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 :

(i) the velocity with which the gun recoils.
(ii) the force exerted on gunman due to recoil of the gun

 Mass of the gun $m_1=3\ kg$

Mass of bullet $m_2=30\ g=0.03\ kg$

Velocity of bullet $v_2=100\ m/s$

(i) Let $v_1$ be the recoil velocity of the gun.

According to the law of conservation of momentum:

$m_1\times v_1=m_2\times v_2$

Or $3\times v_1=0.03\times 100$

Or $v_1=\frac{3}{3}$

Or $v=1\ m/s$

Thus, the gun recoils with a velocity of $1\ m/s$

(ii) Initial velocity of the gun $u=0$

Final velocity $v=1\ m/s$

Time $t=0.003\ s$

Therefore, acceleration $a=\frac{v-u}{t}$

Or $a=\frac{1-0}{0.003}$

Or $a=\frac{1000}{3}\ m/s^2$

Therefore, force $F=ma$    [Newton's second law of motion]

Or $F=3\ kg\times\frac{1000}{3}\ m/s^2$

Or $F=1000\ N$

Updated on: 10-Oct-2022

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