Will a 100MeV muon and a 50MeV muon deposit a different amount of energy in a scintillator? I have a question where I need to calculate how much energy a $100\,\mathrm{MeV}$ muon will deposit in $1\,\mathrm{cm}$ of scintillator. I know that the muons lose energy about $2\,\mathrm{MeV/(g/cm^2)}$, but I would think that regardless of the amount of energy of an incoming muon they will deposit the same amount within $1\,\mathrm{cm}$ of scintillator. The only thing that would change is how far the muon travels through the scintillator. If it loses all of its energy it will never leave, but if it has enough it travels all the way through. Is this incorrect?
 A: The $2\,\mathrm{MeV/(g/cm^2)}$ rule of thumb is a heuristic. 
It is roughly the energy-loss of minimum ionizing particle in a medium made of moderate $A$ atoms. We get away with it for most purposes, because the valley that "minimum ionization" sits at the bottom of is quite broad in energy space and the $A$ dependence is pretty weak (and weaker for muons than for other particles).
A more complete treatment will allow you to calculate the actual energy loss for your two cases and look to see if they are close enough to count as "the same" for your purposes.

The easy, on-line place to find a more complete treatment is in the Particle Data Book (published and maintained by the particle data group). The chapter you want is the one on "Passage of particles through matter" (Chapter 33 in the 2018 edition; and I'm not going to link it directly because they are PDFs and they change regularly).
In the current edition Figure 33.1 is exactly what you are looking for:

(I grabbed that from their downloadable images archive for the 2018 edition.)
We see that $100\,\mathrm{MeV}$ muons are close to the minimum ionization point, but that $50\,\mathrm{MeV}$ muons are on the rising slope.
Notice that this is a log-log figure, so the difference in energy loss between  $50$ and $100\,\mathrm{MeV}$ muons is non-trivial.
