Chapter 9.1, Problem 13SSC

(a)

To determine

To draw: The arrows showing the momentum of the ball before and after bat hits the ball.

Answer to Problem 13SSC

The sketch of the arrows showing the momentum of the ball before and after bat hits the ball is shown in Figure 1.

Explanation of Solution

Given:

The mass of the ball is $m=0.174\text{kg}$ .

The velocity of the ball before hitting the bat is $v_{\text{i}}=26.0\text{m/s}$ .

The velocity of the ball after hitting the bat is

$v_{\text{f}}=-38.0\text{m/s}\left(\begin{array}{l}\text{negative sign }\\ \text{indicates opposite direction}\end{array}\right)$ .

Calculation:

Consider a ball is moving in the positive $x$ direction with the velocity $v_{\text{i}}$ hits the bat and after hitting the bat, the ball moves in the opposite direction with the velocity $v_{\text{f}}$ .

Sketch the vector diagram for the momentum of the ball before and after hitting the ball.

Figure 1

Conclusion:

Thus, the sketch of the arrows showing the momentum of the ball before and after bat hits the ball is shown in Figure 1.

(b)

To determine

The change in the momentum of the ball.

Answer to Problem 13SSC

The change in the momentum of the ball is $11.1\text{kg}\cdot \text{m/s}$ .

Explanation of Solution

Given:

The mass of the ball is $m=0.174\text{kg}$ .

The velocity of the ball before hitting the bat is $v_{\text{i}}=26.0\text{m/s}$ .

The velocity of the ball after hitting the bat is $v_{\text{f}}=-38.0\text{m/s}\left(\begin{array}{l}\text{negative sign }\\ \text{indicates opposite direction}\end{array}\right)$ .

Formula used:

Change in momentum is,

  $\Delta p=m\left(v_{\text{f}}-v_{\text{i}}\right)$

Calculation:

Consider a ball is moving in the positive $x$ direction with the velocity $v_{\text{i}}$ hits the bat and after hitting the bat, the ball moves in the opposite direction with the velocity $v_{\text{f}}$ .

The change in the momentum of the ball is,

$\begin{array}{l}\Delta p=m\left(v_{\text{f}}-v_{\text{i}}\right)\\ \Delta p=\left(0.174\text{kg}\right)\left[-38.0\text{m/s}-26.0\text{m/s}\right]\\ \Delta p=\left(0.174\text{kg}\right)\left(-64\text{m/s}\right)\\ \Delta p=-11.1\text{kg}\cdot \text{m/s}\left(\text{negative sign indicates direction}\right)\\ \Delta p=11.1\text{kg}\cdot \text{m/s}\end{array}$

Conclusion:

Thus, the magnitude of change in the momentum of the ball is $11.1\text{kg}\cdot \text{m/s}$ .

(c)

To determine

The impulse delivered by the bat.

Answer to Problem 13SSC

The impulse delivered by the bat is $11.1\text{kg}\cdot \text{m/s}$ .

Explanation of Solution

Given:

The mass of the ball is $m=0.174\text{kg}$ .

The velocity of the ball before hitting the bat is $v_{\text{i}}=26.0\text{m/s}$ .

The velocity of the ball after hitting the bat is $v_{\text{f}}=-38.0\text{m/s}$ .

Formula used:

Impulse is,

  $I=F\times t$

Calculation:

Consider a ball, moving in the positive $x$ direction and velocity $v_{\text{i}}$ , hits the bat.After hitting the bat, the ball moves in the opposite direction with the velocity $v_{\text{f}}$ . So, the impulse of the bat will be equal to the product of the force $\left(F\right)$ exerted by the bat and the time of contact $\left(\Delta t\right)$ of the bat with the ball. Also, it is equal to the change in the momentum of the bat $\left(\Delta p\right)$ .

The impulse delivered by the bat is,

$\begin{array}{c}\mathrm{Impulse}=\Delta p\\ F\Delta t=m\left(v_{\text{f}}-v_{\text{i}}\right)\\ F\Delta t=\left(0.174\text{kg}\right)\left[38.0\text{m/s}-\left(-26.0\text{m/s}\right)\right]\\ F\Delta t=\left(0.174\text{kg}\right)\left(64\text{m/s}\right)\\ F\Delta t=11.1\text{kg}\cdot \text{m/s}\end{array}$

Conclusion:

Thus, the impulse delivered by the bat is $11.1\text{kg}\cdot \text{m/s}$ .

(d)

To determine

The average force exerted by the bat on the ball.

Answer to Problem 13SSC

The force exerted by the bat on the ball is $F=1.4\times {10}^4\text{N}$ .

Explanation of Solution

Given:

The mass of the ball is $m=0.174\text{kg}$ .

The velocity of the ball before hitting the bat is $v_{\text{i}}=26.0\text{m/s}$ .

The velocity of the ball after hitting the bat is $v_{\text{f}}=-38.0\text{m/s}$ .

The time of contact of the bat and the ball is $\Delta t=0.80\text{ms}$ .

Formula used:

Force in terms of impulse is,

  $F=\frac{I}{t}$

Calculation:

Consider a ball, moving in the positive $x$ direction and velocity $v_{\text{i}}$ , hits the bat.After hitting the bat, the ball moves in the opposite direction with the velocity $v_{\text{f}}$ . So, the impulse of the bat will be equal to the product of the force $\left(F\right)$ exerted by the bat and the time of contact $\left(\Delta t\right)$ of the bat with the ball. Also, it is equal to the change in the momentum of the bat $\left(\Delta p\right)$ .

From part (b), the impulse delivered by the bat to the ball is $F\Delta t=11.1\text{kg}\cdot \text{m/s}$ .

The force exerted by the bat to the ball is,

$\begin{array}{c}F=\frac{m\left(v_{\text{f}}-v_{\text{i}}\right)}{\Delta t}\\ F=\frac{\left(0.174\text{kg}\right)\left[38.0\text{m/s}-\left(-26.0\text{m/s}\right)\right]}{0.08\text{ms}\times \left(\frac{1\text{s}}{1000\text{ms}}\right)}\\ F=\frac{\left(0.174\text{kg}\right)\left(64\text{m/s}\right)}{0.00008\text{s}}\\ F=1.4\times {10}^4\text{kg}\cdot \text{m/s}\times \left(\frac{1\text{N}}{1\text{kg}\cdot \text{m/s}}\right)\\ F=1.4\times {10}^4\text{N}\end{array}$

Conclusion:

Thus, the force exerted by the bat on the ball is $F=1.4\times {10}^4\text{N}$ .