Conservation of momentum / energy problem?

In my answer to a question here, involving firing a gun in space, I am thinking I may have been incorrect about something.

If you were to have a large vacuum chamber and fire a bullet while standing on the floor, and another while sitting on a magnetically levitated platform, would the muzzle velocities of the two bullets be the same (assuming identical mass slugs and identical charges in the cartridges)?

I'm not sure whether kinetic energy imparted to the marksman and the bullet are the same, or if it's momentum. I know that regardless, momentum must be conserved, but I forget how to calculate the amount of momentum in this situation...

I used the following values in my example:
mass of bullet: 115grain =~ 0.00745kg
muzzle velocity of bullet: 360m/s (this is a typical value when fired within the atmosphere, but it doesn't matter for my question)

Anyone less rusty than me in dynamics able to straighten me out on this one?

2007-07-05T15:56:35Z

Forgot to list assumed mass of marksman = 100kg (including the mass of the platform)

My gut tells me that the bullet fired while standing will be much faster than the one fired from the mag-lev platform, but I can't come up with the reasoning for it.

2007-07-05T16:12:58Z

magnetically levitated platform isolates the shooter from the earth (simulating floating in free space). Standing with a solid footing means that the bullet is pressing against the mass of the earth.

none2perdy2007-07-05T15:57:45Z

Favorite Answer

Conservation of momentum does apply

m1v1 = m2v2

1 = bullet after firing
2 = shooter after firing

Their respective kinetic energies after bullet leaves muzzle are not the same ecause of the v^2 factor in the energy equation.

For your two scenarios, the mass of the shooter is the same, and assuming that the platform is much heavier than the shooter, his recoil is the same in both cases, so the bullets muzzel velocity would be the same

Anonymous2007-07-05T15:59:25Z

Momentum is simple: mv=mv
The mass of the bullet multiplied by the velocity of the bullet (the values you have) is the total momentum of the bullet.

I am not quite sure what you mean by a "magnetically levitated" platform, but assuming similar energy expended to accelerate each bullet, momentum would be conserved.