So bullets maim and kill people because they impact with great force?
Small mass but huge acceleration? A bullet at say 10% the speed would hardly do anything I suppose. So what if the mass was increased ten-fold so then the forced stayed the same? Would that much slower yet much larger bullet do the same damage as a traditional bullet?
What if the bullet speed was slowed to a crawl, but the mass adjusted accordingly to maintain the same force?
I can’t picture an incredibly dense bullet traveling at say 10 mph ripping a hole through a person.
I must be not understanding several things, about physics or bullets or both.
Sorry if my question is vague.
Bonus question! (didn’t want to make a new thread. There really should be a fun thread for really quick simple questions like the following.)
It would do a different kind of damage. Just imagine being hit by a car going, say, 30 mph. A much larger object going very slowly would move you without necessarily doing any damage unless you encountered a stationary object like a cow or a building.
But it wouldn’t move you very much, because it would have to be going very, very slowly to have the same energy as a bullet did.
Also, the bullet doesn’t have acceleration once it leaves the gun (except for friction slowing it down) until it hits you, and de-accelerates to zero.
Not sure if you want to keep the kinetic energy the same (mv^2) or the momentum the same (mv).
Well first of all while f=ma is correct, for purposes of discussion probably what you want is e=m(v[sup]2[/sup]). This is the formula for kinetic energy, one measure of how much damage a bullet can inflict, expressed either in Joules (metric) or foot-pounds (English). Note that doubling the speed quadruples the energy, whereas increased mass is linear.
A large mass moving at low speed could have the same total energy, but then momentum becomes a factor: it would tend to impart it’s energy to the target as movement, pushing it rather than damaging it. Of course if you have a very heavy bullet then that much force concentrated on a very small area would be like being impaled.
Ok, so what I’m now wondering is, at what speed would a bullet stop penetrating and start pushing a person? (not looking for concrete numbers, just conceptually)
for example, a sufficiently dense bullet that maintains the same kinetic energy as a regular bullet traveling at 10mph.
I can’t imagine my impossibly dense bullet slowly tearing into someone’s out stretched palm, like a regular bullet. Would the person just be pushed backwards slowly?
I understand that my bizzaro bullet is probably a tiny Blackhole, but let’s ignore those consequences if we can.
I feel that kinetic energy isn’t the only thing factoring into why a bullet rips holes in people. What am I missing?
perhaps, something with the instantaneous nature of energy imparted onto the target at high speeds?
A slow moving mass can certainly tear through someone. Imagine someone sat a very heavy spike on your chest. If the spike is heavy enough it will tear a hole through your chest just by the force of gravity.
But if you’re hit by a very large slow moving object, it doesn’t usually tear through you because instead of the energy ripping your body apart, the energy just moves your body. Imagine standing still while someone opens a door that hits you. You might get injured, but mostly you’re going to be pushed backwards.
There’s a reason a bullet (FMJ, anyway) will bounce off a steel wall but send you to the hospital.
You’re missing pressure and the amounts of different types of strain your skin (and the flesh underneath) can withstand. Basically, physical stress. That’s basically what you said about energy being imparted at high speed but somewhat more formally.
The cross-sectional area is also important in such considerations, which is implicit in the discussion so far. It’s why getting hit by a car won’t necessarily open up a hole in your chest the way a bullet will, even if the car has much more momentum and energy.
We can simplify the impact to compressional, torsional shear, and direct shear stresses. Torsion shouldn’t be a major consideration for a bullet impact, but compression and direct shear certainly are, as your skin will certain deform and then tear at the point of impact.
And that’s where the difference in momentum vs energy will kick in. Even if the momentum is the same in two situations, the amount of energy involved will be different (and vice-versa). Consequently, the local pressures at the point of impact can be quite different.
No numbers on actual wounds, but a bullet 4 times as massive with the same kinetic energy will have 1/2 the velocity and consequently twice the momentum.
but the spike needs you to be perfectly still laying down. So those slow moving things appear to need you to be immovable AND braced against something so it can deliver its kinetic energy and go thru you.
so let me further stipulate that this occurs in the vacuum of space.
Bullets injure and kill due to penetration and disruption of internal organs and structures, not specifically because of their raw kinetic energy. If someone hit you with a giant Stay Puft marshmallow with the kinetic energy of a .45 ACP, you’d be more likely to suffocate rather than suffer blunt or penetrating damage. From this standpoint, the momentum of the bullet and the shape (and tendency to expand controllably within the body so as to cause maximum impulse transfer and resultant disruption) are more significant than the kinetic energy at impact. Really high speed (hypervelocity) rounds will often fragment or gyrate wildly upon impact, doing damage to a larger volume, but this is a coupling between the bullet construction and the hydraulic environment in which it operates.
Basically, terminal ballistics (the behavior of the bullet within a viscous medium) cannot be reduced to a simple energy relationship.
You don’t need to imagine some special unobtanium bullet. Just imagine a steel rod being thrown at you. Muzzle energy of a bullet fired from a pistol, roughly 500 joules. Density of steel, about 8. 9mm diameter rod, cross-section, about 60 sq mm, and a one metre long rod comes in at about half a kilogram. To match the bullet’s energy the rod will be travelling at about 30 m/s. Which is 70 miles per hour. That will go straight through you. But unless it hits something vital won’t kill you, for the reasons Stranger On A Train notes above. Make the rod 4 metres long, and the speed halves. 35 miles an hour. That is more likely to knock you down.
Or consider a baseball. 150 grams. To match the bullet’s energy that will be travelling at 58 m/s or 130 miles per hour. That isn’t a lot faster than the fasted recorded pitch of 105 miles an hour.
That pitch translates to 330 joules, or twice the energy of a .22 rifle bullet. It might hurt, but it you can catch it.
Pah, I messed up the calculations. 1am here, I shouldn’t try this then. Ignore the above.
You don’t need to imagine some special unobtanium bullet. Just imagine a steel rod being thrown at you. Muzzle energy of a bullet fired from a pistol, roughly 500 joules. Density of steel, about 8. 9mm diameter rod, cross-section, about 60 sq mm, and a one metre long rod comes in at about half a kilogram. To match the bullet’s energy the rod will be travelling at about 42 m/s. Which is 100 miles per hour. That will go straight through you. But unless it hits something vital won’t kill you, for the reasons Stranger On A Train notes above. Make the rod 4 metres long, and the speed halves. 50 miles an hour. That is more likely to knock you down. Maybe.
Or consider a baseball. 150 grams. To match the bullet’s energy that will be travelling at 82 m/s or 180 miles per hour. The fasted recorded pitch was 105 miles an hour.
That pitch translates to 165 joules, or about the energy of a .22 rifle bullet. It might hurt, but it you can catch it.
I don’t really know what you want for the bonus question. You could verify it yourself. Find another reputable source that gives a total volume of all water on earth, insert that for V in V=[sup]4[/sup]/[sub]3[/sub]πr[sup]3[/sup] and solve for r to get the radius which you can then use to calculate the size of the requisite sphere.
Another thing to put the bullet energy and momentum into perspective is to remember that at the time of the shooting, the gun gets and equal and opposite push. The fact that the shooter doesn’t take as much damage as the shootee means there is more than just momentum and energy at play.
