Chapter 7, Problem 1SP

(a)

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 $12\text{kg}\cdot \text{m}/\text{s}$.

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 $60\text{m}/\text{s}$. The time of contact of ball and bat is $0.04\text{s}$. The mass of the baseball is $0.120\text{kg}$.

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

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

$p_i=m{v}_i$

Here,

$p_i$ is the initial momentum of the ball

$m$ is the mass of the ball

$v_i$ is the initial velocity of the ball

Substitute $40\text{m}/\text{s}$ for $v_i$ and $0.120\text{kg}$ for $m$ in the above equation to find the initial momentum of the ball.

$\begin{array}{c}{p}_i=\left(0.120\text{kg}\right)\left(40\text{m}/\text{s}\right)\\ =4.8\text{kg}\cdot \text{m}/\text{s}\end{array}$

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

$p_f=m{v}_f$

Here,

$p_f$ is the final momentum of the ball

$v_f$ is the final velocity of the ball

Substitute $60\text{m}/\text{s}$ for $v$ and $0.120\text{kg}$ for $m$ in the above equation to find the final momentum of the ball.

$\begin{array}{c}{p}_f=\left(0.120\text{kg}\right)\left(60\text{m}/\text{s}\right)\\ =-7.2\text{kg}\cdot \text{m}/\text{s}\end{array}$

The negative sign is due to the fact that the final momentum is in negative direction.

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 $-7.2\text{kg}\cdot \text{m}/\text{s}$ for $p_f$ and $4.8\text{kg}\cdot \text{m}/\text{s}$ for $p_i$ in the above equation to find $\Delta p$.

$\begin{array}{c}\Delta p=4.8\text{kg}\cdot \text{m}/\text{s}-\left(-7.2\text{kg}\cdot \text{m}/\text{s}\right)\\ =12\text{kg}\cdot \text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the change in momentum of the baseball during the process is $12\text{kg}\cdot \text{m}/\text{s}$.

(b)

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

Linear momentum is the product of mass of an object and its velocity. It was called as the change in quantity of the motion. It is a vector and it has the same direction as that of velocity.

The change in momentum is found by taking the difference between the initial and the final values of momentum. If the change in momentum is greater than final momentum it must mean that the initial and final momentum have opposite directions.

The change in momentum is found to be $12\text{kg}\cdot \text{m}/\text{s}$ which is evidently greater than the final momentum $7.2\text{kg}\cdot \text{m}/\text{s}$. This implies 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.

(c)

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 $12\text{Ns}$.

Explanation of Solution

Given info: The change in momentum experienced by the baseball is $12\text{kg}\cdot \text{m}/\text{s}$.

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

$\text{impulse}=\Delta p$

Substitute $12\text{Ns}$ for $\Delta p$ in the above equation.

$\text{impulse}=12\text{Ns}$

Conclusion:

Therefore, the magnitude of impulse required to make the change in momentum is $12\text{Ns}$.

(d)

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 $300\text{N}$.

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

Rewrite the above equation for $F$.

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

Substitute $12\text{Ns}$ for $\text{impulse}$ and $0.04\text{ s}$ for $\Delta t$ in the above equation to find the average force acting on the ball.

$\begin{array}{c}F=\frac{12\text{Ns}}{0.04\text{ s}}\\ =300\text{ N}\end{array}$

Conclusion:

Therefore, the magnitude of average force that acts on the baseball to produce the impulse is $300\text{N}$.