A bullet of mass $20 g$ is fired left to right with a velocity of $150 m / s$ from a pistol of mass $2 k g$ . What is the recoil velocity of the pistol?

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Solution

Given,
Mass of bullet = $m_{1} = 20 g = 0.02 k g$
Mass of pistol = $m_{2} = 2 k g$
Initial velocity of bullet = $u_{1} = 0$
Initial velocity of pistol = $u_{2} = 0$
Final velocity of bullet = $v_{1} = 150 m s^{- 1}$
Let final velocity of pistol = $u_{2} = 0$
By conservation of linear momentum, Initial momentum = Final momentum
$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$
$\Rightarrow 0 = 0.02 \times 150 + 2 v_{2}$
$\Rightarrow v_{2} = - 150 m s^{- 1}$
Therefore, the pistol recoils with a velocity of 150 m/s towards the left.

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