Question

A gun is mounted on a railroad car. The mass of the car, the gun, the shells and the operator is 50 m where m is the mass of one shell. If the velocity of the shell with respect to the gun (in its state before firing) is 200 m/s, what is the recoil speed of the car after the second shot? Neglect friction.

Answer 1

It is given that:
Mass of the car, the gun, the shells and the operator = 50m
Mass of one shell = m
Muzzle velocity of the shells, v = 200 m/s

Let the speed of car be v.

On applying the law of conservation of linear momentum, we get:

$$\begin{gathered} 49mV+mv=0 \\ \Rightarrow 49m\times V+m\times 200=0 \\ \Rightarrow V=-\frac{200}{49}\mathrm{m}/\mathrm{s}(\text{'}-\text{'}\mathrm{sign}\mathrm{indicates}\mathrm{direction}\mathrm{towards}\mathrm{left}) \end{gathered}$$

Thus, when another shell is fired, the velocity of the car with respect to the platform is,
$\mathrm{V}=\frac{200}{49}\mathrm{m}/\mathrm{s},\mathrm{towards}\mathrm{left}$

When one more shell is fired, the velocity of the car with respect to the platform is,
${\mathrm{V}}_1=\frac{200}{48}\mathrm{m}/\mathrm{s},\mathrm{towards}\mathrm{left}$
∴ Velocity of the car w.r.t the earth $=200\left(\frac{1}{49}+\frac{1}{48}\right)\mathrm{m}/\mathrm{s}$