|I still think most people are missing the reason for the question about
| exactly 1/2 of the KE being transferred in a perfectly inelastic collision
| between same mass objects, one at rest.
| . . .
If you need to know just where the particular factor 1/2 comes from, it is
traceable to the exponent of the speed in our definition of kinetic energy
(KE=k*V^2). If instead we used KE = k*V^n , then we would deduce Ef =
The numerical results of our calculations/measurements strongly depend upon
just what quantities we chose to calculate/measure (because we found them
useful in our creation of models).
The most interesting fact underlying this discussion is: It is impossible
for an interaction between two point particles to consist soley of an
exchange of (conserved) momentum and kinetic energy. There must be an
additional degree of freedom in the form of a source/sink of
energy/momentum, even if only temporarily active. Even if the overall
interaction conserves both kinetic energy and momentum, intermediate system
states cannot all have done so. The appearance of particular numerical
values is incidental to this over-reaching observation.
A graphical exposition of the above fact is in:
"Mandated Energy Dissipation - e pluribus unum", AJP, 64,10, 1996, pg 1291.