Masses connected to spring momentum

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Homework Help Overview

The problem involves two blocks connected by a spring on a frictionless surface, focusing on the motion of the blocks after the spring is released. The subject area includes concepts from dynamics and momentum conservation.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply a formula related to final velocities and center of mass, but some participants question the validity of this formula in the given context and seek clarification on its derivation.

Discussion Status

The discussion is ongoing, with participants exploring the reasoning behind the original poster's approach and questioning the applicability of the formula used. There is no explicit consensus on the correctness of the method or the final answer.

Contextual Notes

Participants are discussing the assumptions made regarding the system's initial conditions and the effects of the spring's action on both masses.

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.

q17 com.png


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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