Hi Jammonius,
The physics excerpt you posted contains physics errors and numerous unjustified and invalid assumptions.
Excerpt:
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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.
That much is valid so far.
If momentum is conserved it can be used to calculate unknown velocities following a collision.
It can be used that way only under some circumstances. Specifically, it must be known or assumed exactly how elastic the collision is; or, the final masses and velocities of all the bodies except one must already be known.
(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.
The last sentence states that, if we assume the mass of the initially moving colliding object is zero before the collision ((m1)i = 0), the initially stationary object will not be moving after the collision. For what it's worth, this is true. A ghost (assuming ghosts have zero mass) colliding with a pumpkin will not cause the pumpkin to move.
But no real moving object (not even a photon) has zero mass. The moving upper block of the WTC certainly did not have zero mass.
So, for conservation of momentum, there cannot be pulverization.
If pulverization here is referring to phenomena of the collapse of WTC structures on 9/11, this does not follow from the previous argument, which applies only to the collision of an object of zero mass.
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.
That part is essentially correct, describing the characteristics of an entirely inelastic collision.
Conservation of Energy:
In elastic collisions, the sum of kinetic energy before a collision must equal the sum of kinetic energy after the collision.
That is true only for a perfectly
elastic collision, in which no energy is converted into other forms. Note that this means that if any any permanent crushing or breaking of the colliding bodies occur, the collision cannot be perfectly elastic and kinetic energy will not be conserved.
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.
These equations and the ones that follow for kinetic energy are all wrong. Kinetic energy is (1/2)(m * v^2), not (1/2)(m * v).
Even if that error were corrected:
Kinetic energy is not being conserved if it is also being converted into other forms, as it is in an inelastic collision in which pulverization occurs. However, we can accept this equation as describing the overall conservation of energy (rather than "conservation of kinetic energy" which does not apply) in the collision.
If (1/2)(m1 * v21)i + (1/2)(m2 * v22)i = (Pulverize) + (Fail Floor Supports), there well be no momentum transfer.
(Still the wrong equation for kinetic energy.)
But also: momentum is always conserved in any collision, even if some or all of the kinetic energy is converted into other forms such as heat or work done in crushing. So if the momentum of the initially moving body changes in a collision, momentum is always transfered. If the initially moving body comes to a stop, then all its momentum must have been transfered.
What this means is that the scenario the equation describes, in which the pulverization and column breaking exactly absorbs
all of the kinetic energy, is not possible. However, we can accept it as an approximation of an outcome that
is physically possible, which is an inelastic collision in which the velocity of m2 after the collision is negligibly small. In that case, what happens in the collision is that m1 comes momentarily to rest, and m2 is very much larger than m1, which is moved only a miniscule amount as a result.
This can apply to the WTC towers because at the time of the collision under discussion, the bottom block is still rigidly attached to the earth. So the stationary object in the collision is not the bottom block with mass m2, it's the bottom block + the earth with mass m2 + m(earth) which is very many orders of magnitude greater than m1.
However, even that outcome does not arrest the collapse at that point, because another floor's supports have now failed (which is what converted
(Fail Floor Supports) amount of the kinetic energy). So m1 is not supported and will not stay at rest; it will begin falling again.
In reality, (1/2)(m1 * v21)i + (1/2)(m2 * v22)i < (Pulverize) + (Fail Floor Supports),
(Equation is still wrong for kinetic energy.)
But also: this is an unjustified assumption. No actual quantities for m1i, v21i, (Pulverize), or (Fail Floor Supports) have been presented. (Under the reasonable assumption that v22i = 0, we can ignore m2i.) So the "<" equation is a claim with no evidence.
No rationale is given to assume or otherwise believe that the kinetic energy of the moving upper block exceeds the (Fail Floor Supports) and (Pulverize) values.
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.
(Still the wrong equation for kinetic energy.)
But also: no, we do not have to assume that. We can alternatively assume that the earlier assumption made without evidence or justification, that the energy of pulverization and failing floor supports exceeds the kinetic energy of the falling mass, is false.
Appendix B: Assuming elastic collisions:
Assume that the top floor stays intact as a solid block weight, Block-A. Start the collapse timer when the 109th floor fails. At that instant, assume floor 108 miraculously turns to dust and disappears. So, Block-A can drop at free-fall speed until it reaches the 108th floor. After Block-A travels one floor, it now has momentum. If all of the momentum is transferred from Block-A to Block-B, the next floor, Block-A will stop moving momentarily, even if there is no resistance for the next block to start moving.
(m1 * v1)i = (m2 * v2)f
There is no reason to expect Block-A to stop moving even under the silly assumption of an elastic collision, which we know is impossible because crushing is occurring.
The final velocity of a moving mass m1, colliding elastically at velocity v1 into a stationary mass m2, is (m1 - m2)/(m1 + m2) * v1. That means it's only zero if the two masses are equal.
If Block-A stops moving, after triggering the next sequence, the mass of Block-A will not arrive in time to transfer momentum to the next "pancaking" between Block-B and Block-C. In other words, the momentum will not be increased as the "collapse" progresses.
However, as we can observe, the building disintegrated from the top down and there was no block of material.
This is irrelevant, as it depends on two demonstrably false assumptions: that the collision is elastic, and that the masses are equal.
In conclusion, I suggest you inform Judy Wood of these physics errors, so that they can be corrected and the argument -- if any argument remains to be made after the corrections -- can be restated in a more valid way.
Respectfully,
Myriad