We did that very experiment (and calculations) in junior college physics using a bow and arrow, and everything balanced out. What’s-his-name prevailed. 
Oh, yeah. The arrow from the fixed bow had very little additional velocity. I forget why, but recoil goes into the bow (rifle) anyway.
Sorry, but I still disagree. The energy of the bullet is much greater than the kinetic energy received by the gun/user. This is because the gun has much more mass than the bullet. For most gun/bullet combinations, I would estimate the kinetic energy of the bullet is 300 to over 1000 times more than the energy imparted to the gun.
Again, I think you and TBone2 are confusing energy with momentum…
OK, one more time…
No one is confusing energy and momentum. Yes, they are two different quantities, but the point is that they are proportional to one another in this case. At the instant when a bullet exits a gun’s barrel, both the bullet and the gun have received X energy and have gained Y momentum. My point is precisely the same whether we choose to discuss it in terms of energy OR momentum.
I fear that YOU may be confusing momentum with speed or, more precisely, velocity.
Sorry, TBone, but CrafterMan is correct. Momentum is balanced, but energy is not.
Momentum is mass times velocity*. It’s going to be conserved, so the momentum of the gun (and person holding it) moving backwards is the same as the momentum of the bullet going forwards. ** The conservation is easy to prove if you go back to Force=Mass x Acceleration and remember that forces have to balance. At the other end, if the bullet sticks in the target, then the target/bullet combination will now have all the momentum that the bullet had, (again the same momentum imparted to the shooter).
Energy (kinetic) is mass times velocity squared. It does NOT have to be the same for the bullet and gun/shooter. Easy to show if you pick a couple different numbers for velocity and bullet mass, then balance momentums (or better yet, pick a couple sets of masses that push on each other with a constant force for a set time, then calculate the different accelerations imparted, and the resulting momentums of the two masses).
The total energy of the system is conserved, but what that means is that the total chemical energy of the gunpowder is equal to the total kinetic energy of the bullet and gun/shooter (added together – energy doesn’t have a direction, just a total). And when the bullet hits the target, kinetic energy is not conserved there either (unless the bullet bounces off backwards without doing any damage). Most of the bullet’s energy is turned into heat and structural damage to the target – again easy to see if you choose some real numbers and do the momentum balance.
- Momentum is not exactly m*v, but very close at non-relatavistic speeds.
** Momentum also is imparted to gases exiting the muzzle, and to all the air being pushed out of the way of the bullet, so in the real non-idealized physics worls, the target will actually receive less momentum than the gun/shooter.
OK, so you’re telling me that energy can be created, yes? And I’m supposed to imagine a “couple sets of masses that push on each other?” (Those pesky masses! Ya gotta watch 'em all the time!)
Try this: Venture to your local bowling alley, rent some shoes, get a lane, pick out a ball – any ball. Walk up to the foul line with ball in hand(s), stop, and while standing straight – feet together and flat, back straight, no leaning, no bending or twisting or “winding up!” – thrust the ball toward the pins.
When you pick yourself up off your ass, and after the laughter subsides, let us know how it turned out.
TBone2:
Momentum of gun = m[sub]gun[/sub] v[sub]gun[/sub]
Momentum of bullet = m[sub]bullet[/sub] v[sub]bullet[/sub]
Kinetic energy of gun = E[sub]gun[/sub] = 0.5 m[sub]gun[/sub] v[sub]gun[/sub][sup]2[/sup]
Kinetic energy of bullet = E[sub]bullet[/sub] = 0.5 m[sub]bullet[/sub] v[sub]bullet[/sub][sup]2[/sup]
Do you agree with the above?
As a starting point, you must assume the momentum of the gun is the same as the bullet’s momentum. Thus
m[sub]gun[/sub] v[sub]gun[/sub] = m[sub]bullet[/sub] v[sub]bullet[/sub]
solving for m[sub]gun[/sub]:
m[sub]gun[/sub] = m[sub]bullet[/sub] v[sub]bullet[/sub] / v[sub]gun[/sub]
Plugging m[sub]gun[/sub] into energy of gun equation:
E[sub]gun[/sub] = 0.5 m[sub]bullet[/sub] v[sub]bullet[/sub] v[sub]gun[/sub]
From the energy of bullet equation we know that:
m[sub]bullet[/sub] = 2 E[sub]bullet[/sub] / v[sub]bullet[/sub][sup]2[/sup]
Plugging this into the (new) Energy of gun equation gives us:
E[sub]gun[/sub] / E[sub]bullet[/sub] = v[sub]gun[/sub] / v[sub]bullet[/sub]
From the original conservation of momentum equation we know that
v[sub]gun[/sub] / v[sub]bullet[/sub] = m[sub]bullet[/sub] / m[sub]gun[/sub]
Thus
E[sub]gun[/sub] / E[sub]bullet[/sub] = m[sub]bullet[/sub] / m[sub]gun[/sub]
Let’s plug in some numbers.
Let’s say I’m shooting an SA58 FAL ;). The caliber is NATO 7.62 x 51 (.308 WIN). The rifle weighs approximately 9.5 lbs with full magazine. The bullet weight is 147 grains. There are 7000 grains to a pound, thus the bullet weighs 0.021 lbs. (The weight ratio is equivalent to the mass ratio.)
The gun is 452.4 times more massive than the bullet. The kinetic energy the gun receives is:
E[sub]gun[/sub] = E[sub]bullet[/sub] m[sub]bullet[/sub] / m[sub]gun[/sub]
E[sub]gun[/sub] = E[sub]bullet[/sub] / 452.4
E[sub]gun[/sub] = 0.22% E[sub]bullet[/sub]
In the above example, the kinetic energy the gun receives is less than one quarter of one percent of the bullet’s kinetic energy. With all due respect, this is a far cry from the 100% figure you previously stated.
Think about this statement. How can something dependent on velocity (momentum) be proportional to something dependent on velocity squared (kinetic energy)?
momentum/KE = (m v) / (0.5 m v[sup]2[/sup]) = 2/v
The ratio is dependent on velocity so momentum can’t be proportional to kinetic energy. Proportionality would mean
(momentum/KE)[sub]1[/sub] = (momentum/KE)[sub]2[/sub]
which can’t be true unless the velocities are equal. In this case, conservation of momentum determines both the force of the gun’s kick and the bullet’s impact. Energy is still conserved, but in a more complicated way.
Energy cannot be created from nothing, but it can change forms. You start off with chemical energy in the gunpowder. That energy gets turned into kinetic energy, of both the bullet and the gun. Those energies must be added together: The total energy of the system is greater than any individual part.
Momentum also cannot be created from nothing, but it can’t be “hidden” in other forms, either. Momentum is a vector, so the bullet’s momentum forward can (and does) cancel out the gun’s momentum backwards. Since momentum can’t be hidden in other forms, it’s almost always easier to analyze interactions like this one using conservation of momentum. You don’t need to know the energy of the gunpowder to be able to say that the momentum of the bullet must exactly balance the recoil momentum of the gun. Once you know that, you can calculate the vellocities, and once you know the velocities, then you can calculate the kinetic energy.
To state it another way: Energy is always conserved, but kinetic energy is not. If you try to conserve kinetic energy, then you’ll end up with neither the bullet nor the gun ever moving at all (since the system starts with zero kinetic energy, and energy can’t cancel out the way momentum can).
Thanks Chronos. It also goes without saying that a lot of the energy (most?) from the explosion gets turned into low-grade heat.
Would someone please answer this, the old “immovable object” (gun) question? Can you have zero momentum in the gun? I remember the answer to be no, but I don’t remember why that is. I think you actually shake the earth a little, but I’ll wait for an answer from more agile minds. 
Peace,
mangeorge
Momentum must be conserved, so the gun (or the system of gun + whatever it’s attached to) must have momentum. So the gun, or the system of gun and Earth, or whatever, must move. But if we attach the gun to something with enough mass (like the Earth) then it will only need a very extremely small speed to have enough momentum.
Doesn’t the bullet (slug, shot, what have you) also arrive at the target with a compressed column of hot gas in front of it? Not all of the air it encounters on the way to the target is pushed to the side – some must be compressed in front of it. Unless we’re operating in a vacuum, the atmosphere plays an important role in ballistics.
Some of the air that the bullet passes through would also likely be accelerated towards the target, such that there could be some additional force applied after the fact.
Energy - a scalar - is conserved
Kinetic energy - a vector - is not conserved. Not as a vector.
There’s a thread about calibers and stopping power in which Susannann posted a link to http://www.firearmstactical.com/hwfe.htm which is a bit of a long read, but goes into more detail than I’ve seen elsewhere on what actually happens to the victim. It also pretty thoroughly demolishes the whole stopping power'' or throwing victim backwards into the nearest glass object’’ phenomenon in more detail than the bad movie physics page I linked to earlier…
BTW: A lot of this gun/bullet physics stuff was covered here:
http://boards.straightdope.com/sdmb/showthread.php?s=&threadid=75987
Agree. Excellent article. I also like where it says “energy dumping” and hydrostatic shocks” are myths. Very true, but I am still amazed how many shooters believe that stuff…
Thanks, Chronos.
Because I brought it up, and now I’m curious, The physics are the same for gun, bow and arrow, thrown spear, or just about any kind of projectile, right?
The energy source really isn’t relavent, is it.
As long as you keep track of everything that’s flying away. If you fire a rocket launcher, for instance, you won’t feel any (or as much) recoil, but that’s just because the rocket also sends a bunch of hot gas backwards. For a gun, bow, or spear, it all works essentially the same way.