Kvant Physics Problem 33

Verified: yes
Verdicts: PASS + PASS
Solve time: 5m04s
Source on kvant.digital

Problem

Two beads with masses $m_1$ and $m_2$ can slide without friction along a smooth horizontal wire ring. Initially, the beads are connected by a thread, and a compressed spring is located between them. The thread is burned through. After the beads begin to move, the spring is removed. At what point on the ring will the beads collide for the eleventh time? The collisions of the beads are perfectly elastic.

Setup and Assumptions

Two beads of masses $m_1$ and $m_2$ move without friction on a smooth horizontal circular wire ring of circumference $L$. Initially the beads are connected by a thread, and a compressed spring is placed between them. After the thread is burned, the spring pushes the beads apart. When the spring has finished acting, it is removed.

The unknown is the point of the ring at which the beads meet for the eleventh time.

The ring is rigid and fixed in an inertial reference frame. The wire is perfectly smooth, so no external tangential force acts on the system. The spring acts only during the initial release and is removed afterward. The beads are treated as point masses. All subsequent collisions between the beads are perfectly elastic. Gravity does not influence the motion because the ring is horizontal.

Let the point where the spring releases the beads be chosen as the origin on the ring. Let the positive direction along the ring coincide with the initial motion of mass $m_1$.

Physical Principles

The total momentum of the two-bead system along the ring is conserved because the forces exerted by the spring are internal and there are no external tangential forces.

For a perfectly elastic one-dimensional collision of two masses, the collision is equivalent to the particles passing through one another if the particles are regarded as distinguishable only by their trajectories. This standard trick allows the motion to be analyzed without explicitly treating each collision.

The center of mass of an isolated system moves with constant velocity. Since the initial total momentum is zero, the center of mass remains fixed.

Uniform motion occurs between collisions because no forces act after the spring is removed.

Derivation

Immediately after the spring ceases to act, let the velocities of the beads be $v_1$ and $v_2$, directed oppositely around the ring.

Conservation of momentum gives

$$m_1 v_1=m_2 v_2 .$$

The total momentum is zero, so the center of mass remains permanently at its initial position.

To find collision points, it is convenient to use the standard unfolding method. Instead of allowing the beads to collide elastically, imagine that they pass through each other unchanged. The set of actual collision events is identical to the set of crossings of these fictitious trajectories.

Let the coordinate along the ring be measured from the release point. In the fictitious motion,

$$x_1=v_1 t, \qquad x_2=-v_2 t ,$$

with coordinates understood modulo $L$.

A collision occurs whenever the two coordinates on the ring coincide:

$$x_1-x_2=(v_1+v_2)t=nL,$$

where $n=1,2,3,\ldots$.

Thus the time of the $n$-th collision is

$$t_n=\frac{nL}{v_1+v_2}.$$

The position of the collision is the position of bead $1$ at that instant:

$$x_n=v_1 t_n =\frac{nL,v_1}{v_1+v_2}.$$

Using the momentum relation,

$$m_1 v_1=m_2 v_2,$$

we write

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

Substituting into the expression for $x_n$,

$$x_n = \frac{nL,v_1} {v_1!\left(1+\frac{m_1}{m_2}\right)} = \frac{nL} {1+\frac{m_1}{m_2}} = \frac{nLm_2}{m_1+m_2}.$$

Since positions on the ring differing by an integer multiple of $L$ are identical,

$$x_n \equiv \frac{nLm_2}{m_1+m_2} \pmod L.$$

For the eleventh collision, $n=11$:

$$x_{11}\equiv \frac{11Lm_2}{m_1+m_2} \pmod L.$$

This coordinate is measured from the point where the spring released the beads, in the direction of the initial motion of mass $m_1$.

Result

The position of the $n$-th collision is

$$x_n \equiv \frac{nLm_2}{m_1+m_2} \pmod L.$$

For the eleventh collision,

$$x_{11}\equiv \frac{11Lm_2}{m_1+m_2} \pmod L.$$

Hence the collision point is located at the arc distance

$$\boxed{ x_{11}= \left( \frac{11m_2}{m_1+m_2} - \left\lfloor \frac{11m_2}{m_1+m_2} \right\rfloor \right)L }$$

from the release point, measured along the direction of the initial motion of mass $m_1$.

No numerical substitution is possible because the problem statement provides no numerical values for $m_1$, $m_2$, or $L$.

Sanity Checks

The expression has dimensions of length because the factor multiplying $L$ is dimensionless.

If $m_1=m_2$, then

$$x_n=\frac{nL}{2}\pmod L.$$

Successive collisions alternate between diametrically opposite points of the ring, which matches the symmetry of equal masses moving with equal speeds in opposite directions.

If $m_1\gg m_2$, then

$$x_n\approx nL\frac{m_2}{m_1},$$

so the collision points advance only slightly from the release point. This agrees with the fact that the heavy bead moves slowly while the light bead circles the ring rapidly.

If $m_2\gg m_1$, then

$$x_n\approx nL \pmod L,$$

so collisions occur close to the release point, which is consistent with the heavy bead being nearly stationary.

The most common source of error is the use of the momentum relation. Writing $m_1v_1=m_2v_2$ with the wrong mass in the numerator when eliminating one velocity changes the collision position from

$$\frac{nLm_2}{m_1+m_2}$$

to

$$\frac{nLm_1}{m_1+m_2},$$

which places every collision point at the wrong location on the ring.