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A gun is mounted on a railroad car. The mass of the car with all of its components is 80m and mass of each shell to be fired is 5m. If the muzzle velocity of the shells is $100 m s^{- 1}$ horizontally w.r.t gun, then calculate the recoil speed of the car after second shot (in $m s^{- 1}$ ). Consider car to be at rest initially.

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answer is 13.

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Detailed Solution

Mass of gun is $80 m - 5 m = 75 m$

After first short, if speed of car is $v_{1}$ , then we have

$75 m v = 5 m (100 - v)$

$\Rightarrow v_{1} = \frac{500}{80} = \frac{100}{16} m s^{- 1}$

After second shot, if speed of car is $v_{2}$ , then we have

$75 m (\frac{100}{16}) = 70 m v_{2} - 5 m (100 - v_{2})$

$\Rightarrow 75 v_{2} = \frac{75 \times 100}{16} + 500$

$\Rightarrow v_{2} = \frac{100}{16} + \frac{100}{15} m s^{- 1} \Rightarrow v_{2} = 100 (\frac{1}{16} + \frac{1}{15}) m s^{- 1} \approx 13 m s^{- 1}$

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