Question

a gun of mass 3 kg fires a bullet of mass 30gm. The bullet takes 0.003 sec to move through the barrel of the gun and acquires a velocity of 100m/s. Calculate - 1.the velocity with which the gun recoil. 2. the force exerted on the gun man due to recoil of gun.. PLZ solve it

Answer 1

1. velocity with which gun recoils is 1m/s .
mass of gun * velocity of gun = mass of bullet * velocity of bullet
3 * v = 30/1000 *100
3v = 0.03*100
3v = 3/100*100
3v=3
v=3/3
v=1

Answer 2

AnsWer:

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

1. 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.

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

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

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

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

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

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

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

hope help u mate ✌