Question

A gun fires a 0.25 kg projectile which acquires a velocity of 300 m/s. If the projectile takes 0.0050 seconds to travel the length of the barrel, what is the force exerted by the gun on the projectile? (Use impulse-momentum theorem

Answer 1

The impulse-momentum theorem states

$$\rm impulse=change\: in\: momentum$$ $$F\Delta t=\Delta p$$

$$F=\frac{\Delta p}{\Delta t}=\frac{0.25\:\rm kg\times 300\: m/s-0 }{0.005\:\rm s}=15000\:\rm N$$