The fact that the objects are in vacuum has very little to do with Newton's Law.
Instead, as always there will be an exchange of momentum; exactly how much momentum is exchanged depends on whether the collision is elastic or inelastic (most collisions are somewhere in between the two...)
Newton's law can be restated as "the change in momentum of one object is equal and opposite to the change in momentum of the other object". But to get the final velocities, you need to know the mass (and the energy after the collision).
It is usually helpful to analyze collisions in the center of mass frame. Since the two objects in your example have the same mass (you said they are identical), the center of mass moves at half the velocity of the incoming particle. In that frame of reference, one particle appears to come from the left at $v/2$, and the other comes from the right at the same velocity.
In a perfectly inelastic collision, they will hit each other and stick: all relative velocity is gone. So if they have $v'=0$ in a frame moving at $v/2$ then their final velocity is $v/2$ for both.
In a perfectly elastic collision, they will bounce off each other and continue with the speed they started with: so the particle that was initially stationary, and therefore was moving at $v/2$ to the right in the c.o.m. frame, ends up moving with $v/2$ to the left; transforming back to the lab frame, it will have velocity $v$. The other particle started out with $v$, and by a similar argument ends up stationary.
Now if some energy is lost during the collision, then the final velocities in the c.o.m. frame will not be $±v/2$ - they will be slightly less (but still equal and opposite). That means in turn that the velocity of the first particle will not be reduced completely to zero, and the second particle will not get all the energy/momentum of the first.