Question

a 5.0-g bullet leaves the muzzle of a rifle with a speed of 318 m/s. what force (assumed constant) is exerted on the bullet while it is traveling down the 0.91-m-long barrel of the rifle?

Answer 1

F Δt = M ΔV impulse applied to bullet

F Δt = .005 kg * 318 m/s = 1.6 kg-m/sec

Δt = .91 m / 318 m/s = .0029 sec

F = 1.6 kg-m/s / .0029 sec = 560 Newtons