Solving a Problem: What Went Wrong & Velocity for Both Masses

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The discussion centers on a problem involving two connected masses and their velocities. The original poster is confused about their solution and whether the velocity is the same for both masses. Respondents clarify that while the masses are connected and share the same speed, they move in opposite directions. There is also a note that the provided solution incorrectly identifies the hanging mass. The key takeaway is that the velocities are indeed the same, despite the confusion over the mass designation.
Mohmmad Maaitah
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Homework Statement
As in the first provided picture:
Relevant Equations
Newton second law
This is the problem:
1683368077531.png


And this the answer provided by the examiner:
1683367960587.png

And this is my own answer:
IMG_20230506_130351_883.jpg


So what did I get wrong???
Also I want to know if the Velocity is the same for both masses.
 

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Mohmmad Maaitah said:
So what did I get wrong???
The diagram shows m_1 as the hanging mass, but the given solution switches that. Your work is fine.
Mohmmad Maaitah said:
Also I want to know if the Velocity is the same for both masses.
Sure---they are connected. (The speeds are the same, but they move in different directions, of course.)
 
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Doc Al said:
The diagram shows m_1 as the hanging mass, but the given solution switches that. Your work is fine.

Sure---they are connected. (The speeds are the same, but they move in different directions, of course.)
Oh lord, he switched the masses :doh:
 
Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

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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