I was having an argument with another person online a couple of days ago about bullets and the physics the govern how hard they “hit”.
Basically, here’s his argument-
He is convinced that bullets from pistols have more mass, but less acceleration. Bullets from rifles have less mass but more speed and can penetrate objects more easily because they have a small “cross section”:
Since Force = mass x acceleration, the two examples CAN equal the same amount of force produced, but the rifle bullet “hits harder” because of the smaller cross section. (He gave an example of armor piercing rounds that more easily ‘punch through’ a given target)
Here’s my argument:
You’re on crack. F=ma is a measure of work using a vector measurement. Projectile “damage” is normally described by 1/2mv<squared>=Kinetic Energy. (Velocity being a constant). It’s all about the speed…a given projectile going twice as fast has 4 times the energy as one the same mass going slower, and therefore does a lot more damage. The reason that the “smaller cross section” penetrates better (I reason) is because there is a greater amount of kinetic energy being applied to a smaller area of the target. An armor piercing round and a normal round hit with the same amount of energy, it’s just that the AP round focuses it’s energy on a smaller area and therefore penetrates.
After going back and forth with this guy for a while, I started to become unsure of myself. Certainly f=ma can apply to a bullet since it has mass and acceleration. Can the result of f=ma and 1/2mv<squared> equal each other? It’s newtons to what, joules?
I was thinking that the bullet isn’t necessarily doing “work” on the target so much as transferring energy to it…
So who has the better understanding of the physics of a bullet? and why? Basically I need someone to describe why the formula f=ma doesn’t apply to bullet damage…
I leave it in the Dopers capable hands…my brain’s fried trying to remember my high school physics…
I think you’re both correct. You can think of it both ways.
You can think of it in terms of force, as your friend did. Everything being equal, a faster bullet will decelerate at a greater rate when it hits the target, and therefore it applies more force on the target. The faster bullet also takes more time to come to a stop, so it applies force for a longer period of time. As for bullet size, a smaller cross-section bullet has a smaller rate of deceleration and travels deeper into the target.
Or you can think of it in terms of kinetic energy - your analysis is perfectly correct. If you want to equate the two, energy (work) is force times distance. Kinetic energy of the bullet is equal to the force applied by the bullet multiplied by the distance it travels into the target.
No force on the bullet during flight. I think the debate here is over ``terminal ballistics’’, namely when the bullet is slowing down having hit the target, in which case there is work done on the bullet (to slow it down) and therefore there is a force applied. Call it a force applied by the target on the bullet, or vice versa, as you please.
What it really comes down to is how the bullet expands. If it mushrooms nicely, then there’s a big surface area to generate friction which applies a force on the bullet, slows it down and transfers its kinetic energy into damage to the target. Also, as bullets are generally spinning, the edges of the jacket as the bullet opens form these nice blades like on a blender… If it doesn’t open up, it can go straight through the target, slowing down (accelerating/ decelerating) less, and not dumping all its energy. On the other hand, an excessively light-weight bullet might hit a bone and break apart without continuing to penetrate into the vitals. In this case, I guess the kinetic energy would go into spring potential energy as the bone flexes, and if it doesn’t break, would be slowly scrubbed off as the bone vibrates after being struck without doing as much damage.
While an armour-piercing round is heavier (and denser) than a hunting bullet, it is also designed to not deform on impact, being made of a much harder material. This means that the impact is concentrated on a very small piece of the target, and overwhelms the shear strength of the armour.
Yes, but I think they’re both on crack. Hence, the question. Depending upon the rifle or pistol, either bullet could have more mass or speed. A .22 might be shot from either. A 44 magnum bullet might weigh 300 grains, but it can be shot from rifles. So-called elephant rifles might have bullets weighing 500 grains.
I’m not sure I follow the exact problem in the question (ballistics isn’t my thing) but I will say that increasing velocity increases kinetic energy faster than increasing mass. In other words, doubling the velocity will get you more power than doubling the mass of the projectile.
And dum-dum bullets are built to break up as they hit, which causes a lot more damage, even tho’ the kinetic energy transfered is the same as a regular bullet of the same size.
Professional hit men (the “shot to the back of the head” type) use small caliber bullets so the bullet won’t come out the other side. It hits the far inside of the skull and bounces around inside, again doing maximum damage.
Ok…I used the calculator I linked to for the following info. Mass is in grains, velocity is in feet/second. I just made up the numbers so don’t nitpick me if the numbers I used are somehow ridiculous for bullets…just an illustration.
You can see from this that half the mass (300 vs. 600) at twice the velocity (1000 vs. 500 f/s) gets you twice the kinetic energy of the 600 grain bullet (666 ft-lb vs. 333 ft-lb). I can’t say what formulas were used but hopefully that’ll help.
Mass = 300
Velocity = 1000
Kinetic Energy = 666.021 ft-lb
Momentum = 1.332 slug f/s
Mass = 600
Velocity = 1000
Kinetic Energy = 1332.043 ft-lb
Momentum = 2.664 slug f/s
Mass = 300
Velocity = 500
Kinetic Energy = 166.505 ft-lb
Momentum = 0.666 slug f/s
Mass = 600
Velocity = 500
Kinetic Energy = 333.011 ft-lb
Momentum = 1.332 slug f/s
I should note that I stripped out ‘Kinetic Pulse’ and Energy/Momentum’ data from my list that is provided with the calculator. In hindsight maybe I shouldn’t have…I don’t know what it means (which is why I removed it) but that doesn’t mean it wouldn’t be relevant to someone. Check the calculator yourself to see what I mean (it’s simple…only asking for mass and velocity).
A .44 magnum 240 grain bullet has the same mass regardless of whether it is fired out of a 5 inch barrel or a 30 inch barrel. The muzzle velocity of the same bullet is going to be pretty much the same between the two. A longer barrel doesn’t always mean a higher velocity (but it can), but it does increase accuracy. Also, what is your friend describing as an “armor piercing bullet”? A heavy penetrator, or a bullet dipped in teflon?
It seems to me that you are both comparing apples and oranges, without knowing what the difference is.
It depends on what you mean by “it.” (I can’t believe I just said that.)
Kinetic energy is the amount of “work” the bullet is capable of doing. “Work” consists of applying a force over a certain distance. If one bullet has twice the kinetic energy of another (either because it’s heavier or faster, or both), that means it can exert more force on the target and/or do so for over a longer distance. So if you want to describe the amount of damage a certain bullet can do, kinetic energy is the most convenient number to use. Saying “this bullet has 1000 Joules of kinetic energy” is the same thing as saying “if this bullet were to hit a target which decelerates it with a force of 1000 Newtons, it will travel 1 meter into the target.” The exact amount of force depends on the hardness of the target and the speed, mass and shape of the bullet.