Question

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 to keep it in position. The number of bullets fired per second by the machine gun is:

• A. 25
• B. 5
• C. 100
• D. 50

Hint:

The force required to keep the machine gun in position is equal to the rate of change of momentum. Use this principle to find the number of bullets fired per second.

Answer 1

The Correct Option is D

The force exerted by the machine gun is related to the rate of change of momentum of the bullets: \[ F = \frac{d(mv)}{dt} \] Since \( m = 10 \, \text{g} = 0.01 \, \text{kg} \) and \( v = 250 \, \text{m/s} \), the momentum of each bullet is: \[ mv = 0.01 \times 250 = 2.5 \, \text{kg m/s} \] Now, the rate of change of momentum gives the force: \[ F = \frac{d(mv)}{dt} = \frac{10 \times 250}{1000} \, \text{N} \] Thus, the number of bullets fired per second \( n \) is: \[ n = \frac{125}{\frac{10 \times 250}{1000}} = 50 \] Thus, the number of bullets fired per second is 50.