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  #1  
Old 06-14-2012, 08:14 AM
foopyx foopyx is offline
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f=ma question

Good morning you lovely and informative people.

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.)

Is this accurate? http://earthsky.org/earth/if-you-mad...ig-would-it-be
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  #2  
Old 06-14-2012, 08:47 AM
John Mace John Mace is offline
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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).

Last edited by John Mace; 06-14-2012 at 08:51 AM..
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  #3  
Old 06-14-2012, 08:54 AM
Lumpy Lumpy is offline
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Well first of all while f=ma is correct, for purposes of discussion probably what you want is e=m(v2). 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.

Edit: John Mace beat me to it.
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  #4  
Old 06-14-2012, 08:56 AM
John Mace John Mace is offline
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ETA: The car wouldn't be going 30 mph. Much slower, considering it's got so much mass.
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  #5  
Old 06-14-2012, 09:41 AM
foopyx foopyx is offline
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Thanks! starting to fix my stupids.

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?
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  #6  
Old 06-14-2012, 09:48 AM
John Mace John Mace is offline
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What shape is the bullet?
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  #7  
Old 06-14-2012, 09:54 AM
Lemur866 Lemur866 is offline
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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.
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Old 06-14-2012, 09:55 AM
foopyx foopyx is offline
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The bullet is bullet shaped....

Let's say it's a perfect sphere about ten millimeters diameter.
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  #9  
Old 06-14-2012, 09:57 AM
Great Antibob Great Antibob is offline
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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.
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  #10  
Old 06-14-2012, 09:59 AM
foopyx foopyx is offline
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Quote:
Originally Posted by Lemur866 View Post
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.
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.
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  #11  
Old 06-14-2012, 10:11 AM
John Mace John Mace is offline
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In that case, your cells explode long before a bullet comes into play!
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  #12  
Old 06-14-2012, 10:19 AM
california jobcase california jobcase is offline
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I thought it was e= 1/2(mass)(velocity squared).

Last edited by california jobcase; 06-14-2012 at 10:22 AM..
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  #13  
Old 06-14-2012, 10:26 AM
Stranger On A Train Stranger On A Train is offline
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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.

Stranger
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  #14  
Old 06-14-2012, 10:26 AM
Francis Vaughan Francis Vaughan is offline
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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.

Last edited by Francis Vaughan; 06-14-2012 at 10:29 AM..
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  #15  
Old 06-14-2012, 10:35 AM
Francis Vaughan Francis Vaughan is offline
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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.

Last edited by Francis Vaughan; 06-14-2012 at 10:36 AM..
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  #16  
Old 06-14-2012, 10:39 AM
foopyx foopyx is offline
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Quote:
Originally Posted by Stranger On A Train View Post
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


Stranger
Ah, right there was my mistake. Thank you everyone!


(no one addressed my bonus question!)
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  #17  
Old 06-14-2012, 11:02 AM
Inner Stickler Inner Stickler is offline
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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=4/3πr3 and solve for r to get the radius which you can then use to calculate the size of the requisite sphere.
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  #18  
Old 06-14-2012, 11:05 AM
Buck Godot Buck Godot is offline
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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.
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  #19  
Old 06-14-2012, 11:19 AM
Iggy Iggy is offline
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The kinetic energy from a .30-06 is about the same as a Ford Explorer traveling at 8.5mph.

As you have surmised it is not just an issue kinetic energy. The pressure exerted on your flesh at the moment of impact will play a huge role.
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  #20  
Old 06-14-2012, 11:41 AM
Xema Xema is online now
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Quote:
Originally Posted by John Mace View Post
Not sure if you want to keep the kinetic energy the same (mv^2) ...
Quote:
Originally Posted by Lumpy
... what you want is e=m(v2)
It may be a nitpick, but (as CA jobcase notes) E = m * v2 / 2
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  #21  
Old 06-14-2012, 11:57 AM
Lumpy Lumpy is offline
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Quote:
Originally Posted by Xema View Post
It may be a nitpick, but (as CA jobcase notes) E = m * v2 / 2
You're right, the cite I looked at was saying that energy is proportional to m*v2
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  #22  
Old 06-14-2012, 12:00 PM
Chronos Chronos is offline
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Quote:
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.
That proves that there's more than momentum involved, but it doesn't say anything about energy, since the energy is distributed unevenly between the gun and the bullet.
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  #23  
Old 06-14-2012, 12:33 PM
panamajack panamajack is offline
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If you're in free space, then the momentum transfer becomes a bigger factor at slow speed.. It's probably fairly difficult to determine how much you're going to start moving with the incoming object, but obviously it'll need to be going faster if it's going to penetrate than if your body takes longer to be moved.

In contrast a typical fired bullet would hit you almost the same as if you were standing on the ground. Assume the bullet is traveling on the order of the speed of sound. The earliest your body can move on the ground is as long as it takes the impact to travel (at the speed of sound in your body) to your feet. While that speed is faster than some bullets, if you're hit ~1m from the ground the bullet can easily get through you in the millisecond before you could possibly interact with the ground. Thus it's going to be the same as in free space. About the only possible difference is the attitude of the body under its own weight, but I'm guessing that's very sleight.

As an aside, the second question is a good candidate for Wolfram Alpha. The details of how it interpreted your question are provided. Although it doesn't always provide complete references, so the verification part isn't answered. You have to trust it on the numbers it starts with.
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  #24  
Old 06-14-2012, 01:20 PM
foopyx foopyx is offline
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Quote:
Originally Posted by Inner Stickler View Post
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=4/3πr3 and solve for r to get the radius which you can then use to calculate the size of the requisite sphere.
Quote:
Originally Posted by panamajack View Post
As an aside, the second question is a good candidate for Wolfram Alpha. The details of how it interpreted your question are provided. Although it doesn't always provide complete references, so the verification part isn't answered. You have to trust it on the numbers it starts with.
I appreciate the feedback, but usually some smart person comes along and does the math for me. ;-)

Thanks for all the answers!
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  #25  
Old 06-14-2012, 01:35 PM
Inner Stickler Inner Stickler is offline
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Originally Posted by foopyx View Post
I appreciate the feedback, but usually some smart person comes along and does the math for me. ;-)
Tsk, that's no way to learn.
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  #26  
Old 06-14-2012, 06:15 PM
Kevbo Kevbo is offline
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Quote:
Originally Posted by Buck Godot View Post
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.
Two things you need to look at:

Momentum = m*v
Also Momentum = f * t (This is also sometimes called impulse) **
Energy = m * (v**2) / 2

A peculiar thing about explosions and collisions is that the total momentum of the various masses ends up the same after the event as before. If the gun and bullet both start at rest, (total momentum=0) then after firing the bullet and gun end up with equal and opposite momentum, (total momentum = 0) but since the bullet gets much greater velocity, the energy goes mostly to the bullet. So the shooter's shoulder has to soak up the same momentum as the Elk he is shooting at, but the Elk absorbs a heckuva lot more energy, due to the much higher V of the bullet, and that v squared issue.

The force in the OP is hard to define, because A is hard to define. You can stop a bullet all at once (big a) or gradually with air friction alone (very low a) So how much force you apply to the target is very hard to calculate. BUT if you know the mass of the target, and bullet, and assume that the bullet stops inside the target, then it is very easy to calculate how much momentum, or impulse you have applied to the target.

** if F is not constant then you have to integrate f over t

Last edited by Kevbo; 06-14-2012 at 06:17 PM..
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