How is kinetic energy related to momentum in a spring-mass system?

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SUMMARY

The discussion centers on the relationship between kinetic energy and momentum in a spring-mass system involving two blocks with masses m and 3m. When released from a compressed spring, the block of mass 3m has a final kinetic energy of E, while the block of mass m has a final kinetic energy of 3E. This conclusion arises from the understanding that momentum is conserved and that the smaller mass must travel at three times the velocity of the larger mass to maintain momentum equality. The equations governing momentum and kinetic energy are pivotal in deriving these relationships.

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There are two blocks one of mass m and the other of mass 3m. The blocks are both start at rest on a frictionless surface and are connected by a spring which is initially compressed. The blocks are then released. If the final kinetic energy of the block of mass 3m is E what is the final kinetic energy of the other block?

My Understanding

The one object that unites the two masses is the force of the spring. In this situation it would release it's energy equally in both directions at the same time. Given that I assumed that as the force is the same the 3m mass would travel at 1/3 the velocity but both objects would have the same overall kinetic energy that is 'E'.

Now that answer turned out to be incorrect and it also confused me how a kinetic energy question appeared on an Impulse/Momentum quiz so I must be missing how kinetic energy and momentum directly relate to each other. I know that

Momentum = mass x velocity
Kinetic energy = (1/2)mv^2
The force of the spring would cause the two masses to accelerate for a limited time, after which they would maintain a constant velocity with no acceleration.
The spring would impart it's accelerating force on the two masses for an identical amount of time.

The other possible answers were one such as E/3, E/9, 3E, 9 E, sqrt(E) etc.. so I assume it's related to the mv^2 equation but I don't see how.
 
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In the two boxes spring situation, it's not kinetic energy that is equally spread, but momentum. So the smaller mass must have 3x the velocity of the big one to equal out. Then, if K.E. = .5mv2 and momentum =mv then K.E. = .5v*momentum and solving for momentum gives momentum = (K.E.)/(.5v). If E is the kinetic energy of the big mass, and the big mass has velocity v, then the small mass has velocity 3v and the momenta are equal and can be set equal to each other.

Momentum = Momentum
(K.E.)/(.5v) = (K.E.)/(.5*3v)
(E)/(.5v) = (?)/(.5*3v) [multiply both sides by .5v]
E = (?)/3
? = 3*E
 
Thank you for clearing that up.
 

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