Conservation of energy in the breaking of glass 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.
 A: 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.
