Masses connected to spring momentum

  • #1
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.
 

Answers and Replies

  • #2
vela
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Can you explain your reasoning a bit more? Where did vfM = 2vCM - v0M come from, for example?
 
  • #3
That is a formula we were taught
 
  • #4
vela
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Well, it seems wrong to me. How do you know is applies here?
 

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