Editing Elastic collision (section)

===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).