How to calculate the number of glass sheets that will be broken by a falling object? In season 1, episode 7 of King of the nerds the contestants are asked to calculate how many sheets of glass will be broken by a falling object. They are shown 1 example case and then asked to calculate the result for the fallowing 3 test, for which they are given the data. After watching this I wanted to know how to calculate the answer.
A schematic drawing of the experiment (drawn by me):

Screen captures of the experiment from the show:
 
A screen capture with all the data given:

I've translated the data to normal (metric) units:
Ball drop data


*

*The ball weighed 5.44 kg

*The ball diameter was 0.1905 m

*The distance from the bottom of the ball to the ground was 4.26 m

*The distance from the bottom of the ball to the first sheet of glass was 2.38 m

*The sheets of glass were 0.6096 m x 0.6096 m x 0.00635 m thick

*The sheets of glass were spaced 0.0762 m apart

*There were 20 sheets of glass

*The weight of one sheet of glass was 5.21 kg

*4 sheets of glass broke


Challenge Variations
(A)


*

*Ball weight 2.72 kg

*Ball diameter 0.1524 m

*Glass thickness 0.003175 m

*Glass weight, one sheet 2.72 kg

*26 sheets of glass, spaced 0.0762 m apart

*First sheet 1.9304 m from bottom of ball


(B)


*

*Ball weight 3.63 kg

*Ball diameter 0.17145 m

*Glass thickness 0.00635 m

*Glass weight, one sheet 5.21 kg

*13 sheets of glass, spaced 0.1524 m apart

*First sheet 1.9304 m from bottom of ball


(C)


*

*Pig weight 6.804 kg

*Glass thickness 0.00635 m

*Glass weight, one sheet 5.21 kg

*26 sheets of glass, spaced 0.0762 m apart

*First sheet 1.9304 m from bottom of ball


The results of the experiments were:


*

*A: 5 sheets of glass were broken

*B: 2 sheets of glass were broken

*C: 5 sheets of glass were broken


I'm interested in the correct way to calculate the analytically solution to the problem, and what physics is behind it.
 A: First, if you review the image of the whiteboard, the glass in variation A is 1/8" thick, not 1/4", thus the 6 pound sheet vs. 11.5 pound sheet.
Now, divide the sample puzzle into 5 segments:  1) ball only; 2) ball plus sheet 1; 3) ball plus sheets 1 and 2; 4) ball plus sheets 1, 2, and 3; and 5) ball plus sheets 1, 2, 3, and 4.  
Calculate the mass of each section (shown here in kg): 1) 5.44; 2) 10.66; 3) 15.88; 4) 21.09; and 5) 26.31.
Using the equation mass x gravity x height and the gravity constant 9.8 m/s/s, you can then calculate the potential energy at the beginning of each section (when the ball is first released and/or just as a sheet of glass breaks), I have shown my equation along with total potential energy (height in meters):  1)5.44 x 9.8 x 2.39 = 127.36; 2) 10.66 x 9.8 x 0.08 = 7.96; 3) 15.88 x 9.8 x 0.08 = 11.86; 4) 21.09 x 9.8 x 0.08 = 15.75; and 5) 26.31 x 9.8 x 0.08 = 19.65.
Add them all up to find the total potential energy accumulated in the system, 182.57.  As it all falls, potential energy is converted to kinetic energy = to potential energy just before impact.  Since Kinetic energy is used up and everything stops after 4 sheets, we have to assume that each sheet used approximately 1/4 of the total energy to break the sheet, or 45.64.
Before I could use this information to solve the variations, I needed one more piece of data.  Variation A uses 1/8" glass whereas all others use 1/4" glass.  I found it difficult to find the difference in breaking strength between the 2 thicknesses.  However, I did find a table of allowable weight for glass shelving.  Not great, but it worked.  Anyway, the ratio of strength of 1/4" thick to strength of 1/8" thick is just over 2, 2.197.  Therefore, I calculated 20.78 as the required kinetic energy to break 1/8" glass.  I would love a better way to do this part.
Now, for each Variation, I would start in each section.  I would calculate the kinetic energy at impact - energy to break sheet of glass + new potential energy of next section = kinetic energy of next impact.  I would then repeat that process until there is not enough energy to break another sheet, energy goes negative.
I calculated: A) 5 sheets; B) 2 sheets; and C) 4 sheets.  This is identical to the guesses made by the contestant that won the challenge.  My theory is that the extra sheet on Variation C may have broken due to variations in individual panes of glass.
Interestingly enough, I also found that variation A may have broken all sheets if just one more sheet had broken.  New Energy in every following section would have been enough to break the next sheet and energy would be built up rather than lost all the way to the bottom of the tower.
Any other thoughts?
