Chapter 7, Problem 16P

(a)

To determine

The magnitude and direction of momentum of the baseball.

Answer to Problem 16P

The magnitude of momentum of the baseball is $\underset{\_}{3.75\text{kg}\cdot \text{m}/\text{s}}$ and direction is $\underset{\_}{36.87°}$.

Explanation of Solution

Write the expression for the horizontal component of momentum.

  $\Delta p_x=m\Delta v_x$        (I)

Here, $m$ is the mass, $\Delta v_x$ is the change in horizontal momentum.

Write the expression for the vertical component of momentum.

  $\Delta p_y=m\Delta v_y$        (II)

Here, $m$ is the mass, $\Delta v_y$ is the change in vertical momentum.

Write the expression for the magnitude of momentum.

  $\left|p\right|=\sqrt{p_x^2+p_y^2}$        (III)

Write the expression direction of momentum.

  $\theta ={\tan }^{-1}\frac{p_x}{p_y}$        (IV)

Conclusion:

Substitute, $0.15\text{kg}$ for $m$, $20\text{m}/\text{s}$ for $\Delta v_x$ in equation (I) to find the horizontal component of momentum.

  $\begin{array}{c}\Delta p_x=\left(0.15\text{kg}\right)\left(20\text{m}/\text{s}\right)\\ =3\text{kg}\cdot \text{m}/\text{s}\end{array}$

Substitute, $0.15\text{kg}$ for $m$, $15\text{m}/\text{s}$ for $\Delta v_x$ in equation (II) to find the vertical component of momentum.

  $\begin{array}{c}\Delta p_y=\left(0.15\text{kg}\right)\left(15\text{m}/\text{s}\right)\\ =2.25\text{kg}\cdot \text{m}/\text{s}\end{array}$

Substitute, $3\text{kg}\cdot \text{m}/\text{s}$ for $\Delta p_x$, and $2.25\text{kg}\cdot \text{m}/\text{s}$ for $\Delta p_y$ in equation (III) to find the magnitude of momentum.

  $\begin{array}{c}\left|p\right|=\sqrt{{\left(3\text{kg}\cdot \text{m}/\text{s}\right)}^2+{\left(2.25\text{kg}\cdot \text{m}/\text{s}\right)}^2}\\ =3.75\text{kg}\cdot \text{m}/\text{s}\end{array}$

Substitute, $3\text{kg}\cdot \text{m}/\text{s}$ for $\Delta p_x$, and $2.25\text{kg}\cdot \text{m}/\text{s}$ for $\Delta p_y$ in equation (IV) to find the direction of momentum

  $\begin{array}{c}\theta ={\tan }^{-1}\frac{2.25\text{kg}\cdot \text{m}/\text{s}}{3\text{kg}\cdot \text{m}/\text{s}}\\ =36.87°\text{from}\text{negative}x-\text{axis}\end{array}$

Therefore, the magnitude of momentum of the baseball is $\underset{\_}{3.75\text{kg}\cdot \text{m}/\text{s}}$ and direction is $\underset{\_}{36.87°}$.

(b)

To determine

The average force of the bat on the ball.

Answer to Problem 16P

The average force of the bat on the ball is $\underset{\_}{\text{75}\text{N}}$.

Explanation of Solution

Write the expression for momentum in terms of force and time.

  $\Delta p=F_{\text{avg}}\Delta t$        (V)

Here, $F_{\text{avg}}$ is the average force, $\Delta t$ is the time.

Conclusion:

Substitute, $3.75\text{kg}\cdot \text{m}/\text{s}$ for $\Delta p$, and $50\times {10}^{-3}\text{s}$ in equation (V) to find the average force.

$\begin{array}{c}3.75\text{kg}\cdot \text{m}/\text{s}=F_{\text{avg}}\left(50\times {10}^{-3}\text{s}\right)\\ F_{\text{avg}}=\frac{3.75\text{kg}\cdot \text{m}/\text{s}}{50\times {10}^{-3}\text{s}}\\ =75\text{N}\end{array}$

Therefore, the average force of the bat on the ball is $\underset{\_}{\text{75}\text{N}}$.