball is 0.057kg, the initial speed $u$ is $50\mathrm{m}/\mathrm{s}$, the final speed $v$ is 50m/s in the opposite direction of initial speed and the distance ball is crushed $d$ is 0.02m.

The change in momentum of a system over the time is equals to the net external force. The difference between the final and starting values of momentum is the change in momentum.

The net force is given by

$\mathit{F}\mathbf{=}\frac{\mathbf{∆}\mathbf{p}}{\mathbf{∆}\mathbf{t}}$Here $\mathbf{∆}\overrightarrow{\mathbf{p}}$is the change in momentum whereas$\mathbf{∆}\mathit{t}$is the time interval.

The mass $m$of the tennis ball is 0.057kg, the initial speed $u$ is 50 m/s, the final speed v is 50 m/s in the opposite direction of initial speed and the distance ball is crushed d is 0.02m.

The equation for the change in momentum,

$∆P=m\left(v-u\right)$ ........................(1)

Here, $∆P$ is the change in the momentum, $m$ is the mass of the ball, $u$ is the initial speed and $v$is the final speed.

Write equation for the average speed during crush

$v\text{'}=\frac{\left(v-v_1\right)}{2}$ ...........................(2)

Here, $v$ is the average speed during crush, $v$ is the final speed of the tennis ball and $v_1$ is the speed of ball at moment of hit which is zero.

Write the equation for the time taken by the ball to stop

$∆t=\frac{d}{v\text{'}}$ ........................(3)

Here, $∆t$ is the time taken by the ball to crush, $d$is the distance tennis ball has crushed and $v\text{'}$ is the average speed during crush.

Write the equation for the force on the tennis ball by the wall

$F=\frac{∆P}{2\times ∆t}$ ................................(4)

Here, $F$ is the force on the tennis ball due to the wall, $∆P$ is the change in the momentum and $∆t$ is the time taken by the ball to crush.

Substitute 50 m/s for v and 0 for $v_1$ in equation (3)

localid="1668503594971" $$\begin{gathered} v\text{'}=\frac{50}{2}\mathrm{m}/\mathrm{s} \\ =25\mathrm{m}/\mathrm{s} \end{gathered}$$

Substitute 0.02 m for $d$and 25 m/s for $v\text{'}$in equation (3)

$$\begin{gathered} ∆t=\frac{0.02\mathrm{m}}{25\mathrm{m}/\mathrm{s}} \\ =0.0008\mathrm{s} \end{gathered}$$

Substitute 50m/s for $v,-50\mathrm{m}/\mathrm{s}$ for $u$and 0.057kg for $m$ in equation (1)

$$\begin{gathered} ∆P=0.057\left(50--50\right)\mathrm{kg}·\mathrm{m}/\mathrm{s} \\ =5.7\mathrm{kg}·\mathrm{m}/\mathrm{s} \end{gathered}$$

Substitute $5.7\mathrm{kg}·\mathrm{m}/\mathrm{s}$ for $∆P$ and $0.0008\mathrm{s}$for $∆t$in equation (4)

$$\begin{gathered} F=\frac{5.7\mathrm{kg}·\mathrm{m}/\mathrm{s}}{2\times 0.000\mathrm{ss}} \\ =3562.5\mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Therefore, the force on the tennis ball is $3562.5\mathrm{kg},\mathrm{m}/{\mathrm{s}}^2$.