Question

A bullet of mass 5g is fired at velocity 900ms-1 from a rifle of mass 2.5kg. What is the rexoil velocity of the rifle

Answer 1

Here I have used the law of conservation of momentum:
M1×U1+M2×U2=M1V1×M2V2
Please check the answer!:)

Answer 2

Hey dear,

◆ Answer-
v2 = -1.8 m/s

◆ Explanation-
# Given-
m1 = 5 g = 0.005 kg
m2 = 2.5 kg
v1 = 900 m/s
v2 = ?

# Solution-
Initially both gun and bullet are at rest.

Initial momentum is-
p1 = m1u1 + m2u2
p1 = 0.005×0 + 2.5×0
p1 = 0

Final momentum is-
p2 = m1v1 + m2v2
p2 = 0.005×900 + 2.5×v2
p2 = 4.5 + 2.5v2

By law of conservation of momentum-
p1 = p2
0 = 4.5 + 2.5v2
v2 = -4.5/2.5
v2 = -1.8 m/s

Therefore, recoil velocity of the gun is 1.8 m/s in opposite direction as that of bullet.

Hope this helps...