Question

The expanding gases that leave the muzzle of a rifle also contribute to the recoil. A .30 caliber bullet has mass 7.20×10−3 kg and a speed of 601 m/s relative to the muzzle when fired from a rifle that has mass 2.80 kg . The loosely held rifle recoils at a speed of 1.95 m/s relative to the earth.

A.Find the momentum of the propellant gases in a coordinate system attached to the earth as they leave the muzzle of the rifle.

Answer 1

Given that :

mass of rifle, M = 2.80 kg

recoil velocity of rifle relative to earth, V = 1.95 m/s

mass of bullet, m = 7.2 * 10^-3 kg = 0.0072 kg

velocity of bullet relative to muzzle, v = 601 m/s

We know that, momentum of rifle is given by -

p_r = M V = (2.80 kg) (1.95 m/s)

p_r = 5.46 kg.m/s

velocity of bullet relative to earth is given by,

v_b = (601 m/s) - (1.95 m/s)

v_b = 599.05 m/s

We know that, momentum of bullet is given by -

p_b = m v_b = (0.0072 kg) (599.05 m/s)

p_b = 4.31 kg.m/s

Part-A : Momentum of the propellant gases in a coordinate system attached to the earth which is given as -

using a formula, we have

p_g = p_r - p_b = [(5.46kg.m/s) - (4.31 kg.m/s)]

p_g = 1.15 kg.m/s (Answer)

I hope help you.