Collision problem with a spring

[PinkDaisy]

The problem is:

A spring that is compressed 0.5m by a block (block1) are at the top of a 10m tall frictionless hill. The block is released and slides down the hill where it collides with block2, which started at rest. (a) What is the speed of block1 prior to the collision? (b) Find the speeds of both blocks following the collision assuming that the coefficient of restitution is 0.50.

mass of block1 = 10kg, mass of block2=15kg, and spring constant k=500N/m

So for (a) I got:

mgh + .5mv(i)^2 = .5mv(f)^2
(10)(9.8)(10) + (.5)(10)(0) = (.5)(10)(v^2)
v=14.0 m/s

I'm not sure where to even start with part b.

Thanks

[Galileo]

If mass1 if sort of 'fired' away by the spring, then the kinetic energy when it is released is equal to the potential energy from the spring.

[PinkDaisy]

Isn't there also a component due to gravitational potential energy?

Does the equation start out looking like this:

.5m(1)v(1)^2 + m(1)gy + .5kx^2 + .5m(2)v(2)^2 = .5m(1)v(1)^2 + m(1)gy(init) + .5kx^2 + .5m(2)v(2)^2

0 + m(1)gy + .5kx^2 + 0 = .5m(1)v(1)^2 + 0 + 0 + .5m(2)v(2)^2

How does the coefficient of restitution come into play?