The recoil velocity of a gun of $5 k g$ , if a bullet of $25 g$ acquires a velocity of $500 m s^{- 1}$ after firing from the gun, is .

5 m s^{- 1}

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2.5 m s^{- 1}

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2 m s^{- 1}

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Solution

The correct option is B $2.5 m s^{- 1}$
Let: $m_{1}$ = $25 g$ = $0.025 k g$ , $u_{1}$ = 0,
$v_{1} = 500 m s^{- 1}$ , $m_{2}$ = $5 k g$ , $u_{2}$ = 0
Let velocity of gun after firing be $v_{2}$
From law of conservation of momentum, $m_{1}$ $u_{1}$ + $m_{2}$ $u_{2}$ = $m_{1}$ $v_{1}$ + $m_{2}$ $v_{2}$
$\Rightarrow$ $(0.025 \times 0) + (5 \times 0) = (0.025 \times 500) + (5 \times v_{2})$
$\Rightarrow$ $0 = 12.5 + (5 \times v_{2})$
$\Rightarrow$ $5 \times v_{2} = - 12.5$
$\Rightarrow$ $v_{2} = \frac{- 12.5}{5} = - 2.5 m s^{- 1}$

Thus, recoil velocity of gun is equal to $2.5 m s^{- 1}$ . Here, the negative(- ve) sign shows that gun moves in the opposite direction of bullet.

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The recoil velocity of a gun of 5 kg, if a bullet of 25 g acquires a velocity of 500 ms−1 after firing from the gun, is .

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