How Do Kinetic Energies of Particles A and B Compare After Spring Release?

[cd80187]

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 5.0 times the mass of B, and the energy stored in the spring was 87 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 (a) particle A and (b) particle B?


So here is the problem. I know that the final kinetic energies of A + B = 87 J. Next, I solved for Va final, using the momentum equation and got Va final = 5 times Vb final. Next I substituted the values into the kinetic energy equation and ended up with... .5 x (5 x Mb)(5Vb) squared + .5 (Mb)(Vb) squared, where M is the mass of particle b, and V is the final velocity. So i was able to narrow it down to two unknowns, but I am unsure of where to go from here. When I solve for the two unknown, i can finish it, but I am just stuck. Thanks for the help in advance

 

[Doc Al]

So here is the problem. I know that the final kinetic energies of A + B = 87 J. Next, I solved for Va final, using the momentum equation and got Va final = 5 times Vb final.

You might want to redo that. A is 5 times heavier than B.

Next I substituted the values into the kinetic energy equation and ended up with... .5 x (5 x Mb)(5Vb) squared + .5 (Mb)(Vb) squared, where M is the mass of particle b, and V is the final velocity. So i was able to narrow it down to two unknowns, but I am unsure of where to go from here. When I solve for the two unknown, i can finish it, but I am just stuck.

What two unknowns? (The mass drops out and you can solve for the speed.)