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A Physics of bullet impacts, I wonder if anyone can help?

  1. Feb 27, 2016 #1
    If I have a bullet with 700 Joules KE, the mass is 0.008036m, velocity 417.39m/s, momentum p=3.35kg-m/s
    Kinetic energy is the ability to do work, Kinetic energy converts into FORCE...so yes KE will exert a force...which will do work, as to do work requires force and distance. Thus, when the bullet is slowing down, its Kinetic energy is being turn into force....to have exert force you must use energy, when energy is being exerted it is doing work, by exerted said force, again KE is the ability to do work. Thus, net force accelerates things. The bullet is slowing down as the target is exerting force on the bullet but by 3rd law of Newton the bullet is exerting force on the target, the expended KE reduces velocity, but the bullet transfers energy thus transfers FORCE. (Now, force by 3rd law on less rigid material will deform the bullet, thus energy is expended with the bullet breaking apart of being deformed....transferring energy, exerting force on target as well).

    The bullet is made of a material where it absorbs impacts can can continue to exert a force, it doesn't penetrate a 80kg object on a frictionless surface, I did the calculations and......The 700Joules doesn't penetrate the 80kg object but allow the bullet to exert a force of 700N's on the box and it applies it for .47809144437 seconds......the 80kg box is displaced 1 meter....its final velocity is 4.183300132 m/s...its KE=700Joules.....that 80kg now has the ability to do work....

    Lets say the box is flat and we are using a .357 Sig with a flat nose....the boxes side is flat....how would I calculate the time is lasted....like somebody come up the Elasticity and density of our material...just find a numbers...find the equation and get it done...and I'll give you a gold star.

    After which, There will be more questions...so regarding this....
    Last edited: Feb 27, 2016
  2. jcsd
  3. Feb 27, 2016 #2


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    I don't think this is correct. The collision should last a LOT less time than half a second.

    Also, your post is very difficult to understand. I don't really even know what you're trying to say in the first paragraph.
  4. Feb 27, 2016 #3
    I know, we are ASSUMING.....the energy is conserved as it doesn't give it off in deformation and such...it lasts a bit longer, if it penetrated or deformed? the impulse lasts shorter...
  5. Feb 27, 2016 #4


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    Assuming what?

    I'm sorry but I can't understand you.
  6. Feb 27, 2016 #5
    The conservation of energy not in total effect, its energy is used up in converting into work...not deformation, penetration, etc.
  7. Feb 27, 2016 #6
    The bullet isn't deformed, the impulse should last longer.....
  8. Feb 28, 2016 #7


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    Okay, you're saying that the kinetic energy is solely used to perform work to move the target, is that correct?

    That doesn't sound right either. Typically when an object is deformed in a collision the deformation actually serves to increase the time of the collision and reduce the force. This is why automobiles are designed to crumple during a crash. The deformation of the car's bumper and frame increases the duration of the collision, which lessens the force of the collision, which increases occupant survival.
  9. Feb 28, 2016 #8

    Our material of our bullet is a strong material that isn't going to deform easily, it absorbs the impact and doesn't rebound, its our "magic" bullet...

    Second, yes our bullet is made of a very tough material impacting a tough material....so the Kinetic energy is being turned into force....like a hammer and a nail.
  10. Feb 28, 2016 #9


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    If you're trying to answer a physics question then inventing magic bullets which don't obey the laws of physics isn't going to help you. A perfectly rigid bullet can't absorb the impact, as that requires that it deform and absorb energy (that's literally what it means to absorb the impact).


    You've violated conservation of momentum here. The initial momentum of the system is 3.35 kg*m/s but the final momentum is 334.4 kg*m/s.

    Okay, so from your original post it looks like you're asking how to find how long the collision lasts between a bullet and a box? Is that right?
  11. Feb 28, 2016 #10

    okay, no,no you aren't understanding it. The bullet doesn't defy physics, that isn't the point....

    Think of it like a hammer, a hammer does work right? It doesn't shatter or truly rebound....our bullet is made of a durable material which will absorb the impact and won't shatter....like a hammer, our bullet isn't deforming..the bullet maybe is deforming...but not shattering...

    this is a inelastic collision, except our force is directed into work, our energy isn't just entirely taken up, okay? If my bullet impacts the target, its slowing down...its exerting a force, in a typical situation, a bullet penetrating, the impulse if VERY short....the box is moving, force times a distance, it requires the ability to do work something to bullet has by a virtue of motion called KINETIC ENERGY.
  12. Feb 28, 2016 #11


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    Hold on. A hammer absorbs part of the energy of the collision by deforming. A temporary deformation is known as an elastic deformation. The nail and the person swinging the hammer all undergo this kind of deformation too. Not to mention the energy absorbed by the wood as the nail penetrates it. The hammer doesn't rebound because this turns out to be very close to a perfectly inelastic collision. But notice that all of the kinetic energy of the hammer is now gone. The nail and wood are not moving.

    I know what kinetic energy and work are, you don't need to keep trying to explain them. It only serves to clutter up your posts and make them harder to read.

    The problem here is that this isn't an inelastic collision. In an inelastic collision the kinetic energy is turned into another form, usually by deforming the colliding objects, but the momentum is conserved. It isn't possible to transfer all of the kinetic energy from one object to the other in an inelastic collision. Note that this is by definition. There is no way to manipulate your example to be an inelastic collision where all of the KE is transferred to the 2nd object. Nor is there any way to transfer all of the kinetic energy to the 2nd object even in a perfectly elastic collision, as the two masses are not equal. To conserve both energy and momentum in an elastic collision, the less massive object MUST rebound off of the larger.
  13. Feb 28, 2016 #12
    however, you don't take into account KE internal and absorption of force.

    this is an inelastic collision, SO its KE->Force-->Fs->Mechanical ENERGY...
  14. Feb 28, 2016 #13


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    I've already explained that this isn't an inelastic collision. It can't be if your target moves away at 4.18 m/s and has all of the initial 700 joules of KE.

    This doesn't make any sense. Those terms aren't directly related to each other.

    Look, you can continue to assert things which aren't correct, or you can elaborate on what you actually want to know and maybe we can help you.
  15. Feb 28, 2016 #14

    They are related W=∆KE, KE is stored Mechanical energy....

    Second, it must be inelastic because the KE-> force -> work done -> Kinetic energy
    Favg as in Average impact force.

    again, you aren't accounting for KE-internal nor the Work-energy theorem...
  16. Feb 28, 2016 #15

    The linked paper shows a method to experimentally determine the force of a bullet impacting ballistic gelatin.

    Bullets can be made of hard materials (solid copper, brass) that have minimal deformation when impacting tissue or ballistic gelatin.

    However, as the target material becomes harder, the interaction time becomes shorter. At some point, when a bullet impacts at over 1000 feet per second, either the bullet or the target or both will likely experience significant deformation. I don't know of any bullet materials that can impact ceramic armor or hardened steel without significant deformation, and I've done experiments.

    See: http://www.dtic.mil/dtic/tr/fulltext/u2/a567525.pdf

    Even in cases where the armor stops the bullet, both the armor and the bullet experience significant deformation. The bullet is also significantly deformed when it penetrates the armor.
  17. Feb 28, 2016 #16
    When the bullet hits the block, it is not possible for the block to instantly accelerate to the bullet velocity. Either the bullet or the block or both must deform to preserve conservation of momentum and not increase the kinetic energy.
  18. Feb 28, 2016 #17
    Very much true, now it should be noted deformation is occurring, I guess that is a better way of putting this, by better much more accurate.

    Its more of a situation in which any possible scenario which a 700J bullet is capable of exerting 700Newtons and displacing it by a meter, though conservation of energy laws prevent this from happening.....but this is a scenario I just made which it could happen...even though we have no material to do so currently.

    although, Im a HUGE fan of your work and the BTG group....Im a huge fan, in fact...

    though, I was being stupid when I made this post, I should probably remove it to save myself from the embarrassment.

    However, I do know about impact forces and such, regarding Force Average = Ek/d, this I learned from the .30-06 document made by M Courtney a while back.

    again, Im a huge fan and its an honor for you to respond to my post, thank you.
  19. Feb 28, 2016 #18


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    No it can't. You've violated conservation of momentum in your example. There is literally nothing that could ever be made that would allow this to happen. Since this thread apparently has little to do with real physics, thread locked.
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