Chapter 7, Problem 86P

To determine

The average force exerted on the ball.

Answer to Problem 86P

The average force exerted on the ball is $\text{34}\text{N}$.

Explanation of Solution

Write the expression for velocity of the ball before bounce.

  $\overrightarrow{u}=u_x\cos \theta \widehat{i}-u_y\sin \theta \widehat{j}$        (I)

Here, $u$ is the initial velocity of the ball before bounce, $u_x$ is the velocity of the ball strikes the ground below the horizontal position, $u_y$ is the velocity of the ball strikes the ground in vertical position, and $\theta$ is the angle with respect to ground below the horizontal position.

Write the expression for velocity of the ball after bounce.

  $\overrightarrow{v}=v_x\cos \varphi \widehat{i}+v_y\sin \varphi \widehat{j}$        (II)

Here, $v$ is the final velocity of the ball after bounce, $v_x$ is the velocity of the ball strikes the ground above the horizontal position, $v_y$ is the velocity of the ball strikes the ground in vertical position, and $\varphi$ is the angle with respect to ground above the horizontal position.

Write the expression for the change in momentum of the ball.

  $\Delta \overrightarrow{p}={\overrightarrow{p}}_f-{\overrightarrow{p}}_i$        (III)

Here, $\Delta p$ is the change in momentum of the ball, $p_f$ is the final momentum of the ball, and $p_i$ is the initial momentum of the ball.

The expression for initial momentum of the ball is,

  ${\overrightarrow{p}}_i=m\overrightarrow{u}$        (IV)

Here, $m$ is the mass of the tennis ball.

The expression for final momentum of the ball is,

  ${\overrightarrow{p}}_f=m\overrightarrow{v}$        (V)

Write the expression from the relation between force and rate of change of momentum.

  $F_{\text{avg}}=\frac{\Delta \overrightarrow{p}}{\Delta \overrightarrow{t}}$        (VI)

Here, $F_{\text{avg}}$ is the average force exerted on the ball and $\Delta t$ is the interaction time period of the ball with ground.

Conclusion:

Substitute the equation (IV) and (V) in equation (III).

  $\begin{array}{c}\Delta \overrightarrow{p}=m\overrightarrow{v}-m\overrightarrow{u}\\ =m\left(\overrightarrow{v}-\overrightarrow{u}\right)\end{array}$        (VII)

Substitute $54\text{m/s}$ for $u_x$ and $u_y$, $22°$ for $\theta$ in equation (I) to find $\overrightarrow{u}$.

  $\begin{array}{c}\overrightarrow{u}=\left(54\text{m/s}\right)\cos \left(22°\right)\widehat{i}-\left(54\text{m/s}\right)\sin \left(22°\right)\widehat{j}\\ =\left(50.1\text{m/s}\right)\widehat{i}-\left(20.2\text{m/s}\right)\widehat{j}\end{array}$

Substitute $53\text{m/s}$ for $v_x$ and $v_y$, $18°$ for $\varphi$ in equation (II) to find $\overrightarrow{v}$.

  $\begin{array}{c}\overrightarrow{v}=\left(53\text{m/s}\right)\cos \left(18°\right)\widehat{i}+\left(53\text{m/s}\right)\sin \left(18°\right)\widehat{j}\\ =\left(50.4\text{m/s}\right)\widehat{i}+\left(16.4\text{m/s}\right)\widehat{j}\end{array}$

Substitute $\left(50.1\text{m/s}\right)\widehat{i}-\left(20.2\text{m/s}\right)\widehat{j}$ for $\overrightarrow{u}$, $\left(50.4\text{m/s}\right)\widehat{i}+\left(16.4\text{m/s}\right)\widehat{j}$ for $\overrightarrow{v}$, and $0.060\text{kg}$ for $m$ in equation (VII) to find $\Delta p$.

  $\begin{array}{c}\Delta \overrightarrow{p}=\left(0.060\text{kg}\right)\left[\left(50.4\text{m/s}\right)\widehat{i}+\left(16.4\text{m/s}\right)\widehat{j}-\left(50.1\text{m/s}\right)\widehat{i}+\left(20.2\text{m/s}\right)\widehat{j}\right]\\ =\left(0.018\text{kg}\cdot \text{m/s}\right)\widehat{i}+\left(2.196\text{kg}\cdot \text{m/s}\right)\widehat{j}\end{array}$

Substitute $\left(0.018\text{kg}\cdot \text{m/s}\right)\widehat{i}+\left(2.196\text{kg}\cdot \text{m/s}\right)\widehat{j}$ for $\Delta p$ and $0.065\text{s}$ for $\Delta \overrightarrow{t}$ in equation (VI) to find average force.

  $\begin{array}{c}{F}_{\text{avg}}=\frac{\left(0.018\text{kg}\cdot \text{m/s}\right)\widehat{i}+\left(2.196\text{kg}\cdot \text{m/s}\right)\widehat{j}}{0.065\text{s}}\\ =\left(0.277\text{N}\right)\widehat{i}+\left(33.8\text{N}\right)\widehat{j}\end{array}$

The magnitude of the average force exerted on the ball is,

  $\begin{array}{c}{F}_{\text{avg}}=\sqrt{{\left(0.277\text{N}\right)}^2+{\left(33.8\text{N}\right)}^2}\\ =33.8\text{N}\\ \approx \text{34}\text{N}\end{array}$

Therefore, the average force exerted on the ball is $\text{34}\text{N}$.