According to a Newtonian observer, the following formulas are valid:
m.(c - v) = M.v x = v.t L = c.tKnown quantities are : c , m , M , L . The unknowns are : t , x , v . Hence:
t = L/c v = m/(M + m).c x = m/(M + m).LExample for (c - v) = v , hence v = c/2 :
m = M x = L/2In agreement with what might be expected in this case.
Let's proceed with photons instead of cannon balls. Reference (wikipedia) is:
Photon.
According to this reference, the (linear) momentum p and the energy
E of a photon are mutually related as follows:
p = E / cBecause everybody seems to adhere to relativistic mechanics these days, it becomes bit of a guesswork what Newton would have thought of this. Because, in classical Newtonian mechanics, photons do not necessarily travel with "the" speed of light c. We think a good start is to assume that the momentum and the energy of a photon, travelling with speed (c - v) now, obey the following relationship:
p = E / (c - v) instead of p = E / cIf the Cannon on a railroad car model is assumed valid for photons, then it follows that:
m.(c - v) = M.v = E / (c - v)Conclusion:
E = m.(c - v)2In practice it is almost impossible to distinguish this outcome from E = m.c2 .