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Masses connected to spring momentum

  1. Dec 18, 2016 #1
    The problem statement, all variables and given/known data
    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.
     
  2. jcsd
  3. Dec 18, 2016 #2

    vela

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    Can you explain your reasoning a bit more? Where did vfM = 2vCM - v0M come from, for example?
     
  4. Dec 18, 2016 #3
    That is a formula we were taught
     
  5. Dec 18, 2016 #4

    vela

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    Well, it seems wrong to me. How do you know is applies here?
     
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