Chapter 3, Problem 3.1P

To determine

To show the shell’s relative speed to the ground is $\underset{\_}{v_s=\frac{v}{1+\frac{m}{M}}}$.

Answer to Problem 3.1P

The shell’s relative speed to the ground is shown as $\underset{\_}{v_s=\frac{v}{1+\frac{m}{M}}}$.

Explanation of Solution

Write the momentum of the system of gun and shell before the trigger.

    $P_{\text{initial }}=M{v}_{og}+m{v}_{os}$        (I)

Here, $P_{\text{initial }}$ is the initial momentum of the system, $M$ is the mass of the gun, $v_{og}$ is the velocity of the gun before the trigger, $m$ is the mass of the shell, and $v_{os}$ is the velocity of the shell before the trigger.

The velocity of the shell before the trigger relative to the ground is zero since it is initially at the rest, and similarly the velocity of the gun before trigger relative to the ground is zero as it is at the rest position initially.

Apply the above condition in equation (I).

    $\begin{array}{c}{P}_{\text{initial }}=M\left(0\right)+m\left(0\right)\\ =0\end{array}$        (II)

Write the momentum of the system of gun and shell after the trigger.

    $P_{\text{final }}=M{v}_g+m{v}_s$        (III)

Here, $P_{\text{final }}$ is the final momentum of the system, $v_g$ is the velocity of the gun after the trigger, and $v_s$ is the velocity of the shell after trigger.

From conservation of momentum, the initial and final momentum are equal.

Apply the above condition in equation (III).

    $0=M{v}_g+m{v}_s$

Rewrite the above equation, to find $v_g$.

    $\begin{array}{c}-m{v}_s=M{v}_g\\ v_g=-\left(\frac{m}{M}\right)v_s\end{array}$        (IV)

The velocity of the shell relative to the gun is equal to the difference in velocity of the shell and the gun after trigger.

    $v=v_s-v_g$        (V)

Use equation (IV) in (V).

    $\begin{array}{c}v=v_s-\left(-\frac{m}{M}\right)v_s\\ =v_s+\left(\frac{m}{M}\right)v_s\\ =\left(1+\frac{m}{M}\right)v_s\end{array}$

Rearrange the above equation.

    $\begin{array}{c}v=\left(1+\frac{m}{M}\right)v_s\\ v_s=\frac{v}{\left(1+\frac{m}{M}\right)}\end{array}$

Conclusion:

Therefore, the shell’s relative speed to the ground is shown as $\underset{\_}{v_s=\frac{v}{1+\frac{m}{M}}}$.