So AB, what would Newton have to say? I say the perp would be knocked back away. Only the Warren Commission seems to think a round hitting from behind causes the recieving party to move toward the shooter.
You obviously think catching a brick at a couple of hundred feet per second wouldn’t include you in the brick’s momentum?
Your assumption must have the 45 ACPs passing on through and not transferring their energy onto the mass, othewrwise just wanting to be argumentive.
Now: First of all, Mr Newton would tell you that P=mV. Second, he would tell you that momentum is conserved in all interactions. Recall that velocity and momentum are vector quantities; they can have a negative value. Positive or negative value indicates direction.
Two interactons are relevant: First is the interaction of the bullet with the shooter's body (including the pistol). Second is the interaction of the bullet with the perp's body. I shall assume, for simplicity's sake, that at the moment of firing the shooter is not moving, that at the moment of impact the perp is not moving, and that the bullet loses no velocity during its brief movement from muzzle to perp.
Definition of terms:
Ms = Mass of shooter.
Mp = Mass of perp.
Mb = Mass of bullet.
Ps = Momentum of shooter (after shot).
Pp = Momentum of perp (After impact).
Pb = Momentum of bullet (during flight).
Vs = Velocity of shooter (after shot).
Vp = Velocity of perp (After impact).
Vb = Velocity of bullet (during flight).
Prior to the shot,
Ps = Pb = 0.
At the shot (when bullet leaves muzzle),
Ps + Pb = 0 (momentum is conserved)
Ps = -Pb.
MsVs = MbVb
Now, this is all the momentum that bullet will ever have. The interaction of the bullet striking the perp is called an "inelastic collision"; the bullet is completely merged with the perp. Momentum is still conserved. Recall that the perp is stationary prior to impact, therefor his momentum prior to impact Pp is zero.
Pb + Pp = Pbp.
MbVb + 0 = (Mb+Mp)Vpb
So: The momentum of the perp with a bullet in his gut is the same as the momentum of the bullet prior to impact, and in the same direction. It is also equal in magnitude to the momentum of the shooter after the shot, but in opposite direction. If the perp and shooter are the same mass, then they will have the same velocity (actually perp velocity is slightly LESS ... he's got a bullet in his gut and therefore greater mass) but in opposite directions. Since firing a .45ACP pistol (or even a 12ga slug) does not throw me back several feet, it will not do that to the perp. Let's throw some real numbers in here.
Bullet = 230gr or 0.033lb at 950 ft/s. Perp = 200lb. Pb = 0.033 * 950 = 31.35 ft-lb/s
Since momentum is conserved,
Pbp = (200lb + .033lb)Vbp = 31.35 ft-lb/s
Doing some algebra,
Vbp = 31.35/200.033 = 0.157 ft/s
Sorry, perp ain't flying back two or three (or even one) feet due to being shot with a .45ACP. In fact, most of his motion after the shot will be due to physiological reactions and gravity. The bullet's momentum is negligible in comparison. Also, when you shoot the .45, you will be given a (backward) velocity of 0.157 ft/s (it's called recoil). That "feels" right to me.