A bullet has a mass of 0.06 kg. Starting from rest, after the gun's trigger is pulled, a constant force acts on the bullet for the next 0.025 seconds until the bullet leaves the barrel of the gun with a speed of 992 m/s. What is the change in momentum of the bullet?

The change in momentum of the bullet : 59.52 kg m/s

Given

m=0.06 kg

Δt=0.025 s

vo=0(from rest)

vt= 992 m/s

Required

The change in momentum

Solution

The change in momentum  = ΔP

ΔP  =m(vt-vo)

ΔP =0.06(992-0)

ΔP =59.52 kg m/s

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