91Ó°ÊÓ

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 $\mathbf{∆}\stackrel{\mathbf{→}}{\mathbf{p}}$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

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

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

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