Masses connected to spring momentum

AI Thread Summary
In the discussion, a problem involving two blocks connected by a spring on a frictionless table is analyzed. The first scenario has one block stationary while the other moves after the spring is released, achieving a speed of 2.0 m/s. In the second scenario, both blocks are released simultaneously, and the user attempts to calculate the speed of the lighter block using conservation of momentum principles. The user arrives at a speed of 1.79 m/s for the lighter block, but this is identified as incorrect, prompting a request for clarification on the formulas used and their applicability to the scenario. The conversation highlights the importance of correctly applying conservation laws in spring-mass systems.
Raj Kishore
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Homework Statement
An m = 1.3 kg block and an M = 3.0 kg block have a spring compressed between them and rest on a frictionless table. With a stopper in place that prevents m from moving, the spring is compressed and released so that M moves away with a speed 2.0 m/s.

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The spring is compressed exactly as before, but this time without the stopper in place. Both blocks are released at the same time. What is the speed of mass m when the spring stops acting on it?

The attempt at a solution
What I did to solve this was I used the equation vfM = 2vCM - v0M

vCM = (Mv0M + mv0m)/(m+M)

Then I said Mv0M = mvfm + MvfM

So, (Mv0M - MvfM)/m = vfm

I got the final answer to be 1.79 m/s, but it is wrong. Thanks for helping.
 
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Can you explain your reasoning a bit more? Where did vfM = 2vCM - v0M come from, for example?
 
That is a formula we were taught
 
Well, it seems wrong to me. How do you know is applies here?
 

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