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

Answer:

 F=264.045 N

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

$a=\dfrac{v^2-u^2}{2S}$

$a=\dfrac{304^2-0^2}{2\times 0.77}\ m/s^2$                                      

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