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From: Luigi Fortunati <fortunati.luigi@gmail.com>
Newsgroups: sci.physics.research
Subject: Re: The hidden error
Date: Sat, 29 Mar 2025 00:38:53 PDT
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Mikko il 28/03/2025 06:16:16 ha scritto:
> On 2025-03-27 08:26:11 +0000, Luigi Fortunati said:
>
>> I have completed the animation of the elastic collision
>> https://www.geogebra.org/classic/hxvcaphh
>> and the inelastic one
>> https://www.geogebra.org/classic/atdrbrse
>> where, in both cases, I noticed a strange phenomenon.
>> 
>> In the inelastic collision, body A with mass m_A=1 exerts a force
>> F_AB=+v on body B, because it increases its speed from vi_B=-v to
>> vf_B=0.
>
> In a collision the force is not constant in time. It is initilally
> sero and finally sero but if it is always zero there is no collision.
> How the force varies duriong the collision depends on details that
> are not discussed below. In the special case of zero duration of the
> collision the force is infinite.

Zero duration does not exist, the time of the collision is very short 
but it is never zero.

[[Mod. note -- I think Mikko was trying to describe the limit where
the duration goes to zero, with the force scaling proportionally to
1/duration.  In this limit, the force is a Dirac delta function,
  https://en.wikipedia.org/wiki/Dirac_delta_function#History
and isn't actually a real-valued function.  It can be rigorously
defined via the theory of distributions,
  https://en.wikipedia.org/wiki/Distribution_(mathematics)
but the intuitive notion of an infinitely narrow and infintely tall
"spike" of finite area is relatively simple and often sufficient.
Another representation is that a Dirac delta function is the derivative
of a Heaviside step function.  I suspect a Dirac delta function can also
be defined via non-standard analysis, but I'm not sure of this.
-- jt]]

I obtained the force from Newton's second law F=ma, knowing <m> and 
knowing <a>.

When body A has mass m_A=1, in the inelastic collision the acceleration=
 
of body B is a_B=+v, so the force acting on B is equal to +v (the mass 
of B being equal to 1)

And when the mass of body A doubles to m_A=2, the acceleration of body 
B and the force acting on B increase only from +v to +4/3v instead of 
doubling from +v to +2v.

Why does the force on body B increase but not double if the mass A that 
exerts it doubles?

This was the question.

Luigi Fortunati