A machine gun fires 15 bullets per second, each of mass 0.04kg,with a muscle speed of 30m/s. The average force exerted on the gun due to recoil is?

$$M_{total} = \text{number of bullets per second} \times \text{mass of each bullet}$$ $$M_{total} = 15 \text{ bullets/s} \times 0.04 \text{ kg/bullet}$$ $$M_{total} = 0.6 \text{ kg/s}$$

The momentum of the bullets is given by $p = M_{total} \times v$, where $v$ is the muzzle speed.

$$\Delta p = M_{total} \times v$$ $$\Delta p = 0.6 \text{ kg/s} \times 30 \text{ m/s}$$ $$\Delta p = 18 \text{ kg} \cdot \text{m/s}^2$$

According to Newton's second law, the average force is the rate of change of momentum. Since the momentum calculated in Step 2 is per second, this directly gives the force.

$$F_{recoil} = \frac{\Delta p}{\Delta t}$$ $$F_{recoil} = \frac{18 \text{ kg} \cdot \text{m/s}}{1 \text{ s}}$$ $$F_{recoil} = 18 \text{ N}$$

The average force exerted on the gun due to recoil is $\text{18 N}$.