How is energy conserved when I throw a heavy metal ball at a piece of glass and cause it to shatter? The ball starts off with significant kinetic energy and ends up with very little. The glass sees what I think can be a small change in potential energy, depending on how you position things, but otherwise, I still see humongous energy losses. Is it heat energy during the process of shattering? Or am I neglecting something else...

EDIT: according to a comment on one of the answers, I decided to clarify/simplify. I said metal ball because it's not likely a metal ball will absorb much of the energy through a deformation.


1 Answer 1


Chemical bonds should definitely be taken into account.

Let us make an order-of-magnitude estimate: We take a square glass panel with length $L = 1\,\mathrm{m}$ and width $e= 1\, \mathrm{cm}$. We will compute the energy required to break it in two, considering only the chemical bond energy.

What we need is the number of broken bonds and the energy of each bond. We can consider that each molecule occupies a sphere with a radius $$r \simeq 1 \,\mathrm{\dot{A}} = 10^{-10} \,\mathrm{m}.$$

The bond energy should be $$E_{bond} \simeq 1\,\mathrm{eV} = 1.6 10^{-19}\,\mathrm{J}.$$

So to break the panel, one has to break bonds. Their number can be computed by assuming that with a straight separation, we have broken the bonds on a surface $S_\mathrm{glass}= Le$. Each molecule occupies a surface $S_\mathrm{molec} \simeq r^2$, so the number of broken molecules is $$N \simeq \frac{Le}{r^2}.$$

Finally, $$E_\mathrm{broken} \simeq N E_\mathrm{bond} \simeq \frac{eL}{r^2}N.$$

We reach $$E_\mathrm{broken} \simeq \frac {10^{-2}}{10^{-20}} 1.610^{-19} \simeq 0.1 \,\mathrm{J}. $$

This might seem small. However, one never breaks the glass in one straight separation but in a spiderweb shape. The total crack length must be around $10L$. So $$E_\mathrm{breaking} \simeq 1 \,\mathrm{J},$$ which is the equivalent of a 100 g ball travelling at 40 km/h.

  • $\begingroup$ Disagree. Most balls would deformate much more than glass. So ball would soak up most of energy(force is the same, but path is longer). Energy soaked up by glass would be used up mostly for process described by you, but all energy from ball would just dissipate to heat and sound waves(minor part). $\endgroup$
    – Vashu
    Commented May 27, 2018 at 5:04
  • $\begingroup$ For sure. I didn't claim that it was the only phenomenon but that it had to be taken into account. It would be interesting to have an estimate of the heat loss during the process. I think that for an iron ball it should be quite small. However for a soccer-like ball I don't know $\endgroup$
    – Cabirto
    Commented May 27, 2018 at 9:46
  • $\begingroup$ Vashu I'm editing the question to allow the ignoring of the energy absorbed by the ball's deformation. It will make calculations satisfactory and simpler. $\endgroup$
    – user196799
    Commented Jul 24, 2018 at 10:18

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