| Appendix A: For those concerned about momentum.
Conservation of Momentum and Conservation of Energy
Conservation of Momentum:
The amount of momentum (p) that an object has depends on two physical quantities: the mass and the velocity of the moving object.
p = mv
where p is the momentum, m is the mass, and v the velocity.
If momentum is conserved it can be used to calculate unknown velocities following a collision.
(m1 * v1)i + (m2 * v2)i = (m1 * v1)f + (m2 * v2)f
where the subscript i signifies initial, before the collision, and f signifies final, after the collision.
If (m1)i = 0, and (v2)i = 0, then (v2)f must =0.
So, for conservation of momentum, there cannot be pulverization.
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If we assume the second mass is initially at rest [(v2)i = 0], the equation reduces to
(m1 * v1)i = (m1 * v1)f + (m2 * v2)f
As you can see, if mass m1 = m2 and they "stick" together after impact, the equation reduces to ,
(m1 * v1)i = (2m1 * vnew)f
or vnew = (1/2) * v1
If two identical masses colliding and sticking together, they will travel at half the speed as the original single mass.
Conservation of Energy:
In elastic collisions, the sum of kinetic energy before a collision must equal the sum of kinetic energy after the collision. Conservation of kinetic energy is given by the following formula:
(1/2)(m1 * v21)i + (1/2)(m2 * v22)i = (1/2)(m1 * v21)f + (1/2)(m2 * v22)f + (Pulverize) + (Fail Floor Supports)
where (Pulverize) is the energy required to pulverize a floor and (Fail Floor Supports) is the energy required to fail the next floor.
If (1/2)(m1 * v21)i + (1/2)(m2 * v22)i = (Pulverize) + (Fail Floor Supports), there well be no momentum transfer.
In reality, (1/2)(m1 * v21)i + (1/2)(m2 * v22)i < (Pulverize) + (Fail Floor Supports),
So, for conservation of energy, we must assume there is some additional energy such that,
(1/2)(m1 * v21)i + (1/2)(m2 * v22)i + (Additional Energy) = (Pulverize) + (Fail Floor Supports),
where (Additional Energy) is the additional amount of energy needed to have the outcome we observed on 9/11/01.
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