Question

A bullet of mass 0.02 Kg is fired with a speed of 60 m/s from the gun of mass 4.0 Kg, the recoil velocity of the gun is in m/s

Answer 1

Let, mass of bullet (m1)=0.02kg

mass of gun (m2) = 4kg

velocity of bullet(v1) = 60 m/s

velocity of gun(v2) =?

We know, for expolsive forces, initial momentum (m1u1+m2u2)=0.

Now, From conservation of linear momentum,

m1u1+m2u2 = m1v1+m2v2

or, 0 = 0.02×60+4×v2

or, 0 = 1.2 +4v2

or, 1.2 +4v2=0

or, 4v2 = -1.2

or, v2= -1.2/4

= -0.3 m/s

Answer 2

Answer:

hope my attachment answer is correct