I did this for brand W a while back, with squat for a response. Go figure. Anyhoo...here's what you wanted.
[Excerpt from Shooting Times Magazine, September 1991 Article by Lane Pearce "A handful of Fire and Fury!"]
What is recoil energy? Or understanding why guns kick
"There's a big difference between the energy and the momentum of a moving object, as you can see in the following equations:
[a] Kinetic energy = 0.5 x mass x (velocity2)
Momentum = mass x velocity
Because the velocity term in the first equation is squared, velocity influences the bullet's energy much more than momentum.
Further searching led me to Hatcher's Notebook, by Major General Julian Hatcher, a compendium of useful information on firearms-related topics. Two chapters of the book cover Hatcher's observations and theory associated with recoil. He described this complicated phenomena using classical physics and a good measure of empirical calculations based on extensive testing.
Studying Hatcher, I confirmed that for practical purposes, the recoil or "kick" of a firearm can be approximated using the simple equations shown above. For example, let's predict the energy of a S&W Model 29 .44 Magnum revolver firing a 240-grain bullet, propelled by 24 grains of powder, at 1,400 fps [feet per second].
To do this you first convert the weight of the bullet and powder in grains to their equivalent value in terms of mass. In the language of physics, the weight is expressed in pounds-force [lb] and mass is expressed in pounds mass [lbm]. To convert pounds-force to pounds-mass, you devide by the gravitational constant, 23.174. (This conversion is necessary to make the use of physics as difficult as possible, much like Mr. Newton must have intended!)
To find the energy of the bullet, devide the weight of the bullet in grains by 7,000 to convert to pounds, then devide that figure by 32.174. The result of these calculations is 0.001066 [lbm], which is the mass of the bullet. That's a tiny number, but when you multiply by the square of the velocity [1,400 X 1,400] and devide that number in half (look back at equation [a] above) you get 1044. This is the muzzle energy in foot-pounds [ft-lbs] of our sample .44 Magnum bullet.
To calculate the recoil energy of the gun, remember, you must first determine the momentum of the bullet and powder charge. Take the tiny number for the mass of the bullet shown above, and add it to the mas of the powder [24 grains/7000/32.174 = 0.0001065 lbm] and multiply the result by 1,400 (see equation ) and you get 1.641. This number is the momentum, in pounds-mass x fps [lbm-fps] of the bullet and powder gasses exiting the barrel. This value can be used to calculate the recoil velocity and the revolver and, in turn, its recoil energy.
Because the momentum of the gun is equal to the momentum of the bullet and powder gasses, we can assume that the momentum of the gun equals 1.641 lbm-fps. Then, using equation again, we can algebraically calculate the gun's recoil velocity.
The weight of the gun, in pounds, must be devided by the gravitational constant to get its mass. The three pound S&W Model 29 masses 0.093243 lbm. Dividing the momentum of the bullet and powder gasses by the mass of the gun produces a recoil velocity of 17.6 fps. Put the recoil velocity and the revolver's mass into the kinetic energy equation and you get 14.44 ft-lbs [of recoil energy].
The exact mathematical equation is: 
V=[w + (k x c)] x v
---------------------
W
Where:
V= the gun's recoil velocity
w= the bullet weight, in pounds [take the weight in grains and divide by 7,000]
c= the weight of the powder charge, in pounds [again divide the weight in grains by 7,000]
v= the bullet's muzzle velocity
W= the guns weight in pounds [take the weight in ounces and divide by 16]
And finally:
k= an empirically determined constant that approximates the effect of the gasses escaping from the muzze of the firearm just after the bullet exits the barrel. To get this value exactly, you'd have to know the pressure distribution over time in the gun's chamber and barrel, the exit velocity of the gasses, and the effects of gas loss at the barrel/cylinder gap of the specific test revolver and a host of other variables. Again, however, we can approximate this number for any practical application of the equation. Hatcher reported that the value for "k" ranges between 1 and 2 depending on the gun/cartridge interior ballistics. For shotguns and low pressure handguns, 1.25 is a good approximation. For magnum handgun loads, and low-intensity rifle cartridges, e.g., .30-40 Krag, .30-30 WCF, etc... you would use 1.5. For .30-06 and other high-(>45,000 psi) pressure firearms, 1.75 may be used fo predict the recoil velocity and energy."
Nifty huh? heheh
Flakbait
Admin, Delta 6's Flight School
