# A ball inside a cavity

Seems not a trivial problem:

There is a semi-cylindrical cavity, with radius $R$ as shown in the Figure. A small ball (point mass for simplicity) with an initial horizontal velocity $v$ flies into the cavity. Friction is absent, blows against the cavity's wall are absolutely elastic. What should be the value of $v$ so that the ball is spending minimum time inside the cavity before flying off ?

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i'm missing something!!!! the value which comes to my mind is infinity (or for the sake of theory of relativity, $c$). It would just fly off never entering the cavity. What am I missing? –  Vineet Menon Nov 16 '11 at 8:35
I suppose given that relativity enforces a finite velocity, in the limit of a very large cavity (or very strong gravity) one cannot avoid hitting the wall at least once. –  David Z Nov 16 '11 at 8:38
@DavidZaslavsky: but then the higher the velocity of you ball, the lower you would stay inside of cavity, even if the cavity is very large. –  Vineet Menon Nov 16 '11 at 10:04
@Martin: did you intend to require that the ball still be traveling from left to right when it exits the cavity? –  Harry Johnston Nov 17 '11 at 1:46

I don't think there is a lower bound.

Set $x$ horizontal, $y$ vertical, $(0,0)$ at the center of the circle.

We'll find $x$ and $y$ for the point of impact. Say the ball takes time $t$ to cross the cylinder. It falls a distance $\frac{1}{2} g t^2$, so $y = -\frac{1}{2}gt^2$. We also have $x = vt - R$, or $t = (x+R)/v$ . Using the fact that for a circle $dy/dx = -x/y$, the slope of the wall where it hits is $\frac{2x}{gt^2} = \frac{2xv^2}{g(x+R)^2}$.

The y-velocity of the ball at impact is $-gt$. The ball's trajectory's slope at impact is $v_y/v = \frac{-gt}{v} = -\frac{g(x+R)}{v^2}$.

Multiplying the slopes gives $\frac{-2x}{x+R}$.

Since $0<x<R$ for large $v$, we see that the slopes multiply to something $0>product>-1$. That is, the slope of the ball's trajectory is not quite perpendicular to the wall. Instead, it's a little shallower than that. That means the ball bounces back up off the wall at a slightly higher angle than it hit the wall, and therefore bounces right back out almost the way it came, but slightly higher, in the limit of high $v$. Having the ball come off the wall at a higher angle clearly makes it bounce out of the cylinder in this limit. That means there is no minimum time, since the time inside shrinks as $v$ increases.

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Very nice answer! –  Martin Gales Nov 17 '11 at 6:36