Editing Momentum (section)

====Elastic collisions====
{{Main|Elastic collision}}
[[File:Elastischer stoß.gif|thumb|right|Elastic collision of equal masses]]
[[File:Elastischer stoß3.gif|thumb|right|Elastic collision of unequal masses]]
An elastic collision is one in which no [[kinetic energy]] is transformed into heat or some other form of energy. Perfectly elastic collisions can occur when the objects do not touch each other, as for example in atomic or nuclear scattering where electric repulsion keeps the objects apart. A [[gravity assist|slingshot maneuver]] of a satellite around a planet can also be viewed as a perfectly elastic collision. A collision between two [[Pool billiards|pool]] balls is a good example of an ''almost'' totally elastic collision, due to their high [[stiffness|rigidity]], but when bodies come in contact there is always some [[dissipation]].<ref>{{cite web |url=http://hyperphysics.phy-astr.gsu.edu/hbase/elacol.html |title=Elastic and inelastic collisions |work=Hyperphysics |first=Carl |last=Nave |date=2010 |access-date=2 August 2012  |archive-url=https://web.archive.org/web/20120818114930/http://hyperphysics.phy-astr.gsu.edu/hbase/elacol.html |archive-date=18 August 2012 }}</ref>

A head-on elastic collision between two bodies can be represented by velocities in one dimension, along a line passing through the bodies. If the velocities are {{math|{{var|v}}{{sub|A1}}}} and {{math|{{var|v}}{{sub|B1}}}} before the collision and {{math|{{var|v}}{{sub|A2}}}} and {{math|{{var|v}}{{sub|B2}}}} after, the equations expressing conservation of momentum and kinetic energy are:

<math display="block">\begin{align} m_{A} v_{A1} + m_{B} v_{B1} &= m_{A} v_{A2} + m_{B} v_{B2}\\
\tfrac{1}{2} m_{A} v_{A1}^2 + \tfrac{1}{2} m_{B} v_{B1}^2 &= \tfrac{1}{2} m_{A} v_{A2}^2 + \tfrac{1}{2} m_{B} v_{B2}^2\,.\end{align}</math>

A change of reference frame can simplify analysis of a collision. For example, suppose there are two bodies of equal mass {{mvar|m}}, one stationary and one approaching the other at a speed {{mvar|v}} (as in the figure). The center of mass is moving at speed {{math|{{sfrac|{{var|v}}|2}}}} and both bodies are moving towards it at speed {{math|{{sfrac|{{var|v}}|2}}}}. Because of the symmetry, after the collision both must be moving away from the center of mass at the same speed. Adding the speed of the center of mass to both, we find that the body that was moving is now stopped and the other is moving away at speed {{mvar|v}}. The bodies have exchanged their velocities. Regardless of the velocities of the bodies, a switch to the center of mass frame leads us to the same conclusion. Therefore, the final velocities are given by<ref name=FeynmanCh10/>

<math display="block">\begin{align}  v_{A2} &= v_{B1}\\
v_{B2} &= v_{A1}\,. \end{align}</math>

In general, when the initial velocities are known, the final velocities are given by<ref>{{cite book|last1=Serway|first1=Raymond A.|first2=John W. Jr. |last2=Jewett |title=Principles of physics: a calculus-based text|date=2012|publisher=Brooks/Cole, Cengage Learning|location=Boston, Massachusetts|isbn=978-1-133-10426-1|page=245|edition=5th}}</ref>

<math display="block">\begin{align}
 v_{A2} &= \left( \frac{m_{A} - m_{B}}{m_{A} + m_{B}} \right) v_{A1} + \left( \frac{2 m_{B}}{m_{A} + m_{B}} \right) v_{B1} \\
 v_{B2} &= \left( \frac{m_{B} - m_{A}}{m_{A} + m_{B}} \right) v_{B1} + \left( \frac{2 m_{A}}{m_{A} + m_{B}} \right) v_{A1}\,.
\end{align}</math>

If one body has much greater mass than the other, its velocity will be little affected by a collision while the other body will experience a large change.