Re: Fun problem

"Igor Khavkine" <igor.kh@xxxxxxxxx> a écrit dans le message de news:
1145745718.411724.269370@xxxxxxxxxxxxxxxxxxxxxxxxxxxx

I recently ran into a fun problem from an old Russian physics olympiad.
I found the solution a little surprising and intriguing. The problem
requires no more than high school classical mechanics and some
ingenuity. Here it is for your amusement.

Consider a point particle sliding on a flat table (ignore friction).
The table has a cylindrical hole of finite depth (vertical walls, flat
bottom). The particle can approach the hole with different velocities
and with different impact parameters (the particle's motion need not be
directed toward the centre of the hole). As the particle falls into the
hole, it starts bouncing off the walls and the bottom (assume elastic
collisions). Sometimes it gets stuck in the hole forever, sometimes it
escapes (bounces out). Determine the relation between the depth of the
hole, its radius, the particle's initial velocity, and impact parameter
necessary for the particle to escape after it falls in.

It seems not so difficult nor intriguing. We'll use symmetry everywhere.
The level of the table is the maximum height achievable by the particle,
therefore it escapes if and only if both at the table level and at the edge
of the hole. Consider the intersections of the cylinder and the initial
particle trajectory. The particle must bounce just at the vertical of the
middle of the two points. Between the bounce and the second point, the
trajectory is the same as between the first point and the bounce, but time
and space inverted/reflected. Now, given the impact parameter, it remains
to calculate the initial velocity so that the time to get at the middle
point from the first intersection for an horizontal motion is equal to the
time for falling the distance equal to the depth of the hole. The only
possible snag is that friction must be absolutely zero, and not merely
negligible.

Obviously, they aren't the only solutions. There is a denumerable
infinity of solutions because of multiple bouncing. For example, the speed
calculated above divided by two, three...

If the trajectory passes by the axis of symmetry of the hole, we have also
the solutions with each speed multiplied by two, three... for one, two...
bouncing of the wall, some of them having already been found.

There are still more solutions in the non symmetric case when the point
bounces off the wall, but I think it is enough for now.

--
~~~~ clmasse on free F-country
Liberty, Equality, Profitability.

.

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