A machine gun fires a bullet of mass 40 g with a velocity of 1200 m s^-1. The man holding it can exert a maximum force on 144 N on the gum. How many bullets can he fire per second at the most?

3

\(F = \frac{\Delta p}{\Delta t}\)

When the gun applies maximum force : 

\(F_{max} = 144 \) N

\(\Rightarrow \Delta t = \frac{\Delta p}{F_{max}} = \frac{m(v-u)}{F_{max}}\)

\(\Rightarrow \frac{0.04(1200-0)}{144} = \frac{1}{3}\) s

$\therefore$To achieve the same change in momentum, the minimum time will be $\frac{1}{3}$ s. Any lesser time and the force applied will be greater.

$\therefore$ In one second 3 bullets can be fired.