Forgive me if I'm wrong, but the way I understand things it's energy of the CB that breaks up the rack, not the force of your stroke. The correct formula is E=1/2 mv^2. Energy varies as the square of the velocity. Therefore, a CB speed of 15 MPH delivers less than 1/2 the energy to the rack that a CB speed of 22 MPH delivers.
F=ma seems meaningless in determining breaking energy because it fails to account for the the duration of the applied force (your stroke), so the final velocity remains unknown. A short stroke has less distance to accelerate the cue, so greater force is needed to achieve the same speed as a longer stroke (one that has more time to accelerate the cue).
Also, that force must be constant throughout the stroke. An object can only be accelerated smoothly while a steady force is being applied to it.
Think about pushing a car from a standstill. At the beginning, all of your available force is pushing on the car, so acceleration is at a maximum, even though the car seems to be barely moving. As time goes on, the car goes faster and faster, but the faster it goes, the less force you are able to apply.
Once the car's speed hits your maximum running speed, you can no longer apply an accelerating force to it. Any force you can apply at that point will be very light, just enough to maintain its velocity (overcoming rolling frictional forces, air resistance, etc).
The same thing is happening throughout the duration of your stroke (only it's much harder to detect because the time is so short), but the breaker with the steadiest force throughout the stroke will achieve the best cue acceleration and highest final velocity. It's a timing thing.
Since this is a collision, transfer of momentum (P) must be considered here as well. P=mv. Momentum is always conserved as momentum (unlike kinetic energy, which can be transformed into other forms of energy such as heat, light, etc). Heavier cues stroked at the same velocity transfer greater momentum to the CB. If a 24 oz. cue was able to transfer every bit of its momentum to the CB, a 6 MPH tip speed would results in a 24 MPH CB speed, while an 18 oz cue would have to travel at 8 MPH to achieve the same CB velocity. Of course, this is not exactly what is seen in real life because:
1. The COR of the cue is < 1
2. Much of the momentum of the cue is carried forward of the contact point in the form of follow through. The cue would have to stop dead in its tracks for all of the momentum of the cue to be transferred to the CB, and this is not what happens, unless you were to use a 6 oz solid phenolic cue (same mass and material as the CB).
Still, a heavier cue stroked at the same velocity as a lighter cue will transfer more momentum to the CB. Since the CB is always 6 oz, and since momentum is always conserved, the CB must leave the cue tip at a higher velocity than the cue is moving forward.
But, instead of an 18 oz cue, imagine a 1 oz phenolic needle fired at 24 MPH. You will not achieve a 24 MPH break with this setup. The needle will rebound backwards violently, carrying most of its momentum back the way it came from.
Similarly, try using a 16 pound cue to get the CB up to 24 MPH. Almost all of the considerable momentum of that slow moving cue would continue forward long after the CB contact time had elapsed, transferring very little to the CB.
Cue weight is definitely a factor at a purely mechanical level, but the key is in whether or not the player can deliver the heavier cue at the same velocity as the lighter cue. Cue design may enter into this as well, like in those 24 oz Cannon cues that are claiming to achieve greater CB speed through an increase in momentum. Anybody try one of those?
Now, Dr. Dave will come on here and tell me I'm full of it.