Question

A machine gun has a mass of 20 kg. The firing rate of 500 bullets per second and mass of each bullet Is 20 g. If the speed of the bullets 500 ms^-1. Find the force required to keep the gun in its position.

Answer 1

m_gun = 20 kg, m_b = 20 g

v_gun = ? v_b = 500 ms^-1

From Law of conservation of momentum

M_gun V_gun + m_b v_b = 0

⇒ V_gun = - \(\frac {20 \times 10^{-3} \times 500}{20}\)

= – 0.5^-1

∴ Force required to hold its position

F = m \(\frac{v-u}{t}\) = 20 × \(\frac{(0.5-0)}{(\frac{1} {500})s}\)

= 5000 N.