Question

A machine gun fires a bullet of mass 25 g with a velocity of 1000 ms\(^{-1}\). If the man holding the gun can exert a maximum force of 100 N on the gun, the maximum number of bullets that he can fire per second is:

• A. 4
• B. 12
• C. 8
• D. 6
• E. 3

Hint:

The rate of firing is limited by the maximum force the shooter can apply. Use the formula \( {Rate} = \frac{F}{p} \) for momentum-related problems.

Answer 1

The Correct Option is A

Step 1: The impulse imparted to the gun by firing one bullet is equal to the change in momentum of the bullet. 
The momentum of one bullet is: \[ p = m \times v = 0.025 \times 1000 = 25 \, {kg m/s} \] The maximum force on the gun is \( F = 100 \, {N} \), and the time taken to impart this force for each bullet is: \[ \Delta t = \frac{p}{F} = \frac{25}{100} = 0.25 \, {seconds} \] Thus, the maximum number of bullets fired per second is: \[ {Rate} = \frac{1}{\Delta t} = \frac{1}{0.25} = 4 \]