In frictionless rigid-body collision dynamics, what happens when a ball falls into a crevice where two planes intersect?

Question

When a ball bounces on a horizontal plane, its vertical velocity $v$ gets negated and multiplied by a coefficient of restitution $\varepsilon<1$, and its horizontal velocity $u$ is unchanged. The time taken in flight between two bounces is

$$t_n=\frac{2v_n}{g}=\frac{2v_0}{g}\varepsilon^n,$$

so the total time taken, as it comes to rest after infinitely many bounces ("chatter"), is

$$t_0+t_1+t_2+\cdots=\frac{2v_0}{g}(1+\varepsilon+\varepsilon^2+\cdots)=\frac{2v_0}{g}\frac{1}{1-\varepsilon}.$$

If the plane is sloped, then the normal and tangential motions can be analyzed separately; the above equations have $g$ replaced by $g\cos\theta$, and there's a constant tangential acceleration $g\sin\theta$.

Suppose the ball has finished chattering, and is sliding down this plane, and hits a second plane, and tries to settle into the corner. See this picture (but the planes may have different angles). If the planes intersect at a right angle, then the same type of chattering motion ensues. If the angle is obtuse, then the ball is sent into the air, bouncing between the two surfaces, and the motion appears chaotic. (The same happens with an acute angle; though it would deflect the ball downward into the first plane, causing multiple collisions at once.)

What exactly is this motion? If it's too complicated, I want to know at least that it takes a finite amount of time to come to rest. (I tried to use inequalities to bound the time, but ended up with a geometric series of $\sqrt{\varepsilon^2+1}^n$, which diverges.)

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For the case where both planes have the same inclination $\theta$ from the horizontal (so the angle between them is $\pi-2\theta$), I've found a self-similar trajectory; each parabolic arc has $k$ times the speed, $k$ times the time, and $k^2$ times the length of the previous arc. The total time taken is thus

$$t_0+t_1+t_2+\cdots=\frac{t_0}{1-k}.$$

I first guessed that the scale factor would be $k=\varepsilon$, but instead

$$\varepsilon=k\,\frac{k^2+2k\sin^2\theta+(1-2\sin^2\theta)}{1+2k\sin^2\theta+k^2(1-2\sin^2\theta)}.$$

This is a cubic equation if we want to solve for $k$ in terms of $\varepsilon$. I also found the restrictions on the initial conditions required to produce such a trajectory, but those details are irrelevant.

I wonder if there are more complex self-similar trajectories, repeating (and scaling by $k$) after several arcs.

Answer 1

Let's say the inclination of the left plane is $\theta$, and of the right plane is $\phi$, and the angle between them is $\alpha=\pi-\theta-\phi>\pi/2$. For notational convenience let's also define $\beta=\pi-\alpha$. The velocity has components $(u,v)$ tangent and normal (respectively) to the left plane. The process of bouncing off the right plane is described by the equations

$$u'\cos\beta+v'\sin\beta=1\big(u\cos\beta+v\sin\beta\big)$$

$$u'\sin\beta-v'\cos\beta=-\varepsilon\big(u\sin\beta-v\cos\beta\big)$$

which we can solve for the new velocity

$$u'=u\,(\cos^2\beta-\varepsilon\sin^2\beta)+v\,(1+\varepsilon)\sin\beta\cos\beta$$

$$v'=u\,(1+\varepsilon)\sin\beta\cos\beta+v\,(\sin^2\beta-\varepsilon\cos^2\beta).$$

The ball is initially sliding, so $u>0,\,v=0$. We'll consider the case where $\beta$ is close enough to $\pi/2$ that $u'<0$; that is, $\cot^2\beta<\varepsilon$. There is a possibility of chattering on the left surface again, but this would have to end before the tangential acceleration brings the ball back to the right surface. Comparing times,

$$\frac{2v'}{g\cos\theta}\frac{1}{1-\varepsilon}<\frac{-2u'}{g\sin\theta}$$

$$\frac{v'}{-u'}<(1-\varepsilon)\cot\theta$$

$$\frac{v'}{-u'}=\frac{(1+\varepsilon)\sin\beta\cos\beta}{\varepsilon\sin^2\beta-\cos^2\beta}=\frac{(1+\varepsilon)\sin2\beta}{\varepsilon-1-(1+\varepsilon)\cos2\beta}<(1-\varepsilon)\cot\theta$$

This is independent of the initial speed $u$. So when the ball slides down again, the process repeats itself with initial speed $-u'=u(\varepsilon\sin^2\beta-\cos^2\beta)$. Eventually we get infinitely many bounces off the right surface, and infinity-squared many bounces off the left surface. The total time is

$$\frac{2u}{g\sin\theta}\big((\varepsilon\sin^2\beta-\cos^2\beta)+(\varepsilon\sin^2\beta-\cos^2\beta)^2+(\varepsilon\sin^2\beta-\cos^2\beta)^3+\cdots\big)$$

$$=\frac{2u}{g\sin\theta}\frac{(\varepsilon\sin^2\beta-\cos^2\beta)}{1-(\varepsilon\sin^2\beta-\cos^2\beta)}.$$

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So, at least we have an open set of initial conditions and plane angles for which the whole trajectory can be easily calculated.