Editing Collision (section)

==Examples==

===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).

===Perfect inelastic collision===
[[Image:Inelastischer stoß.gif|a completely inelastic collision between equal masses]]

In a perfect [[inelastic collision]], i.e., a zero [[coefficient of restitution]], the colliding particles [[Coalescence (physics)|coalesce]]. Using conservation of momentum:
::<math>m_a \mathbf v_{a1} + m_b \mathbf v_{b1} = \left( m_a + m_b \right) \mathbf v_2,</math>
the final velocity is given by
::<math>\mathbf v_2 = \frac{m_a \mathbf v_{a1} + m_b \mathbf v_{b1}}{m_a + m_b}.</math>
The reduction of total kinetic energy is equal to the total kinetic energy before the collision in a [[center of momentum frame]] with respect to the system of two particles, because in such a frame the kinetic energy after the collision is zero. In this frame most of the kinetic energy before the collision is that of the particle with the smaller mass. In another frame, in addition to the reduction of kinetic energy there may be a transfer of kinetic energy from one particle to the other; the fact that this depends on the frame shows how relative this is.
With time reversed we have the situation of two objects pushed away from each other, e.g. shooting a [[projectile]], or a [[rocket]] applying [[thrust]] (compare the [[Tsiolkovsky rocket equation#Derivation|derivation of the Tsiolkovsky rocket equation]]).

===Animal locomotion===
Collisions of an animal's foot or paw with the underlying substrate are generally termed ground reaction forces. These collisions are inelastic, as kinetic energy is not conserved. An important research topic in [[prosthetics]] is quantifying the forces generated during the foot-ground collisions associated with both disabled and non-disabled gait. This quantification typically requires subjects to walk across a [[force platform]] (sometimes called a "force plate") as well as detailed [[kinematic]] and [[Dynamics (mechanics)|dynamic]] (sometimes termed kinetic) analysis.

===Hypervelocity impacts===
[[File:HRIV Impact.gif|thumb|Video of the hypervelocity impact of NASA’s [[Deep Impact (spacecraft)|Deep Impact probe]] on comet [[Tempel 1]].]]
Hypervelocity is very high [[velocity]], approximately over 3,000 [[metre per second|meters per second]] (11,000&nbsp;km/h, 6,700&nbsp;mph, 10,000&nbsp;ft/s, or [[Mach number|Mach]] 8.8). In particular, hypervelocity is velocity so high that the strength of materials upon impact is very small compared to [[inertia]]l stresses.<ref Name="AIAA">{{cite book
|title= Critical technologies for national defense
|author= Air Force Institute of Technology
|year= 1991
|publisher= AIAA
|location= 
|isbn= 1-56347-009-8
|page= 287
|url= https://books.google.com/books?id=HsEorBWNGWwC&dq=Hypervelocity+3%2C000&pg=PA287}}</ref> Thus, [[metal]]s and [[fluid]]s behave alike under hypervelocity impact. An impact under extreme hypervelocity results in [[vaporize|vaporization]] of the [[impact force|impactor]] and target. For structural metals, hypervelocity is generally considered to be over 2,500&nbsp;m/s (5,600&nbsp;mph, 9,000&nbsp;km/h, 8,200&nbsp;ft/s, or Mach 7.3).  [[Meteorite]] [[impact crater|craters]] are also examples of hypervelocity impacts.