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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: $49mV+mv=0\phantom{\rule{0ex}{0ex}}⇒49m×V+m×200=0\phantom{\rule{0ex}{0ex}}⇒V=-\frac{200}{49}\mathrm{m}/\mathrm{s}\left(\text{'}-\text{'}\mathrm{sign}\mathrm{indicates}\mathrm{direction}\mathrm{towards}\mathrm{left}\right)$ 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 $=200\left(\frac{1}{49}+\frac{1}{48}\right)\mathrm{m}/\mathrm{s}$

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