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What happens to kinetic energy?

Draconis_Aganata

Inactive Member
I am wondering something, what happens to kinetic energy, say from a bullet hitting a metal plate, does it convert into something else (i.e. heat) or does it just stop?
 
Note: bear with me as this isn't my speciality

From my understanding of kinetic energy and such related principles, the energy would neither stop nor be converted when it hit something very hard. The energy would instead be transmitted from the projectile into whatever it struck. Think about pushing a swing - your hand touches swing - and swing moves and it's the same here.

Which is exactly why I stated in the article that Wes put up that even when wearing a nano constructed breastplate I would not want to be shot by something moving at 1100mph+
 
Some energy would be converted into heat/light, but not all (the bullet could bounce off). Because the bullet is smaller, it would keep more kinetic energy than the plate did.

The amount would be dependant on the angle of impact, the bounciness of the bullet and the plate, etc.
 
A part of the bullet's energy would be transferred to the metal, not necessarily in a kinetic form. The metal could even recieve no kinetic energy at all (if it's too heavy and too hard to move).

You could add the weight of the bullet, the hardness and the shape (a round-shaped bullet has more chance of going splat on impact and lose a lot of it's kinetic energy than a pointy-shaped bullet.) of both objects in what Wes said.
 
What occurs on impact depends greatly on the speed and strength of both impactor and impactee. There are several different possible combinations.

=Relative velocity between the two objects are relatively small (only a few hundred mph) and the impactee is strong enough to prevent deformation. In this situation the projectile will plastically deform on impact and transmit its kinetic energy to the target almost entirely. Some will end up as thermal energy within the bullet (*due to friction in the deformation), but this will be a small fraction of the total energy.

=Relative velocity is high (1-2k mph) or the impacteee is not strong enough to prevent deformation. This will result in significantly more energy been transferred into heat due to friction in the deformation. At the extremes of this situation the impactor may have sufficient energy to cause a local failure of the impactee and penetrate, carrying part of its energy through to the target.

=At very high velocities (several thousand mph) a strong, dense impactor (such as the depleted uranium and tungsten kinetic-kill rounds used by today’s militaries) could completely overcome the yield strength of the impactor (such as what the above kinetic-kill rounds do) and punch right through with minimal lose of energy to the impactee. However, there is likely to be a huge amount of heat generated as the impactor travels through the armor. In current KK rounds it is this superheated (near plasma) gas that destroys the interior of a target and kills the personnel therein.
 
You could think about it on a particle level.
  • You've got a bunch of particles that are very strongly held together on one side, and a bunch of particles that are pretty strong and moving very fast on the other. When the fast-moving bullet hits the front of the breastplate or whatever, it pushes on the particles at the point of impact. Those particles (acquiring the momentum of the bullet) are forced backwards, but they are held in place by the particles around them - so the particles around them are forced backwards as well, and the ones around those, and so on. So you have a wave of force travelling all across the plate and either

    1) the bonds holding the molecules of armor together are stronger than the pressure caused by the bullet, and the entire plate is thrown back at a speed proportional to the speed of the bullet, or

    2) the bonds are weaker than the pressure and the bullet pierces through the armor, travelling through at a velocity reduced by an amount proportional to the strength of the bonds, while the plate is thrown back at a speed proportional to the loss of speed in the bullet.
The basic idea for this is given by the formula p=mv, and p1=p2; where p is momentum, m is mass, and v is velocity. So, if a bullet that weighs 50g strikes an extremely hard piece of armour that weighs 5kg, the piece of armour will be thrown back at a velocity ~1/100 of the bullet's (since it weighs 100x as much). If the armour is only hard enough to absorb half the bullet's momentum, the armour will be thrown back at a velocity 1/200 of the bullet, but the bullet will pierce through at 1/2 its original speed; if it absorbs 3/4, the bullet will go at 1/4 the speed and the armour is knocked at 3/400 (or 1/133) the speed. Note that a bullet going at 1/2 the speed will only pierce 1/4 as deep into a substance providing uniform resistance (such as flesh). A bullet going 1/4 the speed will puncture 1/16 as deep.

Addendum: Momentum is conserved, but you will probably notice some loss of kinetic energy if you do calculations. This is what turns into heat (the light is a by-product of the heat).
 
Scribbles said:
Kimura said:
Yangfan said:
Don't forget the case where the bullet bounces back. :)
When is that? >.>

I think its when the bonds are stronger and the material is elastic.
Ah, okay. Yeah, if the bonds are strong but elastic, then the bullet will bounce a bit, and the difference of the initial and final velocities of the bullet (it will be bigger, since the final should be negative) gives you the amount the person feels. It'll be felt more at the epicenter of the collision, though, since the elasticity means it will be temporarily deformed.
 
Newton guys... Newton is what you are looking for. Newtons 3 laws explain all of this.

1: An object at rest will remain at rest unless acted on by an unbalanced force. An object in motion continues in motion with the same speed and in the same direction unless acted upon by an unbalanced force.

So lets take an example of a target shooter shooting at a steel plate. They shoot at the steel plate, hit the plate, and the plate falls over. The bullet was the unbalanced force. It put the plate off balance as started it's motion.

Acceleration is produced when a force acts on a mass. The greater the mass (of the object being accelerated) the greater the amount of force needed (to accelerate the object).

This has everything to do with the actual firing of the bullet. Mass x Acceleration dictates Force. The heavier the bullet is, the more powder is needed to move it forward, but the faster it goes the most damage it does. This can be explained easily. A Single grain of sand moving at the speed of light would hit an object with the same Kenetic Energy as a Humvee travelling at about 200mph. The smaller object, however, would move so fast that whatever it hit would probably not even notice it happened, where they would definitely notice the truck :p


For every action there is an equal and opposite re-action.

By far the most famous of the three. This, with a combination of the first law describes bounce. If I shoot at a plate, and the plate has more resting energy than the bullet can transfer into the plate ((5.56 v 7.62 argument abound)) then the bullet will travel back in the same angle it hit at. Meet the luckiest man in the world

These are just basics though :p



 
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