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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.
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Solution
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:
\[49\text{ mV + mv } = 0\]
\[ \Rightarrow 49m \times V + m \times 200 = 0\]
\[ \Rightarrow V = - \frac{200}{49}\text{m/s} (' - ' \text{ sign indicates direction towards left })\]
Thus, when another shell is fired, the velocity of the car with respect to the platform is,
\[V = \frac{200}{49}\text{ m/s, towards left }\]
When one more shell is fired, the velocity of the car with respect to the platform is,
\[V_1 = \frac{200}{48}\text{ m/s, towards left }\]
When one more shell is fired, the velocity of the car with respect to the platform is,
\[V_1 = \frac{200}{48}\text{ m/s, towards left}\]
∴ Velocity of the car w.r.t the earth \[= 200\left( \frac{1}{49} + \frac{1}{48} \right) \text{m/s}\]
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