Kvant Physics Problem 558

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Solve time: 2m49s
Source on kvant.digital

Problem

F558. A small heavy ball enters through an opening into a smooth sphere of the same mass, passing at a distance $R/2$ from the center of the sphere ($R$ is the radius of the sphere). After the ball enters, the opening closes automatically. Assuming that collisions between the ball and the sphere are perfectly elastic, find the trajectories of the ball and of the center of the sphere in the reference frame in which the sphere was initially at rest. Determine the parameters of these trajectories and mark on them the points at which collisions occur.

Setup and Assumptions

A small heavy ball of mass $m$ enters a hollow smooth spherical shell of radius $R$ and mass $m$. The shell is initially at rest in an inertial reference frame. The ball enters through a hole that subsequently closes, so the ball remains inside the shell.

The initial velocity of the ball is denoted by $\mathbf v_0$, with magnitude $v_0$. The incoming trajectory passes at a distance

$$b=\frac{R}{2}$$

from the center of the shell.

The shell is perfectly smooth, and every collision between the ball and the inner surface is perfectly elastic. External forces are neglected. Gravity is neglected. The shell is rigid.

The unknowns are the trajectory of the ball and the trajectory of the center of the shell in the frame where the shell was initially at rest, together with the positions of all collision points.

Since the masses are equal,

$$M=m.$$

Physical Principles

The system consists only of the ball and the shell. Since no external force acts on the system,

$$\mathbf P=\text{const},$$

where $\mathbf P$ is the total momentum.

During a collision, the force acts along the normal to the spherical surface. The normal passes through the center of the sphere.

For a perfectly elastic collision, the relative velocity of the colliding bodies reverses its normal component while preserving its magnitude. Since the collision is central and the masses are equal, the normal component of the relative velocity is exchanged between the bodies.

It is convenient to describe the motion in the center of mass frame. In that frame the center of mass moves uniformly, and the relative motion is equivalent to the motion of a point particle in a fixed spherical cavity with specular reflections from the wall.

For a spherical reflector, the angle of incidence equals the angle of reflection.

Derivation

Initially the shell is at rest and the ball has velocity $\mathbf v_0$. The center of mass velocity is

$$\mathbf V_{\rm cm} = \frac{m\mathbf v_0+m\cdot0}{2m} = \frac{\mathbf v_0}{2}.$$

The velocity of the ball relative to the center of mass is

$$\mathbf u = \mathbf v_0-\mathbf V_{\rm cm} = \frac{\mathbf v_0}{2}.$$

The shell center moves in the center of mass frame with velocity

$$-\frac{\mathbf v_0}{2}.$$

Consider the relative motion of the ball with respect to the shell center. The relative velocity is

$$\mathbf v_{\rm rel} = \mathbf v_0.$$

Thus, before the first collision, the relative trajectory is a straight chord of the sphere whose impact parameter is

$$b=\frac R2.$$

The chord length between successive reflections is

$$\ell = 2\sqrt{R^2-b^2} = 2\sqrt{R^2-\frac{R^2}{4}} = \sqrt3,R.$$

Let $\alpha$ be the angle between the radius drawn to a collision point and the incoming ray. Then

$$\sin\alpha = \frac bR = \frac12,$$

hence

$$\alpha=30^\circ.$$

The angle between the chord and the tangent plane is

$$90^\circ-\alpha=60^\circ.$$

After reflection, the ray makes the same angle with the normal. Since the sphere is perfectly symmetric, the reflected chord has the same distance $b$ from the center.

Let the first collision point be $A$. The reflected ray reaches a second collision point $B$. The central angle between successive collision points is obtained from the geometry of a chord with distance $R/2$ from the center:

$$\cos\frac{\Delta\varphi}{2} = \frac bR = \frac12.$$

Therefore

$$\frac{\Delta\varphi}{2}=60^\circ, \qquad \Delta\varphi=120^\circ.$$

Thus each collision point is obtained from the preceding one by a rotation of $120^\circ$ about the axis perpendicular to the plane of motion.

After three such steps,

$$3\Delta\varphi = 360^\circ,$$

so the fourth collision coincides with the first:

$$A\to B\to C\to A\to\cdots$$

The collision points form an equilateral triangle inscribed in the great circle lying in the plane of motion.

The time between successive collisions is

$$\tau = \frac{\ell}{v_{\rm rel}} = \frac{\sqrt3,R}{v_0}.$$

In the center of mass frame, the relative trajectory is a periodic broken line joining the vertices of this equilateral triangle.

Now return to the laboratory frame, where the shell was initially at rest.

The center of mass moves uniformly with velocity

$$\mathbf V_{\rm cm} = \frac{\mathbf v_0}{2}.$$

In the center of mass frame, the shell center moves along a straight line with velocity $-\mathbf v_0/2$. Transforming back to the laboratory frame,

$$\mathbf r_{\rm shell}(t) = \mathbf V_{\rm cm}t - \frac{\mathbf v_0}{2}t = 0.$$

Hence the center of the shell remains fixed at its original position for all time.

For the ball,

$$\mathbf r_{\rm ball}(t) = \mathbf r_{\rm shell}(t) + \mathbf r_{\rm rel}(t) = \mathbf r_{\rm rel}(t),$$

because the shell center is stationary.

Consequently, the ball moves along the same periodic broken-line trajectory described above: successive straight segments are chords of length $\sqrt3,R$, and the reflections occur at the three vertices $A$, $B$, and $C$ of the inscribed equilateral triangle.

Result

The center of mass velocity is

$$V_{\rm cm}=\frac{v_0}{2}.$$

The relative trajectory inside the sphere is a straight chord reflected specularly at the wall. Since

$$b=\frac R2,$$

the distance between successive collision points is

$$\ell=\sqrt3,R.$$

The central angle between consecutive collision points is

$$\Delta\varphi=120^\circ.$$

Therefore the collision points are the three vertices of an equilateral triangle inscribed in the circle of motion. The sequence of collisions is

$$A\rightarrow B\rightarrow C\rightarrow A\rightarrow\cdots.$$

The time between collisions is

$$\tau=\frac{\sqrt3,R}{v_0}.$$

The shell center remains at rest in the given reference frame:

$$\boxed{\mathbf r_{\rm shell}(t)=\text{const}}.$$

The ball moves periodically along the sides of the inscribed equilateral triangle, reflecting successively at its three vertices:

$$\boxed{\Delta\varphi=120^\circ,\qquad \ell=\sqrt3,R,\qquad \tau=\frac{\sqrt3,R}{v_0}}.$$

Sanity Checks

The dimensions are consistent. The chord length $\ell$ has units of length, and the collision interval $\tau=\ell/v_0$ has units of time.

If the impact parameter approached zero, the chord would become a diameter and the particle would oscillate along a single straight line through the center. The present result differs because $b=R/2$ produces a nontrivial polygonal orbit.

The relation

$$\cos\frac{\Delta\varphi}{2}=\frac bR$$

gives $\Delta\varphi=120^\circ$ when $b=R/2$, and three such angles exactly fill $360^\circ$. This confirms the periodicity after three reflections.

The step most susceptible to error is the determination of the angular advance between collision points. Using $\sin(\Delta\varphi/2)=b/R$ instead of the correct relation $\cos(\Delta\varphi/2)=b/R$ would lead to an incorrect polygon and an incorrect collision sequence.