Editing Collision (section)

===Billiards===
{{Anchor|Cue sports}}Collisions play an important role in [[cue sports]]. Because the collisions between [[billiard balls]] are nearly [[Elastic collision|elastic]], and the balls roll on a surface that produces low [[rolling friction]], their behavior is often used to illustrate [[Newton's laws of motion]]. After a zero-friction collision of a moving ball with a stationary one of equal mass, the angle between the directions of the two balls is 90 degrees. This is an important fact that professional billiards players take into account,<ref>{{cite web|last=Alciatore |first=David G. |date=January 2006 |url=http://billiards.colostate.edu/technical_proofs/TP_3-1.pdf |archive-url=https://ghostarchive.org/archive/20221009/http://billiards.colostate.edu/technical_proofs/TP_3-1.pdf |archive-date=2022-10-09 |url-status=live |title=TP 3.1 90° rule |access-date=2008-03-08 }}</ref> although it assumes the ball is moving without any impact of friction across the table rather than rolling with friction.
Consider an elastic collision in two dimensions of any two masses ''m''<sub>a</sub> and ''m''<sub>b</sub>, with respective initial velocities '''v'''<sub>a1</sub>  and '''v'''<sub>b1</sub> where '''v'''<sub>b1</sub> = '''0''', and final velocities '''v'''<sub>a2</sub> and '''v'''<sub>b2</sub>.
Conservation of momentum gives ''m''<sub>a</sub>'''v'''<sub>a1</sub> = ''m''<sub>a</sub>'''v'''<sub>a2</sub> + ''m''<sub>b</sub>'''v'''<sub>b2</sub>.
Conservation of energy for an elastic collision gives (1/2)''m''<sub>a</sub>|'''v'''<sub>a1</sub>|<sup>2</sup> = (1/2)''m''<sub>a</sub>|'''v'''<sub>a2</sub>|<sup>2</sup> + (1/2)''m''<sub>b</sub>|'''v'''<sub>b2</sub>|<sup>2</sup>.
Now consider the case ''m''<sub>a</sub> = ''m''<sub>b</sub>: we obtain '''v'''<sub>a1</sub> = '''v'''<sub>a2</sub> + '''v'''<sub>b2</sub> and |'''v'''<sub>a1</sub>|<sup>2</sup> = |'''v'''<sub>a2</sub>|<sup>2</sup> + |'''v'''<sub>b2</sub>|<sup>2</sup>.
Taking the [[dot product]] of each side of the former equation with itself, |'''v'''<sub>a1</sub>|<sup>2</sup> = '''v'''<sub>a1</sub>•'''v'''<sub>a1</sub> = |'''v'''<sub>a2</sub>|<sup>2</sup> + |'''v'''<sub>b2</sub>|<sup>2</sup> + 2'''v'''<sub>a2</sub>•'''v'''<sub>b2</sub>. Comparing this with the latter equation gives '''v'''<sub>a2</sub>•'''v'''<sub>b2</sub> = 0, so they are perpendicular unless '''v'''<sub>a2</sub> is the zero vector (which occurs [[if and only if]] the collision is head-on).