• #### Re: Inertia and third principle

From Mikko@21:1/5 to All on Sun Aug 4 07:53:54 2024
Luigi Fortunati il 30/07/2024 19:05:34 ha scritto:
In my animation https://www.geogebra.org/m/qterew9m ...
If body A passes some momentum to body B, body B must politely return
to body A the *same* momentum it received, no more and no less.
[[Mod. note -- Newton's laws say that (assuming that there are no
external forces acting) the *total* momentum of the system remains
constant.

Sure! The *total* momentum of the system remains (undoubtedly)
constant, this is established by the law of conservation of momentum
which I have never disputed.

If we look at the momentum *changes* during the collision, we have
p_A_after - p_A_before = 3.75 - 15 = -11.25 , while B's momentum change
during the collision is p_B_after - p_B_before = 2.25 - -9 = +11.25
so the momentum *changes* are indeed precisely opposite. -- jt]]

Yes, even the *changes* in momentum during the collision are equal and opposite, otherwise the law of conservation of momentum which I have
never disputed would not be correct.

My objection concerns something else, it concerns the equality between
action and reaction (the third principle).

The law of conservation of momentum is equivalent to lawes of inertia
and reaction together. If conseervation of momentum is violated without interaction the law of inertia is violated. If conservation of momentum
is violated in an interaction the law of reaction is violated.

--
Mikko

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• From Mikko@21:1/5 to Luigi Fortunati on Thu Aug 8 08:56:30 2024
On 2024-08-08 07:03:32 +0000, Luigi Fortunati said:

Luigi Fortunati il 05/08/2024 20:41:20 ha scritto:
[[Mod. note --
To analyze a collision at the level of granularity you're trying to
achieve (i.e., forces acting on the individual mass particles which
make up bodies A and B, you need to include the inter-particle forces
between the particles which make up each body. That is, if we label the
columns of particles in your animations from left to right as A5, A4,
A3, A2, A1 (comprising body A) and B1, B2, B3 (comprising body B),
then we need to take into account that the actual structure is
effectively
A5 <--> A4 <--> A3 <--> A2 <--> A1 B1 <--> B2 <--> B3
where each '<-->' denotes a spring...

Exactly, between the particles there are elastic forces that act like springs.

And as in all springs, if there is compression there are forces, if there is no compression there are no forces.

Before the collision there are no forces because there is no compression, during the collision suddenly the forces are there because the compression is there: this is why I talk about forces that are activated.

To analyze the collision, we really need to write out Newton's 2nd
law for all the particles, taking into account all the forces. It's
a bit of a mess, but it will give a precise answer.
-- jt]]

Great, I use Newton's 2nd law F=ma.

In my animation https://www.geogebra.org/m/qterew9m the force that brakes body A is F=ma=15*-3/4=-11.25.

The one that stops body B and then accelerates it backwards is F=9*(1/4--1)=9*+5/4=+11.25.

Therefore, the forces acting on bodies A and B during the collision are equal and opposite.

This result was tormenting for me because I was certain (and I am certain!) that the action of body A on body B cannot be equal to the reaction of body B on body A.

But my certainty was shattered on the rock of the mathematical result mentioned above.

With these doubts, I spent days and days trying to clarify every little thing, I consumed an entire notepad filling it with notes, numbers, formulas and concepts.

And finally, I understood where the solution was.

The forces acting on bodies A and B are (as I have already said) certainly equal to +11.25 and -11.25.

But how much of that +11.25 is action force exerted by A against B and how much of that -11.25 is reaction force exerted by B against A?

The two forces are equal. If they are not equal to the number shown then
there is no reason to show that numbmer.

Certainly, the force +11.25 on body B is exerted entirely by body A.

However, not all of the -11.25 force on body A is exerted by body B.

Yes, it is. You can prove that from the third law or from the conservation
of momentum. Each momentum is changed by the product of the force and
the duration of the interaction.

--
Mikko

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• From Mikko@21:1/5 to Luigi Fortunati on Thu Aug 15 13:52:30 2024
On 2024-08-13 11:58:42 +0000, Luigi Fortunati said:

There is some error in my previous post, which I have corrected here.

The force compresses if it encounters a reaction and accelerates if it does not.

In a typical collision the interaction first compresses both bodies and
then acclerates them. The amount of compression and acceleration of the
bodies need not be same. In come cases the direction of acceleration can
be opposite to the direction of motion so the speed is reduced.

Sometimes one of the bodies is so massive that the acceleration is too
small to be detected, e.g. when someting falls to the ground.

--
Mikko

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• From Mikko@21:1/5 to Luigi Fortunati on Sat Aug 24 08:46:57 2024
On 2024-08-23 14:34:29 +0000, Luigi Fortunati said:

Mikko il 15/08/2024 15:52:30 ha scritto:
There is some error in my previous post, which I have corrected here.

The force compresses if it encounters a reaction and accelerates if it does not.

In a typical collision the interaction first compresses both bodies and
then acclerates them. The amount of compression and acceleration of the
bodies need not be same. In come cases the direction of acceleration can
be opposite to the direction of motion so the speed is reduced.

Sometimes one of the bodies is so massive that the acceleration is too
small to be detected, e.g. when someting falls to the ground.

These are generic considerations that demonstrate neither equality nor inequality between the action and reaction of the two bodies.

My intent was not to demonstrate what needn't be demostrated, only that
your reasoning involves false assumptions. Those whose opĂ¶inion matters believe in Newtons's third law until otherwise is proven and, so it more
than enough to point out that you have not proven.

--
Mikko

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