A rifle man, who together with his rifle has a mass of $100 kg$ , stands on a smooth surface and fires $10$ shots horizontally. Each bullet has a mass $10 g$ and a muzzle velocity of $800 m/s$ . What velocity does the rifle man acquire at the end of $10$ shots?

0.8 m/s

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0.5 m/s

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0.3 m/s

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1.2 m/s

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Solution

The correct option is A $0.8 m/s$
Let $m_{1}$ and $m_{2}$ be the masses of the bullet and the rifleman and $v_{1}$ and $v_{2}$ be their respective velocities after the first shot.
Initially, the rifleman and bullet are at rest, therefore initial momentum of system $= 0$ .
As external force is zero, momentum of system remains conserved i.e.
Initial momentum = Final momentum
After first shot,
$m_{1} v_{1} + m_{2} v_{2} = 0$
or $v_{2} = \frac{- m_{1} v_{1}}{m_{2}}$
( $∵ m_{1} << m_{2}, m_{2} \approx 100 kg$ )
$= - \frac{(10 \times 10^{- 3} kg) (800 m/s)}{100 kg} = - 0.08 m/s$

After 10 shots,
$10 m_{1} v_{1} + m_{2} v_{2}^{'} = 0$
Therefore, velocity acquired after $10$ shots
$v_{2}^{'} = \frac{- 10 m_{1} v_{1}}{m_{2}} = 10 v_{2} = 10 \times (- 0.08) = - 0.8 m/s$
(-ve sign indicates velocity of man is opposite to bullet's velocity)

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