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I was thinking about why bullet can break molecular bonds when it hits solid target despite the fact that the kinetic energy of the individual atoms that make up the bullet is less that the molecule bond energy threshold of the target.

For example, lets say we fire 300 m/s graphite bullet at teflon target. Carbon fluorine bond is 5.5 eV and carbon atom moving at 300 m/s velocity has 0.005 eV kinetic energy yet the graphite bullet is able to break the teflon carbon fluorine bonds.

How exactly do many carbon atoms with low kinetic energy join their energies together and break those bonds? Is this a phonon thing? Do they create together phonon wave with higher than 5.5 eV energy?

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  • $\begingroup$ Why do you think that it is the kinetic energy that breaks the bonds? You can fracture a solid without giving it any kinetic energy by straining it. $\endgroup$ – lnmaurer Jul 17 at 4:12
  • $\begingroup$ What other energy can it be? The only energy bullet has is kinetic energy. Yes yes I know mass equals energy and if it isnt liquid helium cooled bullet then it has thermal energy but for all practical purposes, bullets do stuff becose of kinetic energy. $\endgroup$ – wav scientist Jul 18 at 0:50
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    $\begingroup$ The bullet-target system can have energy other than kinetic energy. I have posted an answer below. $\endgroup$ – lnmaurer Jul 18 at 4:55
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The bullet-target system only has kinetic energy --- until the bullet collides with the target.

Consider a system that is easier to visualize: instead of a bullet we have a baseball, and we throw it at a flexible rubber sheet. As the baseball collides with the sheet, the baseball causes the sheet to stretch. In the process, the baseball slows down. In other words, the collision converts kinetic energy in the baseball to strain energy in the sheet. If the strain in the sheet becomes to great, the sheet fractures (rips), and the strain energy is dissipated.

The same process happens when a bullet hits a target. In the collision, the kinetic energy of the bullet is converted to strain energy in the target (and in the bullet, for that matter). If the strain becomes large enough, molecular (and intermolecular) bonds break and the target fractures.

In other words, the kinetic energy doesn't directly cause bonds to break; strain from the collision causes the bonds to break.

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  • $\begingroup$ Thanks for clarification. That strain, its mediated by phonons, correct? $\endgroup$ – wav scientist Jul 19 at 17:14
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    $\begingroup$ I'm not sure what that question means. If something is strained, then its constituent atoms are, on average, further apart. There is then a restoring force from the bonds connecting the atoms. Chemical bonds are complicated, but I wouldn't say that phonons have much to do with it. $\endgroup$ – lnmaurer Jul 21 at 17:41
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This is getting into chemistry but anyways, the question is based on physics. enter image description here
You are thinking of intramolecular bond energy, which is about the bond between atoms of a molecule. Breaking these would mean breaking a compound into its elements. For example, breaking the bond you talked about, for water ($H_2O$), would result in separate hydrogen and oxygen gas, instead of splitting up water.

What you need to consider is intermolecular bonds - the bonds between different molecules. This is way weaker. This is what you break when splitting something like a block of wood or boiling some water away.

Search for intermolecular forces for more info.

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  • $\begingroup$ That is only true if you are cutting with the direction of wood grain, if you cut cross the grain, cross the cellulose fibers, then you are breaking these big molecules, these big polymers, and that isnt just van der waals bonds but covalent bonds too. Imagine shooting carbon fiber, you are breaking covalent bonds. Or imagine shooting block of metal, you will be breaking ionic bonds. $\endgroup$ – wav scientist Jul 16 at 22:02
  • $\begingroup$ @TNTCookie strong covalent bonds get broken in many situations. Jewelers break carbon-carbon bonds, which are quite strong, every time they cut a diamond. $\endgroup$ – lnmaurer Jul 17 at 4:23

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