Two masses connected by a rope on a pulley on a ramp

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To solve the problem of two masses connected by a rope on a pulley on a ramp, it's essential to create distinct symbols for each mass to avoid confusion. After establishing the free body diagrams, the next step is to write the force balance equations for each mass. These equations can be organized either by separating horizontal and vertical forces or by considering forces parallel and normal to the slopes. This structured approach simplifies the problem and clarifies the forces acting on each mass. Properly defining the variables and equations is crucial for an accurate solution.
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Homework Statement
If the left-hand slope in the figure makes a 58° angle with the horizontal, and the right-hand slope makes a 32° angle, how should the masses compare if the objects are not to slide along the frictionless slopes?
Relevant Equations
F=ma
I have no clue how to do the problem. I created a free body diagram for each block. I assume that it is simpler than I am making it out to be.
 

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First, use a different symbol for each mass. Calling them both "m" will confuse.

Next, write the force balance equations for each mass.
You can write an equation for each of horizontal and vertical or for parallel to the slope(s) and normal to them. Your choice.
 
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My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

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