2
$\begingroup$

I’m going over a chapter on linear momentum in my physics course right now and am somewhat puzzled with what happens with some of the kinetic energy that is lost in a perfectly inelastic collision.

Imagine a world without sound, heat, or any non-mechanical forms of energy. Now imagine that there are two perfectly square blocks, M1 and M2, in empty space that each have a mass of 1 kilogram. M1 flies toward M2 in a perfectly straight line with a velocity of 1 meters per second. M1 sticks to M2, creating M3, and the new 2 kilogram rectangular block moves with a velocity of 1/2 meters per second. In the collision, 1/4 J of energy was lost.

What happened to that energy, given that no sound or heat was emitted? Does it requires a certain amount of energy to form a single 2 kilogram object out of two 1 kilogram objects?

EDIT:

I was able to figure things out thanks in great part to the posts below. In case this bothers someone else in the future, the way that I think about it is, it requires energy to slow down M1 and speed up M2 to the same velocity (you can imagine M3 as two separate particles). Fundamentally, the energy is lost in that speeding up/slowing down process in a world without friction, heat, etc.

$\endgroup$
0

5 Answers 5

3
$\begingroup$

Your question contains a contradiction. You imagine a world without any forms of non-mechanical energy- in such a world, inelastic collisions could not exist. The whole point of inelastic collisions is that mechanical KE is lost to non-mechanical forms of energy.

$\endgroup$
2
$\begingroup$

Even in a vacuum, the inelastic deformation would result in an increase in the temperatures of the blocks. An increase in temperature means there is an increase in the kinetic energies (KE) of the atoms and molecules of the blocks, i.e., an increase in KE at the microscopic level.

In effect, the macroscopic KE of the blocks associated with their overall motion that was "lost" has simply been converted to microscopic KE.

In so far a heat is concerned, there are three mechanisms: Conduction, convection, and radiation. The first two don't apply to your example (blocks in a vacuum). But there is still can be heat transfer by electromagnetic radiation since that requires no medium to transfer (occurs in a vacuum).

Hope this helps

$\endgroup$
0
1
$\begingroup$

If you have a perfectly inelastic collision, you either are working in a system which does not have conservation of energy, or you have another way of tracking the energy which is "lost" in the collision. If you have neither, then you do indeed have a contradiction.

We have a similar problem in electronics. You can develop idealized circuits which have short-circuits in them. These circuits behave incredibly poorly, and many an entrance-level electrical engineer has asked what happens in these circuits. And we have to go into the details of all of the things we skipped in the model - capacitances and impedances which one usually get to ignore.

At some point they ask "well, what happens in the simplified model?" And the answer is "nothing." The simplified model simply cannot describe any real physics in that particular corner case. And that's okay. We work with such models all the time.

For a real trip, I'll point out a nasty trick you'll learn in the future. When working in the framework of Einstein's relativity, rigid bodies do not exist. If you try to think through how they work, you arrive at all sorts of nasty paradoxes.

As for your example, its worth noting that, when you get far enough into physics, you'll find out that heat is not actually anything special. What it ends up being is a measure of the random element of the motion of particles, jostling around inside the object. When we have an inelastic collision that "turns energy into heat," what we really mean is that the starting objects had nice easy to track energy related to the velocity of the object -- the center of mass of the object. When the collision occurs, some of that energy goes into terribly hard to predict quivvverings of the material which is much harder to track. Instead of trying to track it all with velocities, we choose to lump it together, and measure it with "heat."

$\endgroup$
1
  • $\begingroup$ ”When working in the framework of Einstein's relativity, rigid bodies do not exist.” At nonzero temperatures, even purely elastic bodies don’t exist! Everything is viscous to some degree. +1 $\endgroup$ Commented Nov 12, 2021 at 1:40
0
$\begingroup$

In an inelastic collision, the internal energy of the colliding masses increases. Energy is conserved, but kinetic energy is not conserved; you have to account for the increase in internal energy. For example, the colliding masses can change shape and increase in temperature.

Sometimes this confuses those studying basic physics mechanics, since much of mechanics assumes rigid bodies for which the internal energy cannot change. For example, the mechanics definition of work as equal to the change in kinetic energy assumes no change in internal energy. Where there are significant changes in internal energy, the first law of thermodynamics is needed. See my answer to Positive and negative Work, question on this exchange.

$\endgroup$
0
$\begingroup$
  • The objects might deform. Energy is then trapped in the stresses and strains of the deformed material.
  • There might be glue between the objects. Energy is then trapped chemically in the bonds created that adhere the surfaces together.
  • There might be velcro or similar micro-mechanical adhesion. Energi is then trapped in the rearrangement, stretching and compression of peaks and asparities that interlock.

In a realistic scenario, energy will also always be transformed into thermal energy via heat generation, increasing the temperature slightly and possibly increasing the temperature of the surrounding air.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.