Calculating Momentum of Propellant Gases in a Rifle Recoil

[ooohffff]

Homework Statement

The expanding gases that leave the muzzle of a rifle also contribute to the recoil. A .30 caliber bullet has mass 7.20×10−3kg and a speed of 601 m/s relative to the muzzle when fired from a rifle that has mass 3.10 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.

I'm not totally sure how to set up the equations because I can't really visualize what the problem is asking?

Homework Equations

The Attempt at a Solution

m1=.0072 kg
m2=3.1 kg
v1i= 0 m/s
v1f=601 m/s
v2i = ?
v2f=-1.95 m/s

 

[Daymare]

You need to apply momentum conservation here.From the perspective of the ground both the bullet and the rifle were initially at rest.Now after firing the relative velocity of the bullet is 601m/s relative to the rifle which has a relative velocity of 1.95m/s with respect to the ground.So what is the relative velocity of the bullet with respect to the ground?.Its (601-1.95)m/s.Now apply momentum conservation and see how much momentum is missing from one side.

 

[haruspex]

I can't really visualize what the problem is asking?

It is saying that three things move as a result of the firing of the bullet:

• The bullet
• The gun
• The exhaust gases from the explosive charge

Each of these has momentum. Since they form a closed system, the sum of their momenta is...?

 

[haruspex]

You need to apply momentum conservation here.From the perspective of the ground both the bullet and the rifle were initially at rest.Now after firing the relative velocity of the bullet is 601m/s relative to the rifle which has a relative velocity of 1.95m/s with respect to the ground.So what is the relative velocity of the bullet with respect to the ground?.Its (601-1.95)m/s.Now apply momentum conservation and see how much momentum is missing from one side.

I may be wrong, but I interpreted ooohfff's post as seeking clarification of the question rather than hints on how to solve it.
But maybe it's the relative velocity aspect that was the block.

 

[ooohffff]

You need to apply momentum conservation here.From the perspective of the ground both the bullet and the rifle were initially at rest.Now after firing the relative velocity of the bullet is 601m/s relative to the rifle which has a relative velocity of 1.95m/s with respect to the ground.So what is the relative velocity of the bullet with respect to the ground?.Its (601-1.95)m/s.Now apply momentum conservation and see how much momentum is missing from one side.

Great, thanks...I forgot to subtract the 1.95 but fixed it. The other problem I have is that I get the right number for the missing momentum but I get the negative of it when it should be positive? I plug in mv=(.0072)(599.05)+(3.1)(-1.95)

 

[haruspex]

mv=(.0072)(599.05)+(3.1)(-1.95)

It is easy to confuse yourself when writing balance equations that way. Safer is the form Σm_iv_i=0.