# Reason for different type of energy transfer for two kinds of collisions

According to my physics book, if an electron were accelerated with 15 MeV of (kinetic?) energy and collided into a 100g thermally insulated copper block (not sure if the fact it is thermally insulated is relevant) the energy would transfer into thermal energy on the copper block and the copper block would heat up by a few degrees.

However, if I were to collide into that same copper block with a bowling ball travelling with 15 MeV of kinetic energy then that energy would be transferred into kinetic energy on the copper block and the copper block would move.

It seems to me, that in both cases, there is a thing smashing into another thing with 15 MeV of energy. When the first thing is small (e.g. electron) the energy is transferred into thermal energy. When the first thing is large the energy is transferred into kinetic energy.

Why doesn't the electron's energy get transferred into kinetic energy as well?

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Thermally insulated is important. If it wasn't thermally insulated, then you would have to find out the heat lost to the environment when the energy transfer occurred for the electron case and it's harder (and time dependent). By saying thermally insulated, you can just find the steady-state temperature and ignore the idea that the heat would start at the impact point and spread throughout the block. – tpg2114 Jul 15 '13 at 4:07

Momentum is transferred in both cases (conservation of momentum is absolute), but how much?

Figure the momentum of the two projectiles.

• The electron is fully relativistic, so you get $$p^2 = E^2 - m^2 = \left(15.0^2 - 0.5^2\right)\,\mathrm{MeV}^2$$ or $p \approx 1.5\times 10^{7}\,\mathrm{eV}$ (in $c = 1$ units, of course).

• The bowling ball is fully classical, so you get $$p^2 = 2mE$$ with a mass in the $10^{36}\,\mathrm{eV}$ range. So $p \approx 6 \times 10^{21}\,\mathrm{eV}$.

The bowling ball transfers roughly 14 orders of magnitude more momentum than the electron!

The energy from the electron ending in thermal modes because there aren't any other places for it to go.

Figuring it again in mks units is left as an exercise for the interested student.

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Is momentum the only reason? If I put enough energy into the electron to have the same momentum as the bowling ball (18\times 10^{29} by my calculation) would that cause the electron to transfer kinetic energy to the block? – Pace Jul 15 '13 at 11:57
If you give the electron enough momentum to be the same as the bowling ball you would have to give it enormously more energy. Far more than could be contained in the tiny little beam stop proposed: it would blow right through. – dmckee Jul 15 '13 at 11:59
Then I would argue the answer isn't simply the difference in momentum but also the shape of the impact. If, instead of supercharging one electron, I charged a plane of electrons (equal in dimensions to the copper block) so that the total momentum transferred was equal to that of the bowling ball, but no electron had more than 15 MeV, would that cause the block to move? – Pace Jul 15 '13 at 12:50
Of course the microscopic details of the collision matter (and your hoard of electrons would render the target into plasma), but that is not the point of the exercise which is to show how the relationship between kinetic energy and momentum is mediated by the mass of the projectile and you can't draw trivial conclusions about it if you don't know the masses involved. – dmckee Jul 15 '13 at 15:09

So I will take a stab from a more "big picture" point of view.

When an electron hits the block of copper, it's mass (and physical size) is considerably less than the size of the copper block. So at impact, it's really hitting what could be thought of as a single atom in the block.

Now, temperature is the measure of kinetic energy of molecules/atoms. So when all of the kinetic energy from the electron gets converted (let's say no losses) into kinetic/vibrational energy of the atom it hit, that atom then vibrates in the lattice and this vibration distributes the energy throughout the block. All of the atoms begin to vibrate more and this is measured as a temperature increase in the block.

An example of what this looks like (this is a copper atom hitting a copper block)[from Wikipedia]:

As you can see, the motion of the atoms is not coherent but is random as they bounce around.

When a bowling ball hits the block, it's of the same/bigger scale than the block so it's not transferring energy to the kinetic modes of the atoms. Rather, it's transferring momentum to the entire block and not just to the random vibrations/translations of the atoms itself. So this will cause the kinetic energy of the block to increase, rather than internal energy.

TL;DR: Electron is tiny and hits the atom, converting kinetic energy into random motion/vibration of the atoms in the lattice. Temperature is defined as the average kinetic energy of the atoms. Bowling ball is huge and provides a coherent translation, rather than a random one, to the block. This is kinetic energy of the block rather than internal.

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+1 for animation :) – Chris White Jul 15 '13 at 4:34
Well, at 15 MeV the electron is like to break a few bonds and penetrate a few g/cm^2 into the target, but nice. – dmckee Jul 15 '13 at 5:18
Great animation, but dmckee's answer deals with the crux of the problem, which is momentum conservation rather than energy conservation. – Michael Brown Jul 15 '13 at 5:19
@dmckee I willingly admit it's outside of what I know, I could only give the qualitative description. Quantitative wins out though :) – tpg2114 Jul 15 '13 at 5:19