Kvant Physics Problem 57

Two identical balls, each of mass $m$, are connected by a massless spring of stiffness $k$.

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Problem

Two identical balls connected by a massless spring move along a smooth horizontal floor with the same speed, perpendicular to a vertical wall. Describe how the collision of the system with the wall takes place. How will the balls move after the impact? The collision of a ball with the wall is perfectly elastic, and the collision time may be neglected.

Setup and Assumptions

Two identical balls, each of mass $m$, are connected by a massless spring of stiffness $k$. The system moves on a smooth horizontal floor toward a rigid vertical wall. The common velocity of both balls before any collision is $v_0$, directed perpendicular to the wall.

Let the wall be located at $x=0$, and let the positive $x$ direction point away from the wall. Initially the spring is undeformed, so its length is equal to the natural length $l_0$. The ball closer to the wall will be called ball 1, and the other ball will be called ball 2.

The collision of a ball with the wall is perfectly elastic, and its duration is neglected. Consequently, when a ball strikes the wall, its velocity instantaneously changes sign while retaining the same magnitude.

The floor is smooth, so no external horizontal forces act except during the instantaneous impacts with the wall. The spring obeys Hooke's law, and its mass is neglected.

The task is to describe the entire subsequent motion and determine the motion of the balls after the interaction with the wall has ended.

Physical Principles

The motion between impacts is governed by Newton's second law and Hooke's law.

If $x_1$ and $x_2$ are the coordinates of the balls, the spring force is proportional to the extension

$\xi=x_2-x_1-l_0.$

The equations of motion are

$m\ddot x_1=k\xi,$

$m\ddot x_2=-k\xi.$

The center-of-mass coordinate

$X=\frac{x_1+x_2}{2}$

satisfies

$\ddot X=0$

whenever neither ball is in contact with the wall.

The relative coordinate $\xi$ satisfies

$\ddot \xi+\omega^2\xi=0,$

where

$\omega=\sqrt{\frac{2k}{m}}.$

A perfectly elastic collision with the wall reverses the velocity of the colliding ball:

$u\rightarrow -u.$

During such an instantaneous collision, the position of the ball does not change.

Derivation

Initially both balls move toward the wall with velocity

$\dot x_1=\dot x_2=-v_0,$

and the spring is undeformed:

$\xi=0, \qquad \dot\xi=0.$

Ball 1 reaches the wall first. At the instant immediately after the collision,

$\dot x_1=+v_0, \qquad \dot x_2=-v_0.$

Hence

$\dot X=\frac{\dot x_1+\dot x_2}{2}=0,$

so the center of mass remains at rest for all subsequent motion until any further wall interaction occurs.

For the relative coordinate,

$\dot\xi=\dot x_2-\dot x_1=-2v_0.$

Since $\xi=0$ at that instant, the subsequent relative motion is

$\xi(t)=-\frac{2v_0}{\omega}\sin\omega t.$

The spring is first compressed. The maximum compression is

$|\xi|_{\max} = \frac{2v_0}{\omega},$

which gives

$|\xi|_{\max} = v_0\sqrt{\frac{2m}{k}}.$

Because the center of mass is at rest, the individual velocities are determined by symmetry. From $v_1+v_2=0$ and $\dot\xi=v_2-v_1$, one obtains

$v_2=\frac{\dot\xi}{2}, \qquad v_1=-\frac{\dot\xi}{2},$

hence

$v_2(t)=-v_0\cos\omega t, \qquad v_1(t)=+v_0\cos\omega t.$

After half a period,

$t=\frac{\pi}{\omega},$

the spring again has its natural length:

$\xi=0,$

and the relative velocity becomes

$\dot\xi = -2v_0\cos\omega t = 2v_0.$

At this instant,

$\dot x_1=-v_0, \qquad \dot x_2=+v_0.$

Ball 1 is again at the wall at this moment and undergoes a second perfectly elastic collision, reversing its velocity to $+v_0$, while ball 2 retains velocity $+v_0$. Immediately after this collision,

$\dot x_1=\dot x_2=+v_0, \qquad \xi=0,$

so the system performs uniform translation away from the wall.

Resolution of the collision-ordering issue

After the first collision, the coordinates satisfy

$x_1=X-\frac{l_0+\xi}{2}, \qquad x_2=X+\frac{l_0+\xi}{2},$

and since $X=l_0/2$, this gives

$x_2(t)=l_0+\frac{\xi(t)}{2}.$

Substituting the oscillation,

$x_2(t)=l_0-\frac{v_0}{\omega}\sin\omega t.$

The minimum position of ball 2 occurs when $\sin\omega t=1$, hence at $t=\frac{\pi}{2\omega}$,

$x_{2,\min}=l_0-\frac{v_0}{\omega}.$

If $x_{2,\min}>0$, ball 2 never reaches the wall. If $x_{2,\min}\le 0$, a contact with the wall would occur during the motion of the internal oscillator.

The corresponding velocity is

$v_2(t)=-v_0\cos\omega t,$

so at the instant of minimum position,

$v_2\left(\frac{\pi}{2\omega}\right)=0.$

Thus any possible contact with the wall by ball 2 occurs only at zero velocity. A perfectly elastic collision at zero velocity produces no change in motion, since the velocity reversal $u\to -u$ leaves $u=0$ invariant. Therefore, even in the regime $l_0 \le \frac{v_0}{\omega}$, no additional dynamical stage is generated and the subsequent evolution remains identical.

This removes any ambiguity in collision ordering and shows that the only dynamically relevant wall interaction after the first event is the second impact of ball 1 at $t=\pi/\omega$.

Result

The collision process consists of two instantaneous impacts of the ball nearest the wall.

After the first impact, the center of mass remains at rest and the system performs harmonic oscillations in the relative coordinate with angular frequency $\omega=\sqrt{2k/m}$. The spring reaches maximum compression

$\xi_{\max} = -v_0\sqrt{\frac{2m}{k}}, \qquad |\xi_{\max}| = v_0\sqrt{\frac{2m}{k}}.$

After a time

$t=\frac{\pi}{\omega} = \pi\sqrt{\frac{m}{2k}},$

the spring returns to its natural length, ball 1 collides again with the wall, and both balls acquire the same velocity.

Immediately after the second collision,

$\dot x_1=\dot x_2=v_0, \qquad \xi=0,$

so the final motion is uniform translation away from the wall with speed

$\boxed{v_{\text{final}}=v_0}.$

Sanity Checks

The dimensions of the maximum compression are correct:

$v_0\sqrt{\frac{m}{k}} \sim \frac{\mathrm{m}}{\mathrm{s}} \sqrt{ \frac{\mathrm{kg}} {\mathrm{kg}/\mathrm{s}^2} } = \mathrm{m}.$

If the spring becomes extremely stiff, $k\to\infty$, then

$|\xi|_{\max}\to0,$

so the system approaches rigid-body behavior.

If the spring becomes very soft, $k\to0$, the oscillation period becomes large and energy exchange between kinetic and elastic forms dominates the motion before the second collision.

The total kinetic energy before the first collision equals

$m v_0^2.$

After the first collision, energy is fully contained in the internal oscillation of the spring, while the center of mass remains at rest. After the second collision, the kinetic energy again equals

$m v_0^2,$

confirming complete recovery of the initial translational state and overall energy conservation throughout the process.