Question

A bullet of 20 gm is shot from a gun of 2 kg. at a velocity of 100 m/s.Find out the recoil velocity of the gun????

Answer 1

here use conservation of linear momentum ,
let m is the mass of bullet
M is the mass of gun
initial velocity of bullet is u
initial velocity of gun U
final velocity of bullet v
final velocity of gun V
then according to law of conservation of linear momentum ,
mu + MU = mv + MV

according to above phenomena
u =0
U= 0
v =100 m/sec
V=?

hence,
0 + 0 = (20/1000 )x 100 +2 x V
V =-1 m/s
hence gun speed 1 m/s backward direction .

Answer 2

mass of the gun = m1=2 kg
initial velocity of the gun u1 = 0
mass of the bullet m2 = 20gm= 0.02kg
initial velocity of the bullet u2= 0
final velocity of gun v1= ?
final velocity of bullet v2= 100m/s-1
According to law of conservation of momentum
m1v1+m2v2 = m1v1+m2u2
2×v1+0.02×100 = 20×0 +0.020× 0
2v1 + 2 =0 or 2v1 = -2
v1 =-1m/s
Therefore, the gun will recoil with a velocity of 1 m/s backwards.