25
$\begingroup$

I teach grade 12 physics and am about to introduce collisions. I am explaining that in elastic collisions, kinetic energy is conserved and in inelastic collisions, kinetic energy is not conserved. The kinetic energy in inelastic collisions can be lost to heat, sound and electromagnetic radiation.

As I think about it though, do heat, sound and light not also have kinetic energy? Heat is movement of particles which is kinetic energy. Sound is movement of air particles which is also kinetic energy. Lastly, if I am not mistaken, light (EMR) has a relativistic kinetic energy.

If you were to track every molecule and photon's kinetic energy after a collision, would kinetic energy be conserved in all collisions?

The only thing that makes me think it wouldn't would be somehow if the energy was converted into some form of potential energy. Like a collision where one object causes another object to increase in height therefore increasing the gravitational potential energy of the object.

Edit: Note, I am not teaching my students anything deeper than the basic macroscopic conception of collisions. While I was prepping, it just got me thinking about this.

$\endgroup$
6
  • 1
    $\begingroup$ Kinetic energy in this context refers to the energy inherent to the motion of the objects and does not refer to "energy but not potential energy". $\endgroup$
    – DKNguyen
    Sep 6 at 4:24
  • 2
    $\begingroup$ Just a practical note for something that might trip you up on a test problem. In the world described by introductory physics, an increase in potential energy - which occurs when a field does negative work on an object - would count as kinetic energy for the purposes of determining the elasticity of a collision. Since the field is not part of the colliding object, work done by the field on the colliding objects must have come at the expense of the objects' kinetic energy. Consider a car crash on a slope vs the exact same car crash on level ground. $\endgroup$
    – g s
    Sep 7 at 5:23
  • 1
    $\begingroup$ If you include photon-electron "collisions" you can change electron potentials/orbits which I'd contend cannot reasonably be considered kinetic energy. The sum of "mass" and "energy" will still be constant but now we have long left the explanatory realm of "kinetic energy" ;-). $\endgroup$ Sep 7 at 11:05
  • $\begingroup$ @g s would a more accurate classical description then be "mechanical energy is always conserved in elastic collisions."? Mechanical energy being defined as the potential + kinetic energy of a closed system. $\endgroup$
    – Morphyl
    Sep 7 at 12:37
  • $\begingroup$ @Morphyl I think you just need to be very careful to identify what counts as the object doing the colliding. For a toy example: suppose you have two idealized weights on a spring, and you bounce an idealized ball off of one of the weights. If the system is the ball and the impacted weight, the collision is elastic: the conserved kinetic energy of the system compresses the spring after the collision. If the system is the ball and the object composed of two weights and a spring, the collision is inelastic: the spring was compressed during the collision, so kinetic energy wasn't conserved. $\endgroup$
    – g s
    Sep 7 at 15:49

11 Answers 11

15
$\begingroup$

The definition of an elastic collision between two bodies is one for which the kinetic of energy of the two bodies remains the same after the collision (see https://en.wikipedia.org/wiki/Elastic_collision). A collision can only be perfectly elastic if there is no heat, sound, or light (or anything else) generated as a byproduct of the collision -- any energy devoted into these forms must be subtracted from the total incoming kinetic energy, i.e. the total kinetic energy of the two bodies is not conserved. While light and sound do have kinetic energy, they carry away some fraction of the initial system KE, so the collision is not elastic by definition.

As I see it, the answer to the question "Are all collisions really elastic?", given the definition of elastic, is really no. Not only that, but there is really no example of a perfectly elastic collision on the everyday scale of matter. Even cases where the objects do not touch, for example a fly-by scattering between two stars, generates heat from gravitational tides and would thus be inelastic.

$\endgroup$
5
  • 33
    $\begingroup$ I think this misses OP's point, though, that atomic collisions have no other means to dissipate energy as "heat" since that concept doesn't even apply to single atoms. If an atom bombards a solid and scatters, it deposits some of its kinetic energy into the solid at the point of the collision as kinetic energy. That kinetic energy is confined in the solid as phonons, but each individual interaction is perfectly conserving energy, so the lattice vibrations (ie: heat) spreads out as these "elastic" collisions reverberate through the solid. I think OP is more correct than this answer admits. $\endgroup$
    – J...
    Sep 6 at 16:21
  • 2
    $\begingroup$ That is a fair criticism of my answer. The usual context at the level of OP's physics class would be treating colliding bodies as point masses, so in that sense the translational kinetic energy of the two center of masses is not conserved, even if the total KE is conserved via vibrations in the solids' lattices, etc. Though I suppose that's not the precise definition of "elastic". $\endgroup$
    – zhutchens1
    Sep 6 at 20:23
  • 4
    $\begingroup$ No, and phonons do shed energy as electromagnetic radiation, so there is a sort of "friction", even within the solid, so we really can't say that system is properly elastic in that sense either, but the individual interactions are quantized so, at the lowest level, any individual collision will either lose energy to a photon, etc, or it will not. $\endgroup$
    – J...
    Sep 6 at 20:45
  • 1
    $\begingroup$ @J... : I have deuteron, the fusion remnant of two hydrogen atoms, here... It claims you have oversimplified with "have no other means to dissipate energy", ignoring dissipation by ejecting positrons and (anti-)neutrinos and also ignoring changes in binding energy. $\endgroup$ Sep 7 at 3:14
  • 1
    $\begingroup$ @EricTowers Naturally. The unspoken objective here was to bring OP's oversimplified view together with this answer in a way that demonstrated how close to a reasonable approximation OP's conjecture was. It's trivial to design an experiment that avoids such high energies (and therefore interactions) so we don't need to consider them when we're exploring the idea of whether or not perfectly elastic interactions are possible at all, even in some limited sense. $\endgroup$
    – J...
    Sep 7 at 11:39
11
$\begingroup$

Heat
The difference between the collisions of macroscopic and microscopic objects is that the concept of heat is not applicable to the latter. Indeed, thermodynamic concepts work only for large collections of particles, containing of the order of $N_A\approx 10^{23}$ particles.

Collisions at microscale (compound objects)
So a more refined distinction between the elastic and inelastic collisions, is that in the latter some kinetic energy may be converted into the internal energy of the colliding particles. To give a few examples:

  • Atom absorbs a photon, so that an electron in atom moves to a higher energy state, i.e., the energy of the atom is changed, while photon ceases to exist. Note that in this case the internal energy is partially potential and partially kinetic, but it is internal to the atom, which we treat as a whole.
  • Raman scattering is similar to the example above, but the photon is re-emitted with a lower energy. This is perhaps more similar to the idea of collisions in elementary mechanics.
  • Nuclear systems exhibit similar behavior: a nuclei may absorb a neutron or an alpha particle (or deflect it absorbing only part of its energy), while becoming excited, i.e., increasing its internal energy.

Collisions of elementary particles
In all these examples we deal with objects having internal structure, such as atoms or nuclei. In case of elementary particles the collisions are elastic, even though some particles may cease to to exist and others may appear in their place. Again, this is not completely true, since even the elementary particles, such as protons and electrons, do have inner structure in terms of quarks and could probably be excited to higher energy states.

Energy conservation
As pointed in other answers, whether collision is elastic or inelastic, it does not change the fact that the energy is conserved. However, for macroscopic objects the heat and the kinetic/potential energy of the object as a whole are sharply distinct, whereas for microscopic objects this distinction is less obvious - which is probably the main premise of this question.

        

$\endgroup$
5
  • 1
    $\begingroup$ +1, but FYI protons and neutrons have internal structure. Each contains quarks. Electrons and quarks do not. They are elementary. $\endgroup$
    – mmesser314
    Sep 6 at 15:52
  • $\begingroup$ @mmesser314 Thanks. I did mention the quarks - however, I do not know, if they have been observed in excited states - this is not my field. $\endgroup$ Sep 6 at 16:25
  • 4
    $\begingroup$ Isn't a mole of particles rather a large amount to start seeing thermodynamic concepts? For example, a mole of iron would be about 56 g of material, which is definitely not a microscopic amount. $\endgroup$ Sep 6 at 17:53
  • $\begingroup$ @AndrewMorton indeed, we even know how the error scales with the number of particles. There is no sharp transition between macroscopic and microscopic - what we really talk about are the limiting cases. $\endgroup$ Sep 6 at 19:38
  • 1
6
$\begingroup$

There is actually no such thing as an elastic collision at the atomic (or even subatomic) scale, but that is not necessarily for the reason one might think. The reason is that, when collisions are mediated by electromagnetic forces (as most collisions, from the everyday scale down to the molecular scale), the number of quanta involved never remains constant. Any collision or scattering process in which electromagnetism is involved will entail the emission of an enormous number of very, very soft (that is, low frequency) photons. These carry off very little energy, but they are huge in number, limited only by the macroscopic situation in which a microscopic collision might be set.

In low-energy collisions, these electromagnetic excitations go essentially undetected, but they draw a small amount of energy off the initial particles. In nominally two-body collision, the two principal bodies will have slightly less combined energy after the collision than before, because of the profuse emission of very long-wavelength photons. However, at shorter wavelengths, these photons can be seen experimentally. As collisions become more energetic, this soft photon emission spectrum connects up smoothly to the more energetic bremsstrahlung photons that are emitted in fast collisions.

$\endgroup$
1
  • $\begingroup$ An interesting possibility is to regard these photons as bodies, just like the colliding bodies, and regard their energy as kinetic energy (as OP mentioned). With this choice, all energy after the collision is kinetic energy, so the collision is elastic. (Assuming no change of internal energy of the bodies). Thus, the question of elastic vs non-elastic depends on what one, by choice, calls a body. $\endgroup$
    – Menno
    Sep 9 at 7:00
4
$\begingroup$

Can It Wiggle?

When computing the dynamics of two colliding bodies, this is a good question to ask. For every molecule, the answer is: yes. That's because the bond lengths of the molecule can vary, making them atomic-scale springs.

As far as I know, nucleons cannot "wiggle" (but I am not a physicist, so take that with a grain of salt). So if you bounce two H atoms together, I believe you should see an "elastic collision". But as soon as you upgrade to H2, you now have molecules that can wiggle, and some of the collision energy can go into making the atoms of a single H2 molecule bounce like a spring. Therefore, such collisions will appear inelastic.

Can It Spin?

If you can orient the colliders, then they may also have angular momentum, so some of the collision energy may be converted to spin, rather than whole-body momentum.

Can It Deform?

If a collider can permanently change its shape, then some of the collision energy may go into reconfiguring its internal bonds. You generally need bigger molecules to observe plastic deformation, but I doubt you need more than 100. Chemists could probably tell us the smallest deformable molecule (I'm guessing less than a dozen atoms of a typical metal could be bent into all sorts of shapes).

Conclusion

Of course, some of these modes are just a few components of "heat" which we may think of as part of the "internal energy" of a body. Deformation addresses the fixed structure of a body. In general, elasticity starts with the presumption that the colliding bodies are immutable: they cannot change their internal state, so everything interesting about the collision can be observed externally, as properties of the entire bodies. This assumption fails as soon as the collision products have multiple internal states.

$\endgroup$
7
  • $\begingroup$ I think this is an interesting track of investigation even if I'm not sure how much of it is true. (Also not a physicist.) Let's say the molecule starts wiggling. How long will it wiggle? Macroscopic springs are dampened as wiggling turns into heat. So our intuition is that wiggling is a way to absorb energy. But the same is not obvious on the molecular level. Perhaps Buzz's answer is the key. $\endgroup$ Sep 8 at 12:45
  • $\begingroup$ "nucleons cannot "wiggle"" - oh they can. Given that basically everything can "wiggle" kickstarted the string theory trying to unify the explanations under this "wiggling". $\endgroup$
    – Lodinn
    Sep 8 at 14:15
  • $\begingroup$ @Lodinn interesting! Can atomic collisions transfer energy to nucleon vibration states? $\endgroup$ Sep 8 at 19:06
  • $\begingroup$ @DanielDarabos well, the wiggling is a component of heat, so I think it will continue to wiggle until it reaches 0 K. If the molecule has neighbors, it can transfer some of its wiggling to them, but if its isolated, then I presume it can only cool via blackbody radiation. So the answer is: "Not forever", but possibly: "A very long time". $\endgroup$ Sep 8 at 19:09
  • $\begingroup$ @Lodinn just to be clear, by "nucleon wiggling", do you mean that protons/neutrons can move relative to each other in a nucleus (which I buy), or that a single free proton can vibrate due to the motions of its constituent quarks? Because the latter is what I doubt. $\endgroup$ Sep 8 at 19:31
4
$\begingroup$

The law of conservation of energy is absolute. In a sense, if nature could be successfully modeled with classical mechanics and classical electrodynamics, at the pedantic level, you would have to include potential energy , as you note, in order to fulfill the law. Classically there is only potential and kinetic energy. In the simplest inelastic collision between two particles, the potential energy in the deformations would not allow to talk about conservation of just kinetic energy.

It gets worse at the quantum level, or at the level of large velocities close to the speed of light , where mass is no longer conserved.To get conservation of energy one has to use models with Lorentz invariance and four vectors.

$\endgroup$
5
  • $\begingroup$ Mass isn't conserved even at non-quantum but relativistic level. $\endgroup$
    – Ruslan
    Sep 6 at 20:23
  • $\begingroup$ @Ruslan: At the relativistic level, energy is not conserved either. $\endgroup$ Sep 7 at 3:30
  • $\begingroup$ @Ruslan, true. I kept at the atomic level. Maybe I should edit $\endgroup$
    – anna v
    Sep 7 at 4:01
  • $\begingroup$ Really nice answer. "close to the speed of light , where mass is no longer conserved" Can you please elaborate on this? $\endgroup$ Sep 7 at 4:29
  • $\begingroup$ @ÁrpádSzendrei it is the four vector link, close to the speed of light mass is the "length" of the four vector, and depends on the four vector algebra of the specific interaction. $\endgroup$
    – anna v
    Sep 7 at 5:53
2
$\begingroup$

The problem is essentially one of definition.

At a macro level, the concept of kinetic (vs. say potential) energy is easy. As scale reduces, the definitions used in a classroom become harder and harder to maintain without propping up the definition by new clauses/criteria, redefining it, or abandoning it altogether.

This isnt entirely uncommon. Other macro properties and terms (pressure, volume, particle, colour, position, collision, elastic) also just fail us when scale reduces enough.

Essentially, when we cant average realistically, or take a higher level view of behaviour, the concepts that work at higher level dissolve.

The terms we use so freely, may simply not be applicable "as written and usually used" at those smaller scales.

So I would explain like this:

As scale shrinks, the concepts of "kinetic energy" as well as "collision" and "elastic" simply arent well enough defined to be helpful. And thats okay, they werent created to be.

Its the same as if you go to the doctor, and he wants to know if you have grown in the year (as a child), he doesnt want to be bothered with the height and angle of each hair, or whether one cell is higher than last time. (Pick your own better analogy!) He wants to know a big picture thing of it.

Similarly, yes we can try to probe each micro outcome, and match it to our higher level concepts. That'll maybe help for a while. But even those fail, as you scale down.

Ultimately the answer is.. .

...you cant answer this question, because you'd have to first rigorously define your terms. What counts as kinetic energy? Elastic? Collision? And make these valid at all scales you're asking the question.

Do that and fine. Mostly, we dont bother.

$\endgroup$
1
$\begingroup$

For sound wave, a half of the wave is kinetic energy, and the other half is elastic potential energy. At each fixed position, the kinetic energy and potential energy are constantly converted from one to another. The potential energy is higher in the dense part of the sound wave, and lower in the dilute part.

For heat, it also composed of kinetic and potential energy, unless the system is simple as ideal gas composed of non-interacting mono-atom. For complex system. like solid or liquid, the heat is the oscillation of constituent lattice points, energy in form of potential and kinetic.

For electron magnetic wave, there are no masses, therefore there are no kinetic energy, but a field energy (energy density) in terms electric and magnetic field intensities.

$\endgroup$
3
  • 1
    $\begingroup$ Light has kinetic energy (identical to its total energy $h\nu$) and redshifts as a result of momentum exchanged when scattering. Note that $\gamma m_0 = 0/0$ for $v=c$ and we can therefore not use typical kinetic energy calculation. Note also that $E = T+m_0c^2 = T$ for $m_0 = 0$ $\endgroup$
    – g s
    Sep 6 at 4:34
  • $\begingroup$ It is not called "kinetic energy". $\endgroup$
    – ytlu
    Sep 6 at 4:51
  • $\begingroup$ @gs There were radiation energy density before the appearance of plank constant, and before the establish of special relativity. $\endgroup$
    – ytlu
    Sep 6 at 5:21
1
$\begingroup$

What can happen to two molecules (or atoms, or smaller particles) colliding?

  1. They may separate internally unchanged, having just exchanged some momentum. This is called an elastic or adiabatic interaction.

  2. One or two of them may get excited in rotational or vibrational mode or a constituent particle (e.g. electron in a molecule) may change its state to a higher-energy one, leaving the initial particles with less kinetic energy. (the two variants are not really different when looked closer)

  3. The reverse may also happen - an excitation of one of the particles to be traded for kinetic energy.

  4. A constituent particle (an electron, an atom, a radical or some other particle) can get completely unbound and separate from the one of the initial particles.

  5. A constituent particle can be exchanged between the initial particles - e.g. two neutral molecules may become two ions when one electron moves from one to the other.

  6. An "interaction particle" (e.g. photon) can be radiated.

  7. Particles can fuse together, leaving a single, new, probably excited particle

Did I miss something?

... or any of the above 2..7 combined by whatever means.

$\endgroup$
1
$\begingroup$

In an educational context, I think it's worth highlighting the concept of the domain of applicability of scientific models. Human intuition very often likes to operate via analogies to observed phenomena at the 'human scale' (i.e. macroscopic objects that we 'directly' perceive with our senses and have some evolved intuitive understanding of how they should behave [ignoring for the moment that sometimes these intuitions are quite wrong.]) When this intuitive sense is combined with a more rigorous mathematical modeling of these physical behaviors, it can then also be tempting to extend these same models via analogy to other domains.

This use of analogy can be fruitful in some cases, but it can also cause people to make bad predictions. When we say that a given mathematical model is valid, it's because we have experimental evidence that supports the model. In the case of classical mechanics, this evidence involves human-scale macroscopic objects. Early models of microscopic interactions attempted to apply these mechanics to the microscale (such as the Bohr model of the atom), but we've gathered enough evidence to be confident that classical mechanical models aren't a very good fit for the behaviors that frequently occur at that scale.

I think one of the most important things to learn in early scientific education is that we have models that can be used to make useful predictions, but that these models also have limitations and constraints. This domain of applicability is in turn defined by the accumulated experimental evidence showing both when the models work and when they do not. In my opinion, cultivating this mindset is an important part of developing scientific literacy, as it highlights the fact that it's not about our intuitive sense of what's going on, but rather the models we've built, the evidence we've collected and, in the context of sciences like Physics, the math underlying them.

$\endgroup$
0
$\begingroup$

While it is true that much of the KE lost by two bodies colliding inelastically is converted into KE of individual particles, not all of it is. More importantly, I recommend that you make a clear distinction between the macroscopic KE of the colliding bodies and the KE of the particles that comprise them and the environment with which they interact, otherwise you will raise more questions than you answer. If, for example, you conflate the translational KE of a football with the molecular KE of the air molecules within it, you will make it harder, not easier for students to figure out what is happening.

$\endgroup$
0
$\begingroup$

An inelastic microscopic collision involves the excitation of the internal degrees of freedom in one or both of the objects involved. My go to example is that the collision between two molecules can excite the vibrational modes of the molecules involved. Thus after the collision the net kinetic energy of the two molecules is lower than before they collided.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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