Question

Question 33:
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

Lakhmir Singh Physics Class 9

Answer 1

This is the solution. I hoped that this solution will help you.

Answer 2

Dear Student,

◆ Answers -

1. |v1| = 1 m/s
2. |F| = 1000 N

● Explanation -

(i) According to law of conservation of momentum,

pi = pf

m1u1 + m2u2 = m1v1 + m2v2

3×0 + 0.03×0 = 3×v1 + 0.03×100

0 = 3v1 + 3

v1 = -3/3

v1 = -1 m/s

Therefore, recoil velocity of gun is 1 m/s.

(ii) Force exerted on the gunman -

F = ma

F = m × (v1-u1)/t

F = 3 × (-1-0)/0.003

F = -1000 N

Therefore, force exerted on the gunman by the gun would be 1000 N.

Thanks...