This is for physics geeks only! I'm asking about the falling plates at most pistol ranges. As we all have seen, there is always one that is really hard to knock down. I am trying to figure out if (except for mass and speed, all else equal) a bullet's momentum or it's energy is what knocks the plate down. I think energy is what matters, but I want to be sure. OK, I'm rusty on my physical science. From the get go, I know that the elasticity of a bullet complicates things, particularly when energy is considered as opposed to momentum. I think that the best assumption to make about elasticity to simplify the question is to pretend that the bullet and plate collisions are perfectly elastic, much like billiard balls. (Even though my examples are hollow points, pretend that they don't deform.) We also assume that all shots hit dead center, the distance is negligible, and air resistance is also negligible. As examples, I have Remington's 185g .45+P Golden Saber vs. Remington's 230g .45+P Golden Saber. The speeds and energy are: 230g 875 f/s 391 ft-lbs 185g 1140 f/s 534 ft-lbs If you multiply the grains by f/s to get the momentum, the 185g bullet has around 5% more momentum. But the energy difference is huge... over 1/3 more. I have one final observation/question. Remington has a hot 9mm 115g bullet that goes 1250 ft/s, and has 399 ft-lbs of energy. would it have about the same plate-knocking effectiveness as the example 230g round which has 391 ft-lbs of energy? Wanna kill these ads? We can help!