Editing Elastic collision (section)

===One-dimensional Newtonian===
[[File:Prof Walter Lewin Elastic Collisions.ogv|thumb|Professor [[Walter Lewin]] explaining one-dimensional elastic collisions]]

In any collision without an external force, [[momentum]] is conserved; but in an elastic collision, kinetic energy is also conserved.<ref name=serway257>{{harvnb|Serway|Jewett|2014|p=257}}</ref> Consider particles A and B with masses ''m''<sub>A</sub>, ''m''<sub>B</sub>, and velocities ''v''<sub>A1</sub>, ''v''<sub>B1</sub> before collision, ''v''<sub>A2</sub>, ''v''<sub>B2</sub> after collision. The conservation of momentum before and after the collision is expressed by:<ref name=serway257 />
<math display="block"> m_{A}v_{A1}+m_{B}v_{B1} \ =\ m_{A}v_{A2} + m_{B}v_{B2}.</math>

Likewise, the conservation of the total [[kinetic energy]] is expressed by:<ref name=serway257 />
<math display="block">\tfrac12 m_{A}v_{A1}^2+\tfrac12 m_{B}v_{B1}^2 \ =\ \tfrac12 m_{A}v_{A2}^2 +\tfrac12 m_{B}v_{B2}^2.</math>

These equations may be solved directly to find <math>v_{A2},v_{B2}</math> when <math>v_{A1},v_{B1}</math> are known:<ref name=serway258>{{harvnb|Serway|Jewett|2014|p=258}}</ref>

<math display="block">
\begin{array}{ccc}
v_{A2} &=& \dfrac{m_A-m_B}{m_A+m_B} v_{A1} + \dfrac{2m_B}{m_A+m_B} v_{B1} \\[.5em]
v_{B2} &=& \dfrac{2m_A}{m_A+m_B} v_{A1} + \dfrac{m_B-m_A}{m_A+m_B} v_{B1}.
\end{array}
</math>

Alternatively the final velocity of a particle, v<sub>2</sub> (v<sub>A2</sub> or v<sub>B2</sub>) is expressed by:

<math>v_2 = (1+e)v_{CoM}-ev_1</math>

Where:
*e is the [[coefficient of restitution]].
*v<sub>CoM</sub> is the velocity of the [[center of mass]] of the system of two particles:
<math>v_{CoM} = \dfrac{m_Av_{A1} + m_Bv_{B1}}{m_A+m_B}</math>
*v<sub>1</sub> (v<sub>A1</sub> or v<sub>B1</sub>) is the initial velocity of the particle.

If both masses are the same, we have a trivial solution:
<math display="block">
\begin{align}
v_{A2} &= v_{B1} \\
v_{B2} &= v_{A1}.
\end{align}</math>
This simply corresponds to the bodies exchanging their initial velocities with each other.<ref name=serway258 />

As can be expected, the solution is invariant under adding a constant to all velocities ([[Galilean relativity]]), which is like using a frame of reference with constant translational velocity. Indeed, to derive the equations, one may first change the frame of reference so that one of the known velocities is zero, determine the unknown velocities in the new frame of reference, and convert back to the original frame of reference.

====Examples====

;Before collision:
:Ball A: mass = 3 kg,       velocity = 4 m/s
:Ball B: mass = 5 kg,       velocity = 0 m/s
;After collision:
:Ball A: velocity = −1 m/s
:Ball B: velocity = 3 m/s

Another situation:
[[Image:Elastischer stoß3.gif|frame|center|Elastic collision of unequal masses.]]

The following illustrate the case of equal mass, <math>m_A=m_B</math>.

[[Image:Elastischer stoß.gif|frame|center|Elastic collision of equal masses]]

[[Image:Elastischer stoß2.gif|frame|center|Elastic collision of masses in a system with a moving frame of reference]]

In the limiting case where <math>m_{A}</math> is much larger than <math>m_{B}</math>, such as a ping-pong paddle hitting a ping-pong ball or an SUV hitting a trash can, the heavier mass hardly changes velocity, while the lighter mass bounces off, reversing its velocity plus approximately twice that of the heavy one.<ref name=serway258-9>{{harvnb|Serway|Jewett|2014|pp=258–259}}</ref>

In the case of a large <math>v_{A1}</math>, the value of <math>v_{A2}</math> is small if the masses are approximately the same: hitting a much lighter particle does not change the velocity much, hitting a much heavier particle causes the fast particle to bounce back with high speed. This is why a [[neutron moderator]] (a medium which slows down [[fast neutron]]s, thereby turning them into [[thermal neutron]]s capable of sustaining a [[chain reaction]]) is a material full of atoms with light nuclei which do not easily absorb neutrons: the lightest nuclei have about the same mass as a [[neutron]].

====Derivation of solution====
To derive the above equations for <math>v_{A2},v_{B2},</math> rearrange the kinetic energy and momentum equations:

<math display="block">\begin{align}
m_A(v_{A2}^2-v_{A1}^2) &= m_B(v_{B1}^2-v_{B2}^2) \\
m_A(v_{A2}-v_{A1}) &= m_B(v_{B1}-v_{B2})
\end{align}</math>

Dividing each side of the top equation by each side of the bottom equation, and using <math>\tfrac{a^2-b^2}{(a-b)} = a+b,</math> gives:

<math display=block> v_{A2}+v_{A1}=v_{B1}+v_{B2}  \quad\Rightarrow\quad v_{A2}-v_{B2} = v_{B1}-v_{A1}</math>

That is, the relative velocity of one particle with respect to the other is reversed by the collision. 

Now the above formulas follow from solving a system of linear equations for <math>v_{A2},v_{B2},</math> regarding <math>m_A,m_B,v_{A1},v_{B1}</math> as constants:
<math display="block">\left\{\begin{array}{rcrcc}
v_{A2} & - & v_{B2} &=& v_{B1}-v_{A1} \\
m_Av_{A1}&+&m_Bv_{B1} &=& m_Av_{A2}+m_Bv_{B2}.
\end{array}\right.</math>
Once <math>v_{A2}</math> is determined, <math>v_{B2}</math> can be found by symmetry.

====Center of mass frame====
With respect to the center of mass, both velocities are reversed by the collision: a heavy particle moves slowly toward the center of mass, and bounces back with the same low speed, and a light particle moves fast toward the center of mass, and bounces back with the same high speed.

The velocity of the [[center of mass]] does not change by the collision. To see this, consider the center of mass at time <math> t </math> before collision and time <math> t' </math> after collision:
<math display="block">\begin{align}
\bar{x}(t) &= \frac{m_{A} x_{A}(t)+m_{B} x_{B}(t)}{m_{A}+m_{B}} \\
\bar{x}(t') &= \frac{m_{A} x_{A}(t')+m_{B} x_{B}(t')}{m_{A}+m_{B}}.
\end{align}</math>

Hence, the velocities of the center of mass before and after collision are:
<math display="block">\begin{align}
v_{ \bar{x} } &= \frac{m_{A}v_{A1}+m_{B}v_{B1}}{m_{A}+m_{B}} \\
v_{ \bar{x} }' &= \frac{m_{A}v_{A2}+m_{B}v_{B2}}{m_{A}+m_{B}}.
\end{align}</math>

The numerators of <math> v_{ \bar{x} } </math> and  <math> v_{ \bar{x} }' </math> are the total momenta before and after collision. Since momentum is conserved, we have <math> v_{ \bar{x} } = v_{ \bar{x} }' \,.</math>