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-----Original Message-----
From: Carl E. Mungan [mailto:mungan@USNA.EDU]
Sent: Sunday, October 27, 2002 3:05 PM
[snip]
Note that if you apply energy and momentum conservation in this same
way, it clears up the apparent violation of energy conservation in
the Appendix of my previous message about a rock dropped 5 m on earth
as viewed by someone traveling upward at 5 m/s:
cons. of E => v^2 + 2vu = 2gh + M/m*(u^2 - u'^2)
where v = 10 m/s of rock relative to earth after dropping 5 m, u = 5
m/s of earth initially relative to the observer, and u' = speed of
earth finally relative to the observer.
cons. of p => v = M/m*(u - u')
Substitute the second into the first to find v = sqrt(2*g*h). Sorry,
no Nobel prize.