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!

They are directly related, so it is both. Momentum = mv Kinetic energy = (1/2) m (v squared). m = mass v = velocity

A factor in knocking over plates with a bullet: The impact of the bullet does not really knock the plate FLAT. Rather, it shifts the plate enough so that the plate center of mass is no longer over the support holding the plate upright. When that shift is achieved, the plate will continue to fall based on the torque exerted by its own weight "twisting" around the support that is no longer directly underneath the plate. A comparable issue: the intellectual "conflict" between bullet energy and bullet momentum is a classic one. Quantitative "disagreement" about bullet "stopping power" among calibers/loads usually comes down to deciding what emphasis is placed on energy vs. momentum.

Momentum is what knocks targets around, even when energy is a lot lower than other rounds. Watch hickok45's swinging targets for examples of this (especially the "dueling tree").

It is almost entirely momentum and the part that energy plays is insignificant at bullet velocities. English

Took this from a post on physicsforum: "Kinetic energy is a magnitude. It only describes how hard an object in motion will another object and is expressed in foot pounds. Momentum is the application of kinetic energy and is expressed in joules, foot pounds per second, etc..... When comparing moving objects that are similar (IE, bullets of different weights and velocities), the projectile with more momentum would tend to move its target father regardless of kinetic energy. Example.... A rimfire rifle is fired at a steel plate at 50 yards. If you hit it, it flips over. You have a choice of using a rifle that will propel a 40 grain bullet at 1249 f/p/s, or a 17 grain bullet at 2199 f/p/s at 50 yards (these are actual figures). If you only used the formula for KE, you would find the bullets would provide 139 ft-lbs & 183 ft-lbs respectively, so you would likely pick the bullet with the higher KE for the task at hand. Looking into the figures a little more, by figuring momentum.......the 40 grain bullet would provide 7.14 ft-lbs/second, and the 17 grain bullet would provide 5.34 ft-lbs/second (weight/7000*velocity). The heavier, slower bullet would be more likely to flip the plate because it has more momentum even though the lighter faster bullet has 44 more ft-lbs of kinetic energy." This is why the .45 knocks bowling pins around like nothing, despite having less energy than some 9mm offerings.

Permit me to expand on Arclight610's comments. Momentum and kinetic energy are both "qualities of motion"; the difference between these two qualities is NOT intuitively apparent. How can an explanation of that difference be stated to someone WITHOUT invoking the mathematical velocity versus velocity-squared mantra? I offer this angle: An object is moving, and I exert force on it until it stops. How LONG in TIME it takes me to stop the object is the measure of the object's momentum. How FAR in DISTANCE the object travels until I stop it is the measure of the object's kinetic energy. The physics concepts of "impulse" and "work" also work into this comparison.

[ame="http://www.youtube.com/watch?v=RHlLmYVCzKY"]Professor Irwin Corey[/ame] has the answer to this recurring question.

The Taylor Knockout Factor method of handgun round effectiveness to incapacitate a person also goes with momentum ... but with an additional factor. TKO takes the product of mass, velocity, (thus using momentum) AND bullet DIAMETER as its computational basis. Thus, the wide bullet picks up rating over a narrow bullet of equal mass and velocity.

It seems to be a momentum issue. Slow-motion video shows that faster bullets disintegrate before the plate starts moving. Thus, much of the energy is dispersed in other directions. A heavier bullet seems to dump more of the energy into the target over a longer period, allowing the plate to be knocked down more effectively. My experience mirrors this. In subgun matches, the 124 and 147gr 9mm rounds seem to be more reliable at putting down targets than 115gr bullets at higher velocities. In fact, most times, the fast 90gr loads barely rock the plates. Another factor is that jacketed bullets work better than un-jacketed bullets. The copper jacket probably slows down the deformation of the bullet, allowing more energy to be directed into the plate.