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From: Luigi Fortunati <fortunati.luigi@gmail.com>
Newsgroups: sci.physics.research
Subject: Re: inelastic collision (was: Re: Newton e Hooke)
Date: 3 Mar 2025 08:09:36 GMT
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Jonathan Thornburg [remove -color to reply] il 01/03/2025 10:18:54 ha 
scritto:
> ...

I improved my animation https://www.geogebra.org/classic/c4ruteax where 
the particles are atoms that compress in the collision and then expand 
again without detaching and the central dots are the atomic nuclei.

> ...

> In article <vps7vq$3laqc$1@dont-email.me>, Luigi Fortunati replied
>> Body B has a momentum -3 and, therefore, cannot transfer -3.75 to body 
>> A because it does not have it.
>
> This is mistaken.  (Linear) momentum doesn't have an inherent zero point,
> so there's never a case where one body doesn't have enough momentum to
> transfer some to another body.  Rather, momentum is analogous to position
> on a number line, where being at position -3 doesn't prevent you from moving
> a distance 3.75 either to the right or to the left.
>
> One way to "see this in action" is to consider what the collision would
> look like if analyzed in a different inertial reference frame (IRF).  For
> example, let's consider an IRF which is moving to with a velocity v=-10
> (i.e., moving the left at a speed of 10) with respect to Luigi's original
> IRF.  In this new IRF, each velocity is the velocity in Luigi's original
> IRF + 10.
>
> In this new IRF, the speeds and momenta before the collision are
>    v_A_before = +11 --> p_A_before = +55
>    v_B_before =  +9 --> p_B_before = +27
>    p_total_before = p_A_before+p_B_before = +82
> so that after the collision, the total momentum must also be p=+82.  Hence
> the common body of mass 8 must be moving at a speed of p/m = +10.25 after
> the collision, and A and B's speeds and momenta after the collision must be
>    v_A_after = +10.25 --> p_A_after = +51.25
>    v_B_after = +10.25 --> p_B_after = +30.75
>    p_total_after = p_A_after+p_B_after = +82
> The velocity changes during the collision are thyus
>    Delta_v_A = v_A_after - v_A_before = +10.25 - +11 = -0.75
>    Delta_v_B = v_B_after - v_B_before = +10.25 -  +9 = +1.25
> and the momentum changes during the collision are
>    Delta_p_A = p_A_after-p_A_before = +51.25 - +55 = -3.75
>    Delta_p_B = p_B_after-p_B_before = +30.75 - +27 = +3.75
>
> Notice how the velocity changes during the collision, AND the momentum
> changes and A <--> B transfers during the collision, are exactly the same
> as when we analyzed the collision in Luigi's original IRF.

We are in complete agreement on this, I also say that the variation of 
the momentum of body A is
Delta_p_A=p_A_after-p_A_before=-3.75
and the variation of the momentum of body B is
Delta_p_B=p_B_after-p_B_before=+3.75

So, on the variations of the momentum quantities we both say the same 
thing.

What differentiates us is the *force* FB_A (of body B on body A) which 
for you is -3.75 and for me is -3.

From a physical point of view, the force FA_B (that body A exerts on 
body B) is nothing other than the force that particle A1 exerts on 
particle B1 because A1 and B1 are the only particles of the 2 bodies 
that touch.

And the force FB_A (that body B exerts on body A) is nothing other than 
the force that particle B1 exerts on particle A1.

If only particles A1 and B1 existed, they could only exchange their 
small opposing forces +1 and -1, and nothing else.

Since they exchange much larger forces, it means that they receive some 
help to increase their paltry forces.

Obviously, particle A1 is reinforced by the particles A2-A5 behind it, 
which, inside body A, transmit their positive forces forward to A1 
where they accumulate, to then discharge themselves on particle B1.

And in body B, the particles B2 and B3 behind it do the same, which 
transmit their negative forces forward to the left to B1 where they 
accumulate, to then discharge themselves on particle A1.

These are physical considerations and not just mathematical ones.

Could you explain to me how it is possible that the particle B1 where 
the negative forces of only 2 particles (B2 and B3) accumulate, can 
exert the *same* force as the particle A1 where the positive forces 
that accumulate (and push) are exactly double since they come from 4 
particles behind (A2, A3, A5 and A5) and not from 2?

Ciao.