1. The problem statement, all variables and given/known data 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? 2. Relevant equations Us = (1/2)kx2 K = (1/2)mv2 ∑pi = ∑pi 3. The attempt at a solution Using conservation of mechanical energy, I get mbva2 + (1/2)mbvb2 = 60 J Setting the initial momentum equal to the final momentum, I get 2mbva + mbvb = 0 Solving for vb I get vb = -2va Substituting that into the energy equation mbva2 + 2mbvb2 = 60 J Substituting k for mbva2 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.