Question

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

Answer 1

Answer:

The bullet takes 0.003 seconds to move through the barrel of the gun, and acquires a velocity of 100 m/s. Calculate:a) the velocity at which the gun recoil b) the force exerted on the gunman due to recoil of the gun.

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.

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

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

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

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

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

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

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

hope help u mate ✌