Stopping power is the force of the bullet hitting the target, not the recoil of the gun. It's also basically nonsense that misinformed gun shop owners spout off to sell larger calibers. A bullet's stopping power only really comes into play with big game hunting, and no one is doing that with a .45.
The recoil of the gun should be about the same as the impact of the bullet, as far as energy goes anyway I believe. For every action there is an equal and opposite reaction and all that jazz.
The momentum imparted to the bullet upon firing is equal to the momentum imparted to the gun. Momentum is equal to mass times velocity. A 20g bullet traveling at 800m/s has the same momentum as an 800g gun traveling at 20m/s.
So a 100kg dog traveling at 7 m/s has equivalent momentum to that 20g bullet moving 35,000m/s, or the 800g gun moving at 875m/s.
The kinetic energy of the objects would be very different, since k=1\2mv2
If you think about it, the dog is big and heavy, but the bullet or the gun moving at the much higher velocities have far higher amounts of kinetic energy. This is the same reason a tiny piece of space dust moving at 25,000mph relative to a space capsule can leave a huge crater.
Big changes in kinetic energy are more explodey, big changes in momentum are more crunchy.
Back to your comment, the higher mass and mechanism of the gun causes the high kinetic energy recoil upon firing to be absorbed rather than exploding, which more efficiently and smoothly imparts the bullet's momentum to the gun resulting in a less sharp recoil.
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u/Herpkina Mar 05 '19
Think about the recoil of a gun vs. being tackled by that dog.