Chapter 7, Problem 24P

(a)

To determine

The speed of bullet-baseball system just after collision.

Answer to Problem 24P

The speed of bullet-baseball system just after collision is $33.3\text{m}/\text{s}$.

Explanation of Solution

According to momentum conservation law, sum of momentum bullet and baseball before and after collision will be same.

    $m_{\text{bullet}}{v}_{\text{i}\text{bullet}}+m_{\text{ball}}{v}_{\text{i}\text{ball}}=\left(m_{\text{bullet}}+m_{\text{ball}}\right)v$

Here, $m_{\text{bullet}}$ is the mass of bullet, $m_{\text{ball}}$ is the mass of ball, , $v_{\text{i}\text{bullet}}$ is the velocity of bullet before collision, $v_{\text{i}\text{ball}}$ is the velocity of bullet after collision and $v$ speed of bullet-baseball system just after collision.

Conclusion:

Substitute 0.030 kg for $m_{\text{bullet}}$, 0.15 kg for $m_{\text{ball}}$, $200\text{m}/\text{s}$ for $v_{\text{i}\text{bullet}}$ and $0\text{m}/\text{s}$ for $v_{\text{i}\text{ball}}$ in the above equation to find $v$.

    $\begin{array}{c}\left(0.030\text{ kg}\right)\left(200\text{m}/\text{s}\right)+\left(0.15\text{ kg}\right)\left(0\text{m}/\text{s}\right)=\left(0.030\text{ kg}+0.15\text{ kg}\right)v\\ v=\frac{\left(0.030\text{ kg}\right)\left(200\text{m}/\text{s}\right)}{0.18\text{kg}}\\ =33.3\text{m}/\text{s}\end{array}$

Thus, the speed of bullet-baseball system just after collision is $33.3\text{m}/\text{s}$.

(b)

To determine

The average force due to air friction during the upward-motion of ball-bullet system.

Answer to Problem 24P

The average force due to air friction is $0.94\text{N}$.

Explanation of Solution

Write the equation for work done by friction from the difference between potential energies at height on considering and neglecting the air resistance.

    $\begin{array}{c}{W}_{\text{air-res}}=\left[\left(m_{\text{bullet}}+m_{\text{ball}}\right)gH\right]-\left[\left(m_{\text{bullet}}+m_{\text{ball}}\right)gh\right]\\ =\left(m_{\text{bullet}}+m_{\text{ball}}\right)g\left[H-h\right]\end{array}$        (I)

Here, $W_{\text{air-res}}$ is the work done by air resistance, $g$ is the gravitational acceleration, $H$ is the height reached by ball-bullet system in the absence of air resistance and $h$ is the is the height reached by ball-bullet system in the presence of air resistance.

According energy conservation law, potential energy at height $H$ will be equal to kinetic energy of system.

    $\begin{array}{c}\left(m_{\text{bullet}}+m_{\text{ball}}\right)gH=\frac{1}{2}\left(m_{\text{bullet}}+m_{\text{ball}}\right)v^2\\ H=\frac{v^2}{2g}\end{array}$        (II)

Write the equation for average force of air resistance from $W_{\text{air-res}}$.

    $F=\frac{W_{\text{air-res}}}{h}$        (III)

Here, $F$ is the average force of air resistance.

Conclusion:

Substitute 0.030 kg for $m_{\text{bullet}}$, 0.15 kg for $m_{\text{ball}}$, $33.3\text{m}/\text{s}$ for $v$ and $9.8{\text{m}/\text{s}}^2$ for $g$ in equation (II) to find $H$.

    $\begin{array}{c}H=\frac{{\left(33.3\text{m}/\text{s}\right)}^2}{2\left(9.8{\text{m}/\text{s}}^2\right)}\\ =56.6\text{m}\end{array}$

Substitute 0.030 kg for $m_{\text{bullet}}$, 0.15 kg for $m_{\text{ball}}$, $56.6\text{m}$ for $H$, 37 m for $h$ and $9.8{\text{m}/\text{s}}^2$ for $g$ in equation (I) to find $W_{\text{air-res}}$.

    $\begin{array}{c}{W}_{\text{air-res}}=\left[\left(0.030\text{ kg}+0.15\text{ kg}\right)\left(9.8{\text{m}/\text{s}}^2\right)\right]\left(56.6\text{m}-37\text{m}\right)\\ =34.6\text{J}\end{array}$

Substitute $34.6\text{J}$ for $W_{\text{air-res}}$ and 37 m for $h$ in equation (III) to find $F$.

    $\begin{array}{c}F=\frac{34.6\text{J}}{37\text{m}}\\ =0.94\text{N}\end{array}$

Thus, the average force due to air friction is $0.94\text{N}$.