Steve Landsburg posted about this very problem today at http://www.thebigquestions.com/2014/06/18/follow-the-bouncing-ball/ He references the paper http://ics.org.ru/doc?pdf=440&dir=e by Gregory Galperin which does indeed prove that the number of collisions does give the first N digits of pi. -Thomas C On Fri, Nov 26, 2010 at 11:48 PM, Gary Antonick <gantonick@post.harvard.edu> wrote:
Gareth,
At the same time I think a reasonable argument can be made for the final velocities to be -V at the limit. I think this is what the circle and tangent line are suggesting.
Which means, if I've got it right, that the wall (the third ball) would have to be very, very heavy.
- Gary
On Fri, Nov 26, 2010 at 4:05 PM, Gareth McCaughan <gareth.mccaughan@pobox.com> wrote:
On Friday 26 November 2010 22:17:54 Gary Antonick wrote:
As M/m approaches infinity the size of the extra small wedge approaches zero. Does this mean at the limit the two balls will have the same final velocity? If so, what is this final velocity? (And how is this scenario reversible?)
In the limit, the final velocity of both balls is zero. The bigger M/m is, the greater the fraction of the energy and momentum that ends up belonging to the bigger ball; as its mass tends to infinity, the corresponding velocity goes to zero.
Of course a final velocity *equal* to zero makes for an irreversible scenario, but that's only the limit. In any actual case with finite M, the balls end up with unequal, nonzero final velocities.
-- g
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