Sujet : Re: inelastic collision (was: Re: Newton e Hooke)
De : fortunati.luigi (at) *nospam* gmail.com (Luigi Fortunati)
Groupes : sci.physics.researchDate : 03. Mar 2025, 09:09:36
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <vq2ekr$u7jg$1@dont-email.me>
References : 1 2 3 4 5 6 7 8 9 10 11 12
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.