Give Peace A Chance

[W. Niederhut]

5 minutes ago, Bill Brown said:

 

Again, you said: "V1 is the velocity of JFK's head before impact".  That is a direct quote.

That is why you weren't making any sense.

 

I'll re-post my equation for you, while you work on distinguishing your right from your left hand. Deal?

[W. Niederhut]

On 7/24/2024 at 6:05 PM, W. Niederhut said:

How about an old-fashioned physics equation, Jean?

The Lone Nutters can try to twist that around, rhetorically, as much as they want, but it doesn't get any more "self-explanatory" than a mathematical equation expressing Newton's Law of Conservation of Momentum.

It doesn't matter if M2 is a bullet, a baseball bat, or a frying pan.

The equation still holds.

There was no momentum vector originating from the TSBD that could have slammed JFK's head violently backward.

(And it wasn't de-cerebrate posturing.  The backward motion of the head was instantaneous.)

If M1 is the mass of JFK's head and M2 is the mass of the bullet, and

V1 is the velocity of JFK's head before impact, and V2 is the velocity of the bullet before impact, and

V1' is the velocity of JFK's head after impact, and V2' is the velocity of the bullet after impact, then,

M1V1 + M2V2 = M1V1' + M2V2' (+M3V3)

(Where M3 and V3 are the mass and velocity of the displaced skull and brain matter)

 

Re-posting my momentum equation for Bill Brown.

[Bill Brown]

5 minutes ago, W. Niederhut said:

I'll re-post my equation for you, while you work on distinguishing your right from your left hand. Deal?

 

Again, you said: "V1 is the velocity of JFK's head before impact".  That is a direct quote.  YOU were confused by the terms "before" and "after".  It has nothing to do with my determination of right and left.

That is why you weren't making any sense.

[W. Niederhut]

Just now, Bill Brown said:

 

Again, you said: "V1 is the velocity of JFK's head before impact".  That is a direct quote.  YOU were confused by the terms "before" and "after".  It has nothing to do with my determination of right and left.

That is why you weren't making any sense.

We're talking about V1'-- not V1.

Go back and study my equation, Bill.

[Bill Brown]

19 minutes ago, W. Niederhut said:

We're talking about V1'-- not V1.

Go back and study my equation, Bill.

 

Okay.  So V1 is before impact.  V1' is after impact.

But this is where you get it wrong...

You said: "It doesn't matter if M2 is a bullet, a baseball bat, or a frying pan."

But it does matter.  A bullet will not transfer the same amount of kinetic energy to the head that a baseball bat or a frying pan will.

Edited August 14 by Bill Brown

[Bill Brown]

19 minutes ago, W. Niederhut said:

We're talking about V1'-- not V1.

Go back and study my equation, Bill.

 

What is your response to this...?

In the Zapruder Film frame of reference, V1 = 0, M2 << M1, and V2' is also small. So we can simplify as:

M2*V2 = M1*V1' + M3*V3

M1, M2, V2, V1', and V3 are either known or may be reasonably estimated.

In the Zapruder Film, we *observe* both V1' and V3 to be left-to-right (> 0). Thus, V2 > 0, and the bullet is moving left-to-right.

[W. Niederhut]

44 minutes ago, Bill Brown said:

 

Okay.  So V1 is before impact.  V1' is after impact.

But this is where you get it wrong...

You said: "It doesn't matter if M2 is a bullet, a baseball bat, or a frying pan."

But it does matter.  A bullet will not transfer the same amount of kinetic energy to the head that a baseball bat or a frying pan will.

Correct.

That's why I added the M3 and V3 variables to the equation.

But you still need to work on distinguishing right from left.

Edited August 14 by W. Niederhut

[Bill Brown]

2 minutes ago, W. Niederhut said:

Correct.

That's why I added the M3 and V3 variables to the equation.

But you still need to work on distinguishing right from left.

 

"But you still need to work on distinguishing right from left."

No.  I need to work on distinguishing V1 from V1'.

[Sandy Larsen]

On 8/14/2024 at 10:35 AM, Bill Brown said:

You said: "It doesn't matter if M2 is a bullet, a baseball bat, or a frying pan."

But it does matter.  A bullet will not transfer the same amount of kinetic energy to the head that a baseball bat or a frying pan will.

 

What you say (in red) is true. The reason being that some of the kinetic energy will be transformed into another form of energy... namely heat. And the amount of heat lost (dissipated) will depend upon the materials that collide.

HOWEVER...

One of the beauties of using the law of conservation of momentum is that energy is irrelevant! It need not be accounted for whatsoever... even if it is lost. All that matters are the initial and final values of momentum in the system under study. The total momentum will remain the same over time, as long as no external force acts upon any object in the system.

 

[Kevin Balch]

On 8/17/2024 at 5:03 AM, Sandy Larsen said:

 

What you say (in red) is true. The reason being that some of the kinetic energy will be transformed into another form of energy... namely heat. And the amount of heat lost (dissipated) will depend upon the materials that collide.

HOWEVER...

One of the beauties of using the law of conservation of momentum is that energy is irrelevant! It need not be accounted for whatsoever... even if it is lost. All that matters are the initial and final values of momentum in the system under study. The total momentum will remain the same over time, as long as no external force acts upon any object in the system.

 

In textbook problems which are either perfectly elastic collisions or perfectly inelastic collisions that’s true. Not so when there is deformation of materials, fracture of materials, fluids etc.

Even in idealized rigid bodies the moment of inertia needs to be considered. Everyone considers the head mass of 5 kg. It was attached to a body of 70 kg.

I tried to explain what I was getting at but this Wikipedia article explains it better than I could.

https://en.wikipedia.org/wiki/Coefficient_of_restitution

Edited Monday at 07:27 PM by Kevin Balch