Kinetic Energy Loss in Head-On Elastic Collision: Mass Ratios

mangafan
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in a head-on elastic collision between two masses, one of which is stationary, a experimenter would like the incident mass to lose quarter of its kinetic energy in the collision. What must be the ratio of the masses of the two objects for this to be the case?
 
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Thread 'Need help understanding this figure on energy levels'

Thread 'Need help understanding this figure on energy levels'

This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
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Thread 'Understanding how to "tack on" the time wiggle factor'

Thread 'Understanding how to "tack on" the time wiggle factor'

The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
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