Re: The hidden error

On 2025-04-01 12:27:23 +0000, Luigi Fortunati said:

Mikko il 29/03/2025 11:48:23 ha scritto:

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.

>
Acceleration is given by the final velocity (which is vf_B=0) minus the initial velocity (which is vi_B=-v) and,

No, it is not. That is the change of the velocity. Mean acceleration is
the ratio of the chenge of velcity to the duration of the change.
Acceleration is the short duration limit of the acceleration.

Mass in some unit of measurement can be expressed with just one number,
since it is a scalar and does not need anything else.

But the mass is not that number. It is the number times the unit.

--
Mikko