Energy loss in a collision

[kamenges]

I am having trouble reconciling conservation of momentum and conservation of energy. If two bodies of equal mass, one stationary and one moving, collide and both bodies exhibit perfect elastic deformation at the contact point (infering no loss of energy due to plastic deformation of damping) momentum must be conserved. So the sum of the mass/velocity products of the two bodies after the collision must equal the mass/velocity product of the initially moving body alone. But if both bodies are moving after the collision then the total energy of the system must be less than the energy of the initial moving body alone. So assuming no losses due to deformation, where does the energy go?

Thaks in advance for any help.

 

[GregLocock]

WHat makes you think both bodies will be moving after the collision? Ever played pool?

Cheers

Greg Locock

 

[kamenges]

Hi Greg-
Granted, I'm a lousy pool player. But I've had more cue balls follow targets into the pockets than stop on the table. I think the mechanism that stops the cue ball is tangential motion at the bottom of the cue ball opposing the direction of travel (back english). This removes energy from the ball and, as a friction component, can be modelled and accounted for.

 

[prex]

What Greg probably meant is that in a centered collision of perfect spherical bodies the first one stops, of course.
However in an uncentered collision both bodies will be moving after.
The answer to your question is simple (though too much time passed from my courses in mechanics...): both the momentum and the energy are conserved in frictionless and perfectly elastic collision. prex

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[jabonet]

well, besides the conservation of energy, you also have conservation of momentum, MV, that means that if the system is elastic, and has no loss of energy, the center of gravity before and after will be moving at the same speed, this happens ever if the impact is not centered.
In your case, the center of gravity is moving at half the speed of the ball, and when it hits the second one, the ball will stop and the second one will move at the same speed, thus leaving the CG moving at a contant speed. and making the total energy of the two ball system constant.
In pool tables, you also have a spin motion of the ball, that is not always trasfered to the other ball, so the impact is not plastic, and there is no sure way to know what happens.
This looks like a Homework problem, let me remember you that this forum is not for that... just in case.

 

[GregLocock]

Sorry, yes I was being one-dimensional. If you go to two dimensions then of course KE and mv are conserved, even if both balls are moving, because the lateral velocity of the two balls, and so their momentum in that direction, will be of opposite signs.

But it does look like a homework problem. Cheers

Greg Locock

 

[brengine]

kamenges,
I don't know why I'm working out your homework problem, but I am concerned with your statement:
"But if both bodies are moving after the collision then the total energy of the system must be less than the energy of the initial moving body alone. So assuming no losses due to deformation, where does the energy go?"
-->Just because both bodies are moving does NOT mean energy is lost. Could they possibly be moving a different exiting speeds, different than the incoming ball? How would this affect things?
-->Did you actually plug some numbers in too see what would happen?

Ken

 

[Biggadike]

I don't think anyone has mentioned this so far so forgive me if I'm repeating...

If you take the combined centre of gravity for as many balls as you have in the system, its motion vector remains unchanged before and after the collision. It is this vector which quantifies the kinetic energy of the sytem. (not allowing for friction and other losses)

This takes a bit of getting your head around but it does explain a lot I think.

 

[MASSEY]

just say it's due to entropy and take a 75

 

[vtl]

The energy is lost do to frictions. Surfaces friction + aerodynamics friction.

 

[kamenges]

My, my. Someone tosses that dirty 'h'-word out there and people get all bent out of shape.
Just to quell some fears, I'm not a student and this is not a homework problem. I'm the result of a very 'focused' (read 'limited') formal education. I'm a tech scool product so I haven't had any college physics. I haven't needed to look into impact dynamics for work yet, so this is just curiosity.
I saw the folly in my base statement about shared final velocity when I remembered the multiple ball pedellum office toy (ball hits row to cause far ball to move). So at the risk of being further chastised for being something I'm not in pursuit of something that is not what it appears, I'll ask what is really the source of my confusion. What is the energy transfer mechanism in the collision? More correctly, how can a body that is approaching zero velocity (the originally moving body) continue to transfer energy to the body it hit right up until the originally moving body stops completely? Why don't they just separate when they both reach the same velocity?

Thanks,
Keith

 

[MASSEY]

(mAvA) + (mBvB) = (mAv'A) + (mBv'B)

and in your case: e = 1 so that
(v'B-v'A)/(vA-vB)

if it is frictionless and perfectly elastic (as nothing really is) then energy and momentum are both conserved.

 

[GregLocock]

Yeah, we get bent at homework posters because it is specifically forbidden.

But you raise a great question, now.", how can a body that is approaching zero velocity (the originally moving body) continue to transfer energy to the body it hit right up until the originally moving body stops completely? Why don't they just separate when they both reach the same velocity?"

It's a bit like Zeno's paradox.

My guess is that the elastic deformation of the point of contact in either, or both, bodies allows local velocities in excess of the CofGs, so that one body can push the other even though it is 'slower'. Perhaps one way of analysing this would be to say that the potential energy stored in the spring rate of the bodies during the collision gets turned back into KE later on.

ie

KE1 at t= -1

= KE1+KE2+PE1+PE2 during collision

then we have two separate systems

PE1+KE1 at the moment of separation
= KE1 after the PE in 1 has dissipated back into KE

and
PE2+KE2 at the moment of separation
= KE2 after the PE in 2 has dissipated back into KE

One way of examining this would be to simulate it using a spring coupling a large mass to a small mass to represent each body, and the impact site on each body.



Cheers

Greg Locock

 

[vtl]

another loss that would likely involved are the elastic strain energy.

 

[prex]

I think that a simple answer to the question raised by kamenges is that the mechanism of transferring the momentum and the kinetic energy from one body to the other requires as a physical condition that part of the kinetic energy is transferred to elastic energy during the contact phase.
This comes from the fact that momentum must be conserved even during the contact: so, restricting ourselves to the simple case of two equal masses with one at rest and centered collision, during the contact one must have v_i1=v₁+v₂ and it is easy to check that this is incompatible with a conservation of kinetic energy that would require v_i1²=v₁²+v₂².
So some transfer of energy into an elastic deformation is necessary, and this deformation (just like in a squeezed rubber ball) explains why the forces exherted by the first body onto the other one are not strictly related to their relative velocity.
Now to go into more detail, it would be necessary to write down some equations... prex

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[austim]

Biggadike, your argument surprises me considerably.

I read your post to suggest that the KE that we are interested in is 0.5*total mass*mean velocity^2.

What if you consider the collision to be between two identical cars approaching each other at identical speeds? (Any sensitive automotive engineers should read something else while I continue this ;-)).

The CG of this system is stationary, and therefore your argument would have us believe that the KE of the system is zero. Where then does all the strain energy come from when they impact?

Surely the "kinetic energy of the system" = 0.5*sum(MV^2), which will rarely be the same as 0.5*sum(M)*Vmean^2.

Thus in my simple example, Vmean is zero whatever the V for each car may be.

Even for one car striking a second stationary car, the total KE in the system is 0.5*MV^2, but consideration of the movement of the CG would give you only half as much [0.5*(2*M)*(V/2)^2=0.25*MV^2].

 

[Speedy]

If you consider the real effect of the compressive forces at collision, i.e. acceleration, the rate change of velocity and not velocity itselt, then it becomes clearer.

Collision consists of 2 stages, compression and de-compression. Compression will occur up to the point where both bodies have equal velocity. During both stages though, the forces always act in the same direction. This would explain why the impacting body will continue to transfer momentum after the point where both bodies have equal velocity. Tell a man there are 300 billion stars in the universe and he'll believe you. Tell him a bench has wet paint on it and he'll have to touch to be sure.

 

[kamenges]

Thanks for the responses, everyone. Elastic compression is something I should have intuitively considered since I made it an initial requirement. But I can certainly see how this would allow for full energy/momentum transfer. I'm still trying to get my mind around the non-centered collision case since both bodies will have an ending velocity not equal to zero. But I suspect I am missing a signing convention or getting a vector addition wrong or something like that.

Thanks again for the info.
Keith