Question

A bullet of mass 0.01 kg is fired from a gun weighing 5.0 kg.if the initial speed of the bullet is 250 m/s,calculate the speed with which the gun recoils.

Answer 1

According to momentum conservation :
M1V1 = M2V2
So, 0.01×250 = 5×V2 ( V2 is speed of gun recoil)
so V2 = 2.5/5 = 0.5 m/s

Answer 2

Dear Student,

◆ Answer -

vg = -0.5 m/s

● Explanation -

# Given -

mb = 0.01 kg

mg = 5 kg

vb = 250 m/s

vg = ?

# Solution -

According to law of conservation of momentum,

0 = mb.vb + mg.vg

mb.vb = -mg.vg

vg = -mb.vb/mg

Substitute values,

vg = -0.01 × 250 / 5

vg = -0.5 m/s

Hence, the gun recoils with speed of 0.5 m/s.

Thanks dear. Hope this is helpful to you...