Thanks, very informative! Also, don't certain factors play a part in effecting the force felt by the shooter vs the victim. Like the shotgun absorbing some force itself (recoil?), the stock/padding, etc?
This will get long, so hang in there.
Yes, certain factors DO play a very important part in affecting the force felt by the shooter vs. the victim.
Always remember: momentum transfer is conserved in a closed system.
What affects the force felt by the shooter?
All other things being constant, the mass of the shotgun and the length of travel of the shotgun in recoil over time.
If you shoot a heaver shotgun, the force felt by the shooter "feels" less than for a lighter shotgun.
If you use a padded butt on the stock, you can reduce the felt recoil because it effectively lengthens the recoil distance traveled, as well as the time in which the distance is traveled. Remember...force is mass times acceleration. Acceleration is change in velocity divided by change in time. If you take a longer time to drop the recoil velocity to zero, then acceleration goes down and the force felt decreases.
What affects the force felt by the shooter?
Again, all other things being being constant, it's force which is equal to mass times acceleration. (And in closed system, momentum is conserved. I keep saying that, it seems!) So what affects the force felt by the victim?
The same things that affect the force felt by the shooter. Mass of the slug and how quickly the slug velocity drops to zero, because that affects the acceleration of the slug...and therefore the acceleration of the combined mass of the victim and the slug (because momentum is conserved). The velocity question is where it gets difficult because you have to make some assumptions.
Here's an apt analogy that's very easy to visualize:
Ever play pool or billiards? All the billiard balls (including the cue ball) are of the same mass. If you line up a perfectly straight (head on) shot with your cue ball and the 8-ball, what happens to the two balls when they collide?
The cue ball stops and the 8-ball moves off at the same velocity the cue ball had initially.
If the two ball could, somehow, "stick together" on impact and the table was a frictionless surface, what would happen? The combined mass of both the balls stuck together would move at 1/2 the initial velocity of the cue ball. (Because momentum is conserved.)
Now...imagine what would happen if you had a billiard ball that was 2881 times the mass of the cue ball. For the same scenario:
Head on shot: The cue ball hits the far more massive 8-ball and bounces back with ALMOST the same velocity it had initially. Some momentum would transfer and the 8-ball would move with a very tiny fraction of the initial velocity of the cue ball.
If they stuck together on a frictionless surface...the combined mass of both balls would move together at a tiny fraction of the initial velocity of the cue ball.
Remember...momentum is mass times velocity, and momentum is always conserved in a closed system.
So, using the same numbers above slug/victim scenario ("1" for mass of the cue ball and "2880" for mass of the 8-ball):
Initial momentum of the 8-ball is zero because it's initial velocity is zero. (m8v8 = 2880 times zero)
Initial momentum of the cue ball is mcvc, where it's initial mass is "1" and initial velocity is whatever you choose it to be.
Final momentum is m8v8 plus mcvc. Now the final mass will be 2881 and their final velocity will be the equal (because they're stuck together and therefore traveling at the same velocity). We can say the equation for this is mfvf.
Therefore the final momentum is equal to the sum of the two initial momentums (again...this is where it's important to remember that momentum is conserved in a closed system):
mfvf = m8v8 + mcvc
Remember...m8v8 is zero because the massive 8-ball is stationary. So it drops out.
mfvf = mcvc
Therefore final velocity (vf) would be:
vf = (mcvc) divided by mf.
OR
vf = (mc/mf) times vc = (1/2881) times the initial velocity of the cue ball.
BACK TO YOUR SCENARIO:
Since I conveniently used the same mass, let's assume the initial velocity of your 1 ounce slug is 1,760 fps.
Dividing that by the total mass of the victim plus the slug (2881), you get a combined final velocity of the victim being 0.61 fps (which is about 0.4 mph).
BUT...FORCE is equal to mass times acceleration. This is what the victim would "feel". Acceleration is change in velocity divided by the time it takes for that change to happen.
For simplicity, we can assume that 0.61 fps is effectively zero compared to the initial slug velocity of 1,760 fps. So the change in velocity will be 1,760 - zero = 1,760 fps.
How long does it take for the slug to decelerate to zero fps? I really don't have a clue. And therein lies the problem. Objectively speaking, it's definitely not a long time. A fraction of a second, to be sure. It would be interesting to see if anybody has done any experiments for something like this.
But the bottom line the force is directly proportional to how fast that 2881 ounces is "pushed" to a velocity of 0.61 fps. The less time it takes to accelerate the victim to 0.61 fps, the more force the victim will feel.