Re: The hidden error

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Sujet : Re: The hidden error
De : mikko.levanto (at) *nospam* iki.fi (Mikko)
Groupes : sci.physics.research
Date : 29. Mar 2025, 11:48:23
Autres entêtes
Organisation : -
Message-ID : <vs8i6m$144gl$1@dont-email.me>
References : 1 2
On 2025-03-29 00:38:53 +0000, Luigi Fortunati said:

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)

The acceleration cannot be v. The symbol v is reserved for velocity.
The mass of B should not be 1 because 1 is a number, not a mass.

--
Mikko

Date Sujet#  Auteur
27 Mar 25 * The hidden error6Luigi Fortunati
28 Mar 25 +* Re: The hidden error2Mikko
29 Mar 25 i`- Re: The hidden error1Luigi Fortunati
29 Mar 25 `* Re: The hidden error3Mikko
1 Apr 25  `* Re: The hidden error2Luigi Fortunati
2 Apr 25   `- Re: The hidden error1Mikko

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