Infinite sequence of nested, falling, colliding spheres 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...?
 A:             
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 $v'_{i+1} = \frac{4}{3} v_i$.
A: I don't think your system is well-specified. Consider the following simpler variant on the same theme:
Infinitely many ideal, identical billiard balls occupy a 1D world. The initial distance between successive balls (excluding the diameter of the balls themselves) is 1, 1/2, 1/4, 1/8 and so forth.
At time t=0, we give the first ball a push so it starts moving with speed 1 towards the infinite line of balls. A cascade of collisions happen, each transferring 100% of the energy and momentum from one ball to the next. By time 2, infinitely many collisions have happened, and all balls are now at rest and never move again! You can write down parametric equations for the position of each ball at any time, and see that locally the rules for ideal billiard balls are satisfied at every time. But somehow the energy and momentum we put into the system has disappeared.
Even more disturbing, the standard billiard-ball rules are time symmetric. So we can run the experiment backwards -- if we have an infinite line of balls sitting, all at rest, it is completely according to the rules if they spontaneously start an infinite cascade of collisions that end with ball 0 being ejected at some unpredictable speed, at an unpredictable moment in time.
It's not clear that your system of spheres has a failure mode that's quite as dramatic as this one. The fact that total mass is finite may conceivably prevent any momentum shenanigans. But according to statistical mechanics one should expect that the available energy will eventually drain away into micro-vibrations of ever smaller spheres as it tries to distribute itself evenly over all degrees of freedom. And the rules are still time-symmetric, so there are probably valid histories where energy arises spontaneously and ends up affecting the larger shells.
A: Making many extremely unrealistic assumptions:
After a collision, the second sphere will move 4 times as fast as the first (1/4 the mass) and the distance between it and the next sphere wil be halved, hence the time between each consecutive collision will be an eighth the time for the previous collision. It both spheres can feel gravity once released, both will accelerate downwards at the same rate, so the relative speeds will be the same as if there were no gravity and the first sphere were set off at an initial speed x m/s. $\sum_0^\infty 1/(8^{n})=1/9 $ so it will take r/9 seconds for the collision to reach the centre (r is the radius of the first sphere), and the 'centre' will now be a distance of r/2 from the top so it will take 2r/9 seconds for the outer sphere to be hit again. Now the 'centre' will be 1/4 of the distance from the bottom. So first it takes r/3 seconds between when the outer sphere is hit, and after that the intervals between when the outer sphere is hit are 4r/9 seconds. In addition, the spheres will be constantly accelerating downwards. 
