You’re right, the cite I looked at was saying that energy is proportional to m*v[sup]2[/sup]
That proves that there’s more than momentum involved, but it doesn’t say anything about energy, since the energy is distributed unevenly between the gun and the bullet.
If you’re in free space, then the momentum transfer becomes a bigger factor at slow speed… It’s probably fairly difficult to determine how much you’re going to start moving with the incoming object, but obviously it’ll need to be going faster if it’s going to penetrate than if your body takes longer to be moved.
In contrast a typical fired bullet would hit you almost the same as if you were standing on the ground. Assume the bullet is traveling on the order of the speed of sound. The earliest your body can move on the ground is as long as it takes the impact to travel (at the speed of sound in your body) to your feet. While that speed is faster than some bullets, if you’re hit ~1m from the ground the bullet can easily get through you in the millisecond before you could possibly interact with the ground. Thus it’s going to be the same as in free space. About the only possible difference is the attitude of the body under its own weight, but I’m guessing that’s very sleight.
As an aside, the second question is a good candidate for Wolfram Alpha. The details of how it interpreted your question are provided. Although it doesn’t always provide complete references, so the verification part isn’t answered. You have to trust it on the numbers it starts with.
I appreciate the feedback, but usually some smart person comes along and does the math for me. 
Thanks for all the answers!
Tsk, that’s no way to learn.
Two things you need to look at:
Momentum = m*v
Also Momentum = f * t (This is also sometimes called impulse) **
Energy = m * (v**2) / 2
A peculiar thing about explosions and collisions is that the total momentum of the various masses ends up the same after the event as before. If the gun and bullet both start at rest, (total momentum=0) then after firing the bullet and gun end up with equal and opposite momentum, (total momentum = 0) but since the bullet gets much greater velocity, the energy goes mostly to the bullet. So the shooter’s shoulder has to soak up the same momentum as the Elk he is shooting at, but the Elk absorbs a heckuva lot more energy, due to the much higher V of the bullet, and that v squared issue.
The force in the OP is hard to define, because A is hard to define. You can stop a bullet all at once (big a) or gradually with air friction alone (very low a) So how much force you apply to the target is very hard to calculate. BUT if you know the mass of the target, and bullet, and assume that the bullet stops inside the target, then it is very easy to calculate how much momentum, or impulse you have applied to the target.
** if F is not constant then you have to integrate f over t