Velocity from spring pushing on 2 masses

[JoeyBob]

Homework Statement

See attached

Relevant Equations

Ep=1/2kx^2, W=Ek

So i know Ek=123.48 from the potential energy that converts into kinetic energy (Ep=1/2kx^2).

Now by conservation of momentum, m1v1=m2v2

So m2Sqrt(2Ek/(m1+m2))=m1v1

This is where I am making a mistake I think, but not sure how.

Answer is suppose to be 4.44

 

[JoeyBob]

So I tried something else

Ek=Ek1+Ek2

2Ek-m1/m2 * v1^2=v2^2

And substitute this into m1v1^2=m2v2^2, but this didn't give the right answer either.

 

[Delta2]

Not sure how you arrive at some of the equations in post #1 and #2, but let's just reset:
Using conservation of energy (potential energy of spring=sum of kinetic energies) gives you one equation. I don't see this equation in post #1, I see some other equation which doesn't look right.
Using conservation of momentum you get the other equation $m_1v_1=m_2v_2$ which is correct from your post #1.

SO , you will have two equations with two unknowns the velocities of the two bodies $v_1,v_2$.

 

[haruspex]

So m2Sqrt(2Ek/(m1+m2))=m1v1

I read that as $v_1=\frac{m_2}{m_1}\sqrt{\frac{kx^2}{m_1+m_2}}$.
But that's not what I get. Please post your working.