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 68 J. Assume that 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 each particle?
@Physics

Question

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 68 J. Assume that 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 each particle?
   @Physics

anonymous · Answer

$$m_A=2m_B$$
according to the conservation of momentum:
$$v_Am_A=v_Bm_B$$
so by substituting
$$2 \times v_Am_B=v_B\times  m_B$$
cancelling mB
$$2v_A=v_B$$
now square both sides
$$4v_A^2=v_B^2$$
Now we look at the energy in the system
$$68=\frac{1}{2}m_Av_A^2+\frac{1}{2}m_Bv_B^2$$
substituting and factoring out the kinetic energy
$$68=\frac{1}{2}m_Av_A^2(1+2)$$
Therefore the kinetic energy of A is 68/3 or 22.667.
Now the energy of B will be
68-68/3=B
B=45.33.
Therefore two thirds of the energy from the spring is going to B