A bullet of mass $10 g$ is fired horizontally with a velocity of $150 m s^{- 1}$ from a pistol of mass $2 k g$ . Find the recoil of the pistol.

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Solution

Given:
Mas of bullet, $M_{b} = 10 g$
Velocity of bullet, $V_{b} = 150 m s^{- 1}$
Mass of pistol, $M_{g} = 2 k g$
Due to the law of conservation of momentum, the momentum of the bullet is equal to the recoil of the gun. So the sum of the momentum of both the objects is equal to zero.
$M_{b} V_{b} = - M_{g} V_{g}$ .
$⟹ V_{g} = - \frac{M_{b} V_{b}}{M_{g}} = - \frac{0.01 \times 150}{2} = 0.75 m s^{- 1}$ .
$-$ ve sign is ignored in the final answer as it denotes the direction of the velocity of the pistol.

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