A man fires a bullet of mass $200 g$ at a speed of $5 m / s$ . The gun is of mass $1 k g$ . The velocity with which the gun will recoil backward is:

3 m s^{- 1}

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2 m s^{- 1}

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1 m s^{- 1}

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0.01 m s^{- 1}

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Solution

The correct option is A $1 m s^{- 1}$
Given:- $m_{B} = 200 g, V_{B} = 5 m / s, m_{G} = 1 k g, V_{G} = ?$
from the law of conservation of momentum
$0$ = $m_{G} V_{G} + m_{B} V_{B}$
$V_{G} = - \frac{m_{B} v_{B}}{m_{G}} = \frac{0.2 \times 5}{1} = - 1 m / s$
-ve sign represent that gun will recoil backward.

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Similar questions

What is the velocity of recoil of a gun, if a bullet of mass 100 g is fired from a gun of mass 20 kg with a velocity fo 100 ms^–1?

10 g

1 k g

5 m / s

10

1

5 m / s

5 k g

25 g

500 m s^{- 1}

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