Question

A gun of mass 3 kg fires a bullet of mass 30 g. The bullet takes 0.003 s to move through the barrel of the gun and acquires a velocity of 100 m/s. Calculate the velocity by which the gun recoils.

Answer 1

Step 1: Given that:

Mass(M) of the gun = $3kg$

Mass(m) of bullet = $30g=\frac{30}{1000}kg=0.03kg$

Time taken by bullet to move through the barrel of the gun = 0.003sec

The velocity of the bullet(v) = $100m{s}^{-1}$

Step 2: Calculation of recoil velocity of gun:

According to the law of conservation of momentum;

The initial momentum of the gun and bullet = Final momentum of the gun and bullet

Initially, both the gun and the bullet are at rest;

Therefore

Let the recoil velocity of the gun is V.

$M\times 0+m\times 0=M\times (-V)+m\times v$

The direction of recoil is opposite to the direction of motion of the bullet, therefore, the recoil velocity of the gun is taken negative with respect to the velocity of the bullet.

$MV=mv$

$3kg\times V=0.03kg\times 100m{s}^{-1}$

$3V=3$

$V=1m{s}^{-1}$

Thus,

The recoil velocity of the gun is 1m/sec.