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From: Mikko <mikko.levanto@iki.fi>
Newsgroups: sci.physics.research
Subject: Re: The hidden error
Date: 29 Mar 2025 10:48:23 GMT
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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