Let's analyze what happens between each pair of points.
Between A and B:
The block is accelerated at a constant acceleration by the force imparted on it by the mechanical plunger. Constant acceleration gives the block a linearly increasing velocity
Between B and C:
The block is moving at the velocity it left the plunger at, and encounters no external forces. No forces means no acceleration, meaning no change in velocity.
Between C and D:
The block encounters the rough surface, which imparts a frictional force onto the block. The force acts in the direction opposite the block's movement, so it creates a constant negative acceleration, causing the block to decrease in velocity linearly.
Between D and E:
Same as between B and C, just with a lower velocity.
Here is where we need to start considering the center of mass. Before this point, only the first block was moving while the second stayed stationary, so the center of mass had just half the speed of the moving block.
Between E and F:
The blocks collide, and momentum is conserved.
Let's say both blocks have a unit mass of 1, and the moving block has a velocity of 1.
The total momentum of the system is the momentum of the moving block and that of the stationary.
(mv)moving + (mv)stationary = 1*1 + 0 = 1
After the collision, the blocks stick together, so they need to be considered a single object.
(mv)combined = 2*v = 1, since momentum must be conserved
v = 1/2
However, since before the collision, the center of mass simply moved at half the speed of block 1, the speed of the center of mass does not change at E.
Throughout the simulation, acceleration is mostly constant. This is because acceleration is directly proportional to force, and any forces encountered by the block are constant. The forces at which the mechanical arm pushes on block 1 and friction acts on it are constant, never changing in their time period, therefore acceleration is constant.
When we added the momenta of the two blocks separately, the total ended up being 1. Momentum must be conserved. The momentum of the system after the collision can be represented by the combined mass of the two blocks times the velocity of the two blocks. The combined mass is 2. Since momentum must be conserved, the combined mass times the velocity of the combined system must equal 1.
m*v = 1; 2*v = 1; we solve for v, v = 1/2.
Conservation of momentum:
Momentum is equal to the mass of each separate part times the velocity. The total momentum of everything in the system must stay constant, if no external forces act on the system.
During the collision, there are no forces acting on the two blocks combined. They only interface with each other, and with nothing else in the world at the time of the collision.
Draw the center of mass at any point between the two blocks. It's exactly halfway between the two blocks. That point stays halfway between the two blocks at all times.
When one is moving and the other is stationary, we just take the average of their velocities.
Imagine a line connecting the two blocks at all times. Imagine it shrinking as the first block approaches the second block in real time. Now imagine the midpoint of that line throughout the entire simulation. It always stays in the middle, even when it shrinks. This means that the midpoint of the line (the center of mass) moves at half the speed of the first block, when the second is stationary.
The center of mass in this situation must always be halfway between the two blocks. The ratio of the distances between the two blocks and the center of mass must be equal to the ratio of the masses. The blocks have a ratio of 1:1 in mass, so therefore the ratio of the distance between block 1 and the CoM and block 2 and the CoM must also be 1:1.
Say block 2 has double the mass of block 1. The block1:block2 mass ratio is then 1:2. Let's look at the ratio of the distances between block1 and the CoM and block2 and the CoM. That would be 2:1, or the inverse of the mass ratio. This means that the CoM is twice closer to block 2, the heavier block, than to block 1, when block 2 is twice heavier.