Kvant Physics Problem 33

Two beads of masses $m_1$ and $m_2$ move without friction on a fixed rigid circular wire of circumference $L$.

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Solve time: 1m20s
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Problem

On a smooth horizontal wire ring, two beads of masses $m_1$ and $m_2$ can slide without friction. Initially, the beads were connected by a thread, and a compressed spring was located between them. The thread is burned through. After the beads begin to move, the spring is removed. At which point on the ring will the beads collide for the eleventh time? The beads collide perfectly elastically.

Setup and Assumptions

Two beads of masses $m_1$ and $m_2$ move without friction on a fixed rigid circular wire of circumference $L$. Their positions along the ring are described by arc coordinates $x_1(t)$ and $x_2(t)$ measured modulo $L$. A compressed spring is initially placed between the beads while a thread keeps the configuration fixed so that no motion occurs before release.

At the initial moment the thread is burned. The spring then exerts internal forces and gives the beads opposite impulses along the tangent to the ring. After a short interaction time the spring is removed completely and no further external or internal forces act. The beads subsequently move freely along the ring and undergo perfectly elastic collisions when they meet.

The unknown is the geometric location on the ring where the beads collide for the eleventh time.

The system is analyzed in an inertial frame fixed to the ring. The ring is assumed perfectly rigid, and the motion is strictly one dimensional along the circumference.

Physical Principles

The motion between collisions is uniform rectilinear motion along the ring, so each bead satisfies

$x_i(t) = x_i(0) + v_i t \quad \text{mod } L.$

During collisions, conservation of linear momentum and kinetic energy apply. For a one dimensional perfectly elastic collision,

$v_1' = \frac{m_1 - m_2}{m_1 + m_2}v_1 + \frac{2m_2}{m_1 + m_2}v_2,$

$v_2' = \frac{2m_1}{m_1 + m_2}v_1 + \frac{m_2 - m_1}{m_1 + m_2}v_2.$

A standard reduction for one dimensional elastic collisions is the equivalence to a model in which the particles pass through each other without interaction while preserving their velocities. The collision events are then interpreted as crossings of trajectories on the circle.

The center of mass coordinate

$X_{\mathrm{cm}} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$

moves uniformly because the net external force on the system is zero. Hence

$\frac{d^2 X_{\mathrm{cm}}}{dt^2} = 0.$

Derivation

The spring force is internal and acts only during a short interval after release. Because no external force acts, the total momentum immediately after release is conserved and remains constant for all subsequent motion. The symmetry of the initial configuration implies that the spring produces equal and opposite impulses along the tangent direction of the ring, so the total momentum of the two beads is zero. This yields

$m_1 v_1 + m_2 v_2 = 0,$

and therefore

$v_2 = -\frac{m_1}{m_2} v_1.$

The center of mass velocity is

$V_{\mathrm{cm}} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2} = 0,$

so the center of mass remains fixed on the ring at a point $X_{\mathrm{cm}}$ determined entirely by the initial positions:

$X_{\mathrm{cm}} = \frac{m_1 x_1(0) + m_2 x_2(0)}{m_1 + m_2}.$

After the spring is removed, the beads move freely and collide elastically. Replacing elastic collisions by the equivalent picture of particles passing through each other shows that each bead continues uniform motion with constant velocity on the circle. A collision occurs precisely when

$x_1(t) \equiv x_2(t) \pmod{L}.$

In the passing through representation, both particles traverse the circle while the center of mass remains fixed. The only point on the ring that is shared by both trajectories at collision instants is the center of mass point. At each encounter the positions of the beads coincide with the center of mass coordinate, because the symmetry of motion about a fixed center forces the meeting point to remain invariant under exchange of velocities during elastic scattering.

Thus every collision occurs at the same geometric location $X_{\mathrm{cm}}$. Since the collision point does not depend on the collision number, the eleventh collision occurs at the same point as the first collision.

Result

The collision point for the eleventh collision coincides with the center of mass position on the ring,

$X_{\mathrm{cm}} = \frac{m_1 x_1(0) + m_2 x_2(0)}{m_1 + m_2} \quad \text{mod } L.$

No additional numerical substitution is possible without specified initial coordinates. The final answer is therefore

$\boxed{X_{\mathrm{cm}} \text{ on the ring, defined by } \frac{m_1 x_1(0) + m_2 x_2(0)}{m_1 + m_2} \ (\mathrm{mod}\ L)}.$

Sanity Checks

Dimensional consistency holds because $X_{\mathrm{cm}}$ is a linear combination of arc lengths $x_1$ and $x_2$, so it has units of length.

In the limit $m_1 \gg m_2$, the center of mass approaches $x_1(0)$, meaning the heavier bead dominates the collision location, consistent with the inertia of the system.

In the limit $m_1 = m_2$, the center of mass reduces to the midpoint of the initial arc separation, consistent with symmetry of equal masses.

Any incorrect treatment would most likely arise from assuming that successive elastic collisions shift the meeting point along the ring, which would violate the invariance of the center of mass and the equivalence between elastic collisions and free passage in one dimension.