Kvant Physics Problem 373

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Problem

Two billiard balls, one of which is initially at rest, undergo an elastic “oblique” collision. The line passing through the centers of the balls at the moment of collision makes an angle $60^\circ$ with the direction of the initial motion of the incoming ball. During the collision, the balls deform, and part of the kinetic energy of the incoming ball is converted into potential energy of elastic deformation of the balls, which, upon separation of the balls, is again converted into kinetic energy. Determine the maximum fraction of the energy of the balls that is converted into elastic deformation energy during the impact. The balls are assumed to be perfectly smooth.

Setup and Assumptions

Two identical billiard balls of mass $m$ move on a horizontal frictionless table. Ball 1 approaches with initial speed $v$ along a straight line, and ball 2 is initially at rest. The collision is perfectly elastic, meaning total kinetic energy of the two-ball system is conserved over the entire interaction.

The balls are perfectly smooth, so the impulse during collision acts only along the common normal at the point of contact, which coincides with the line joining the centers at the instant of impact. The line of centers at contact makes an angle $\theta = 60^\circ$ with the initial velocity of ball 1.

The task is to determine the maximum fraction of the total kinetic energy of the two-ball system that is stored as elastic deformation energy at any instant during the collision process.

The analysis is performed in the laboratory frame, with no external forces acting on the system during the collision.

Physical Principles

The total kinetic energy of the two-ball system is conserved for a perfectly elastic collision, while during the collision part of the translational kinetic energy is temporarily converted into elastic potential energy associated with compression along the normal direction.

The impulse acts only along the line of centers, so the tangential component of velocity of each ball remains unchanged throughout the collision.

The normal component of the relative motion behaves as a one-dimensional collision characterized by the reduced mass

$\mu = \frac{m_1 m_2}{m_1 + m_2} = \frac{m}{2}.$

For the relative motion along the line of centers, the kinetic energy associated with the normal component of relative velocity is temporarily stored as elastic deformation energy at the instant of maximum compression, when the normal relative velocity becomes zero.

The normal component of the initial relative velocity is

$v_n = v \cos\theta.$

The elastic energy at maximum compression equals the initial kinetic energy of the reduced-mass system associated with this normal relative motion:

$E_{\max} = \frac{1}{2}\mu v_n^2.$

The total initial kinetic energy of the system is

$E_0 = \frac{1}{2} m v^2.$

Derivation

The initial velocity of ball 1 is decomposed into components relative to the line of centers. The normal component is

$v_n = v \cos 60^\circ = \frac{v}{2}.$

Only this normal component contributes to compression because the tangential component does not change during collision and does not store elastic energy.

The relative motion along the normal direction is equivalent to a one-dimensional collision of two masses with reduced mass

$\mu = \frac{m}{2}.$

The kinetic energy associated with this relative normal motion at the beginning of impact is

$E_{\text{rel},n} = \frac{1}{2}\mu v_n^2.$

Substituting $\mu = \frac{m}{2}$ and $v_n = \frac{v}{2}$ gives

$E_{\text{rel},n} = \frac{1}{2}\cdot \frac{m}{2}\cdot \frac{v^2}{4} = \frac{m v^2}{16}.$

At maximum compression, all of this normal relative kinetic energy is converted into elastic deformation energy, so

$E_{\max} = \frac{m v^2}{16}.$

The initial total kinetic energy of the system is

$E_0 = \frac{1}{2} m v^2.$

The required fraction is

$\frac{E_{\max}}{E_0} = \frac{\frac{m v^2}{16}}{\frac{1}{2} m v^2} = \frac{1}{8}.$

Result

The maximum fraction of the total kinetic energy stored as elastic deformation energy is

$\frac{E_{\max}}{E_0} = \frac{1}{8}.$

Substituting units consistently confirms cancellation of mass and velocity:

$\frac{[ \text{kg} \cdot \text{m}^2 \cdot \text{s}^{-2} ]}{[ \text{kg} \cdot \text{m}^2 \cdot \text{s}^{-2} ]} = 1.$

$\boxed{0.125}$

Sanity Checks

The expression depends only on $\cos^2 60^\circ$, consistent with the fact that only the normal velocity component contributes to compression energy. The result scales quadratically with the normal projection, so replacing $60^\circ$ by $0^\circ$ would yield full conversion into normal relative motion and increase the fraction to $\frac{1}{2}$, consistent with a head-on collision of equal masses where the reduced mass energy is half the total initial kinetic energy.

Dimensional consistency holds since both numerator and denominator represent energy, ensuring a dimensionless ratio.

The most error-prone step is the decomposition of kinetic energy into the reduced-mass normal-motion form, where omitting the reduced mass factor $\mu = \frac{m}{2}$ would incorrectly overestimate the stored energy by a factor of two.