Masses connected to spring momentum

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1. Dec 18, 2016

Raj Kishore

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.

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. Dec 18, 2016

vela

Staff Emeritus
Can you explain your reasoning a bit more? Where did vfM = 2vCM - v0M come from, for example?

3. Dec 18, 2016

Raj Kishore

That is a formula we were taught

4. Dec 18, 2016

vela

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