Chapter 7, Problem 16P

(a)

To determine

The change in momentum (magnitude and direction) of the baseball.

Answer to Problem 16P

The magnitude of momentum of baseball is $3.8\text{kg}\cdot \text{m}/\text{s}$ and it directed $37°$ above the horizontal direction opposite ${\overrightarrow{v}}_{\text{i}}$.

Explanation of Solution

Take direction opposite to initial velocity of baseball as $+x$ direction and upward direction as $+y$ direction.

Write the expression for the magnitude of change in momentum along $x$ direction.

$\Delta p_x=m\left(v_{\text{fx}}-v_{\text{ix}}\right)$

Here, $\Delta p_x$ is the change in momentum along $x$ direction, $m$ is the mass of baseball, $v_{\text{fx}}$ is the final velocity of baseball along $x$ direction and $v_{\text{ix}}$ is the initial velocity of baseball along $x$ direction.

Write the expression for the magnitude of change in momentum along $y$ direction.

$\Delta p_y=m\left(v_{\text{fy}}-v_{\text{iy}}\right)$

Here, $\Delta p_y$ is the change in momentum along $y$ direction, $m$ is the mass of baseball, $v_{\text{fy}}$ is the final velocity of baseball along $y$ direction and $v_{\text{iy}}$ is the initial velocity of baseball along $y$ direction.

Write the expression for the magnitude of total change in momentum.

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

Here, $\left|\Delta \overrightarrow{p}\right|$ is the magnitude of change in momentum of total momentum, $\Delta p_x$ is the magnitude of change in momentum along $x$ direction and $\Delta p_y$ is the magnitude of change in momentum along $y$ direction.

Substitute $m\left(v_{\text{fx}}-v_{\text{ix}}\right)$ for $\Delta p_x$ and $m\left(v_{\text{fy}}-v_{\text{iy}}\right)$ for $\Delta p_y$ in above equation to get $\left|\Delta \overrightarrow{p}\right|$.

$\left|\Delta \overrightarrow{p}\right|=m\sqrt{{\left(v_{\text{fx}}-v_{\text{ix}}\right)}^2+{\left(v_{\text{fy}}-v_{\text{iy}}\right)}^2}$ (I)

Write the expression for direction of momentum.

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{m\left(v_{\text{fy}}-v_{\text{iy}}\right)}{m\left(v_{\text{fx}}-v_{\text{ix}}\right)}\right)\\ ={\tan }^{-1}\left(\frac{\left(v_{\text{fy}}-v_{\text{iy}}\right)}{\left(v_{\text{fx}}-v_{\text{ix}}\right)}\right)\end{array}$ (II)

Here, $\theta$ is the angel that momentum vector makes with $x$ axis.

Conclusion:

Substitute $0.15\text{kg}$ for $m$ , $0\text{m}/\text{s}$ for $v_{\text{fx}}$, $-20\text{m}/\text{s}$ for $v_{\text{ix}}$ , $15\text{m}/\text{s}$ for $v_{\text{fy}}$ and $0\text{m}/\text{s}$ for $v_{\text{fy}}$ in equation (I) to get

$\begin{array}{c}\left|\Delta \overrightarrow{p}\right|=0.15\text{kg}\sqrt{{\left(\left(0\text{m}/\text{s}\right)-\left(-20\text{m}/\text{s}\right)\right)}^2+{\left(15\text{m}/\text{s}-0\text{m}/\text{s}\right)}^2}\\ =3.8\text{kg}\cdot \text{m}/\text{s}\end{array}$

Substitute $0.15\text{kg}$ for $m$ , $0\text{m}/\text{s}$ for $v_{\text{fx}}$, $-20\text{m}/\text{s}$ for $v_{\text{ix}}$ , $15\text{m}/\text{s}$ for $v_{\text{fy}}$ and $0\text{m}/\text{s}$ for $v_{\text{fy}}$ in equation (II) to get $\theta$.

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{\left(15\text{m}/\text{s}-0\text{m}/\text{s}\right)}{\left(0\text{m}/\text{s}\right)-\left(-20\text{m}/\text{s}\right)}\right)\\ =37°\end{array}$

Therefore, the magnitude of momentum of baseball is $3.8\text{kg}\cdot \text{m}/\text{s}$ and it directed $37°$ above the horizontal direction opposite ${\overrightarrow{v}}_{\text{i}}$.

(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 $75\text{N}$ in the same direction as $\Delta \overrightarrow{p}$.

Explanation of Solution

Write the expression for the average force of the bat on the ball.

$F_{\text{av}}=\frac{\Delta p}{\Delta t}$

Here, $F_{\text{av}}$ is the average force of the bat on the ball, $\Delta p$ is the momentum change

Conclusion:

Substitute $3.75\text{kg}\text{m}/\text{s}$ for $\Delta p$ and $50\text{ms}$ for $\Delta t$ in above equation to get $F_{\text{av}}$.

$\begin{array}{c}{F}_{\text{av}}=\frac{3.75\text{kg}\text{m}/\text{s}}{50\text{ms}\times \frac{{10}^{-3}\text{s}}{1\text{ms}}}\\ =75\text{N}\end{array}$

Therefore, the average force of the bat on the ball is $75\text{N}$ in the same direction as $\Delta \overrightarrow{p}$.