Question

A bullet is fired by a gun. The force acting on the bullet is given by F = (600 - 2 x 10⁵ t). Where F is in ‘N’ and t is in 's'. As the bullet leaves the barrel of the gun, force acting on it becomes zero. What will be average impulse of the bullet?

Answer 1

Given: F = 600 - 2 x 10⁵ t

∵ The force becomes zero as the bullet leaves the barrel.

∴ 600 - 2 x 10⁵ t = 0

∴ \(t = \frac{600}{2\times10^5} = 3\times10^{-3}\ s\)

Based on Equilibrium of Concurrent Forces and Pulley