A billiard ball rolling without slipping hits an identical, stationary billiard ball in a head-on collision. Describe the motion of the balls after the collision. Prove that the final state does not depend on the coefficient of sliding friction between the balls and the billiard table. (Rolling

Question

A billiard ball rolling without slipping hits an identical, stationary billiard ball in a head-on collision. Describe the motion of the balls after the collision. Prove that the final state does not depend on the coefficient of sliding friction between the balls and the billiard table. (Rolling friction is negligible.)

Accepted Answer

Friction between the two balls is negligible, and so, during the collision, they can only exert forces normal to their surfaces. Thus, the first ball stops after the collision, while the second acquires the first's initial speed [latex] v_0 [/latex]. The rotation of the balls, however, does not change, and so, immediately after the collision, the first ball rotates on one spot and the second slides without rotation at speed [latex] v_o.[/latex]

The friction between the balls and the table is of course important and affects the motion of the balls. The first is accelerated forward by the force of kinetic friction [latex] F_{fr} = μmg,[/latex] whilst the second is slowed by the same force, as shown in Fig. S94.1. The rotation of the first ball is reduced by friction, that of the second one is increased. The part played by the frictional force lasts until both balls reach the state of rolling without slipping. After that their motion is unchanged.

It will be shown that the final motion of the balls depends neither on the frictional coefficient, nor on the possible variation of it with position. After the collision, the initially moving ball rotates with an angular velocity [latex] w = v_o/r[/latex]. Its angular momentum about its axis is therefore [latex] I w = \frac{2}{5}mr^2 (v_o/r) = \frac{2}{5}mv_or[/latex]. The angular momentum about the point of contact with the table [latex] P,[/latex] must be the same, since the centre of mass of the ball is at rest, i.e. the angular momentum attributable to translation is zero. The angular momentum of the ball about [latex] P[/latex] cannot be changed by friction any more, as the line of action of this force runs through [latex] P[/latex]. (The sum of the gravitational force and the reaction of the table is zero, and so they can produce no net torque either.)

The angular momentum of a ball, rolling without slipping at speed [latex] v,[/latex] is the sum of its own angular momentum [latex] \frac{2}{5}mvr [/latex] and the angular momentum [latex] mvr [/latex] due to the motion of its centre of mass. Figure S94.2 shows (on the left) the initially moving ball and the forces acting on it shortly after the collision. On the right of the figure, the ball is shown ultimately rolling without slipping. According to the law of conservation of momentum [latex] \frac{2}{5}mv_0 r = \frac{2}{5}mvr + mvr,[/latex] which yields [latex] v = \frac{2}{7}v_o.[/latex] Similar reasoning shows that the final speed of the other ball has to be [latex] \frac{5}{7}v_o[/latex], regardless of the magnitude of the coefficient of friction.

Question P.94: A billiard ball rolling without slipping hits an identical, ...

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