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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 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.

Respuesta :

davidlcastiblanco

Answer:

Mg= 1.3418 kg*m/s

Explanation:

Mass of rifle

[tex]m_{r}=2.90 kg[/tex]

Recoil velocity of rifle

[tex]v_{r}=1.95 m/s[/tex]

Momentum of rifle

[tex]m*v_{r}=3.0kg*1.95\frac{m}{s}==5.85 \frac{kg*m}{s}[/tex]

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

[tex]m_{b}*m_{b}=7.2x10^{-3}kg*599.05\frac{m}{s}=4.3132 \frac{kg*m}{s}[/tex]

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

The momentum of the propellant gas is 1.537 kgm/s

Data;

  • Mass of bullet = 7.20*10^-3kg
  • speed of bullet = 601m/s
  • mass of rifle = 3.0kg
  • recoil speed = 1.95m/s

Conservation of Linear Momentum

The momentum of the rifle must be equal to the momentum of the bullet.

Momentum of the rifle is

[tex]M = 3.0* 1.95 = 5.85 kg m/s[/tex]

The momentum of the bullet is ?

The velocity of the bullet relative to earth is

[tex]601 - 1.95 = 599.05m/s[/tex]

The momentum of the bullet can be calculated as

[tex]M = 0.0072 * 599.05 = 4.31316 kg m/s[/tex]

Using the law of conservation of momentum

[tex]M = 5.85 - 4.313 = 1.537 kg m/s\\[/tex]

The momentum of the propellant gas is 1.537 kgm/s

Learn more on law of conservation of momentum here;

https://brainly.com/question/25976777

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