[chroma]

Your general point about the slide recoil is correct, but your comparison isn't. Kinetic energy is proportional to the square of velocity. k = 1/2 * mv^2

In your example, the projectile has...

    1/2 * 20g * (365 m/s)^2 = 1330 joules

...of kinetic energy. But your slide has...

    1/2 * 600g * (12.2 m/s)^2 = 44 joules

...of kinetic energy. Solve for the slide velocity...

    1/2 * 0.600kg * (x m/s)^2 = 1330 joules
          0.300kg * (x m/s)^2 = 1330 joules
                    (x m/s)^2 = 4433 m^2/s^2
                            x = 66.58 m/s

So the slide is initially traveling backwards at around 65 meters per second. Probably faster, since only a fraction of the deflagration is transferred to the bullet.

Another nitpick: I think your example is not representative. A 20 gram bullet and 600 gram slide are both very heavy for a handgun. Using typical numbers: Hydra-shok 9mm +P has a bullet weighing 8 grams, which exits the muzzle at 350 meters per second. That gives 490 joules of kinetic energy. I weighed the slide on my Sig Sauer P239 at 300 grams. Plug those numbers in, and you get a slide traveling backwards at 57 meters per second. Not too different, but worth mentioning.

That said, I completely agree with everything else in your comment. Thank you for explaining the intricacies of handgun and ammunition metallurgy.

[icegreentea]

Parent was invoking conservation of linear momentum not an energy balance. There's no requirement that both the bullet and slide have the same kinetic energy, only that the sum of energy between the two bodies remains constant.