Luigi Fortunati il 12/03/2025 13:37:09 ha scritto:
The animation https://www.geogebra.org/classic/hxvcaphh shows how two
rigid bodies interact when they collide head-on.
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When body A comes into contact with body B, an often imperceptible
contraction occurs (first phase) and a subsequent elastic return to the
initial form (second phase).
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The quantitative model is the following, before the collision the
masses are both equal to 1 (m_A=1 and m_B=1), the initial velocities
are vi_A=+1 and vi_B=-1, and the momentum quantities are pi_A=+1 and
p_Bi=-1.
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In the first phase of the collision, a compression zone and a pair of
action and reaction forces F1 and F2 are activated simultaneously in
the contact area (the compression is the cause, the opposite forces are
the effects).
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The force F1 slows the motion of body B to the left until it stops and
the force F2 does the same, slowing the motion of body A to the right
until it stops (the forces are the cause, the decelerations are the
effects).
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During this phase, the changes in the value of the two opposing blue
and red action and reaction forces are displayed.
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When the two bodies have stopped, the contraction is at maximum (1),
the velocities have become zero and the opposing forces measure F1=+1
and F2=-1.
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You can stop the motion at this moment with the "Max compression"
button.
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In the second phase of the impact, the opposing forces F1 and F2
accelerate the two bodies A and B in the opposite direction until the
initial position of contact where both the compression, the forces and
the acceleration cease to exist.
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From this point on, the motion of the two bodies returns to being
inertial, the velocities have been inverted:
body A from vi_A=+1 to vf_A=-1 and body B from vi_B=-1 to vf_B=+1
and the quantities of motion as well:
body A from pi_A=+1 to pf_A=-1 and body B from pi_B=-1 to pf_B=+1.
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All this is very simple because the 2 bodies have the same mass but
things get complicated when we choose to increase (with the appropriate
button) the mass of body A from m_A=1 to m_A=2 because, to the main
contraction between the particles A1 and B1, the secondary compression
between A2 and A1 is added whose action on the left concerns
exclusively body A, while that on the right concerns both body A and
body B.
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But I will talk about this in the next post because here I have gone on
too long.
In the case of body A in my animation with mass m_A=2, the forces
generated by the main compression are F1 and F2, and those generated by
the secondary compression are the forces F3 and F4, both are pairs of
equal and opposite forces that guarantee the conservation of the
quantity of motion.
Therefore, the forces that push to the right (F1+F3) are exactly equal
and opposite to the forces that push to the left (F2+F4).
But Newton's third law does not say what we all know, that is, that the
forces that push to the right are equal and opposite to the forces that
push to the left, but says something completely different: it says that
the forces that body A exerts on body B are equal and opposite to those
that body B exerts on body A.
So I ask: is it body B that exerts the force F4 that pushes body A to
the left? No!
The force F4 is exerted by particle A1 against particle A2, that is, it
is exerted by body A against itself, it is an *internal* force that
body A uses to give itself a push to the right against body B: the
force F3 directed towards body B that would not exist without the
internal force F4!
The force F3 is released to the right against body B and the force F4
is released to the left against the same body A.
Body A exerts two pushes on body B (F1+F3) and body B exerts only one
push on body A (F2), while the other push F4 remains inside body A.
Therefore, the action of body A on body B is *greater* than the
reaction of body B on body A.
But greater than how much? I had already asked myself this for a long
time but I never dwelt on it too much and, instead, it was essential to
do so.
Now, on the Italian physics newsgroup, an interlocutor (rightly)
pointed out to me that if a law is not good, it must be replaced by a
new one that takes its place.
And so, I said to myself that, if the action is not equal to the
reaction, I absolutely had to find what relationship links the action
to the reaction, knowing full well that it had to depend on the two
masses.
I did some calculations and found that the force F_AB exerted by body A
on body B and the force F_BA exerted by body B on body A are linked by
this equation:
FA_B=-F_BA(1+(mA-mB)^2/(mA+mB))
I verified its accuracy on the numerical data of my animation with body
A of mass m_A=1 and then with that of mass M_A=2, and it works
perfectly, just as it works with all the other combinations of masses
of bodies A and B.
So, the new law is defined as follows: "For every action there is a
corresponding opposite reaction depending on the ratio between the two
masses expressed by the equation FA_B=-F_BA(1+(mA-mB)^2/(mA+mB))".
Luigi Fortunati
Newton e Hooke
Re: Newton e Hooke