Question

A bullet of mass 30 g is horizontally fired with a velocity 140m s-¹ from a pistol of mass 3kg, What is the recoil velocity of the pistol?​

Answer 1

m1=30g=0.03kg
m2=3kg
v1=140m/s
v2=?
m1v1=m2v2
0.03*140=3*v2
0.01*140=v2
v2=14m/s
Since the motion of gun is opposite to the motion of bulllet
Value of recoil velocity will be negative
v2=-14m/s

Answer 2

Given:-

•Mass of the pistol(M)=3kg

•Mass of the bullet(m)=30g

=>30/1000

=>0.03kg

•Velocity of the bullet(v)=140m/s

To find:-

•Recoil velocity of the pistol(V)

Solution:-

By applying law of conservation of momentum,we get:-

=>M(-V)+mv=0

=>-MV= -mv

=>V= mv/M

=>V= (0.03×140)/3

=>V=1.4m/s

Thus,recoil velocity of the pistol is 1.4m/s.