How much velocity is lost due to incomplete powder burn prior to the bullet leaving the the muzzle? In other words, what is the velocity difference between a 4" semiauto pistol and a 20" bolt action rifle? The energy to work the action has to come from somewhere. The only energy input is the explosive evolution of gas from the burning powder. This energy can be used to propel the bullet OR work the action- in a semiauto some propels the bullet, some works the action. Again, you don't get something for nothing. A high school physics student (you don't have to be a court testimony expert) could tell you that, but thanks anyway for your input. By the way, I've been shooting for a while, myself.
Well, since high school physics has been mentioned and since I like to kid myself that I remember a lot of mine, I'll chime in.
First of all, you are confusing conservation of momentum and conservation of energy. Disregarding any quantum effects: momentum in a closed system stays constant. Similarly (but via totally different means/theories) total energy (chemical, gravitational potential, heat et al) in a system stays constant.
When the primer and powder in a cartridge is ignited, chemical energy is released in the form of heat and pressure. That increases the pressure in the cartridge which acts upon the base of the bullet as a force, causing the bullet to move forward, through the barrel and out the muzzle. Similarly, the pressure in the chamber pushes on the back of the cartridge which in a bolt action firearm is solidly attached to the frame and in the case of a blowback operated firearm, is attached to (or pushes against if you prefer) the mass of the bolt or slide. In the case of the Buckmark, the two are the same, in the case of the Ruger Mark II series, there is no slide, just a bolt which extends through the back of the frame. In either case, the mass is much much larger than the mass of the bullet and only a tiny fraction of the energy is converted into kinetic energy of the rearward moving mass.
Aside from the bullet, the vast majority of the rest of the energy is used in heating/pressurizing/accelerating the gases which result from the combustion. This comes in the form of the loud report of the gun fire and the blast of air from the muzzle. The muzzle blast gases move at a much higher velocity than the bullet (and since energy = 1/2 mv^2) carry a lot of energy despite the low mass of the gases. Incidentally, this is also why a contact wound from a firearm is many times more devastating than just the bullet from even a short distance. These gases can be redirected after the bullet exits the barrel to reduce the recoil experienced by the shooter by a muzzle brake. Note, any porting *before* the crown of the muzzle *will* reduce muzzle velocity slightly, but porting *after* the crown has practically no effect on muzzle velocity.
The *only* reduction in the kinetic energy of the bullet occurs by changing/reducing the force upon the bullet, force which is exerted by the pressure of the combustion gases on the base of the bullet. Yes, in a blowback operated 22lr firearm, pressure will drop very slightly because the size of the cylinder increases slightly due to the rearward movement of the bolt/slide, but given the extremely short amount of time it takes for the bullet to leave the muzzle, this reduction is negligible and many orders of magnitude less than 30%. A simple test for this theory would be to take a 22lr blowback pistol with a lockable slide or bolt and (with several identical batches of ammunition) fire rounds with the slide locked and the slide unlocked through a chronometer.
Given the above, a reduction in recoil does not necessarily mean that there is a corresponding reduction in muzzle velocity.
http://en.wikipedia.org/wiki/Physics_of_firearms is a good place for the bare basics of the physics of firearms. This page may also be useful:
http://www.bsharp.org/physics/stuff/recoil.html .