Velocity of separation and velocity of approach Can I have a clear cut dimension or difference between velocity of approach and velocity of separation? 
In just simple 1D motion.
Considering two  rigid spherical masses of different masses and moving with different velocities.
 A: I have not seen any formal definition, but this might work.
Let's assume that two bodies, left and right, move along $x$, with positive direction to the right.
Let's define the velocity of approach as positive and the velocity of separation as negative.
Then, the velocity of approach or separation could be determine as $\vec v=\vec v_{left}-\vec v_{right}$. 
For instance, if the bodies move toward each other, the velocity of the left body will be positive, the velocity of the right body will be negative, so the difference will be positive, showing that the bodies are approaching each other.
Another example, when the left body is moving to the left at $1$m/s, while the right body is moving to the left at $3$m/s. The difference will be $-1-(-3)=+2$, indicating the the bodies are approaching at $2$m/s.
A: The problem is that the velocity of approach is a term which introduces ambiguity.  
In one dimension, if bodies $A$ and $B$ are travelling at velocities $\vec v_{\rm A}$ and $\vec v_{\rm B}$ the velocity of approach can be said to be either $\vec v_{\rm A} - \vec v_{\rm B}$ or $\vec v_{\rm B}-\vec v_{\rm A}$.
Taking the magnitude of the subtraction does not work as one could envisage a negative velocity of approach where the bodies are moving away from one another.  
The approach which removes these ambiguities is as follows.
Let $\vec v_{\rm A} = a\hat i$ and $\vec v_{\rm B} = b\hat i$ where $a$ and $b$ are components of the velocity vectors in the $\hat i$ direction as shown in the diagram below.  
 
As drawn $A$ is trying to get closer to $B$ and $B$ is trying to get further away from $B$ so the "velocity" of approach is $(+a) + (-b) = a-b$ and using similar ideas the "velocity" of separation is $(-a) + (+b) = b-a$.
You will note that $(a-b)=-(b-a)$ which is to be expected as the velocity of approach is the opposite of the velocity of separation.  

As an example consider an elastic collision where it can be shown that instead of using the conservation of kinetic energy on can say that for such a collision the "velocity" of approach is equal to the "velocity" of separation.  
The example I have chosen is Worked example 6.5: Elastic collision where the solution is obtained via the direct application of the conservation of kinetic energy.  

An object of mass $2 \,{\rm kg}$, moving with speed of $12\,{\rm m/s}$, collides head-on with a stationary object whose mass is $6 \,{\rm kg}$. Given that the collision is elastic, what are the final velocities of the two objects? Neglect friction.  

Using the velocities of approach and separation one might proceed as follows.
 
$x$ and $y$ are the components of the final velocities in the $\hat i$ direction.  
Equating approach and separation velocities gives $(+12) + (+0) = (-x)+(+y) \Rightarrow 12 = -x+y$
Conservation of momentum gives $2\times 12\hat i + 6 \times 0\hat i = 2 \times x \hat i + 6 \times y \hat i \Rightarrow 24 = 2\,x+3\, y$  
Solving these simultaneous equation gives $x=-6$ and $y=+6$ and so the final velocities as $-6 \hat i$ and $+6 \hat i$.

However you may have used the following diagram for this collision "knowing" that the velocity of the $2\,\rm kg$ mass will be reversed after the collision.  
 
All you need to do is set up two equations in a consistent manner.  
Velocities gives $(+12) + (+0) = (+x)+(+y) \Rightarrow 12 = +x+y$
Momentum gives $2\times 12\hat i + 6 \times 0\hat i = 2 \times x (-\hat i) + 6 \times y \hat i \Rightarrow 24 = -2\,x+6\, y$
Solving these simultaneous equation gives $x=+6$ and $y=+6$ and so the final velocities $+6 (-\hat i)+ = - 6 \,\hat i$ and $+6 \hat i$ as before.
