TITLE: Infinite sequence of nested, falling, colliding spheres
QUESTION [9 upvotes]: Imagine an infinite collection of nested, concentric spheres, of radius 1, $\frac{1}{2}$,
$\frac{1}{4}$, $\frac{1}{8}$, and so on.  Suppose they are somehow suspended in space, fixed on their
common center $x$.  Then the outermost sphere is "released" from its center, and falls
vertically under the influence of gravity, while all the other spheres remain "pinned"
with their centers on $x$.  Next, the top interior of the $r=1$ sphere collides with the top exterior of the $r=\frac{1}{2}$
sphere, knocking it loose from $x$ via a perfectly elastic collision, sending it
downward.  And so on.


Essentially my question is: What happens?  It would be pleasing to understand
the behavior of this system without resorting to explicit calculation of
all the interactions.  Assume the spheres are made of some homogenous, thin
material so that their weight is proportional to their surface area
(or circumference if you'd prefer to drop down to $\mathbb{R}^2$).
I cannot see intuitively the sequence of collisions and overall behavior,
and I have not yet tried careful calculations.  Perhaps there is a line
of reasoning that demystifies the apparent complexities...?

REPLY [4 votes]: I worked out the details for the 2D circles case, where the mass of ring $i$ is twice the mass of ring $i+1$ (because the latter is half the radius).
When ring $i$, traveling at velocity $v_i$, collides with stationary ring $i+1$,
their velocities after the collision are $v'_i= \frac{1}{3} v_i$ and $v'_{i+1}= \frac{4}{3} v_i$, a simple consequence of conservation of momentum and energy.†
(This incidentally corrects some miscalculations in the original answers.) 
Because $v_{i+1}$ then increases a bit under acceleration, we know that $v_k$ grows at least as
fast as $\left( \frac{4}{3} \right)^k$.
So, if I've calculated correctly, assuming a 1-meter radius outer ring, it would take  fewer than 64 nested rings to reach the speed of light. :-)
Which means that, answering my original question ("What happens?") requires
an analysis that incorporates the laws of special relativity.

† Just for the record, here is the derivation of the velocities
just before and just after a collision.
Let the $i$-th ring have mass $m$, and so the $(i+1)$-st ring has mass $\frac{1}{2}m$.
Conservation of momentum:
$$
m v_i + \frac{1}{2} m v_{i+1} = m v'_i + \frac{1}{2} m v'_{i+1} \;.
$$
Because $v_{i+1} = 0$, this reduces to: $v_i=v'_i+\frac{1}{2} v'_{i+1}$.
Conservation of (kinetic) energy:
$$
\frac{1}{2} m v_i^2 + \frac{1}{4} m v_{i+1}^2 =
\frac{1}{2} m (v'_i)^2 + \frac{1}{4} m (v'_{i+1})^2 \;,
$$
which reduces to
$v_i^2 = (v'_i)^2 + \frac{1}{2} (v'_{i+1})^2$.
Solving these two equations simultaneously gives
$v'_i = \frac{1}{3} v_i$
and $v'_{i+1} = \frac{4}{3} v_i$.