Newton's law of motion look the same to all observers in inertial frames of reference.
It is equally true that if momentum is conserved in one inertial reference frame, it is conserved in all inertial frames.

This java applet apply the above concept to one dimensional collision problem.
Two circular objects are confined to move in one diminution (between two gray blocks).
Press start button to run the animation.
Circular objects will move with some predefined velocity (yellow arrow).

Click the mouse button to pause. Click it again to resume the animation.

While the animation is suspended:
Click near the arrow of the velocity vector and drag it left/right to change the initial velocity.
Click at the center of the circle to move it : drag the object left <--> right.
Click within the circle to change the mass of that object.
Click right mouse button to increase mass by one unit.
Click left mouse button to decrease mass one unit.

Press Reset to reset most parameters to default values.
eta is the coefficient of restitution
eta = | relative velocity just after collision/relative velocity just before collision |
for elastic collision eta=1., for perfectly inelastic collision eta=0.

You can select different frame of reference to view the relative motion of all the objects.
lab is a laboratory inertial frame.
$m_1, m_2$and CM are frame of reference with respect to left circular object $m_1$,  right circular object $m_2$ and center of mass for $m_1$ and $m_2$.

The velocity of two objects after collision ($V'_1,V'_2$)can be calculated from velocity before collisions ($V_1,V_2$) and mass of two objects ($m_1,m_2$).

From conservation of momentum $m_1 V_1+m_2 V_2=$ $m_1 V'_1+m_2 V'_2$,
and conservation of energy $\tfrac{1}{2}m_1V_1^2+\tfrac{1}{2}m_2V_2^2=\tfrac{1}{2}m_1V_1'^2+\tfrac{1}{2}m_2V_2'^2$
So $m_1 (V_1-V_1')=m_2(V_2'-V_2)$
and $\tfrac{1}{2}m_1 (V_1^2-V_1'^2)=\tfrac{1}{2}m_2 (V_2'^2-V_2^2)$, which means $\tfrac{1}{2}m_1 (V_1-V_1')(V_1+V_1')=\tfrac{1}{2}m_2 (V_2'-V_2)(V_2'+V_2)$
So $V_1+V_1'=V_2'+V_2$

i.e. The equation need to be solved are
$m_1 V_1'+m_2 V_2'=$ $m_1 V_1+m_2V_2$ and $V_2'-V_1'=V_2-V_1$

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  $V_{cm}=\frac{m_1V_1+m_2V_2}{m_1+m_2}$

It means that $V'_1-V_{cm} = - (V_1-V_{cm})$ and $V'_2-V_{cm}= - (V_2-V_{cm})$
or $V'_{1cm}= -V_{1cm}$ and $V'_{1cm}= -V_{1cm}$ where $V'_{1cm}=V'_1-V_{cm}$ ...etc.
From the point of center of mass coordinate system: both particles bounce back with the same speed (relative to center of mass).

-*-

You are welcomed to check out collision in 2 dimension.