Editing Elastic collision (section)

==Equations==

===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>

===One-dimensional relativistic===
According to [[special relativity]],
<math display="block">p = \frac{mv}{\sqrt{1-\frac{v^2}{c^2}}}</math>
where ''p'' denotes momentum of any particle with mass, ''v'' denotes velocity, and ''c'' is the speed of light.

In the [[center of momentum frame]] where the total momentum equals zero, 
<math display="block">\begin{align}
p_1 &= - p_2 \\
p_1^2 &= p_2^2 \\
E &= \sqrt {m_1^2c^4 + p_1^2c^2} + \sqrt {m_2^2c^4 + p_2^2c^2} = E \\
p_1 &= \pm \frac{\sqrt{E^4 - 2E^2m_1^2c^4 - 2E^2m_2^2c^4 + m_1^4c^8 - 2m_1^2m_2^2c^8 + m_2^4c^8}}{2cE} \\
u_1 &= -v_1.
\end{align}</math>

Here <math>m_1, m_2</math> represent the [[rest mass]]es of the two colliding bodies, <math>u_1, u_2</math> represent their velocities before collision, <math>v_1, v_2</math> their velocities after collision, <math>p_1, p_2</math> their momenta, <math>c</math> is the [[speed of light]] in vacuum, and <math>E</math> denotes the total energy, the sum of rest masses and kinetic energies of the two bodies.

Since the total energy and momentum of the system are conserved and their rest masses do not change, it is shown that the momentum of the colliding body is decided by the rest masses of the colliding bodies, total energy and the total momentum.  Relative to the [[center of momentum frame]], the momentum of each colliding body does not change magnitude after collision, but reverses its direction of movement.

Comparing with [[Classical Mechanics|classical mechanics]], which gives accurate results dealing with macroscopic objects moving much slower than the [[speed of light]], total momentum of the two colliding bodies is frame-dependent. In the [[center of momentum frame]], according to classical mechanics,

<math display="block">\begin{align}
m_1u_1 + m_2u_2 &= m_1v_1 + m_2v_2 = 0 \\
m_1u_1^2 + m_2u_2^2 &= m_1v_1^2 + m_2v_2^2 \\
\frac{(m_2u_2)^2}{2m_1} + \frac{(m_2u_2)^2}{2m_2} &= \frac{(m_2v_2)^2}{2m_1} + \frac{(m_2v_2)^2}{2m_2} \\
(m_1 + m_2)(m_2u_2)^2 &= (m_1 + m_2)(m_2v_2)^2 \\
u_2 &= -v_2 \\
\frac{(m_1u_1)^2}{2m_1} + \frac{(m_1u_1)^2}{2m_2} &= 
\frac{(m_1v_1)^2}{2m_1} + \frac{(m_1v_1)^2}{2m_2} \\
(m_1 + m_2)(m_1u_1)^2 &= (m_1 + m_2)(m_1v_1)^2 \\
u_1 &= -v_1\,.
\end{align}
</math>

This agrees with the relativistic calculation <math>u_1 = -v_1,</math> despite other differences. 

One of the postulates in Special Relativity states that the laws of physics, such as conservation of momentum, should be invariant in all inertial frames of reference. In a general inertial frame where the total momentum could be arbitrary,

<math display="block">\begin{align}
\frac{m_1\;u_1}{\sqrt{1-u_1^2/c^2}} +
\frac{m_2\;u_2}{\sqrt{1-u_2^2/c^2}} &= 
\frac{m_1\;v_1}{\sqrt{1-v_1^2/c^2}} +
\frac{m_2\;v_2}{\sqrt{1-v_2^2/c^2}}=p_T \\
\frac{m_1c^2}{\sqrt{1-u_1^2/c^2}} +
\frac{m_2c^2}{\sqrt{1-u_2^2/c^2}} &=
\frac{m_1c^2}{\sqrt{1-v_1^2/c^2}} +
\frac{m_2c^2}{\sqrt{1-v_2^2/c^2}}=E
\end{align}</math>
We can look at the two moving bodies as one system of which the total momentum is <math>p_T,</math> the total energy is <math>E</math> and its velocity <math>v_c</math> is the velocity of its center of mass. Relative to the center of momentum frame the total momentum equals zero. It can be shown that <math>v_c</math> is given by:
<math display="block">v_c = \frac{p_T c^2}{E}</math>
Now the velocities before the collision in the center of momentum frame <math>u_1 '</math> and <math>u_2 '</math> are:
<math display="block">\begin{align}
u_1' &= \frac{u_1 - v_c}{1- \frac{u_1 v_c}{c^2}} \\
u_2' &= \frac{u_2 - v_c}{1- \frac{u_2 v_c}{c^2}} \\
v_1' &= -u_1' \\
v_2' &= -u_2' \\
v_1 &= \frac{v_1' + v_c}{1+ \frac{v_1' v_c}{c^2}} \\
v_2 &= \frac{v_2' + v_c}{1+ \frac{v_2' v_c}{c^2}}
\end{align}</math>

When <math>u_1 \ll c</math> and <math>u_2 \ll c\,, </math>
<math display="block">\begin{align}
p_T  &\approx m_1 u_1 + m_2 u_2 \\
v_c  &\approx \frac{m_1 u_1 + m_2 u_2}{m_1 + m_2} \\
u_1' &\approx u_1 - v_c \approx \frac {m_1 u_1 + m_2 u_1 - m_1 u_1 - m_2 u_2}{m_1 + m_2} = \frac {m_2 (u_1 - u_2)}{m_1 + m_2} \\
u_2' &\approx \frac {m_1 (u_2 - u_1)}{m_1 + m_2} \\
v_1' &\approx \frac {m_2 (u_2 - u_1)}{m_1 + m_2} \\
v_2' &\approx \frac {m_1 (u_1 - u_2)}{m_1 + m_2} \\
v_1  &\approx v_1' + v_c \approx \frac {m_2 u_2 - m_2 u_1 + m_1 u_1 + m_2 u_2}{m_1 + m_2} = \frac{u_1 (m_1 - m_2) + 2m_2 u_2}{m_1 + m_2} \\
v_2  &\approx \frac{u_2 (m_2 - m_1) + 2m_1 u_1}{m_1 + m_2}
\end{align}</math>

Therefore, the classical calculation holds true when the speed of both colliding bodies is much lower than the speed of light (~300,000 kilometres per second).

===Relativistic derivation using hyperbolic functions===

Using the so-called ''parameter of velocity'' <math>s</math> (usually called the [[rapidity]]),

<math display="block">\frac{v}{c}=\tanh(s),</math>
we get
<math display="block">\sqrt{1-\frac{v^2}{c^2}}=\operatorname{sech}(s).</math>

Relativistic energy and momentum are expressed as follows:
<math display="block">\begin{align}
E &= \frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}} = m c^2 \cosh(s) \\
p &= \frac{mv}{\sqrt{1-\frac{v^2}{c^2}}}=m c \sinh(s)
\end{align}</math>

Equations sum of energy and momentum colliding masses <math>m_1</math> and <math>m_2,</math> (velocities <math>v_1, v_2, u_1, u_2</math> correspond to the velocity parameters <math>s_1, s_2, s_3, s_4</math>), after dividing by adequate power <math>c</math> are as follows:
<math display="block">\begin{align}
m_1 \cosh(s_1)+m_2 \cosh(s_2) &= m_1 \cosh(s_3)+m_2 \cosh(s_4) \\
m_1 \sinh(s_1)+m_2 \sinh(s_2) &= m_1 \sinh(s_3)+m_2 \sinh(s_4)
\end{align}</math>

and dependent equation, the sum of above equations:
<math display="block">m_1 e^{s_1}+m_2 e^{s_2}=m_1 e^{s_3}+m_2 e^{s_4}</math>

subtract squares both sides equations "momentum" from "energy" and use the identity <math display="inline">\cosh^2(s)-\sinh^2(s)=1,</math> after simplifying we get:
<math display="block">2 m_1 m_2 (\cosh(s_1) \cosh(s_2)-\sinh(s_2) \sinh(s_1)) = 2 m_1 m_2 (\cosh(s_3) \cosh(s_4)-\sinh(s_4) \sinh(s_3))</math>

for non-zero mass, using the hyperbolic trigonometric identity <math display="inline">\cosh(a-b)=\cosh(a)\cosh(b)-\sinh(b)\sinh(a),</math> we get:
<math display="block">\cosh(s_1-s_2) = \cosh(s_3-s_4)</math>

as functions <math>\cosh(s)</math> is even we get two solutions:
<math display="block">\begin{align}
s_1-s_2 &= s_3-s_4 \\
s_1-s_2 &= -s_3+s_4
\end{align}</math>
from the last equation, leading to a non-trivial solution, we solve <math>s_2</math> and substitute into the dependent equation, we obtain <math>e^{s_1}</math> and then <math>e^{s_2},</math> we have:
<math display="block">\begin{align}
e^{s_1} &= e^{s_4}{\frac{m_1 e^{s_3}+m_2 e^{s_4}} {m_1 e^{s_4}+m_2 e^{s_3}}} \\
e^{s_2} &= e^{s_3}{\frac{m_1 e^{s_3}+m_2 e^{s_4}} {m_1 e^{s_4}+m_2 e^{s_3}}}
\end{align}</math>

It is a solution to the problem, but expressed by the parameters of velocity. Return substitution to get the solution for velocities is:
<math display="block">\begin{align}
v_1/c &= \tanh(s_1) = {\frac{e^{s_1}-e^{-s_1}} {e^{s_1}+e^{-s_1}}} \\
v_2/c &= \tanh(s_2) = {\frac{e^{s_2}-e^{-s_2}} {e^{s_2}+e^{-s_2}}}
\end{align}</math>

Substitute the previous solutions and replace:
<math>e^{s_3}=\sqrt{\frac{c+u_1} {c-u_1}}</math> and <math>e^{s_4}=\sqrt{\frac{c+u_2}{c-u_2}}, </math> after long transformation, with substituting:
<math display="inline"> Z=\sqrt{\left(1-u_1^2/c^2\right) \left(1-u_2^2/c^2\right)} </math>
we get:
<math display="block">\begin{align}
v_1 &= \frac{2 m_1 m_2 c^2 u_2 Z+2 m_2^2 c^2 u_2-(m_1^2+m_2^2) u_1 u_2^2+(m_1^2-m_2^2) c^2 u_1} {2 m_1 m_2 c^2 Z-2 m_2^2 u_1 u_2-(m_1^2-m_2^2) u_2^2+(m_1^2+m_2^2) c^2} \\
v_2 &= \frac{2 m_1 m_2 c^2 u_1 Z+2 m_1^2 c^2 u_1-(m_1^2+m_2^2) u_1^2 u_2+(m_2^2-m_1^2) c^2 u_2} {2 m_1 m_2 c^2 Z-2 m_1^2 u_1 u_2-(m_2^2-m_1^2) u_1^2+(m_1^2+m_2^2) c^2}\,.
\end{align}</math>