Chapter 7, Problem 1SP

A fast ball thrown with a velocity of 40 m/s (approximately 90 MPH) is struck by a baseball bat, and a line drive comes back toward the pitcher with a velocity of 65 m/s. The ball is in contact with the bat for a time of just 0.005 s. The baseball has a mass of 142 g (0.142 kg).

1. a. What is the change in momentum of the baseball during this process?
2. b. Is the change in momentum greater than the final momentum? Explain.
3. c. What is the magnitude of the impulse required to produce this change in momentum?
4. d. What is the magnitude of the average force that acts on the baseball to produce this impulse?

To determine

The change in momentum of the baseball during the process.

Answer to Problem 1SP

The change in momentum of the baseball during the process is $14.9\text{kg}\cdot \text{m}/\text{s}$.

Explanation of Solution

Given info: The velocity of the ball before coming in contact with the bat is $40\text{m}/\text{s}$ and after striking the ball has velocity $65\text{m}/\text{s}$. The time of contact of ball and bat is $0.005\text{s}$. The mass of the baseball is $0.142\text{kg}$.

Take initial direction of motion of the baseball to be positive.

Write the expression to find the initial momentum of the baseball.

$p_i=mv$

Here,

$p_i$ is the initial momentum of the ball

$m$ is the mass of the ball

$v$ is the velocity of the ball

Substitute $40\text{m}/\text{s}$ for $m$ and $0.142\text{kg}$ for $v$ in the above equation to find the initial momentum of the ball.

$\begin{array}{c}{p}_i=\left(0.142\text{kg}\right)\left(40\text{m}/\text{s}\right)\\ =5.68\text{kg}\cdot \text{m}/\text{s}\end{array}$

Write the expression to find the final momentum of the baseball.

$p_f=mv$

Substitute $65\text{m}/\text{s}$ for $m$ and $0.142\text{kg}$ for $v$ in the above equation to find the final momentum of the ball.

$\begin{array}{c}{p}_f=\left(0.142\text{kg}\right)\left(65\text{m}/\text{s}\right)\\ =9.23\text{kg}\cdot \text{m}/\text{s}\end{array}$

Write the expression for change in momentum of the baseball.

$\Delta p=p_i-p_f$

Here,

$\Delta p$ is the change in momentum

Substitute $-9.23\text{kg}\cdot \text{m}/\text{s}$ for $p_f$ and $5.68\text{kg}\cdot \text{m}/\text{s}$ for $p_i$, since the direction of momentum in both cases are opposite, to find the change in momentum.

$\begin{array}{c}\Delta p=5.68\text{kg}\cdot \text{m}/\text{s}-\left(-9.23\text{kg}\cdot \text{m}/\text{s}\right)\\ =14.9\text{kg}\cdot \text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the change in momentum of the baseball during the process is $14.9\text{kg}\cdot \text{m}/\text{s}$.

To determine

Whether the change in momentum is greater than the final momentum.

Answer to Problem 1SP

Yes, the change in momentum is greater than the final momentum.

Explanation of Solution

The change in momentum is found to be $14.9\text{kg}\cdot \text{m}/\text{s}$ which is evidently greater than the final momentum $9.23\text{kg}\cdot \text{m}/\text{s}$. This is due to fact that the initial and final momentum has opposite directions. This makes the change in momentum a greater quantity than the final momentum.

Conclusion:

Therefore, the change in momentum is greater than the final momentum.

To determine

The magnitude of impulse required to make the change in momentum.

Answer to Problem 1SP

The magnitude of impulse required to make the change in momentum is $14.9\text{Ns}$.

Explanation of Solution

Given info: The change in momentum experienced by the baseball is $14.9\text{kg}\cdot \text{m}/\text{s}$.

Write the expression for relation connecting the impulse and change in momentum of the baseball.

$\text{impulse}=\Delta p$

Here,

$\Delta p$ is the change in momentum

Substitute $14.9\text{Ns}$ for $\Delta p$ in the above equation.

$\text{impulse}=14.9\text{Ns}$

Conclusion:

Therefore, the magnitude of impulse required to make the change in momentum is $14.9\text{Ns}$.

To determine

The magnitude of average force that acts on the baseball to produce the impulse.

Answer to Problem 1SP

The magnitude of average force that acts on the baseball to produce the impulse is $2980\text{N}$.

Explanation of Solution

Write the expression for the impulse associated with the ball.

$\text{impulse}=F\Delta t$

Here,

$F$ is the average force acting on the ball

$\Delta t$ is the time of interaction

Substitute $14.9\text{Ns}$ for $\text{impulse}$ and $0.005\text{ s}$ for $\Delta t$ and rearrange the above equation to find the average force acting on the ball.

$\begin{array}{c}F=\frac{\text{impulse}}{\Delta t}\\ =\frac{14.9\text{Ns}}{0.005\text{ s}}\\ =2980\text{ N}\end{array}$

Conclusion:

Therefore, the magnitude of average force that acts on the baseball to produce the impulse is $2980\text{N}$.