Conservation of Linear Momentum and spring compression

[G-reg]

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 5.00 times the mass of B, and the energy stored in the spring was 111 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?

Homework Equations

mass of particle A = 5 times mass of particle B

KE(before) = KE(after)
(1/2)mv^2 = (1/2)mv^2
The sum of KE(before and after) = 111J

The Attempt at a Solution

(1/2)mv^2 + (1/2)mv^2 = 111J
mv^2 + mv^2 = 222J
5(mass of A)v^2 + (mass of B)v^2 = 222J

and I'm not sure how I can find the individual masses from there
any suggestions on how I might be able to do that?

 

[rl.bhat]

Hi G-reg, welcome to PF.
Before separation, both particles are at rest. So after separation momentum must be conserved.
Hence m1v1 = m1v2.
Now you have two unknowns and two equations. Solve them.

 

[G-reg]

Okay, thanks I got it!
Here is another one that I think is similar but it's still giving me trouble..

A space vehicle is traveling at 6000 km/h relative to the Earth when the exhausted rocket motor is disengaged and sent backward with a speed of 86 km/h relative to the command module. The mass of the motor is four times the mass of the module. What is the speed of the command module relative to Earth just after the separation?

so I know that m1v1=m2v2
being that m1(6000)=m2(86)
and then I would substitute 4m2 for m1 to get m2

am I going about this the right way?