a 7.0kg block and a 12kg block sit on a frictionless table. Between and touching, but not attached to, the blocks is a spring that is compressed by 13.5 cm. the spring has a spring constant of 1200 N/m. if the spring were released at what speed would each block move away.

Answer 1

The energy in the spring results in KE in the blocks.

1/2 k x^2=1/2 m1*v1^2 + 1/2 m2*v2^2
and conservation of momentum applies, so
0=m1*V1 + m2*V2 or
v1=-m2/m1 * v2

put that into the first energy equation, and solve. x=.135m

Answer 2

So I got a little confused. I got stuck on this equation at:

1/2kx^2=1/2 m1* (m2/m1*v^2)+1/2 m2v2^2

Then:

kx^2=m2*v2+1/2(m2v2^2)

Then what?