Chapter 7, Problem 24P

(a)

To determine

The speed of the basketball.

Answer to Problem 24P

The speed of the bullet and baseball combination is $33\text{m/s}$.

Explanation of Solution

Since after the collision, the bullet and baseball combination must have the same momentum as the bullet had just before it stuck the baseball.

Write the equation for law of conservation of momentum.

$p_{\text{f}}=p_{\text{i}}$ (I)

Here, $p_{\text{i}}$ is the initial momentum of the bullet before collision and $p_{\text{f}}$ is the final momentum of the bullet and baseball combination after the collision.

Write the equation for the momentum of the bullet before the collision.

$p_{\text{i}}=m_{\text{bullet}}{v}_{\text{i}}$ (II)

Here, $p_{\text{i}}$ is the initial momentum of the bullet before collision, $m_{\text{bullet}}$ is the mass of the bullet, and $v_{\text{i}}$ is the initial velocity of the bullet.

Write the equation for the momentum of bullet and baseball combination after the collision.

$p_{\text{f}}=\left(m_{\text{bullet}}+m_{\text{baseball}}\right)v_{\text{f}}$ (III)

Here, $p_{\text{f}}$ is the final momentum of the bullet and baseball combination after the collision, $m_{\text{baseball}}$ is the mass of the baseball, and $v_{\text{f}}$ is the final velocity of the bullet and baseball combination.

Conclusion:

Substitute equation (III) and (II) in equation (I) and solve for $v_{\text{f}}$.

$\begin{array}{c}\left(m_{\text{bullet}}+m_{\text{baseball}}\right)v_{\text{f}}=m_{\text{bullet}}{v}_{\text{i}}\\ v_{\text{f}}=\frac{m_{\text{bullet}}{v}_{\text{i}}}{\left(m_{\text{bullet}}+m_{\text{baseball}}\right)}\end{array}$

Substitute $0.030\text{kg}$ for $m_{\text{bullet}}$, $200\text{m/s}$ for $v_{\text{i}}$, and $0.15\text{kg}$ for $m_{\text{baseball}}$ in above equation.

$\begin{array}{c}{v}_{\text{f}}=\frac{\left(0.030\text{kg}\right)\left(200\text{m/s}\right)}{\left(0.030\text{kg}+0.15\text{kg}\right)}\\ =\frac{6\text{kg}\cdot \text{m/s}}{0.18\text{kg}}\\ =33.3\text{m/s}\\ =33\text{m/s}\end{array}$

Therefore, speed of the bullet and baseball combination is $33\text{m/s}$.

(b)

To determine

The average force of baseball and bullet rising.

Answer to Problem 24P

The average force of the baseball and bullet combination is $0.93\text{N}$ acting downward.

Explanation of Solution

Write the equation for the conservation of energy.

$K_{\text{i}}+U_{\text{i}}+W=K_{\text{f}}+U_{\text{f}}$ (IV)

Here, $K_{\text{i}}$ is the initial kinetic energy of the system, $U_{\text{i}}$ is the initial potential enegy of the system, $K_{\text{f}}$ is the final kinetic energy of the system, $U_{\text{f}}$ is the final potential enegy of the system and $W$ is the work done by the air resistance on the bullet and baseball combination.

Write the equation for kinetic energy.

$K=\frac{1}{2}m{v}^2$ (V)

Here, $m$ is the mass of the bullet and baseball combination and $v$ is the velocity of the baseball and bullet combination.

Write the equation for potential energy.

$U=mg{y}_{\text{f}}$ (VI)

Here, $g$ is the gravitational acceleration and $y_{\text{f}}$ is the height of the baseball and bullet rises.

Write the equation for work done by air resistance.

$W=F_{\text{avg}}\Delta y_{\text{f}}$ (VII)

Here, $F_{\text{avg}}$ is the average force of air resistance.

Conclusion:

Substitute $0$ for $U_{\text{i}}$ and $0$ for $K_{\text{f}}$ in equation (IV).

$\begin{array}{c}{K}_{\text{i}}+0+W=0+U_{\text{f}}\\ K_{\text{i}}+W=U_{\text{f}}\end{array}$ (VIII)

Substitute equation (V) and (VI) in equation (VII) to find $W$.

$\begin{array}{c}\frac{1}{2}m{v}^2+W=mg{y}_{\text{f}}\\ W=mg{y}_{\text{f}}-\frac{1}{2}m{v}^2\\ =m\left(g{y}_{\text{f}}-\frac{1}{2}{v}^2\right)\end{array}$

Substitute $0.030\text{kg}+0.15\text{kg}=0.18\text{kg}$ for $m$, $9.80{\text{m/s}}^2$ for $g$, $37\text{m}$ for $y_{\text{f}}$, and $33.3\text{m/s}$ for $v$ in above equation.

$\begin{array}{c}W=0.18\text{kg}\left(\left(9.80{\text{m/s}}^2\right)\left(37\text{m}\right)-\frac{1}{2}{\left(33.3\text{m/s}\right)}^2\right)\\ =0.18\text{kg}\left(362.6{\text{m}}^2/{\text{s}}^2-554.4{\text{m}}^2/{\text{s}}^2\right)\\ =-34.5\text{J}\end{array}$

Rearrange equation (VII) for average force.

$F_{\text{avg}}=\frac{W}{\Delta y_{\text{f}}}$

Substitute $-34.5\text{J}$ for $W$ and $37\text{m}$ for $\Delta y_{\text{f}}$,

$\begin{array}{c}{F}_{\text{avg}}=\frac{-34.5\text{J}}{37\text{m}}\\ =-0.93\text{N}\end{array}$

Therefore, the average force of the baseball and bullet combination is $0.93\text{N}$ acting downward.