Conservation of energy in bouncing ball

[Delta2]

Very confusing for me at least.

I ll define two different (i think) frames:

Frame A: The frame where the ground is stationary before the collision.
Frame B: The ground frame where ground is stationary, before, during and after the collision.

I hope frames A and B are well defined (i sense there might be a problem with the way i define frame A, if so explain to me). Also i think Frame A is inertial before, during and after the collision, while Frame B is inertial before and after, but not during, the collision. We "have the right" to remain in frame A even after the collision right?

What i really want to ask is this:
What are the velocities of the ball and the ground , before and after the collision, in Frame A?
Same question for Frame B.

 

[theodoros.mihos]

For the CM system:$$ mv + MV = 0 ,\, E_{cm} = \frac{1}{2}mv^2 + \frac{1}{2}MV^2 $$For V=0 we have another frame witch move by V before collision and -V after collision.
For what reason a system that in time dt change his velocity from V to -V must conserve any kinetic energy?

 

[A.T.]

What are the velocities of the ball and the ground , before and after the collision, in Frame A?
Same question for Frame B.

http://en.wikipedia.org/wiki/Elastic_collision#One-dimensional_Newtonian

For the inertial frame A set u₁ to zero. For frame B same as above, then subtract v₁ from the final velocities.

 

[Delta2]

Ok it is now clear to me that in Frame B it will be $v_2=-u_2$ for some reason i couldn't see that, thanks. However in Frame A it is $|v_2|<|u_2|$ correct?