Question

A man fires a bullet of mass 200 gm at a speed of 5 m/s.The gun is of one Kg mass.By what velocity the gun rebounds backward?

Answer 1

M1 = 200 g = 0.200 kg

M2 = 1 kg

U1 = 0

U2 = 0

V1 = 5 m/s

V2 = ?

As we know ,

M1U1+M2U2= M1V1+ M2V2

0 = 0.200 × 5 + 1× V2

0 = 1 + 1× V2

-1 = V2

Answer 2

Answer:Mass of bullet (m1)=200gm=0.2kg

Speed of bullet(v1)=5m/sec.

Mass of gun(m2)=1kg

From the law of conservation of momentum

m1v1+m2v2=0

v2=-m1v1/m2=-0.2×5/1=-1m/sec.

Explanation: