Find final speeds of two carts pushed apart by compressed spring

[huskydc]

A massless spring of spring constant 20 N/m is placed between two carts. Cart 1 has a mass M1 = 5 kg and Cart 2 has a mass M2 = 3.5 kg. The carts are pushed toward one another until the spring is compressed a distance 1.8 m. The carts are then released and the spring pushes them apart. After the carts are free of the spring, what are their speeds?

i know both momentum and energy conservation applies here, but don't know where to start...
both carts are initially at rest,

so I'm guessing : PE(initial) = PE(final) + KE (final)

 

[Galileo]

You are correct: Conservation of energy and momentum is the way to deal with this problem. So your starting point should be obvious: Write down the initial energy and momentum (these are known) and equate them to the final energy and momentum respectively. This will give you 2 equations in 2 unknowns (v1 and v2).

 

[huskydc]

also, wouldn't PE final be zero too? after the carts are released, the spring would be relaxed again, thus no compression, thus zero..

that'll make it: initial PE = final KE

but i don't know where to go from here

 

[Galileo]

That's right, the final potential energy of the spring is zero.
What are the expressions for the initial energy, initial momentum, final momentum and final energy?

 

[huskydc]

now we have initial PE = final KE,

it goes...

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

does it make sense? but the problem now is i have two variables to solve...one equation...

 

[Doc Al]

Don't forget conservation of momentum. That will give you the second equation that you need.