Question

A shotgun of mass 3kg fires a bullet of 0.1kg at a velocity of 250m/s. What is the recoil velocity of the gun.

Answer 1

The momentum is equal for both. So

M(gun) * V(gun) = M(bullet) * V(bullet)

So

V(gun)=M(bullet) / M(gun) * V(bullet)

= .01 / 40 * 400 = 1/10 m/s = 0.1 m/s

A 1 m (39 inch) barrel length implies an acceleration of 80,000 m/s2

D=V^2 / (2*a) → a = V^2 / (2*D)

= 160,000 / 2 = 80,000 m/s2

V = a*t → t = V / a

= 400 / 80,000 = 1 / 200 second = 0.005 s

This means the gun is accelerated at

V = a * t → a = V / t

= 0.1 / 0.005 = 20 m/s2

And is displaced by

D = 1/2 * a * t^2 = 0.5 * 20 * 0.005^2

= 0.00025 m = 0.25 mm
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