Compressed spring between two particles

[AlephNumbers]

Homework Statement

Particle A and Particle B are held together with a compressed spring between them. When they are released, the spring pushes them apart, and they then fly off in opposite directions, free of the spring. The mass of A is 2.00 times the mass of B, and the energy stored in the spring was 60 J. Assume the spring has negligible mass and that all its stored energy is transferred to the particles. Once that transfer is complete, what are the kinetic energies of (a) particle A and (b) particle B?

Homework Equations

U_s = (1/2)kx²
K = (1/2)mv²
∑p_i = ∑p_i

The Attempt at a Solution

Using conservation of mechanical energy, I get
m_bv_a² + (1/2)m_bv_b² = 60 J

Setting the initial momentum equal to the final momentum, I get
2m_bv_a + m_bv_b = 0

Solving for v_b I get
v_b = -2va

Substituting that into the energy equation
m_bv_a² + 2m_bv_b² = 60 J

Substituting k for m_bv_a²
k + 2k = 60 J
Particle A has one third of the kinetic energy and particle B has two thirds of the kinetic energy. This seems to make sense since particle B has less mass than A.

 

[haruspex]

Looks right. The body with half the mass will have twice the velocity so twice the energy.

 

[AlephNumbers]

The body with half the mass will have twice the velocity so twice the energy.

That is what my intuition told me, but I just wanted to make sure. Thank you!