This is a discussion of elastic collision in one dimension.
Before collision: two  particles with mass and velocity as  m_1,\vec{v_1} and m_2,\vec{v_1}
After collision: the velocity have been changed to \vec{v_1}' and \vec{v_2}'

Assume there is no external force or the interval is very short, then
total linear momentum is conserved: i.e. m_1\vec{v_1}+m_2\vec{v_2}=m_1\vec{v_1}'+m_2\vec{v_2}'
so m_1(\vec{v_1}-\vec{v_1}')+m_2(\vec{v_2}-\vec{v_2}')=0

For elastic collision, the total energy is also conserved. \frac{1}{2}m_1 v_1^2+\frac{1}{2}m_2 v_2^2==\frac{1}{2}m_1 v_1^{'2}+\frac{1}{2}m_2 v_2^{'2}
It can be re-write as  \frac{1}{2}m_1(v_1^2-v_1^{'2})= -\frac{1}{2}m_2(v_2^2-v_2^{'2})
It is the same as m_1(v_1-v_1')(v_1+v_1')= -m_2 (v_2-v_2')(v_2+v_2')
Since m_1(v_1-v_1')=-m_2 (v_2-v_2'), so (v_1+v_1')=(v_2+v_2')  or v_1-v_2=v_2'-v_1'
The result is
v'_1= \frac{m_1-m_2}{m_1+m_2} v_1 +\frac{2m_2}{m_1+m_2}v_2=V_{cm}+\frac{m_2}{m_1+m_2}(v_2-v_1)=2V_{cm}-v_1
and v'_2=\frac{2m_1}{m_1+m_2}v_1+\frac{m_2-m_1}{m_2+m_1}v_2=V_{cm}+\frac{m_1}{m_1+m_2}(v_1-v_2)=2V_{cm}-v_2
where  V_{cm}=\frac{m_1V_1+m_2V_2}{m_1+m_2}  
and V_{cm}=\frac{v_1+v_1'}{2}=\frac{v_2+v_2'}{2} or v_1+v_1'=v_2+v_2'=2V_{cm}

It means that from the coordinate of center of mass: V_{cm}=0, it reduced to
v_1'=-v_1 and v_2'=-v_2

Define \rho=\frac{m_2}{m_1}, the above equations can be re-write as
\frac{v'_1}{v_1}=\frac{1-\rho}{1+\rho}+ \frac{2\rho}{1+\rho}\frac{v_2}{v_1}=2\frac{V_{cm}}{v_1}-1

The following simulation plot the above two functions.
The X-axis is \frac{v_2}{v_1}, it range from Vscale*xmin to 1. (There is no collision if v_2>v_1)
The blue curve is \frac{v_1^'}{v_1} and red curve is \frac{v_2^'}{v_1}
You can change the ratio of \frac{m_2}{m_1} with slider.

The default value is \frac{m_2}{m_1}=1, so
v_2^'= v_1 , so \frac{v_2^'}{v_1}=1 is a horizontal line
v_1^'=v_2  , so v_2 is a straight line with slope 1 (function of \frac{v_2}{v_1})

Special case:
if m_1<<m_2, v_1'=-v_1+2v_2 and v_2'=v_2 ,-*-
if v_2=0, then v_1'=-v_1 and v_2'=0
   e.g.  a ball hit the wall, it will biunced back with almost the same speed (but in oppositive direction).
if m_1>>m_2, v_1'=v_1 and v_2' =2 v_1-v_2,
if v_2=0, then v_1'=v_1 and v_2'=2 v_1,
  e.g. a speedy car hit you while you stand still, you will be kicked by twice the velocity of the car.

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