Conservation of Momentum for Pulley System
Consider a pulley system
Thank you to a user below for this clarification of what the question is:
Why is the momentum of a pully with two masses moving with velocity $v$ equal to $(m_1+m_2)v$ and not $m_1v−m_2v$ (as it would by vector addition of the momenta of independent/unconnected masses)?
Background
In a textbook I am looking at it says that
The Principle of Conservation of Momentum may be applied to bodies at the ends of a piece of string. We must regard the string-particle system as one body, like a long train going around the corner.
The text then goes on to use this principle to solve problems such as the following:
1. Two particles of masses 4 kg and 3 kg hang from a pulley at the ends of a light inextensible string. The system is released from rest. After 2 seconds, the 3 kg pass picks up a particle of mass 2 kg. How much further will the 4 kg mass move downward before it stops?
2. Two masses of 5 kg and 1 kg hang from a smooth pulley at the ends of a light inextensible string. The system is released from rest. After 2 seconds, the 5 kg mass hits a horizontal table: i. How much further will the 1 kg mass rise? ii. 512
The 1 kg mass then falls and the 5 kg mass is jolted off the table. With what speed will the 5 kg mass begin to rise?
I cannot reconcile the vector nature of velocity and hence momentum with this "going around the corner" business. Perhaps someone can explain this better than the textbook... or perhaps is the textbook incorrect?
Edit: For example, consider problem 1 here. We can show that after two seconds the speed of each particle is $\frac27 g$ m s$^{-1}$. Now I quote:
At this point the 3 kg mass picks up a 2 kg mass, to become a new mass of 5 kg. The system will immediately be jolted to a slower speed. This speed can be found by applying the Principle of Conservation to the particle-string system. The mass of this whole system was 7 kg before the jolt and 9 kg after the jolt. Let $v$ be the new speed.
The textbook then uses:
$$m_1u=m_2v$$ $$\Rightarrow 7\left(\frac{2}{7}g\right)=9v$$ $$\Rightarrow v=\frac29 g\text{ m s}^{-1}.$$
I don't understand how this is a valid application of the conservation of momentum.
My understanding of momentum (it is a vector) would be that the equation is (choosing down as positive):
$$4\cdot \frac{2}{7}g-3\cdot \frac{2}{7}g=4\cdot v-5\cdot v.$$ 512