Question

A 4.4 g bullet leaves the muzzle of a rifle with a speed of 304 m/s. What force (assumed constant) is exerted on the bullet while it is traveling down the 0.77 m long barrel of the rifle?

Answer 1

F=264.045 N

Step-by-step explanation:

Initial speed u = 0

Final speed v = 304 m/s

Distance covered S = 0.77 m

Mass of the bullet m = 4.4 g = 4.4 x 10⁻³ Kg

Acceleration of the bullet a

`a=(v^2-u^2)/(2S)`

`a=(304^2-0^2)/(2* 0.77)\ m/s^2`

a=60010.38 m/s²

Force applied on the bullet F = ma

F=4.4 x 10⁻³ x 60010.38 N

F=264.045 N