From Newsgroup: sci.physics.research

In the previous discussion, Mikko wrote on 05/07/2025 11:42:33:

The equality of action and reaction means that the sum of all forces
in an isolated system (i.e., one with no external interactions) is
zero.

The history of the isolated system deserves a separate discussion.

In my animations https://www.geogebra.org/classic/gb6fye39 and https://www.geogebra.org/classic/atdrbrse, where is the isolated
system?

Body B is not an isolated system because it experiences the external
force F1, pushed to the right by body A (action).

Body A is not an isolated system because it experiences the external
force F2, pushed to the left by body B (reaction).

Mass A2 is not an isolated system because it experiences the external
force F4, pushed to the left by mass A1.

Mass A1 is not an isolated system because it experiences the external
force F3 of the rightward push of mass A2 and the leftward push of
mass B1.

Therefore, there is only one isolated body, and it is the set of
bodies A and B.

We know the masses (m_A=2, m_B=1, m_A1=1, m_A2=1, m_B1=1) and the
initial and final velocities (vi_A=+v, vf_A=+1/3v, vi_B=-v,
vf_B=+1/3v).

Therefore, we can calculate the accelerations (a_A=-2/3, a_B=+4/3) and
the forces acting on bodies A and B (F_A=m_A*a_A=2*-2/3=-4/3,
F_B=1*+4/3=+4/3).

The forces acting on body A are equal and opposite to those acting on
body B.

But there aren't just two forces acting (they're not just F_A and
F_B): there are four forces acting! They are F1, F2, F3, and F4.

Well, in both animations, the sum of all the forces is equal to zero (F1+F2+F3+F4=0), but only one animation has the correct forces: which
one is it?

Is the animation where the forces F1-F4 are F1=+1, F2=-1, F3=+0.33,
and F4=-0.33 correct, or is the one where the forces are F1=+1.33,
F2=-1, F3=+0.33, and F4=-0.67 correct?

[[Mod. note --
Both the accelerations and the forces are (highly) time-dependent.
As you say, we know the initial and final velocities, and hence the
velocity changes, but velocity changes aren't accelerations. Rather,
velocity changes are the time integral of accelerations.

Unfortunately, there's no easy way to calculate the individual
(time-dependent) forces and accelerations. In fact, because the
collision is inelastic, the time profiles of the forces and accelerations
may be different between the different bodies.

I have several comments about the animations:
* Both animations appear to show each inter-body force as increasing
linearly with distance moved after the collision, but the bodies
themselves seem to keep moving at constant speed, with no actual
velocity changes from the "collision". For a completely inelastic
collision, the final end state has all the bodies "stuck together"
all at rest with each other, and hence all the inter-body forces
must be zero in the final end state. That is, the actual time
dependence must be that during the collision the inter-body forces
first increase, but eventually they must decrease back to zero and
stay zero thereafter.
* The animation https://www.geogebra.org/classic/atdrbrse
shows forces violating Newton's 3rd law, e.g., once the collision
starts
(a) this animation shows the force F4 exerted on A2 by A1 as NOT
equal in magnitude and opposite in direction to the force F3
exerted on A1 by A2, and
(b) this animation shows the force F2 exerted on A1 by B1 as NOT
equal in magnitude and opposite in direction to the force F1
exerted on B1 by A1.
* In contast, the animation https://www.geogebra.org/classic/gb6fye39
shows forces satisfying Newton's 3rd law, e.g., at each time
(a) this animation shows the force F4 exerted on A2 by A1 as
equal in magnitude and opposite in direction to the force F3
exerted on A1 by A2, and
(b) this animation shows the force F2 exerted on A1 by B1 as
equal in magnitude and opposite in direction to the force F1
exerted on B1 by A1.

To actually work out the (time-dependent) inter-body forces F1, F2, F3,
and F4, and the corresponding accelerations, we would need to explicitly
model the inelastic-collision behavior of each body, i.e., we'd need to
write out how a mathematical model of how much force it takes to crush
each body by a given amount with a given speed,
F_crush = F_crush(d, v)
where d = the crush distance, measured from 0 at initial contact,
with (say) positive meaning crushing,
and v = the crush velocity = time derivative of d

For example, the animation https://www.geogebra.org/classic/gb6fye39
is roughly consistent with a model

{ k1*d if the two bodies are moving towards each other
F_crush(d, v) = {
{ 0 otherwise

Let's describe the motion of the bodies by the quantities

x_A1(t) = time-dependent position of A1
x_A2(t) = time-dependent position of A2
x_B1(t) = time-dependent position of B1

(all positions relative to 0 at the moment of first collision), and
the corresponding velocities and accelerations

v_A1(t) = time-dependent velocity of A1 = time derivative of x_A1(t)
v_A2(t) = time-dependent velocity of A2 = time derivative of x_A2(t)
v_B1(t) = time-dependent velocity of B1 = time derivative of x_B1(t)
a_A1(t) = time-dependent acceleration of A1 = time derivative of v_A1(t)
a_A2(t) = time-dependent acceleration of A2 = time derivative of v_A2(t)
a_B1(t) = time-dependent acceleration of B1 = time derivative of v_B1(t)

The 4 forces are then given by

F1(t) = -F2(t) = F_crush(x_A1(t)-x_B1(t), v_A1(t)-v_B1(t))
F3(t) = -F4(t) = F_crush(x_A2(t)-x_A1(t), v_A2(t)-v_A1(t))

Now we can apply Newton's 2nd law to each body to get

m_A2 a_A2(t) = F_net_on_A2(t) = F4(t)
m_A1 a_A1(t) = F_net_on_A1(t) = F3(t) + F2(t)
m_B1 a_B1(t) = F_net_on_B1(t) = F1(t)

where m_A1, m_A2, and m_B1 are the masses of the 3 bodies.

With some cleverness one could probably solve these equations analytically,
but it's likely easier to do it numerically using your favorite software library for ODE integration.
-- jt]]
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From Newsgroup: sci.physics.research

On Fri, 25 Jul 2025 14:36:16 PDT, Luigi Fortunati
<fortunati.luigi@gmail.com> wrote:

...
[[Mod. note --
Both the accelerations and the forces are (highly) time-dependent.
As you say, we know the initial and final velocities, and hence the
velocity changes, but velocity changes aren't accelerations. Rather, >velocity changes are the time integral of accelerations.

Unfortunately, there's no easy way to calculate the individual >(time-dependent) forces and accelerations. In fact, because the
collision is inelastic, the time profiles of the forces and accelerations
may be different between the different bodies.

I have several comments about the animations:
* Both animations appear to show each inter-body force as increasing
linearly with distance moved after the collision, but the bodies
themselves seem to keep moving at constant speed, with no actual
velocity changes from the "collision". For a completely inelastic
collision, the final end state has all the bodies "stuck together"
all at rest with each other, and hence all the inter-body forces
must be zero in the final end state. That is, the actual time
dependence must be that during the collision the inter-body forces
first increase, but eventually they must decrease back to zero and
stay zero thereafter.

I thank the moderator for highlighting the imperfections in my
animations, which allowed me to correct them in the new, corrected
animation https://www.geogebra.org/classic/f6bpjatj , where the only
remaining inaccuracy (because I couldn't correct it) is the constant
velocity during the collision.

[[Mod. note -- If this animation doesn't work for you, the author
has an alternate version which may work better:
https://www.geogebra.org/classic/zedqvfcx
-- jt]]

Instead, I completed the collision with compression and subsequent decompression (after which the forces disappear), and now I can give
you a precise calculation of the total forces acting during the
collision (the time integral of the accelerations).

These are the calculations.

The duration of the collision is one second.

During this second, the velocity of body A changes from +1 m/s to +1/3
m/s, and that of body B changes from -1 m/s to +1/3 m/s.

The average acceleration of body A during the collision is equal to _A
= -2/3 = -0.67 m/s^2, and that of body B is _B = +4/3 = +1.33 m/s^2.

The masses are: m_A = 2 kg and m_B = m_A1 = m_A2 = m_B1 = 1 kg.

By Newton's second law (F=ma), the force that accelerated mass A2 is F_A2=1*-0.67=-0.67N (force F4 in the animation), the force that
accelerated mass B1 is F_B1=1*+1.33=+1.33 (force F1 in the animation),
and the forces that accelerated mass A1 are F_A1=1*+0.33+1*-1=-0.67N
(forces F3 and F2 in the animation).

All this demonstrates (if my calculations are correct) that Newton's
third law is incorrect because the action of body A on body B
(F1=+1.33N) is not equal to the reaction of body B on body A (F2=-1N).

Note: The forces acting on the isolated system BodyA+BodyB are exactly
equal and opposite, as they should be (F1+F3=+1.33+0.33=+1.67N to the
right and F2+F4=-1-0.67=-1.67N to the left), but they are not actions
and reactions between body A and body B because the forces F3 and F4
are *internal* to body A and do not affect body B.

Inside body A, the leftward thrust F4=-67N of mass A1 on mass A2 is
greater than the rightward thrust F3=+0.33N of mass A2 on mass A1 (the
internal forces F3 and F4 are not in equilibrium because body A is not
an isolated system).

Unfortunately, there's no easy way to calculate the individual >(time-dependent) forces and accelerations.

Above, I demonstrated that the calculation is simple if you define
time and give it a value: one second, in this case.

Unless you prove my numbers wrong.
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