Question

a bullet of mass 40g fired from a gun of mass 10kg. If velocity of bullet is 400mpersec then recoil velocity of the gun

Answer 1

from conservation of momentum

mass of bullet * velocity of bullet = -mass of gun*velocity of recoil

0.04*400=10*V

V=40*0.04

hence Recoil of gun equals 1.6 m/s

Answer 2

Hey dear,

◆ Answer-
v2 = -1.6 m/s

◆ Explanation-
# Given-
m1 = 40 g = 0.04 kg
m2 = 10 kg
v1 = 400 m/s
v2 = ?

# Solution-
Initially both gun and bullet are at rest.
Initial momentum is-
p1 = m1u1 + m2u2
p1 = 0.04×0 + 10×0
p1 = 0

Final momentum is-
p2 = m1v1 + m2v2
p2 = 0.04×400 + 10×v2
p2 = 16 + 10v2

By law of conservation of momentum-
p1 = p2
0 = 16 + 10v2
v2 = -16/10
v2 = -1.6 m/s

Therefore, recoil velocity of the gun is 1.6 m/s in opposite direction as that of bullet.

Hope this helps...