I didn't redo your calculations and I assume that they are correct, which actually doesn't play any role in what I'll describe now. Notice that in the second scenario the 2kg ball will inevitably start to move. By keeping it still you change the reference frame one more time, which invalidates the use of conservation laws.
You cannot use the conservation of energy or momentum in two reference frames$^*$. It is pretty straight forward to see why this must be the case. Assume there is a ball with mass $m$ and from the reference frame of the stationary observer it is moving with velocity $v$. Clearly the ball has a kinetic energy $mv^2/2$ and a momentum $mv$. However from the reference frame of the ball, it is not moving therefore its momentum and energy is zero. If you were to use the conservation of energy or momentum, you would see that it is violated. Thus you conclude that you cannot use the conservation laws for two different reference frames.
I am not sure I understand what your last question is.
$^*$I am assuming that you don't know about momentum-energy four vector, which combines these two and is in fact conserved.
The pulley (and the attachment to the ceiling) are part of the system here. Because of this, you cannot simply use conservation of momentum on the three given masses.
If the final velocity were $v$, then the total energy of the system would have increased since both the pan and counterweight would be moving and the other mass would not have slowed.
You can't use conservation of momentum equations on only a portion of a system. If you imagine a ball bouncing on the floor, you can't say the momentum of the ball is conserved before and after the bounce. You have to consider the change in momentum of the floor as well.
In your problem, the change in momentum of the ceiling will be small, but relevant.
$$\Delta p_{m1} + \Delta p_{m2} + \Delta p_{pan} + \Delta p_{ceiling} = 0$$
As you do not know the change in this final component, you can't use conservation to solve for the remaining momentum of the other three masses.
Let's change the situation to make this more explicit. Instead of a counterweight, consider two pans and two weights.
Let's imagine the pulley and string to be massless, so the two pans and two weights have a total mass of $4m$. If the balls have a velocity $v$, then the total momentum of the pulley inside the room is $1mv + 1mv = 2mv$.
But by symmetry, we can see that the pulley isn't going to turn. If we imagine the pans at rest after the collision, we find the momentum is now $0mv$.
If the pulley's connection to the ceiling/room/earth is not part of the system, then we say that forces from that connection were external and changed the total momentum. We cannot use conservation of momentum due to external forces.
If the ceiling/room/earth are part of the system, then after the collision, they have gained $2mv$ downward momentum, so the total system does not change. If we picture the box as nearly massless and in a spaceship instead of earth, the entire box would be moving downward at $v/2$ after the balls hit the pans. (assuming completely inelastic collision). The more massive the box, the slower it moves to retain the velocity. Consider it attached to a building/earth, and the momentum still changes, but the velocity change is no longer measurable.
Best Answer
Objects are not damaged by momentum $\vec{p}$ they are damaged by force $\vec{F}$. When two objects collide their momenta changes because of forces they apply each other while being in contact. According to 2nd Newton's law the force can be calculated as follows: $$ \vec{F} = \frac{d\vec{p}}{dt} $$
So the force is determined by the time of interaction. When two objects contact their surfaces are flexed. The bigger is the flex the higher is the force which changes the momentum. At some moment of time the relative normal momentum (and velocity) becomes zero and the objects start move backward.
At this moment the force and the flex is maximal. If the surface (armor) is not strong enough for this flex the object is damaged. This can happen long before the relative normal velocity become zero. In that case the interaction between internal parts of the objects started.
Momentum
Hence damage is not determined only by momentum. It is determined by the force and the ability of the object resist it. The force is not constant during the collision and its maximal value depends on momentum and the time of interaction of the surfaces. The time of interaction depends on both flexibility and strength.
When a basketball hits the floor its momentum changes from $\vec{p}$ to almost $-\vec{p}$ so the total change is almost $2p$. When a glass hits the floor its momentum becomes zero so the change is $p$. The ball has higher flexibility and strength and is not destroyed even though the momentum change is two times higher.
Another example is a bullet that hits a door. It makes hole before the door opens. The force is huge but the momentum of the door is almost not changed because the time necessary to reach critical flex is too short. When one pushes the door with his finger the force is small and does not destroy the surface before the flex stop the finger. The door gets enough momentum to start move.
Energy
When the surface is flexed in irreversible way or damaged some part of kinetic energy of the objects turns into kinetic energy of their parts, heat, sound, light etc. This is called inelastic collision. Elastic collision means no damage. If you need a good model this should be taken into account.