F = ma = d/dt(mv)
KE = (1/2)mV2
I think what you're trying to say is:
m1v1 + m2v2 = m1V1 + m2V2
Given that m1 and m2 (bullet and rifle) are both at rest before firing (ie: v1=0 and v2=0) then
m1V1 = -m2V2
Which is to say that after firing, the momentum of the gun and of the bullet are equal in magnitude but opposite in direction.
Therefore, in general recoil varies linearly (not geometrically) with bullet mass and bullet velocity. It is left as an exercise to the reader that for a given cartridge and load, heavier guns will have a smaller V2 and consequently less felt recoil.
Put the gas driven moving mass (BCG) and recoil spring into that system. Its not as simple as Newtons action/reaction experiment, from the perspective of the felt recoil of the operator.