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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?

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So this is theoretical question, and we should not think that bullet and rod will behave like a liquid? –  BarsMonster Jun 29 '11 at 8:54
As far as I can tell, the only relevant parameters are $M/m$ and $x/H$. –  Scott Carnahan Jun 29 '11 at 9:37
@Scott and BarsMonster: Exactly! –  Martin Gales Jun 29 '11 at 9:39
Interesting question. My 'hunch' would be that if the rod is indeed considered "rigid" (which is obviously not physical) the maximum number of collisions would be one. That said, I'm not sure I personally am capable of showing this using conservation of linear and angular momentum. I'll try a little back-of-an-envelope calculation later this afternoon.. –  qftme Jun 29 '11 at 11:48
@qftime: it is possible to arrange it such that in the center of mass frame, after the first collision the particle is stationary. (In other words, after the first collision the particle and the rod have the same linear velocity.) In which case the total number of collision must necessarily be 2. (One can consider this also by initially having a spinning rod and a particle at rest, make impact, time reverse the movie, and translating the frame.) –  Willie Wong Jun 29 '11 at 13:04
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1 Answer

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.)

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Note that $T_B|\Omega|$ only depends on $\frac{m}{M}$, $\frac{x}{H}$, and $\frac{|v'_0|}{|v_m|}$, the last of which depends only on the previous two, in accordance with Scott's comment above. –  Willie Wong Jun 29 '11 at 13:44
On top of rotation of the bar, it will vibrate in one or several of the transversal vibrational modes. The math of such vibrations is rather ugly :=( (compared to string vibrations). –  Georg Jun 29 '11 at 15:46
I wonder what drives you people to write such informative and long answers for no personal profit of yours. Its really admirable but I feel I have very little patience to do so. –  bubble Jun 30 '11 at 12:59
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