Question 4

An average force of 125 N is applied on a machine gun firing bullets each of mass 10 g at the speed of 250 m s$$^{-1}$$ to keep it in position. The number of bullets fired per second by the machine gun is:

Average force exerted by a machine gun equals the rate of change of momentum of the fired bullets.

$$F = n \cdot m \cdot v$$

$$125 = n \cdot (10 \times 10^{-3}) \cdot 250$$

$$125 = n \cdot 2.5 \implies n = \frac{125}{2.5} = 50$$

Create a FREE account and get:

Other Cracku’s Useful Resources:

JEE Questions JEE Question Bank JEE Mains Previous Papers JEE Advanced Previous Papers JEE Study Material PDF JEE Mains Mock Test JEE Advanced Mock Test JEE Daily Target

JEE Physics Questions JEE Maths Questions JEE Chemistry Questions