Kinetic Energy in Inelastic Collision

[pluspolaritons]

Hi, first post on the forum!

We know that kinetic energy is not conserved in an inelastic collision, and how much of the kinetic energy is lost we can find it out through applying conservation of momentum.

But one thing that I don't understand is how does the system know exactly how much kinetic energy to lose? I understand that the kinetic energy is lost to heat, the internal energy of the objects that collided etc. But I find it strange that the before and after collision kinetic energy is not the same, and 'magically' a chunk of kinetic energy has been taken out from the system without accounting for where the kinetic energy goes directly.

It's like A and B with mass M_a and M_b collide, then there must be a total kinetic energy E_1 lost in this system. How the energy is distributed to internal potential energy, heat, sound etc does not matter as long as it adds up to E_1. Isn't that peculiar? Shouldn't it be that how the energy is lost to internal potential energy, heat etc be different from system to system?

I hope you understand what I'm asking here.

 

[Simon Bridge]

how does the system know exactly how much kinetic energy to lose?

Through the exact process of the collision. The "magic" is because the before-and-after calculations you are doing at this stage hide a lot of the things that are happening "during" the collision.
How much energy goes into making the sound of the objects banging together, for example, is determined by the surfaces of the objects colliding, how they deform (and, maybe, break) and the nature of the fluid (i.e. air) surrounding them. It would be a bit much to ask students to go around calculating all these.

But it is no different from your knee "magically" knowing how much to hurt when you skin it.

 

[A.T.]

We know that kinetic energy is not conserved in an inelastic collision, and how much of the kinetic energy is lost we can find it out through applying conservation of momentum.

But one thing that I don't understand is how does the system know exactly how much kinetic energy to lose?

You could just as well ask: How does a system "know" that it has to conserve momentum or total energy?

 

[Chestermiller]

Hi, first post on the forum!

We know that kinetic energy is not conserved in an inelastic collision, and how much of the kinetic energy is lost we can find it out through applying conservation of momentum.


I hope you understand what I'm asking here.

You can't find it out through conservation of momentum. You need to specify what is happening in advance. If the two bodies stick together, then that is one case, and corresponds to a certain amount of KE loss. If the two bodies collide elastically, then the amount of KE loss is zero. Both of these are equivalent to specifying the amount of KE loss in advance.

Chet

 

[sophiecentaur]

The first stab at this is to use the concept of 'coefficient of restitution', which for students, is usually given. This is how they introduce collisions in school Physics. It's the ratio of rebound velocities to approach velocities but, of course, it depends upon the details of the two objects involved - shape, complex modulus and the hysteresis during deformation. It can be found experimentally and then applied to many situations involving the same two bodies, for instance.
Coefficient of restitution allows you to produce a pair of simulatneous equations - one describing the momentum situations and the other relating the velocities. You can then solve for the KE before and after. It's not a complete answer but may be enough to satisfy you initially.

 

[pluspolaritons]

Through the exact process of the collision. The "magic" is because the before-and-after calculations you are doing at this stage hide a lot of the things that are happening "during" the collision.
How much energy goes into making the sound of the objects banging together, for example, is determined by the surfaces of the objects colliding, how they deform (and, maybe, break) and the nature of the fluid (i.e. air) surrounding them. It would be a bit much to ask students to go around calculating all these.

But it is no different from your knee "magically" knowing how much to hurt when you skin it.

I agree we can find out more of how the kinetic energy is distributed if more conditions were given such as the shape and sizes of the objects etc.

But the knee analogy does not seem quite right. How much the knee hurts really depend on how you cut it, the angle you cut it etc. But in the collision case, how much kinetic energy is lost is completely determined when the parameters mass and velocity are specified.

 

[Simon Bridge]

But the knee analogy does not seem quite right. How much the knee hurts really depend on how you cut it, the angle you cut it etc. But in the collision case, how much kinetic energy is lost is completely determined when the parameters mass and velocity are specified.

1. all analogies have limits, if they didn't, then it wouldn;t be an analogy - it would be the thing itself.
2. there is more to the energy transfer in a collision than the mass and velocty, otherwise a head-on collision of 1kg of lead and 1kg stuffed elephant would behave the same as 1kg of steel hitting 1kg of stone, head on, same speeds.
... there is the mass distribution, geometry, density, and material properties all come to play. Is it a glancing blow that puts some spin on? Is the collision in a fluid medium that creates turbulence and carries energy off as sound and heat? Collision can be very complicated.

The point of the analogy is the bit about magic.

 

[alphali]

1. all analogies have limits, if they didn't, then it wouldn;t be an analogy - it would be the thing itself.
2. there is more to the energy transfer in a collision than the mass and velocty, otherwise a head-on collision of 1kg of lead and 1kg stuffed elephant would behave the same as 1kg of steel hitting 1kg of stone, head on, same speeds.
... there is the mass distribution, geometry, density, and material properties all come to play. Is it a glancing blow that puts some spin on? Is the collision in a fluid medium that creates turbulence and carries energy off as sound and heat? Collision can be very complicated.

The point of the analogy is the bit about magic.

So because we can't include all the parameters the equations we use are not fully accurate?

 

[Simon Bridge]

No - it means that in order to get accurate results we need to have some way to account for our lack of knowledge of the fine details of any interaction. Same with anything in life.

In the case of inelastic collisions, we need to account for kinetic energy that gets "lost" as something else. This can be done very accurately by including other parameters, like the total momentum. i.e. you make more measurements.

The Universe knows where the energy goes and how it gets there... so the collisions take care of themselves.

I believe the original questions have been answered.

 

[nasu]

But in the collision case, how much kinetic energy is lost is completely determined when the parameters mass and velocity are specified.

No, this not true.
I suppose this may be the source of your confusion.
It was already said that the amount of kinetic energy "lost" depends on the specifics of the materials and shapes of the bodies.
If you only know masses and and velocities you cannot determine the outcome of the collision.

 

[Chestermiller]

No, this not true.
I suppose this may be the source of your confusion.
It was already said that the amount of kinetic energy "lost" depends on the specifics of the materials and shapes of the bodies.
If you only know masses and and velocities you cannot determine the outcome of the collision.

I would like to strongly second this observation. It is possible to model the detailed deformations and mechanics during a collision, if you are able to describe the mechanical constitutive behavior of the materials, and are prepared to solve sets of partial differential equations describing the time- and spatial variations in the displacements, stresses, and strains in the colliding bodies. If you have described the mechanical constitutive behavior of the materials correctly, then the constitutive equations will be able to describe yield strain and viscoelasticity. These effects are what give rise to the conversion of mechanical energy to thermal energy (i.e., the loss of kinetic energy).

Chet

 

[Khashishi]

On a microscopic scale, there are no inelastic collisions. So, an inelastic collision is simply repartitioning the kinetic energies of the atoms into other places. Each microscopic collision conserves energy, so the whole big collision must conserve energy.

 

[nasu]

On a microscopic scale, there are no inelastic collisions. So, an inelastic collision is simply repartitioning the kinetic energies of the atoms into other places. Each microscopic collision conserves energy, so the whole big collision must conserve energy.

Yes, but not necessarily the macroscopic kinetic energy.

 

[sophiecentaur]

On a microscopic scale, there are no inelastic collisions. So, an inelastic collision is simply repartitioning the kinetic energies of the atoms into other places. Each microscopic collision conserves energy, so the whole big collision must conserve energy.

Yes, but not necessarily the macroscopic kinetic energy.

These two posts say it all, almost - except that it's "never".
And, of course, there can be collisions in which the KE transfers to 'permanent' internal PE in a molecule or between molecules, if they change configuration or stick together.

 

[e.chaniotakis]

dear Khashishi, nasu , we have had some similar discussion in this tread about the kinetic energy-momentum relationship in the inelastic collisions https://www.physicsforums.com/showthread.php?t=670810