Question

A shell of mass 0.02kg is fired by a gun mass 100kg. if the muzzle speed of the shell is 80m/s.what is the recoil speed of the gun?

Answer 1

by universal law of gravitation , maua + mbub = mava + mbvb

lets take m , v , u of shell as ma , va , ua

lets take m , v , u of gun as mb , vb , ub

ma = 0.02kg

mb = 100kg

ua = ub = 0 (starts from rest)

va = 80 m/s

vb = ?

maua + mbub = mava + mbvb

(0.02*0) + (100*0) = (0.02*80) + (100*vb)

0 = 1.6 + 100vb

-1.6 = 100vb

-1.6/100 = vb

-0.016 m/s is the recoil velocity of he gun

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Answer 2

Explanation:

Mass of the gun, M = 100 kg

Mass of the shell, m = 0.020 kg

Muzzle speed of the shell, v = 80 m/s

Recoil speed of the gun = V

Both the gun and the shell are at rest initially.

Initial momentum of the system = 0

Final momentum of the system = mv – MV

Here, the negative sign appears because the directions of the shell and the gun are opposite to each other.

According to the law of conservation of momentum:

Final momentum = Initial momentum

mv – MV = 0

∴ V = mv / M

= 0.020 × 80 / (100 × 1000) = 0.016 m/s