Zeroth law of thermodynamics and related discrepancy? Consider the case:
Assume molecules to sphere and their |velocity| as an indication of their temperature. Now for head on elastic collision (i.e transfer of heat), assume the momentum is conserved
Therefore doing math,
$m₁$ moving in $+x$ and $m₂$ in $−x$
$$(m₁×v₁)−(m₂×v₂) = (m₁−m₂)×v$$
 Where v is common speed (or temperature)
now if $m₁=m₂=m$ , 
$m×(v₁-v₂)=0$
or $v₁=v₂$
but in the beginning i didn't put any restriction on initial speed, this bizarre answer is coming only because of having common speed but that follow from the Zeroth law of thermodynamics.
 A: Zeroth law of thermodynamics?  Zeroth law of thermodynamics is across systems.  If they are colliding they are in the same system.  
Temperate is not a common velocity.  It is a measure of the average kinetic energy.  
Common velocity? (m₁−m₂) × v  is neither a proper application of zeroth law or temperature.  You asserted a constraint of a common speed and the the result is  v₁=v₂.   No surprise.
The proper equation is  
(m₁×U₁) + (m₂×U₂) = (m₁×v₁) + (m₂×v₂) 
If m₁=m₂ then  
U₁ + U₂ = v₁ + v₂ 
A: Temperature is a statistical property of a large collection of particles moving and interacting at random. Talking about temperature for a pair particles colliding doesn't really make any sense. Your particles simply don't have a temperature and your result doesn't have anything to do with the zeroth law. 
What you seem to have done is calculated the momentum for two particles and then required that at some point (the final state?) they must be moving with equal velocities in opposite directions. So, whether this is what you were trying to do or not, you have effectively asked "what do the initial velocities of the particles have to be to get these final velocities?" and so it is not surprising that you get co condition on the initial velocities. 
A: This has nothing do to with thermodynamics.
Your momentum equation is wrong. Where did you get your $(m₁−m₂)×v$  from? You have presupposed that your final velocity is the same.
It should be:
$$(m_1×v_{1,i}))+(m_2×v_{2,i}) = (m_1×v_{1,f})+(m_2×v_{2,f}) $$  
Where i is initial and f is final. The directions are included in the velocities.
There's also the conservation of energy relation,
$$\frac12 m_1{v_{1,i}}^2 + \frac12 m_2{v_{2,i}}^2 = \frac12 m_1{v_{1,f}}^2 + \frac12 m_2{v_{2,f}}^2 $$ 
Together with conservation of momentum, you get the velocity relation: (This is independent of mass. See comments. Strictly speaking, deriving this is not necessary, but it makes calculation easier.)
$$(v_{1,i}−v_{2,i}) = -(v_{1,f}−v_{2,f}) $$
Assuming that $m_1 = m_2$, you get this from the momentum relation:
$$v_{1,i} + v_{2,i} = v_{1,f} + v_{2,f} $$ 
You can find that $v_{1,f} = v_{2,i} $ and  $v_{2,f} = v_{1,i} $. They aren't the same.
I think I need to reference the question more. This collision has absolutely nothing to do with thermodynamics. If you want to consider collisions for a boundary between 2 systems at the same temperature, you can't just consider a single collision. You need to consider all the collisions at the boundary. On average, you will get no change in average speed of particles in the system.
