Chapter 8, Problem 24P

To determine

The depth of the bullet’s penetration into the block.

Answer to Problem 24P

The depth of the bullet’s penetration into the block is $7.94\text{cm}$.

Explanation of Solution

Assume equal firing speeds and equal forces required for the two bullets to push wood fibers apart. This equal forces act backward on two bullets.

Write the expression for law of conservation of energy for first bullet.

  $K_i+\Delta E_{\text{mech}}=K_f$        (I)

Here, $K_i$ is the initial kinetic energy of the first bullet, $\Delta E_{\text{mech}}$ is the mechanical energy of the system, and $K_f$ is the final kinetic energy of the first bullet.

Write the expression for initial kinetic energy of the first bullet.

  $K_i=\frac{1}{2}{m}_1{v}^2$        (II)

Here, $m_1$ is the mass of the first bullet and $v$ is the velocity of the first bullet.

Write the expression for mechanical energy of the system.

  $\Delta E_{\text{mech}}=-F{d}_1$        (III)

Here, $F$ is the force exerted by the bullet and $d_1$ is the depth of the block penetrates by first bullet.

Write the expression for law of conservation of momentum for second bullet.

  $p_i=p_f$

Here, $p_i$ is the initial momentum and $p_f$ is the final momentum.

Rewrite the above expression for mass and velocity.

  $m_{2}v=\left(m_{\text{block}}+m_1+m_2\right)v_f$        (IV)

Here, $m_2$ is the mass of the second bullet, $v$ is the aped of the second bullet, $m_{\text{block}}$ is the mass of the block, and $v_f$ is the final velocity of bullet fired from gun into the block.

Write the expression for law of conservation of energy for second bullet.

  ${K^{\prime }}_i+\Delta {E^{\prime }}_{\text{mech}}={K^{\prime }}_f$        (V)

Here, ${K^{\prime }}_i$ is the initial kinetic energy of the second bullet, $\Delta {E^{\prime }}_{\text{mech}}$ is the mechanical energy of the system, and ${K^{\prime }}_f$ is the final kinetic energy of the second bullet.

Write the expression for initial kinetic energy of the second bullet.

  ${K^{\prime }}_i=\frac{1}{2}{m}_2{v}^2$        (VI)

Here, $m_2$ is the mass of the second bullet.

Write the expression for mechanical energy of the system.

  $\Delta {E^{\prime }}_{\text{mech}}=F{d}_2$        (VII)

Here, $d_2$ is the depth of the block penetrates by second bullet.

Write the expression for final kinetic energy of the second bullet.

  ${K^{\prime }}_f=\frac{1}{2}\left(m_{\text{block}}+m_1+m_2\right)v_f^2$        (VIII)

Here, $v_f$ is the final speed.

Conclusion:

Substitute the equations (II) and (III) in equation (I).

  $\frac{1}{2}{m}_1{v}^2-F{d}_1=0$

Substitute $7.00\text{g}$ for $m_1$ and $8.00\text{cm}$ for $d_1$ in above relation.

  $\begin{array}{c}\frac{1}{2}\left(7.00\text{g}\right)\left(\frac{{10}^{-3}\text{kg}}{1\text{g}}\right)v^2-F\left(8.00\text{cm}\right)\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)=0\\ \frac{1}{2}\left(7.00\times {10}^{-3}\text{kg}\right)v^2-F\left(8.00\times {10}^{-2}\text{m}\right)=0\end{array}$        (IX)

Rewrite the above expression for $v^2$.

  $v^2=\frac{F\left(8.00\times {10}^{-2}\text{m}\right)}{\frac{1}{2}\left(7.00\times {10}^{-3}\text{kg}\right)}$        (X)

Substitute $7.00\text{g}$ for $m_2$, $7.00\text{g}$ for $m_1$ and $1.00\text{kg}$ for $m_{\text{block}}$ in equation (IV) to find $v_f$.

  $\begin{array}{c}{v}_f=\frac{\left(7.00\text{g}\right)\left(\frac{{10}^{-3}\text{kg}}{1\text{g}}\right)v}{\left(1.00\text{kg}+7.00\text{g}\times \left(\frac{0.001\text{kg}}{1\text{g}}\right)+7.00\text{g}\times \left(\frac{0.001\text{kg}}{1\text{g}}\right)\right)}\\ =\frac{\left(7.00\times {10}^{-3}\text{kg}\right)v}{\left(1.014\text{kg}\right)}\end{array}$        (XI)

Substitute the equations (VI), (VII) and (VIII) in equation (V).

  $\frac{1}{2}{m}_2{v}^2+F{d}_2=\frac{1}{2}\left(m_{\text{block}}+m_1+m_2\right)v_f^2$

Substitute $7.00\text{g}$ for $m_2$, $7.00\text{g}$ for $m_1$ and $1.00\text{kg}$ for $m_{\text{block}}$ in above relation.

  $\begin{array}{c}\frac{1}{2}\left(7.00\text{g}\right)\left(\frac{{10}^{-3}\text{kg}}{1\text{g}}\right)v^2-F{d}_2=\frac{1}{2}\left(1.00\text{kg}+7.00\text{g}\times \left(\frac{0.001\text{kg}}{1\text{g}}\right)+7.00\text{g}\times \left(\frac{0.001\text{kg}}{1\text{g}}\right)\right)v_f^2\\ \frac{1}{2}\left(7.00\times {10}^{-3}\text{kg}\right)v^2-F{d}_2=\frac{1}{2}\left(1.014\text{kg}\right)v_f^2\end{array}$

Substitute equation (XI) in above relation.

  $\begin{array}{c}\frac{1}{2}\left(7.00\times {10}^{-3}\text{kg}\right)v^2-F{d}_2=\frac{1}{2}\left(1.014\text{kg}\right){\left(\frac{\left(7.00\times {10}^{-3}\text{kg}\right)v}{\left(1.014\text{kg}\right)}\right)}^2\\ F{d}_2=\frac{1}{2}\left(7.00\times {10}^{-3}\text{kg}\right)v^2-\frac{1}{2}\frac{{\left(7.00\times {10}^{-3}\text{kg}\right)}^2}{\left(1.014\text{kg}\right)}{v}^2\end{array}$

Substitute equation (X) in above relation.

  $\begin{array}{c}F{d}_2=\frac{1}{2}\left[\left(7.00\times {10}^{-3}\text{kg}\right)-\frac{{\left(7.00\times {10}^{-3}\text{kg}\right)}^2}{\left(1.014\text{kg}\right)}\right]\frac{F\left(8.00\times {10}^{-2}\text{m}\right)}{\frac{1}{2}\left(7.00\times {10}^{-3}\text{kg}\right)}\\ F{d}_2=F\left(8.00\times {10}^{-2}\text{m}\right)\left(1-\frac{\left(7.00\times {10}^{-3}\text{kg}\right)}{\left(1.014\text{kg}\right)}\right)\\ d_2=7.94\text{cm}\end{array}$

Therefore, the depth of the bullet’s penetration into the block is $7.94\text{cm}$.