How to calculate the kinetic energy of supernova ejecta? I believed I could use $0.5\times M_{ej} V_{ej}^2$, with $M_{ej}$ being the ejected mass and $V_{ej}$ being the velocity of the ejected mass. But I noticed in this and this that the the mean velocity is somehow calculated as following : $V_{mean}=(3/5)^{0.5}\times V_{ej}$.
And so the kinetic energy was calculated as : $(3/5)(0.5)\times M_{ej} V_{mean}^2 =(3/10)\times M_{ej} V_{mean}^2$.
So is there any explanation for how this factor of $0.3$ or the mean velocity was calculated ?
 A: When observing supernovae, often the bulk of the material is optically thick, so we only see the surface layers. These are moving at some speed $v_\mathrm{surf}$.
When modeling supernovae, the very simplistic model everyone loves to use is that of homologous expansion of a uniformly dense (density $\rho$) sphere. That is, velocity of the material increases linearly with radius from $0$ at the center to $v_\mathrm{surf}$ at the surface (radius $R$).
The kinetic energy is then
\begin{align}
E & = \int\limits_\mathrm{ejecta} \frac{1}{2} \rho v^2 \,\mathrm{d}V \\
& = \frac{2\pi\rho}{R^2} v_\mathrm{surf}^2 \int_0^R r^4 \,\mathrm{d}r \\
& = \frac{2\pi}{5} \rho R^3 v_\mathrm{surf}^2 \\
& = \frac{3}{10} M_\mathrm{ej} v_\mathrm{surf}^2,
\end{align}
where the second line uses the assumptions of uniform density, spherical symmetry, and homologous expansion ($v = (r/R) v_\mathrm{surf}$), and the fourth line uses the total ejecta mass $M_\mathrm{ej} = (4\pi/3) R^3 \rho$.
The surface velocity is perhaps more properly called the photospheric velocity, since it is at about the depth of the photosphere that one is actually measuring velocity. Of course, looking at different lines of different elements can change one's definition of the photosphere. One may hope this isn't much of a problem in the sense that the vast majority of the mass lies within any reasonable definition of the photosphere. Even then, one would be assuming $\rho$ is constant all the way out to the photospheric radius in order to do the above calculation, and this clearly can't hold for all photospheres.
One also sees what I've written as $v_\mathrm{surf}$ called the "ejecta velocity." It is simply understood that this is the velocity only of the outer layers of the ejecta.
If you want an "effective velocity" then you can define it by
$$ \frac{1}{2} M_\mathrm{ej} v_\mathrm{eff}^2 = E, $$
leading to $v_\mathrm{eff} = \sqrt{3/5} v_\mathrm{surf}$. Note that this is only the average in the squared sense:
$$ v_\mathrm{eff}^2 = \frac{1}{M_\mathrm{ej}} \int\limits_\mathrm{ejecta} v^2 \rho \,\mathrm{d}V. $$
The mean in the linear sense (useful for considering momentum rather than energy) is
$$ v_\mathrm{ave} = \frac{1}{M_\mathrm{ej}} \int\limits_\mathrm{ejecta} v \rho \,\mathrm{d}V = \frac{3}{4} v_\mathrm{surf}. $$
