Shaft said:
I posted all my calculations in Post #17 so anyone can check me.
Hey Shaft, I haven't really checked all of your calculations, but on first glance, they look good. I do have to take exception with one thing, though.
The coefficient of restitution is not actually the main culprit at work here. The coefficient of restitution, by definition, is a ratio of the velocity of two objects involved in the collision prior to, and after impact (more about that later). According to the
Handbook of Chemistry and Physics, the actual formula for the coefficient of restitution is as follows:
C= (v4-v3)/(v2-v1)
where two bodies moving in a straight line with velocities v1 and v2 respectively, collide, and after the impact have velocities v3 and v4.
Now, because we are assuming the collision is perfectly elastic (that is, no energy is lost to heat, sound, etc.), the coefficient of restitution is going to be "1" for both collisions. If we plug the actual velocities of the cues and cue balls that you calculated into the formula, we will indeed get a C.O.R. ratio of "1" for each collision. And since the coefficient of restitution is the same in both cases, it is safe to say that it is not responsible for the difference in energy transfer, because both collisions are 100% efficient (only for the purpose of these calculations, of course).
As you pointed out,
both cues have the same energy prior to the collision with the cue ball, but the heavier cue has more momentum. Also, the lighter cue has a greater velocity prior to the collision. Because of the heavier cue's momentum, an interesting happens after the collision. Things turn around, and after the impact, and the heavier cue now has a greater velocity than the lighter cue. Since the heavier cue has a greater velocity after impact, it obviously has retained more of it's kinetic energy than the light cue. This, of course, means that the heavier cue transfered less energy to the cue ball. And since less energy is transfered to the cue ball, it will have a lower velocity. The bottom line, as you said, is that the lighter cue will impart more velocity to the cue ball due to a greater transfer of energy, assuming all other things are equal.
Shaft's calculations appear to be dead nuts on, and I thank him for taking the time to grind them out for us.
One final note on the coefficient of restitution. The golf world has perverted the definition of the the coefficient of restitution. The C.O.R. is actually the property of two colliding objects, and IS NOT a property that can be assigned to one particular object or material. Golfers have been fooled into thinking a club or ball can be assigned a C.O.R. and that is simply wrong. For instance, if you drop a golf ball onto a concrete walkway, it will bounce back up with an initial velocity that is very close to the final velocity it had when it hit the walkway. Therefore, the collision of those two objects would result in a C.O.R of close to "1". But if you drop the same golf ball into a sand trap, it will not bounce back up much, if at all, and thus the C.O.R would be near "0". So you must take both objects into account when calculating the actual coefficient of restitution. In other words, what did you bounce the golf ball off of? It makes a difference.