Center of Mass and Linear Momentum Question?

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 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 that the spring has negligible mass and that its stored energy is transfered to the particles. Once the transfer is complete, what are the kinetic energies of (a) particle A and (b) particle B

Answer 1

E = ke + KE, the kinetic energy of the spring E is converted into the kinetic energy ke of the small mass m at velocity V and KE of the large mass M = 2m at velocity v. What we are looking for is rho = ke/E and Rho = KE/E the fraction of the spring energy E that gets converted to the small and large particle.

To do this, find KE in terms of ke. KE/ke = 1/2 Mv^2/1/2 mV^2 = 2mv^2/mV^2 = 2v^2/V^2; so KE = ke 2(v^2/V^2). Then from conservation of momentum, we have mV = Mv. This results because the initial momentum of the two particles before the spring is released is zero; so the sum of the two momenta after release has to be 0 = Mv - mV.
Therefore, v/V = m/M and v = V(m/2m) = V/2; V = 2v.

Putting this into the kinetic energy ratio, we have KE = ke 2 (v/V)^2 = ke 2 (v/2v)^2 = ke 2(1/2)^2 = ke/2. Thus 1.0 = (ke/2)/E + ke/E; so that E = 3/2 ke and ke = 2/3 E = 40 J, the kinetic energy of the smaller particle (B). KE = 1/3 E = 20 J, the kinetic energy of the larger particle (A)..

Answer 2

Two principles apply:

1) Conservation of momentum applies.

Since the initial momentum was 0, the final momentum must be 0. Therefore:
M-a * v-a = M-b * v-b (in magnitude, I'm not worrying about the sign)
Since M-a = 2 * M-b, this implies that:
2 * v-a = v-b
v-a = v-b/2

Therefore,
KE-a = M-a * v-a^2/2
= (2 * M-b) * (v-a/v-b)^2 *(v-b)^2
= 2*(M-b)*(v-b)^2 * (1/2)^2
= KE-b * (2/4)
= KE-b/2

2) Conservation of energy:
The energy of the spring, 60 J, must have been converted into the total KE.
But total KE = KE-a + KE-b
= ((1/2) + 1)*KE-b
= (3/2) KE-b

Therefore,
60 = (3/2) KE-b
so KE-b = 40 J
and
KE-a = KE-b/2 = 20 J

a) KE-a = 20 J.
b) KE-b = 40 J.

Answer 3

The kinetic energies of the two masses will be proportional to the percent of the system they make up. Assuming the spring transfers all of it's potential to the two masses, mass A will take 40J of the energy as it makes up 2/3 of the system and 40J is 2/3 of the energy, likewise mass B will take 1/3 of the energy (20J) because it makes up 1/3 of the mass of the system.