Question

A bullet of mass 40gm is horizontally fired with velocity 150m/s from a pistol of
mass 2kg. what is the recoil velocity of the pistol?
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Answer 1

Given :

Mass of Bullet, m = 40 grams=0.04 kg

Mass of Pistol, M = 2 kg

Initial Velocity of Bullet, u = 0 m/s

Initial Velocity of Pistol u₁ = 0 m/s

Final Velocity of Bullet, v = 150 m/s

To find :

The recoil velocity of the pistol, v₁

Solution :

According to the Law of Conservation of Momentum,

$\leadsto\sf mu + Mu_1 = mv + Mv_1$

$\leadsto \sf (0.04)(0) + (2)(0) = (0.04)(150) + (2)(v_1)$

$\leadsto\sf 0 + 0 = 0.6 + 2v_1$

$\leadsto \sf v_1 = \dfrac{-0.6}{2}$

$\leadsto \sf v_1 = - 0.3 \: {ms}^{ - 1}$

thus, the recoil velocity of the pistol will be 0.3 m/s in the opposite direction.