Kvant Physics Problem 12

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Verdicts: SKIP + SKIP
Solve time: 1m11s
Source on kvant.digital

Problem

Two identical heavy steel balls rotate on light rods of length $l$ and $2l$ about points $O_1$ and $O$, the distance between which is equal to $3l$ (Fig. 3). At the initial moment, the balls are located at points $A$ and $B$, having velocities $v$ and $2v$ respectively. How many times will the balls collide over a time $t$? Over what time will the balls collide $k$ times? The collisions are assumed to be perfectly elastic.

Fig. 3

G. L. Kotkin

Setup and Assumptions

Two identical steel balls of mass $m$ move on rigid, massless rods that constrain motion to circular trajectories of fixed radii $l$ and $2l$. The centers of rotation are points $O_1$ and $O$, separated by a fixed distance $O_1O=3l$. The two circular paths are externally tangent at a single point $C$, since $l+2l=3l$.

The balls are initially located at points $A$ and $B$ on their respective circles with tangential speeds $v$ and $2v$. The motion is confined to each circle, so the speeds correspond to angular motion about fixed centers. Collisions occur only at the common tangency point $C$. All collisions are perfectly elastic and instantaneous, and the rods constrain motion without energy loss. Gravitational effects are neglected in the tangential dynamics.

The unknown is the number of collisions in a time interval $t$ and the time of the $k$-th collision.

Physical Principles

Uniform circular motion implies the relation between tangential speed $u$, angular velocity $\omega$, and radius $R$,

$$u=\omega R.$$

At a perfectly elastic collision between two identical masses moving along the same line, velocities are exchanged while conserving total momentum and kinetic energy. For identical masses this exchange preserves the set of speeds while possibly reversing directions.

Since motion is constrained to rigid circles, the angular speed magnitude remains constant between collisions. The dynamics reduces to periodic arrival of both balls at the common point $C$ with a fixed recurrence time determined by angular motion.

For a full circular revolution, the period is

$$T=\frac{2\pi}{\omega}.$$

Derivation

For the first ball on radius $l$, the angular velocity is determined by

$$\omega_1=\frac{v}{l}.$$

For the second ball on radius $2l$, the angular velocity is

$$\omega_2=\frac{2v}{2l}=\frac{v}{l}.$$

Both balls therefore have the same angular speed magnitude

$$\omega=\frac{v}{l}.$$

The geometry of external tangency implies that collisions occur only at point $C$, which lies on both circles. The balls meet at $C$ when their angular displacements relative to the line $O_1O$ and $OO_1$ reach the same geometric direction corresponding to the tangency point.

At the initial moment both balls are located at the diametrically opposite points $A$ and $B$ relative to $C$. The angular distance from either initial point to $C$ equals $\pi$. The time required for each ball to reach $C$ is determined by uniform angular motion,

$$t_1=\frac{\pi}{\omega}=\frac{\pi l}{v}.$$

At $C$ the tangents to both circles coincide. The velocities are collinear and opposite due to opposite directions of motion, so the collision is head-on along a single line. For identical masses in a perfectly elastic collision, velocities are exchanged, which reverses the roles of the balls but does not change the magnitude of their angular velocities. The system therefore repeats the same approach process after each encounter.

The angular configuration is restored after each interval $\pi/\omega$, since after collision each ball continues along the circle with unchanged speed magnitude but reversed direction relative to the next approach toward $C$. Thus the collision events are periodic with interval

$$\Delta t=\frac{\pi}{\omega}=\frac{\pi l}{v}.$$

The number of collisions in time $t$ equals the number of such intervals completed:

$$N(t)=\frac{t}{\Delta t}=\frac{vt}{\pi l}.$$

For the $k$-th collision time,

$$t_k=k,\Delta t=k\frac{\pi l}{v}.$$

Result

The number of collisions in time $t$ is

$$N(t)=\frac{vt}{\pi l}.$$

The time at which the $k$-th collision occurs is

$$t_k=k\frac{\pi l}{v}.$$

Boxed results:

$$\boxed{N(t)=\frac{vt}{\pi l}}$$

$$\boxed{t_k=\frac{k\pi l}{v}}$$

Sanity Checks

The quantity $vt/(\pi l)$ is dimensionless since $v t$ has units of length and division by $l$ removes length, confirming consistency for a collision count.

The collision interval $\pi l/v$ has units of time, matching a half-period of circular motion since $T=2\pi/\omega=2\pi l/v$.

If $v$ increases, the collision rate increases proportionally, consistent with faster angular motion. If $l$ increases, the collision rate decreases due to slower angular speed for fixed tangential velocity.

A sign error in the angular displacement from the initial positions would shift the first collision time by a factor of two, since the relevant arc length to point $C$ is exactly $\pi l$ rather than $2\pi l$.