Help: Where has the Kinetic Energy Gone?

I'm looking to fill some apparent gap in knowledge I have. I may have a bit of difficulty in properly conveying my issue, so I hope a discussion can clean it up. Looking at perfectly inelastic collisions, I am somewhat taken back by the fact that kinetic energy is not conserved? Looking at a simple model. Two bodies, one of mass $m$ and the other $M$ are bound for collision with velocities $v$ , and 0, respectively. after the collision they join together and continue with velocity $v'$

From conservation of momentum we have: $mv + 0 = (m+M)v'$ , which means that $v' = \frac {m}{( m+M)}v$

At this point lets look at the kinetic energy before and after the collision.

Before Collision: $KE_{before} =\frac {1}{2}mv^2$

After Collision: $KE_{after} = \frac {1}{2}(m+M){v'}^2 = \frac {1}{2}(m+M)(\frac {m}{( m+M)}v)^2 = \frac {1}{2}\frac {m^2}{( m+M)}v^2$

Then taking the ratio of after to before $\frac {KE_{after}}{KE_{before}} = \frac {m}{( m+M)} < 1$ , meaning $KE_{before} > KE_{after}$

So is someone going to tell me the quantity of energy $\frac {1}{2}mv^2 - \frac {1}{2}\frac {m^2}{( m+M)}v^2$ has been lost as heat ( some seemingly divine intervention of the Second Law of Thermodynamics )? I wasn't aware that I was operating under the pretext of entropy in this idealized case? We know absolutely nothing of the internal collision mechaincs that would generate said heat and potential energies of various forms, yet some how this example dictates a quantity of transformed kinetic energy in this collision? For instance, In Newtons Laws ( and even conservation of energy - as far as the models are concerned ) you can simply "turn off" entropy, and the mechanics will yield a limiting case. But here, the limiting case of conserved kinetic energy seems to be impossible to show. So what is going on here?

Any help greatly appreciated.

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Comments

I think your interpretation is correct. The "lost" kinetic energy is transferred to heat, sound, vibrational modes, etc. It does seem perplexingly convenient, though. Similarly, if you drop a ball from a certain height, you can be confident that ultimately, the change in gravitational potential energy has resulted in an equivalent amount of heat energy added to the local environment. It would be interesting to try to devise a calorimeter machine to confirm these notions empirically.

Steven Chase - 3 years, 2 months ago

Thank you for your response Steven. I see where I was getting mixed up with this concept. I hadn't realized that I was actively putting constraints on the kinetic energy by joining the masses. The simplest of concepts elude my grasp sometimes... I think what was going through my head was something like: calculating a block sliding down a ramp ignoring friction, only to be told by the math at the end that a certain amount of friction was present. I've obviously never encountered that, so it threw me off.

"Similarly, if you drop a ball from a certain height, you can be confident that ultimately, the change in gravitational potential energy has resulted in an equivalent amount of heat energy added to the local environment."

I hadn't actually ever given that concept much thought, thanks for pointing it out!

Eric Roberts - 3 years, 2 months ago

I would say that much of the lost energy is converted into work causing deformation of the objects. In the case of an ideal spring, at the point when two objects are moving at the same speed (instant of minimum kinetic energy), the WORK has resulted in maximum spring compression (a deformation - which we interpret as elastic potential energy). In this case, the spring is able to return that energy (again through doing work on the objects) and they regain the original amount - an elastic collision. In most cases, though, the objects will deform (requiring work), but do not return that energy as they are now “permanently” deformed. I think of the damage done in an automobile crash. Heat, sound, etc. play a role, but I believe in most macroscopic cases, the deformation is where most of the kinetic that was “lost” will be “found”.

Scott Wiley - 10 months, 1 week ago

Thanks for the reply Scott. This is what I have realized since asking. My gripe was ( as Steven Chase noted above ) the convenience of it seemed perplexing.

Eric Roberts - 10 months, 1 week ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
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`2^{34}`	$2^{34}$
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`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$