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Re: momentum transfer



Hi all-
Rick sez-
************************************************************
While we've been beating this about (incorrectly on my part), no one has
stated just what is correct. Let me try again
Considering the energy aspect, assuming that the collision shares energy
between the two cars equally (using a parked car as the fixed object), then
the head-on at v should be energetically equivalent to a collision at 1.414v-
**********************************
But you've forgotten to account for the energy of the c.m. motion.
If you mean that the change in kinetic energy of each car is the same, then
you are implying an elastic collision (where nobody gets hurt)
Y'all seem to be ignoring the inelasticity, which is crucial to the
analysis.
Regards,
Jack

"I scored the next great triumph for science myself,
to wit, how the milk gets into the cow. Both of us
had marveled over that mystery a long time. We had
followed the cows around for years - that is, in the
daytime - but had never caught them drinking fluid of
that color."
Mark Twain, Extract from Eve's
Autobiography