• Re: The hidden error

    From Mikko@21:1/5 to Luigi Fortunati on Fri Mar 28 12:16:16 2025
    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 Mikko@21:1/5 to Luigi Fortunati on Sat Mar 29 10:48:23 2025
    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 Mikko@21:1/5 to Luigi Fortunati on Wed Apr 2 12:52:36 2025
    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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