PE of the top bead is the "Total E" for this isolated situation. When it falls this PE is converted to KE. Upon collision the PE is converted to KE of the bottom bead, but not all of it. A portion of the E is necessary to bring that top bead back up .1 meters.
So now we have
PE (initially) of the top bead=KE of the bottom bead at collision + PE (final) of the top bead at .1m
PE (Etot)=.0497*9.8*.3=KE + .0497*9.8*.1
KE=mg(.3-.1)=.0497*9.8*.2= 97.4 mJ
Now if it is kJ I have no idea how the magnitude could even be that big is we are talking about small beads falling short distances.
Suerte!
I understand the math you are using where Total PE change must equal KE change but I don't believe that applies to this system because in the time required to elevate the top bead back up to 10cm, the lower bead would have fallen and lost potential energy while gaining kinetic energy after the impact. You can't simply say that the top bead returns to 10 cm instantly after the impact so to solve for KE the moment after impact you need to factor in conservation of momentum at the time of impact.
At the exact instant of impact the top bead had lost a PE of mgh (30cm) and converted 100% into KE.
If after the impact the top bead did not bounce but remained motionless, it would have given 100% of it's KE (and velocity) to the lower bead because they are equal in mass.
Based on conservation of momentum knowing that after the impact the top bead has an upward velocity vector, you know that the bottom bead must have a vector pointed down. That would be in addition to the downward KE that would have been transferred if the top bead was motionless after impact.
I really try not to claim texts solutions are wrong but in this situation I believe they were just throwing numbers out so you can plug and chug without taking into consideration small details, like the fact that the top bead would never bounce in the first place.,, (claim validation quoted below)
If you want a very basic explanation of why, read examkrackers physics page 56.
The table shows that at best with a perfectly elastic collision, if two objects of equal mass collide where M1 has initial velocity Vo, and M2 is stationary before the collision, M1 will be stationary after the collision and M2 will have a new velocity equal to the original Vo.
Worst case scenario the collision is inelastic and balls stick together and fall down together.
All figures and quote from ExamKrackers Physics page 56.
"The table above is based upon a collison between a mass m1 moving at vo and a stationary mass m2. The velocities v1 and v2 represent the respective velocities of the masses after the collision. The velocity vc represents the velocity of the combined mass after a fully inelastic collision."
I hate to call kaplan wrong in their methodologies but the problem appears to have some gaping holes. Apologies if I am just misinterpreting the question.
All that said, JMMTB probably has the right equation to get the answer they provided.
You have to question what the solution would be if the top bead bounced back to it's original height. With that equation the KE of bead 2 after the impact, would be zero...
