Redline. Take brand "X" bow and two arrows. One that weighs 350 grains and one that weighs 500 grains. Shoot them both through the chronograph and calculate the KE. The heavier arrow will have more KE and momentum. This is directly related to the ammount of time that forces are being applied to the arrow.
but when the string is released, (excluding the losses due to bow efficency) the bow will transfer all of its potential energy to the arrow.
Wrong. All of the bows stored energy is not transfered to the arrow. It is also transfered to the riser, limbs, the cams, the string and cables, your hand,etc... etc...
Itisn't as simple as you are trying to make it. The ammount of time that "X" ammount of forces are being applied to the projectileabsolutely plays a part in available KE and momentum.
As the string is released the stored energy is being transfered to the arrow. The longer that the projectile is having the available energy being transfered to it the more energy it is absorbing. This is true in rifles as well. Or any moveable object for that matter.Given the same cartridge being fired out of the same gun a heavier projectile remains in the barrel longer and the pressures (say 60,000 psi) insidethe barrel are able to exert their forces upon the bullet for a longer period of time which will give the heavier projectile more KE and momentum.
KE = ½ mv^2
There is no variable for time in that equation. The amount of time it takes is irrelevant.
That's becasue you are not using the correct equation. I don't know exactlywhat the equation would be but in your equation it only accounts for the velocity after the available energy has been transfered to the projectile.