NTA Abhyas JEE Main2020PhysicsLaws of MotionPractice
A bullet is fired from a gun. The force on the bullet is given by F = 600 - 2 × 10 5 t where F is in newton and t in second. The force on the bullet becomes zero as soon as it leaves the barrel. What is the average impulse imparted to the bullet?
Options
- A9 Ns
- B1 .8 Ns
- C0 .9 Ns
- D0 .3 Ns
Correct answer
C. 0 .9 Ns
Step-by-step solution
Given, F = 600 - 2 × 10 5 t = 0 ⟹ t = 3 × 10 - 3 s Impulese I = ∫ 0 t F ⋅ d t = ∫ 0 3 × 10 - 3 600 - 2 × 10 5 t d t = [ 600 t - 10 5 t 2 ] 0 3 × 10 - 3 = 0.9 N s