You can take this even further: imagine a batter whose arms are amazingly strong, who will always hit uniformly. Now let’s line up our pitches:
- a 100m/s fastball pitched by a military-grade cannon
- a 50m/s “fastball” pitched by Nolan Ryan
- a regulation baseball sitting on a tee in front of the batter
- a ball thrown by a person standing behind the catcher, moving at 20m/s
Assume a near-elastic collision (because losses due to heat or air friction in the immediate vicinity of the collision are small) and use
Conservation of Momentum (mass x velocity):
m[sub]1[/sub]v[sub]1i[/sub] + m[sub]2[/sub]v[sub]2i[/sub] = m[sub]1[/sub]v[sub]1f[/sub] + m[sub]2[/sub]v[sub]2f[/sub]
where “m” is mass, “v” is velocity, subscripts “1” and “2” are “bat” and “ball”, and subscripts “i” and “f” are “initial” and “final”, and assuming no mass change in masses 1 or 2
and
Conservation of Kinetic Energy
[sup]1[/sup]/[sub]2[/sub] m[sub]1[/sub]v[sub]1i[/sub][sup]2[/sup] + [sup]1[/sup]/[sub]2[/sub] m[sub]2[/sub]v[sub]2i[/sub][sup]2[/sup] = [sup]1[/sup]/[sub]2[/sub] m[sub]1[/sub]v[sub]1f[/sub][sup]2[/sup] + [sup]1[/sup]/[sub]2[/sub] m[sub]2[/sub]v[sub]2f[/sub][sup]2[/sup]
I assume a bat of mass 5 kilograms (probably too light) and a ball of .5kg (probably too heavy), with a swing speed of 100m/s (probably too fast) coming into the ball. The ball speeds are in cases 1-4 above, but realize that for the first two, we prepend a negative sign to indicate that they’re moving in a direction opposite that of the bat.
Conservation of Momentum:
500 kg*m/s + {-50, -25, 0, or 10} = 5v[sub]1f[/sub] + .5v[sub]2f[/sub]
and so for Case I: v[sub]1f[/sub] = (450 - .5v[sub]2f[/sub]) / 5
Case II: v[sub]1f[/sub] = (475 - .5v[sub]2f[/sub])/5
Case III: v[sub]1f[/sub] = (500 - .5v[sub]2f[/sub])/5
Case IV: v[sub]1f[/sub] = (510 - .5v[sub]2f[/sub])/5
For each case, we’ve now expressed v[sub]1f[/sub] as a function of v[sub]2f[/sub], which we can substitute into the equations for
Conservation of Mass:
25,000 + .25 * {10,000 ; 2,500 ; 0 ; or 400} = 2.5 * (v[sub]1f[/sub])[sup]2[/sup] + .25 * (v[sub]2f[/sub])[sup]2[/sup]
Substituting from above, we have for Case I:
25,000 + 2,500 = 2.5 * {(475 - .5v[sub]2f[/sub])/5}[sup]2[/sup] + .25 * (v[sub]2f[/sub])[sup]2[/sup]
Case II: 25,000 + 625 = 2.5 * {(450 - .5v[sub]2f[/sub])/5}[sup]2[/sup] + .25 * (v[sub]2f[/sub])[sup]2[/sup]
Case III: 25,000 = 2.5 * {(500 - .5v[sub]2f[/sub])/5}[sup]2[/sup] + .25 * (v[sub]2f[/sub])[sup]2[/sup]
Case IV: 25,000 + 100 = 2.5 * {(510 - .5v[sub]2f[/sub])/5}[sup]2[/sup] + .25 * (v[sub]2f[/sub])[sup]2[/sup]
So now we’ve reduced to solvable quadratics. I dump these into Excel (my trusty graphing calculator being otherwise occupied) and get… a bunch of gibberish. I’m going to leave this posted and try a different solution.
When come back, bring roots.