Conservation of momentum of a rifle

[Edwardo_Elric]

Homework Statement

THe expanding gases that leave the muzzle of a rifle also contribute to the recoil. A .30 caliber bullet has a mass of 0.00720kg and a speed of 601 m/s relative to the muzzle when fired from a rifle that has a mass of 2.50kg. The loosely held rifle recoils at a speed of 1.85m/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.

Homework Equations

If the external forces is zero The Total momentum of the system is constant:
P = p_{A} + p_{B} ...

The Attempt at a Solution

i just used the formula:
R = rifle; B = bullet
Px = mRVR + mBVB
Px = (2.50kg)(-1.85m/s) + (0.00720kg)(601m/s)
Px = -0.2978kg m/s

 

[learningphysics]

What does that tell you about the momentum of the gases?

 

[Edwardo_Elric]

The gas is the total momentum of the system? and there are no external forces... it is conserved

 

[learningphysics]

yes momentum is conserved... momentum before = 0... momentum after must also equal 0. ie: momentum of rifle + momentum of bullet + momentum of gases = 0

 

[Edwardo_Elric]

okay... ty so much

Kindly check if the signs are correct
G = gas
So what ill find here is the momentum of the gases:
0 is the momentum of the rifle before; same as after
0 = mRVR + mBVB + mGVG
P(gas) = -mRVR - mBVB
P(gas) = -(2.50kg)(-1.85m/s) - (0.00720kg)(601m/s)
P(gas) = 0.2978 kg * m/s

 

[learningphysics]

okay... ty so much

Kindly check if the signs are correct
G = gas
So what ill find here is the momentum of the gases:
0 is the momentum of the rifle before; same as after
0 = mRVR + mBVB + mGVG
P(gas) = -mRVR - mBVB
P(gas) = -(2.50kg)(-1.85m/s) - (0.00720kg)(601m/s)
P(gas) = 0.2978 kg * m/s

Looks good to me!