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hiddentrue

This problem presents a valuable shortcut for dealing with a specific kind of elastic collision.


The game of pool is an excellent showcase of interesting physics. One of the most common occurences in a game of pool is an almost perfectly elastic collision between a moving cue ball and a stationary ball of equal mass. These collisions occur in two dimensions.

One very useful rule of thumb for pool players is the right angle rule: after the collision the cue ball's final velocity and the final velocity of the struck ball will make a right angle (assuming that both are moving).

Note

Note that this rule is only valid in the limit that ball spin can be ignored. If the cue ball has significant spin, then the rule will be violated.

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The rule is generically valid for *elastic* collisions between *equal mass* objects when *one is stationary* before the collision.  A short proof is to square the magnitude of each side of the vector version of the equation of momentum conservation:\\ \\
{latex}\begin{large} \[ m^{2}v_{1,i}^{2} = m^{2}(v_{1,f}^{2} + 2\vec{v}_{1,f}\cdot \vec{v}_{2,f} + v_{2,f}^{2}) \]\end{large}{latex} \\

Cancelling the masses and comparing to the equation of kinetic energy conservation will immediately yield the result that \\
{latex}\begin{large}\[ \vec{v}_{1,f}\cdot\vec{v}_{2,f} = 0 \]\end{large}{latex}
which implies that (a.) one of the objects has zero final velocity or else (b.) the objects move at right angles to one another after the collision.

When this rule applies, it is very powerful. The following examples showcase the utility of the right angle rule.

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