Question

a gun of mass 3 kg fires a bullet of mass 30g the bullet takes 0.003 s to move through the barrel of gun and acquires a velocity of 100m/s .calculate the velocity with which the gun recoils and the force exerted on gun man due to recoil of the gun

Answer 1

law of conservation of linear momentum.
m1u1+m2u2=m1v1+m2v2

Answer 2

ANSWER :

The velocity with which the gun recoil is$\sf - 1 \: m/s^{2}$

The force exerted on gun man due to recoil of the gun is 1000 N.

GIVEN :

Mass of gun = 3 kg.

Mass of bullet = 30 g.

Time, t = 0.003 sec.

Velocity, v = 100 m/s.

TO CALCULATE :

The velocity with which the gun recoils.

The force exerted by the gun man due to recoil of the gun.

FORMULA :

$\sf Force \: = \: \dfrac {\Delta p}{\Delta t}$

SOLUTION :

Net force on gun and bullet, system is zero.

$\hookrightarrow \sf 3 \: \times \: v \: (30 \times 10^{-3}) \: \times 100 \: = \: 0$

$\hookrightarrow \sf 3v \: = \: - (30 \: \times \: 10^{-1}$

$\therefore \boxed{\sf V \: = \: - 1 \: m/s^{2}}$

$\bold\green{Force \: = \: \dfrac {\Delta p}{\Delta t}}$

$\hookrightarrow↪ \sf \dfrac {m(v \: - \: 0)}{0.003}$

$\hookrightarrow↪ \sf \dfrac {3 \: \times \: 1}{0.003}$

$\therefore \boxed{\sf F \: = \: 1000 \: N}$

hope help u mate ✌