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 metre per second. Calculate the velocity with which the gun recoils and the force exerted on gunman due to recoil of the gun

Answer 1

let mass of gun and bullet be m1 and m2
let the velocity of gun and bullet be v1 and v2
let the acceleration of gun and bullet be a1 and a2

According to the law of conservation of momentum - 
 m1v1 = m2v2 
v1 = (0.03) 100 / 3 = 1 m/s 
∴ velocity of recoil of gun = 1m/s
Since the bullet applies equal and opposite force on the gun - 
 m1a1 = m2a2 
force on gunman = (0.03 kg) (( 100 - 0 )/ 0.003 m/s^2) = 0.03 * 100000 / 3 = 1000 N (ANS)

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.

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

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

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

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

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

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

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

hope help u mate ✌