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3 years ago

8

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 a

nd a speed of 601 m/s relative to the muzzle when fired from a rifle that has mass 3.00 kg. The loosely held rifle recoils at a speed of 1.95 m/s relative to the earth.
Find the momentum of the propellant gases in a coordinate system attached to the earth as they leave the muzzle of the rifle.

1 answer:

katen-ka-za [31]3 years ago

5 0

Answer:

Mg= 1.3418 kg*m/s

Explanation:

Mass of rifle

$m_{r}=2.90 kg$

Recoil velocity of rifle

$v_{r}=1.95 m/s$

Momentum of rifle

$m*v_{r}=3.0kg*1.95\frac{m}{s}==5.85 \frac{kg*m}{s}$

mass of bullet

mb=7.2x10^{-3} kg=0.0072 kg

velocity of bullet relative to muzzel

v_{b} = 601 m/s

velocity of bullet relative to earth= 601 - 1.95=599.05 m/s

momentum of bullet

$m_{b}*m_{b}=7.2x10^{-3}kg*599.05\frac{m}{s}=4.3132 \frac{kg*m}{s}$

the momentum of the propellant gases = momentum of rifle - momentum of bullet

the momentum of the propellant gases

Mg=5.655 -4.3132

Mg= 1.3418 kg*m/s

the momentum of the propellant gases in a coordinate system attached to the earth as they leave the muzzle of the rifle is 1.3418 kgm/s

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