From Newsgroup: sci.physics.research

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.

Wanting to double the force from F_AB=+v, to F_AB=+2v, I would
spontaneously do so by doubling the mass of body A from m_A=1 to m_A=2
(which in the animation I can vary with the appropriate button).

And instead, this doubling of the force does not occur because with the
mass m_A=2, the force does not double from F_AB=+v to F_AB=+2v but
limits itself to increasing only up to F_AB=+4/3v, which is less than
+2.

And the same thing happens in the elastic collision where, when the
mass of A doubles from 1 to 2, the force increases only from F_AB=+2v
to F_AB=+8/3v and does not double until F_AB=+4 (which is double
F_AB=+2v).

Why does the doubling of the mass not also correspond to the doubling
of the force, if the speed is the same?

Evidently, in all this there is some hidden error: what is it?

Luigi Fortunati
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From Newsgroup: sci.physics.research

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.
--
Mikko
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From Newsgroup: sci.physics.research

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
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From Newsgroup: sci.physics.research

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
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From Newsgroup: sci.physics.research

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, therefore, in our case the acceleration of B is equal to its initial velocity with the sign changed because it is a_B=vf_B-vi_B=0-(-v)=+v.

That is why I wrote that the acceleration of B is a_B=+v.

The mass of B should not be 1 because 1 is a number, not a mass.

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

Luigi Fortunati
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From Newsgroup: sci.physics.research

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
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