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From: Luigi Fortunati <fortunati.luigi@gmail.com>
Newsgroups: sci.physics.research
Subject: Re: Inertia and third principle
Date: Wed, 31 Jul 2024 00:05:34 PDT
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Luigi Fortunati il 22/07/2024 04:56:42 ha scritto:
> In my animation https://www.geogebra.org/m/qterew9m the momentum of 
> body A before the collision is p=mv=15v.
>
> After the collision, the momentum of body A is reduced to 
> p=m*(1/4)v=15*(1/4)v=(15/4)v=3.75v
>
> During the collision, body A lost a momentum equal to 15v-3.75v=11.25v
>
> Where did this momentum lost by body A during the collision go?
>
> Luigi Fortunati
>
> [[Mod. note -- 
> Hint: What is body B's mass?  What was its velocity before the collision?
> What is its velocity after the collision?
> -- jt]]

Thank you for your suggestion.

It's true (and it's also obvious): there are only two bodies and what 
one loses can only be gained by the other and vice versa.

However, Newton's 3rd law dictates that there are no gains or losses, 
otherwise the equality between action and reaction would fail.

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.  In
this case, that's saying that
  p_A + p_B = p_total = constant .
In other words, if we define
  p_total_before = p_A_before + p_B_before
  p_total_after  = p_A_after  + p_B_after
then we have
  p_total_after = p_total_before .
This statement holds in any inertial reference frame.

This statement implies that
  (p_A_after - p_A_before) = - (p_B_after - p_B_before) ,
i.e., the *change* in A's momentum during the collision is precisely
the opposite of the *change* in B's momentum during the collision.
-- jt]]

And instead, in the case of my animation this is precisely what doesn't 
happen and I demonstrate it.

Before the collision, body A had 6 more momentum than body B (+15 for 
body A and -9 for body B), after the collision it has only one and a 
half points more momentum than body B ( body A +3.75, body B +2.25).

Body A gave (and received nothing), body B received (and gave nothing).

[[Mod. note -- I don't see any problem here.
Before the collision,
  p_A_before = +15
  p_B_before = -9 ,
hence 
  p_total_before = p_A_before + p_B_before = +6 ,
and after the collision
  p_A_after = +3.75
  p_B_after = +2.25 ,
hence
  p_total_after = p_A_after + p_B_after = +6 ,
so we do indeed have
  p_total_after = p_total_before .

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_A_after = 2.25 - -9 = +11.25
so the momentum *changes* are indeed precisely opposite.
-- jt]]

The action was greater than the reaction and not equal: body A won and 
moved forward, body B lost and went back.

[[Mod. note -- What you're describing is consistent with the mass of A
being larger than the mass of B.  -- jt]]

My animation https://www.geogebra.org/m/tgtrjjuw clearly demonstrates 
that if the action and reaction were *always* equal and opposite, the 
rods would never tilt.

Luigi Fortunati