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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:

49mV+mv=0 49m × V + m ×200 = 0 V= - 20049m/s ('-' sign indicates direction towards left)

Thus, when another shell is fired, the velocity of the car with respect to the platform is,
V= 20049m/s, towards left

When one more shell is fired, the velocity of the car with respect to the platform is,
V1 = 20048m/s, towards left
∴ Velocity of the car w.r.t the earth = 200149+148 m/s

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