Lead Bullet vs. Rubber Bullet

If you look at the two examples...

The metal bullet embeds itself into the wood with heat and damage to the target... Thus the total kinetic energy is dispersed in the target.

For the rubber bullet, the same thing happens, but as the kinetic energy is stored in the bullet by compression of itself, it is released by the rubber returning to its actual size and rebounding off the target...

So.... errrrrmmmmm the energy is dissipated more quickly by the metal bullet than by the rubber bullet...???

I can't think why there should be any difference!!!

John?

When the metal bullet strikes the target some of its kinetic energy is immediatly transfered to the target and is either converted into heat or is used to accelerate the target, and part of the kinetic energy is converted into heat in the bullet and is eventualy shared with the target.

in the case of the rubber bullet the former part applies but now the part of the kinetic energy that is converted when the bullet rebounds go's to heating and acceleraing the bullet in the opposite direction and is no longer shared with the target.

Hence the target acquires less energy but may well sustain a similar amount of damage depending on the relative speed of sound in the target compared to that in the bullet

The question gets to the root of what force is. See the equations below

F=ma where m is mass and a is acceleration

a=dv/dt where v is velocity and t is time, so;

F=mdv/dt

Now mdv=dp where p is momentum so that gives us

F=dp/dt

for our purposes we can write this expression as

F=Δp/Δt

where Δp means "change in momentum (impulse)" and Δt means "change in time"

Taking a look at that equation we see that the size of a force depends on the change of momentum over the change in time. So lets go back to the problem.

A rubber bullet hits the log and reverses direction, whereas the metal bullet simply comes to a stop in the log. If we say that both have an initial momentum of 1.0 kg-m/s and the rubber bullet has a momentum of -.5 kgm/s after bouncing off the log whereas the metal bullet has a momentum of 0 kg-m/s when it comes to a stop in the log. The change in momentum of each is;

Δp _{rubber bullet} = 10 kg-m/s - (-5 kg-m/s) = 1.5 kg-m/s

Δp _{metal bullet} = 10 kg-m/s - 0 kg-m/s = 1.0 kg-m/s

You can see that the Δp is larger in the case of the rubber bullet which means that the force is larger and thus it is more likely to knock over the log.

Basically it takes more force to turn something completely around than it does to stop it. That's why Ray Lewis is a great linebacker and Antonio Pierce is just average ;)

Thank you, guys, for your inputs, so the equation of force is based on momentum kgm/s rather than kinetic energy kg(m/s)^2.

Can we now safely say that a rubber bullet has more force than a metal bullet fired with the same velocity. (both having same mass). But what makes them different?

Neither bullet has more force than the other, the force is just distributed differently. And I don't totally agree with the wording of the solution, that the rubber bullet applies more force to the wood, it's just that some of the force from the lead bullet is destructive (penetrating the wood), and some deflective, where almost all of the force from the rubber bullet is deflective (likely only leaving a dent). It's like if I poke you repeatedly with a sharpened pencil, with equal force, the effect can be quite different, depending on which end I poke you with!

You wrote: "Neither bullet has more force than the other"

This is wrong for two reasons. The first is that the bullets don't "have force", they apply a force. What the bullets have is kinetic energy. The second reason your statement is incorrect is because the rubber bullet applies more force to the log (and vice-versa) than the metal bullet. Please notice in the equation;

F=Δp/Δt

That the magnitude of a force is given by the change in momentum over the change in time. Since the rubber bullet reverses direction, it's change in momentum is larger. Not only that, because the metal bullet is slowed to a stop by the friction of the wood, it's change in momentum takes longer than the change for the rubber bullet whic just bounces backwards off of it.

So Δp is larger and Δt is smaller for the rubber bullet, thus it applies a larger force to the wood.

I gathered your explanation in your previous entry. And "has more force" was a direct response to the last statement in #3. I still have a problem with some of this. First off, your "let's say" grants 5 units of reverse momentum to the 10 units of forward momentum, which strikes me (pun intended) as ridiculously disproportionate. .05 units maybe. Secondly, I question whether there is a gain in total force, or simply a conversion, as in 10 -.05 rather than 10 - (-5) as you put it, and that the force applied by the rubber bullet is in fact less. The difference is, as I proposed earlier, is what that force results in. I would concede that the rubber bullet applies more force to the surface of the wood, hence more readily knocking the wood over, but that the actual force applied is greater with the lead bullet, as displayed by the destruction, and that no conserved energy was converted to the rebound. I also think the belief that the lead bullet takes longer to come to a stop vs. the rubber bullet takes to bounce may well be an erroneous assumption.

Here's the problem. It's hard to counter "beliefs". All I can do is provide the physical formulas that tells you what it will do. For instance, I have the formula:

F=Δp/Δt

where Δp=p_initial - p_final (by definition)

and p=mv

where v is velocity which returns in the opposite direction will have a negative sign which will make the rebounding momentum negative, which when you put that in the formula;

Δp=p_initial - p_final

Will give you a larger change in momentum rather than if it just stopped. This isn't an interpretation, its a fact.

You wrote: "First off, your "let's say" grants 5 units of reverse momentum to the 10 units of forward momentum, which strikes me (pun intended) as ridiculously disproportionate. .05 units maybe"

Then how about this, get a gun that shoots rubber bullets, stand five feet from a concrete wall and fire straight ahead. I'm sure you'll have nothing to worry about since the rubber bullets will (according to you) come off the wall at 1/20th the speed. Personally I'd advise against it since I think they may come off the wall faster than you believe. After all, riot police have been taught to fire rubber bullets at the ground in order for the rubber bullet ricochet to apply nonlethal pain deterrent to combatants.

http://en.wikipedia.org/wiki/Rubber_bullet

Hmmm Yes I can see what you're getting at now Roger.

Of course as the metal bullet is just stopped by the target the energy is all used up...

But with the rubber bullet it is not only used up but extra energy is given (imparted) to the bullet to enable it to reverse direction and still have an opposite force...

Thus the rubber bullet must exert more force to the target...

I guess the 'trick' is realising that the force still available in the rubber bullet is in the reverse direction and so must have been obtained from the target...

I like it! Tricky one... John.

Thank you again , gentlemen, so Roger is right and Munky is wrong. Thats what I thought too; so too is the EDITOR of the Globalspec.

Actually, my question is " why we have a MORE TOTAL FORCE in a rubber bullet fired than a lead bullet fired ?" afterall, they both have the same Kinetic energy and momentum initially. what makes them have different effects or results afterwards?

Hellooooo! is the EDITOR IN THE HOUSE!

You wrote: "Actually, my question is " why we have a MORE TOTAL FORCE in a rubber bullet fired than a lead bullet fired ?"

The question states that the rubber bullet and the lead bullet are the same mass. I think part of the problem here is people are failing to visualize what this means. Lead is typically 10x more dense than rubber. This means that in order for the bullets to be the same mass, the rubber bullet must be much larger in size. If the bullets were the same size, naturally the lead bullet would exert more force, this is because in that case the lead bullet would have more mass.

I've read with interest the debate on kinetic energy here with great interest, but as is usual with an engineering issue, the most important fact is neglected. Bullets are designed to kill people. Baton rounds (rubber bullets) are supposed not to kill people, but they do. The difference between the two is the PR of the agency using them. Any 'police force' that's fires a baton round into a crowd of people is prepared to kill. Don't believe otherwise. After all, the proof has just been shown on this thread!

Agreed Plbmk (?) Trouble is that most engineering is associated with weapons or the ability to cause harm to others...

I worked for BAe for only 6 months to see the futility of it all... my group would design ever more sophisticated armour piercing missiles for tank busting only for the tank designers to design their armour to resist our best efforts and so we had to redesign our missiles to .... etc... etc...

A never ending circle, quite pointless for everyday use and such an expenditure of tax payers money!!

I'm not a pacifist, but I do try to limit my expertise to more useful applications...

John.

I wouldn't call myself a pacifist either mate, but I would like a basic honesty of purpose. As to the arms race, I can't help thinking of the sabre tooth tiger. Bigger and more impressive prey = bigger more impressive predator = bigger and more impressive prey = bigger more impressive predator. Then a small shift in climate = no prey OR predator. I wouldn't like to draw too much of a parallel……but……

Let's go back to the " hows and whys " of academic physics.

My answer to the " MORE TOTAL FORCE " phenomenon is as follows:

1. Because of the inherent property of the substances,i.e., their relative densities:

Lead > wood > rubber.

2. The k.e. of the lead bullet is all used up in penetrating the wood. Action is linear.

3. The k.e. of the rubber bullet is absorbed by the whole piece of wood without pene-

trating it; a part of the k.e. ends up as a compression energy-- the surface of

the wood serves as the stop point where the bullet is compressed; wherein the

loose molecular spaces are compacted.

4. An instant later, this compressed energy is released as recoil force " kicking the

wood " and ricocheting.

5. The " RELEASING OF THE COMPRESSED ENERGY " is the source of the added total

force. This is missing in the lead bullet.

Correct me if I am wrong in this interpretation.

P.S. The reason why I started this discussion is because I have a project that deals

with modulus of elasticity of substances.

I feel your interpretation is correct i.e some of the kinetic energy carried by the bullet instead of being converted to heat in the wood and bullet as is the case with the metal bullet is stored up in the compression of the rubber bullet and is recycled to accelerate the wood forward and the bullet back.

I think most of the energy lost by the lead bullet comes from breaking the bonds in the wood. Before the bullet there is only solid wood, after the bullet there is a hole. This is because the bonds holding the wood together where the hole was were ripped forcibly apart.

Picture trying trying to tear a piece of wood in half. It would take a lot of energy, but it could be done. That's what the bullet does, on a smaller scale, when it imbeds itself in the wood.

Anyway, I think we are all on the same page with the other stuff.