Collision between a rod and a bullet There lies a homogeneous rigid rod of mass $M$ and of length $H$ on a frictionless table at rest. A small bullet of mass $m$ moves toward the rod with velocity $v_0$, perpendicular to the rod and collides absolutely elastically with the rod at a point, located at a distance $x$ from the center of the rod. Before they finally move away from each other, they may still collide each other a few times.   
What is the maximum number of possible collisions of the bullet with the rod before they leave each other forever?
 A: I don't know if there can be a closed form solution. Here's an estimate:
Note that in the centre of mass frame, the radial distance to the particle is hyperbolic in the sense that once the radial velocity of the particle is positive, it can not decrease any more (between collisions this is obvious from the conversion between Cartesian and polar coordinates, and at collisions this follows from the conservation of linear momentum). So after the initial impact (which is perpendicular to the rod, and hence starts with zero radial velocity), the radial velocity will be no less than the equivalent linear motion with no collisions. This allows us to get an estimate on the maximum time the particle remains in "batting distance". 
On the other hand, by conservation of energy, we can get an absolute upper bound on the relative angular velocity of the rod versus the particle, which combined with the maximum time we have, we can get a bound of the maximum number of collisions (between collisions, the angular position of the rod versus the particle in the frame of the centre of rod's mass must change by at least $\pi$). 
Now we implement this (very crude) estimate. 
First we compute the situation after the first collision. For convenience we re-scale $M+m = 1$, and we re-define the radius of the rod to be $H$. Let $v'_0$ be the initial velocity of the particle in the centre-of-mass frame (so $v_0' = Mv_0$). In the COM frame, letting $v_m$ be the velocity of the particle, the total translational kinetic energy is $T = \frac12 \frac{m}{M} v_m^2$. Letting $\omega$ be the angular velocity of the rod about its mass centre, the rotational kinetic energy is $R = \frac16 MH^2\omega^2$. 
In the COM frame, at the moment of the perpendicular collision (when the COM is on the rod), the total angular momentum is given by
$$ L  = m v_m x + \frac{1}{3}M\omega H^2$$
So conservation of energy and conservation of momentum says that after the initial impact, $v_m$ satisfies
$$ \frac{m}{M}(1 + \frac{m}{M}) \left( v'_0^2 - v_m^2\right) = \frac13 \left(\frac{3mx}{MH}\right)^2 \left(v'_0 - v_m\right)^2 $$
which one can in principle solve using the quadratic formula. The total time within batting range would be bounded above by 
$$ T_B = \frac{\sqrt{H^2 - x^2}}{v_m} $$
Using just the conservation of energy, a very crude upper bound for the relative angular speed between the spinning rod and the particle (using the fact that after the first collision, the centres of masses of the particle and the rod only increase in distance, so subsequent impact spots will get further and further away from the centre of the rod) can be written as 
$$ |\Omega| \leq \sqrt{2} \frac{|v'_0|}{Mx} \max (1, 2 \frac{x}{H}) $$
And the bound on the maximum number of collisions will be
$$ T_B |\Omega| \pi^{-1} $$
Note that this estimate is very wasteful when $x\to 0$, since in that limit, we expect the elastic collision to send the rod off not spinning, and there would be only one collision. This estimate is also completely inadequate to treat the edge case where $v_m = 0$ is the solution to the first collision, in which case the total number of collisions should be two. 
In principle someone more clever can take this and get a much more refined estimate. (Or perhaps "intuit" an answer otherwise.)
