Question

A shell of mass 0.020 kg is fired by a gun of mass 100 kg. If the muzzle speed of the shell is 80 m s⁻¹. What is the recoil speed of the gun?

Answer 1

Initially velocity of gun and shell is zero,
so initial momentum is zero.

And according to conservation of momentum,
initial momentum = final momentum
0 = m1 v1 - m2 v2

m1 & v2 are the mass and velocity of gun.
m2 & v2 are the mass and velocity of shell.

100 (v1) = 0.020 (v2)
v1 = (0.020)(80)/(100)
v1 = 0.016 m per s

Answer 2

Hii dear,

# Answer- v2 = -0.016 m/s

## Explaination-
# Given-
m1 = 0.02 kg
v1 = 80 m/s
m2 = 100 kg
v2 = ?

# Solution-
As both gun & bullet were initially at rest, their initial momentum would be zero.

By law of conservation of momentum,
Initial momentum = Final momentum
0 = m1v1 + m2v2
Putting values,
0 = 0.02×80 + 100v2
v2 = -1.6/100
v2 = -0.016 m/s

Thus, gun will recoil backwards with speed of 0.016 m/s.

Hope that was useful...