Masses connected to spring momentum

[Raj Kishore]

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.

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 v_fM = 2v_CM - v_0M

v_CM = (Mv_0M + mv_0m)/(m+M)

Then I said Mv_0M = mv_fm + Mv_fM

So, (Mv_0M - Mv_fM)/m = v_fm

I got the final answer to be 1.79 m/s, but it is wrong. Thanks for helping.

 

[vela]

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

 

[Raj Kishore]

That is a formula we were taught

 

[vela]

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