Let's assume a setup with a static linear molecule with three identical atoms connected by bonds and a single atom, identical to the other three, being shot at the molecule. Let's also assume that everything happens in 1D and can therefore be illustrated by this simple scheme:

(projectile) A ->    A-A-A (molecule)

Where the direction of the arrow is the direction of motion of the projectile and the dash signs in the molecule are the bonds between atoms.

Assuming a low-energy, non-relativistic, perfectly elastic collision, the projectile will stop dead in its track while the molecule will be set in motion at 1/3rd of the projectile's speed, as it has 3 times the projectile's mass (conservation of Momentum). Balancing also the equations for the Kinetic Energy easily reveals that 2/3rds of projectile's kinetic energy must have been locked away in internal vibrations of the molecule.

Now, it's easy for us to calculate these numbers by considering the projectile and the molecule as just two separate systems and conveniently attaching a total mass to each. But in reality the universe doesn't group atoms into molecules, does it? Each bond in the setup doesn't "know" about the existence of the other bond and each atom could be said to be unaware of all but the atoms bonded or colliding with it.

In this context, not "knowing" total masses, how does the universe "calculates" that in this case the molecule must store 2/3rds of the projectile kinetic energy into internal vibrations, no more, no less?

  • $\begingroup$ Welcome to physics.SE! Material like salutations, etc., are discouraged in questions on SE, so I've deleted that material from your question. $\endgroup$
    – user4552
    Aug 17 '13 at 23:17

If I'm understanding the question correctly, then it really doesn't have to be about individual atoms. You could replace the word "atom" with the word "billiard ball" everywhere in the question, and nothing fundamental would change. The fundamental issue being raised is that interactions propagate at a finite speed, so the ball being hit can't instantly respond in a way that implies that it "knows" the existence of the other 9 balls. In a stack of balls, the speed at which the interaction propagates is limited by the speed of sound, and may be much lower if there are gaps between the balls. So really this is just a paradox about the standard "Newton's cradle."

The resolution of the paradox is that the ball being hit may very well respond by flying off at speed $s$. This is exactly what it does if there are gaps between the balls.

However, conservation of momentum is observed at every moment in the process, and therefore the final result of the collision must obey conservation of momentum.

Given that this last atom is attached to the previous one, the momentum should return backward all the way to the first atom and then to the static projectile which would then restart in the opposite direction with its initial speed.

It would violate conservation of momentum if the last atom were to reverse its direction of motion.

Thinking in terms of classical mechanics it's easy: we are dealing with a mass 1m with speed s heading toward a static mass 10m. With a perfectly elastic collision the outcome is a static Projectile and a Rod moving at 1/10th of the Projectile's speed.

This is not physically possible, and it's not what you'll observe in a Newton's cradle, which is constructed so as to be nearly elastic. The scenario you describe would violate conservation of energy if the collisions were all elastic.

  • $\begingroup$ Ben, thanks for your reply. Allow me to address some of the issues you mention. With your feedback I might be able to further improve/clarify the question. Billiard balls is something I definitely had in mind when thinking about the issue, but it's not a perfect analogy because billiard balls cannot normally be linked together in a single object. The problem as I posed it involves "a static object consisting of 10 identical atoms arranged in a line." In this context I should perhaps edit my question to make it explicit that we are dealing with a 10-atom molecule, the atoms linked by bonds. $\endgroup$
    – manu3d
    Aug 18 '13 at 8:26
  • $\begingroup$ Just modified the question to make it explicit we are dealing with a molecule of 10 atoms bonded in a linear chain. $\endgroup$
    – manu3d
    Aug 18 '13 at 8:39
  • $\begingroup$ Ben, I've drastically re-edited the question to be more concise, to the point, and also in light of some new knowledge. You might want to edit your answer to be consistent with the changes! $\endgroup$
    – manu3d
    Aug 20 '13 at 12:49

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