Trizebulax Answers
 
Hi All,
Once
         again, I awoke thinking of Trizebulax, where Spatz and Felicia live under a sky-green sky, subsisting on cat food and table
         scraps. I find it hard to shake off Trizebulax. What a strange and beautiful place. Somewhat like Earth.
It
         seems we need to complete yesterday's quiz. Henceforth we will refer to Trizebulax as T.
We want to know the speed of the Delrin
         hockey puck just before the next-to-final rebound. 
We  begin by formulating an expression
         for the kinetic energy just after the final rebound. The kinetic energy of the final rebound was dissipated as
         heat in sliding friction during the 1/2 foot slide to the final resting place of the puck.  By definition of the coefficient
         of friction, the drag force, D, during the final slide is equal to the coefficient of friction, F, multiplied by
         the weight, W,  of the puck on T. D = F x W. W equals the mass of the puck, m, times g, the gravitational acceleration
         on T. W = m x g. So the kinetic energy just after the final rebound = force times distance = F x W x .5 feet = .5
         x m x g x .5 feet. 
The coefficient of restitution, which was given as .5, is defined as the kinetic energy after the rebound, divided
         by the kinetic energy before the rebound. Because half the kinetic  energy was lost in the final rebound, the kinetic
         energy just before the final rebound is double the kinetic energy just after the final rebound. Therefore the kinetic
         energy just prior to the final rebound  = 2 x .5 x m x g x .5 feet = m x g x .5 feet. 
The kinetic energy after the next-to-final
         rebound equals the kinetic energy prior to the final rebound plus the energy consumed by drag in the slide between the two
         walls. The slide distance between the two walls is double the distance of the final slide, consuming twice the energy
         consumed by the final slide. Therefore the energy consumed sliding between the walls = 2 x .5 x m x g x .5 feet = m x g x
         .5 feet. Thus the energy just after the next-to-final rebound is (m x g x .5 feet) + (m x g x .5 feet) = m x g x 1 foot. 
We
         have seen that, with a coefficient of restitution of .5, the energy just before a rebound is double the energy just following
         a rebound. Therefore the kinetic energy just before the next-to-final rebound = 2 x m x g x 1 foot. Also, from Newton,
         we know that the kinetic energy is 1/2 x  m x V x V , where V is the speed of the puck. So we have 1/2 x m x V x V
         = 2 x m x g x 1 foot. This leaves us with V x V equal to 4 x g x 1 foot. The acceleration of gravity, g , on T,  was
         defined one half of the gravitational acceleration  on earth. Thus g = 1/2 x 32.2 feet per square second = 16.1
         feet per square second. In conclusion, V x V = 4 x 16.1 feet per square second x 1 foot. Taking the square root we have
         V, the speed prior to the next-to-final rebound  = 8.03 feet per second. So you see, we actually did not need to know
         the weight of the puck to solve this problem.
I have no opinion on who will be president. These are strange times.
The
         sky of Trizebulax was given as green, not blue. The sky of Trizebulax is green because all other colors of the spectrum are
         reflected back into outer space.
The combined number of whiskers of Spatz and Felicia is 19.
A
         coefficient of restitution of 1.5 would violate conservation of energy, and might result in the next big bang.
I
         like the names Felicia and Spatz, but wonder if Spats should not have been named Spats for his white rear paws.
I
         hope there are no errors in my calculation of speed. 
Have a great day.
Pete