JEE MainPhysicsCenter of Mass, Momentum and Collision
A heavy machine gun of mass 40 kg is placed on a rough horizontal surface having a coefficient of static friction 0.5 . It fires bullets of mass 50 g horizontally at a speed of 800 m/s. The minimum number of bullets that must be fired per second so that the gun just starts to slide is : (Take g = 10 m/s ^2 )
Options
- A300
- B5
- C10
- D50
Correct answer
B. 5
Step-by-step solution
The maximum static friction force on the machine gun is: f_ max = _s M g f_ max = 0.5 40 10 = 200 N The average recoil force exerted by the bullets on the gun is: F = n m v where n is the number of bullets fired per second. For the gun to just start sliding, the recoil force must equal the maximum static friction: n m v = f_ max n 0.05 800 = 200 n 40 = 200 n = 5 Answer: 5