Impact force of Rigid objects and no displacement

  • #1
I was wondering something, I dropped a box onto my glass desk but it didn't bounce back, I also hit a hammer against a rigid wall, the matter smacked into the wall, it has KE. It never bounced back, I didn't see any deformation or rebound in either situation, I wasn't hitting too hard nor dropped the box to hard.

However, the KE of the hammer will be turned into force, now the wall didn't move nor did the hammer cause deformation nor penetration, it didn't rebound so how do I calculate the impact force?
KE is converted to force but no displacement ore rebound, it must exert a force but how do I calculate it?

If I fire a bullet at tank armor, it absorbs the impact, you wont feel it on the other side(the impact force)...right?

How would I calculate impact force on non-penetrating projectiles? the wall must have absorbed all the force and the hammer didn't rebound...but how can we calculate the force?
 

Answers and Replies

  • #2
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Try to use the concept of momentum change and impulse to your situation.

Edit: This needs to be posted in the correct section.
 
Last edited:
  • #3
Try to use the concept of momentum change and impulse to your situation.

Edit: This needs to be posted in the correct section.
I say, the KE converts to force, thus force is applied for a time. How do I know the impact time? what about material absorption?
Is their a situation where a bullet with 500Joules(assuming impulse lasts long enough, like .451 Seconds) that hits a target, it doesn't penetrate but the bullet exerts 500 Newtons and displaces the target by 1 meter?
 
  • #4
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Is their a situation where a bullet with 500Joules(assuming impulse lasts long enough, like .451 Seconds) that hits a target, it doesn't penetrate but the bullet exerts 500 Newtons and displaces the target by 1 meter?
This situation does seem to be possible (theoretically at least), you can also find out the initial velocity and mass of the bullet from the given data. But whether it can be actually done, I'm uncertain. You would have to take into account the elasticity/rigidity of the target.
 
  • #5
This situation does seem to be possible (theoretically at least), you can also find out the initial velocity and mass of the bullet from the given data. But whether it can be actually done, I'm uncertain. You would have to take into account the elasticity/rigidity of the target.
Okay, so like a bullet of 500Joules hitting a target, it absorbs the impact, lets assume its made of that material capable of absorbing that impact, so it doesn't rebound it exerts a force on the block, its SHOULD turn 500Joules, it will push that 70kg target with 500 Newtons of force, it will displace it by one meter, transferring energy.

the time is ∆t=.529150 seconds....so doing all the equation I don't want to write here...i got....
d=1 meter
Vf=3.779644m/s
Object of 70kg's.....KE: 500 Joules
 
  • #6
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Okay, so like a bullet of 500Joules hitting a target, it absorbs the impact, lets assume its made of that material capable of absorbing that impact, so it doesn't rebound it exerts a force on the block, its SHOULD turn 500Joules, it will push that 70kg target with 500 Newtons of force, it will displace it by one meter, transferring energy.

the time is ∆t=.529150 seconds....so doing all the equation I don't want to write here...i got....
d=1 meter
Vf=3.779644m/s
Object of 70kg's.....KE: 500 Joules
I'm sorry, I don't quite understand what you've done here. You want to find the force on the bullet (required to stop it) right? Then use the impulse equation delta(p)*delta(t)= F
You don't need to involve the concept of kinetic energy at all ( you can, but that'll be unnecessary).
 
  • #7
I'm sorry, I don't quite understand what you've done here. You want to find the force on the bullet (required to stop it) right? Then use the impulse equation delta(p)*delta(t)= F
You don't need to involve the concept of kinetic energy at all ( you can, but that'll be unnecessary).
not the force to stop the bullet but the force a bullet exerts on target....but i've gotten my answer.
 
  • #8
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not the force to stop the bullet but the force a bullet exerts on target....but i've gotten my answer.
That's the same magnitude in the opposite direction.
 
  • #9
That's the same magnitude in the opposite direction.
Indeed, Newtons 3rd law, which with the KE turned into force, the force may be able to do work on material.
 
  • #10
Drakkith
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Indeed, Newtons 3rd law, which with the KE turned into force, the force may be able to do work on material.
Force and kinetic energy aren't directly related to each other in this manner. You don't even need to know the kinetic energy of an object to figure out the force applied during a collision. Indeed, it doesn't even help you in many cases since it isn't a conserved quantity (total energy is, but not kinetic energy). Also note that neither KE nor force are turned into one another. That's perhaps a bit nitpicky with the terminology on my part, but these things tend to be important when trying to understand physics.

Okay, so like a bullet of 500Joules hitting a target, it absorbs the impact, lets assume its made of that material capable of absorbing that impact, so it doesn't rebound it exerts a force on the block, its SHOULD turn 500Joules, it will push that 70kg target with 500 Newtons of force, it will displace it by one meter, transferring energy.
As I explained in your other thread, this isn't the correct way to find the force exerted on the target or bullet. Not to mention the fact that your calculations here also violate conservation of momentum just like the one in your other thread. If the answer you've gotten is based on this method, then it is wrong and will not help you in any way.
 
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  • #11
Force and kinetic energy aren't directly related to each other in this manner. You don't even need to know the kinetic energy of an object to figure out the force applied during a collision. Indeed, it doesn't even help you in many cases since it isn't a conserved quantity (total energy is, but not kinetic energy). Also note that neither KE nor force are turned into one another. That's perhaps a bit nitpicky with the terminology on my part, but these things tend to be important when trying to understand physics.



As I explained in your other thread, this isn't the correct way to find the force exerted on the target or bullet. Not to mention the fact that your calculations here also violate conservation of momentum just like the one in your other thread. If the answer you've gotten is based on this method, then it is wrong and will not help you in any way.

Well, KE will exert a force that much is true but I was wondering for what situation a object could exert 500N's without penetrating
 
  • #12
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Well, KE will exert a force that much is true
What we're saying is that this sentence doesn't make much sense. KE doesn't exert force, they are different concepts. You're trying to make orange juice from apples, in a manner of saying.
The idea of KE is not helpful in this case. It's not even conserved. The KE of the bullet gets converted to other forms of energy such as heat and sound, NOT to a force.
 
  • #13
What we're saying is that this sentence doesn't make much sense. KE doesn't exert force, they are different concepts. You're trying to make orange juice from apples, in a manner of saying.
The idea of KE is not helpful in this case. It's not even conserved. The KE of the bullet gets converted to other forms of energy such as heat and sound, NOT to a force.
I know that, the conservation of energy....and No, KE and force are not apples and oranges...but they aren't similar->object with KE when slowing down is exerting a force, thus work.
 
  • #14
Drakkith
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I know that, the conservation of energy....and No, KE and force are not apples and oranges...but they aren't similar->object with KE when slowing down is exerting a force, thus work.
What you've said here is true, but it is not very useful in understanding force, KE, and work. If you want to gain a better understanding of this whole situation you're better off just forgetting about the kinetic energy of the projectile for the moment and looking at it in terms of momentum. Since momentum is conserved, it places constraints on what can happen to the projectile and target.

Well, KE will exert a force that much is true but I was wondering for what situation a object could exert 500N's without penetrating
If we're talking about bullets, then it's not possible. A bullet exerts far more than 500 newtons of force because the collision time is a tiny fraction of a second. You literally cannot change this without drastically changing your setup.
 
  • #15
What you've said here is true, but it is not very useful in understanding force, KE, and work. If you want to gain a better understanding of this whole situation you're better off just forgetting about the kinetic energy of the projectile for the moment and looking at it in terms of momentum. Since momentum is conserved, it places constraints on what can happen to the projectile and target.



If we're talking about bullets, then it's not possible. A bullet exerts far more than 500 newtons of force because the collision time is a tiny fraction of a second. You literally cannot change this without drastically changing your setup.
That is very true, this by impact it would be greater then 500N's....

The original question is being lost but...

How does a hammer work then by energy conservation laws? I used to know all of this but now Im confused and I just cant seem to understand anything anymore

The hammer converts KE to TME, like how would that work?
 
  • #16
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That is very true, this by impact it would be greater then 500N's....

The original question is being lost but...

How does a hammer work then by energy conservation laws? I used to know all of this but now Im confused and I just cant seem to understand anything anymore

The hammer converts KE to TME, like how would that work?
The initial impact of hammer with nail is inelastic. However, all of the kinetic energy is in the moving hammer and as long as the nail is much less massive than the hammer, the resulting reduction in total kinetic energy is small.

The subsequent driving of the nail into the wood is a classic example of work = force times distance. The kinetic energy of the hammer head is slowly transferred to the nail over the distance that the nail is driven into the wood. This energy transfer is given by the force of the hammer head on the nail times the distance moved by the nail. Of course, this transferred energy is lost almost immediately by the nail as the frictional force of wood on nail does very nearly equal and opposite work on the nail.

The delta between the force of hammer on nail head and the force of wood on nail shaft is just enough to bring the nail to a halt, deeper in the wood than it started.
 
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  • #17
The initial impact of hammer with nail is inelastic. However, all of the kinetic energy is in the moving hammer and as long as the nail is much less massive than the hammer, the resulting reduction in total kinetic energy is small.

The subsequent driving of the nail into the wood is a classic example of work = force times distance. The kinetic energy of the hammer head is slowly transferred to the nail over the distance that the nail is driven into the wood. This energy transfer is given by the force of the hammer head on the nail times the distance moved by the nail. Of course, this transferred energy is lost almost immediately by the nail as the frictional force of wood on nail does very nearly equal and opposite work on the nail.

The delta between the force of hammer on nail head and the force of wood on nail shaft is just enough to bring the nail to a halt, deeper in the wood than it started.
So, YES....the KE of the hammer is turned into MECHANICAL ENERGY...THUS EXERTING A FORCE, THAT IS ONLY WAY IT CAN DO WORK, that is the only way to displace a nail.....because if NOT that is a violation of Newtons 2nd Law

My point being, if a bullet deforms a little but doesn't penetrate...so it presses against the block, like when I apply my hand....Im asking, if the bullet managed to stay on the surface of the object, like if I were to apply my hand, wouldn't the force allow it do work?
 
  • #18
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My point being, if a bullet deforms a little but doesn't penetrate...so it presses against the block, like when I apply my hand....Im asking, if the bullet managed to stay on the surface of the object, like if I were to apply my hand, wouldn't the force allow it do work?
No.
 
  • #20
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That makes no sense, that is a violation.
No. It is not.

Work is computed as the product of force multiplied by distance moved [in the direction of the force] by the object to which the force is applied. If the surface does not move, no work is done on it.
 
  • #21
No. It is not.

Work is computed as the product of force multiplied by distance moved by the object to which the force is applied. If the surface does not move, no work is done on it.
yes, but if we calculate IMPULSE or impact force truly, then we can use dynamics, as we can gather a NET force
 
  • #22
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yes, but if we calculate IMPULSE or impact force truly, then we can use dynamics, as we can gather a NET force
That is just word salad.
 
  • #23
That is just word salad.
ugh, okay, so we know somethings absorb force, like when I hold a hammer really tight, it barely rebounds....

okay, just...think of it like how a hammer pounds in a nail or a hand pushes a box....
 
  • #24
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ugh, okay, so we know somethings absorb force, like when I hold a hammer really tight, it barely rebounds....
Right. Because your hand is absorbing the recoil -- both momentum and energy.

okay, just...think of it like how a hammer pounds in a nail or a hand pushes a box....
When a hammer pounds in a nail, the nail moves. When a hand pushes a box across the floor, the box moves. When a bullet strikes a surface without making a mark, the surface does not move. No movement = no work done.

That does not mean that there is no force. It typically means that there will be a LOT of force. But it does mean that the force does no work -- transfers no energy to the surface.
 
  • #25
Right. Because your hand is absorbing the recoil -- both momentum and energy.


When a hammer pounds in a nail, the nail moves. When a hand pushes a box across the floor, the box moves. When a bullet strikes a surface without making a mark, the surface does not move. No movement = no work done.

That does not mean that there is no force. It typically means that there will be a LOT of force. But it does mean that the force does no work -- transfers no energy to the surface.
exactly

Im asking, if a bullet deformed a little bit but managed to stick....but it falls down after it has used up all of its energy, ifs slowing down its using its energy, thus exerting force W=Fs
 

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