Inertia and third principle

[Luigi Fortunati]

In my animation https://www.geogebra.org/m/gxqwmxah at the moment of
the collision the F1 and F2 forces of action and reaction between the
two bodies A and B are activated.

Is it true that force F1 is the sum of the blue forces of the particles
of body A and force F2 is the sum of the red forces of the particles of
body B?

Could the forces F1 and F2 exist if the blue and red forces of the
particles of bodies A and B did not exist?

Luigi Fortunati

[[Mod. note -- At the moment of collision, the only impact forces are
those on the parts of the bodies which are in contact (labelled as points
A and B in your very nice animamation). These then propagate pressure
waves into the bodies, which apply forces on the rest of the bodies.
So, the small red and blue arrows which your (very nice!) animation
shows as being applied to each mass point of bodies A and B at the
moment of collision are wrong -- the actual forces are very non-uniform
across each body (as well as being highly time-dependent).

Here's an article with actual data of this sort:
Ross, Patel, and Wenzel
"Vehicle Design and the Physics of Traffic Safety"
Physics Today vol 49 (Jan 2006), pages 49-54
The /Physics Today/ article is behind a paywall, but as of a few minutes
ago google scholar says there's a free copy at
https://people.cas.uab.edu/~nordlund/Courses/PH201/Links/vol59no1p49_54.pdf
The article has two graphs showing velocity as a function of time for
various parts of cars during car crashes. You can clearly see how
different parts of a car have different velocity profiles, and hence
different accelerations.

But, I think the answer to your first question may still be "yes". After
all, ultimately the net force on body A is the sum of all the forces on
A's individual mass points, and similarly for B.

As for your second question, suppose that instead of two solid bodies,
you just had a pair of clouds of dust moving towards each other. Then
they'd interpenetrate and basically go right through each other, so
the "impact" forces F1 and F2 would both be zero.
-- jt]]

[Luigi Fortunati]

In my animation https://www.geogebra.org/m/gxqwmxah at the moment of the collision the F1 and F2 forces of action and reaction between the two bodies A and B are activated.
Is it true that force F1 is the sum of the blue forces of the particles of body A and force F2 is the sum of the red forces of the particles of body B?
Could the forces F1 and F2 exist if the blue and red forces of the particles of bodies A and B did not exist?
Luigi Fortunati
[[Mod. note -- At the moment of collision, the only impact forces are
those on the parts of the bodies which are in contact (labelled as points
A and B in your very nice animamation). These then propagate pressure
waves into the bodies, which apply forces on the rest of the bodies.
So, the small red and blue arrows which your (very nice!) animation
shows as being applied to each mass point of bodies A and B at the
moment of collision are wrong -- the actual forces are very non-uniform
across each body (as well as being highly time-dependent).
Ross, Patel, and Wenzel
"Vehicle Design and the Physics of Traffic Safety"
Physics Today vol 49 (Jan 2006), pages 49-54
The /Physics Today/ article is behind a paywall, but as of a few minutes
ago google scholar says there's a free copy at
https://people.cas.uab.edu/~nordlund/Courses/PH201/Links/vol59no1p49_54.pdf
The article has two graphs showing velocity as a function of time for
various parts of cars during car crashes. You can clearly see how
different parts of a car have different velocity profiles, and hence
different accelerations.
But, I think the answer to your first question may still be "yes". After
all, ultimately the net force on body A is the sum of all the forces on
A's individual mass points, and similarly for B.

The net force on body A is the vector sum of the endless blue internal
forces of its own particles (which push body A to the right) and the
single external red force F2 (which pushes body A to the left).

Luigi Fortunati

[Luigi Fortunati]

As was pointed out to me, what happens in my animation https://www.geogebra.org/m/gxqwmxah is rather complex because it varies over time and generates pressures in the two bodies.

All true, but the impact time is very short and the forces acting in that brief instant can be considered as a whole.

The first overall consideration is that the forces of the collision are of inertial origin, the blue forces are activated due to the inertia of the particles of body A which would like to go to the right and cannot do so because the inertia of the particles of body B opposes (red forces).

And the inertia (red forces) of the particles of body B that would like to go to the left cannot do so because the inertia of the particles of body A opposes it (blue forces).

In this way, only compression is generated and no acceleration is generated because there is no net force.

As for this animation, there is nothing else to say.

Things get complicated when the masses of the two bodies A and B are not equal, as happens in my animation https://www.geogebra.org/m/tmwxph9z where the mass of body A is double that of body B ( mA=2mB).

In this case, the compression is not due to all the blue forces versus the red forces, because the blue forces are double the red forces!

And then, half of the blue forces (on one side) and all the red forces (on the other) generate compression, while the remaining half of the blue forces (which has no reaction forces) generates acceleration towards the right of the body AB (the two bodies A and B are now unified after the collision).

All this can be seen better in my animation https://www.geogebra.org/m/bxndbjwp where I added what happens after the collision: bodies A and B remain stationary, bodies C and D go to the right (a due to the net force), the bodies E and F go towards the left.

The action and reaction forces of the third principle are those where the reaction exists.

Instead, the forces that accelerate without compressing are those where there is no reaction and, therefore, these forces must be excluded from the third law.

Luigi Fortunati

Ps. The velocity v=1/3 of the unified body CD after the collision, calculated with half the blue force (FC/2=1) divided by the unified mass of the body CD (m=3) is perfectly identical to the velocity after the collision calculated by conservation of momentum.

[Luigi Fortunati]

Discovering the existence of an error should be stimulating and don't
disturb.

Maybe what I'm trying to say hasn't been understood or, maybe, I
haven't explained myself well.

And then I propose a clarifying example.

Yesterday you were walking calmly but distractedly and you bumped into
a person with the same size as you who was walking as distractedly as
you.

You exerted your force F=10 (in some unit of measurement) and he
reacted with his force F=-10.

You don't fall backwards and he doesn't fall either because neither
force prevails.

Today, again, you go back to walking distractedly and you collide with
another person who is twice your size and you end up on the ground,
why?

Certainly, the force that you exerted on him is the same as yesterday
(i.e. F=10) and if he also had only exerted a force F=-10 on you (as
the third law prescribes), you would not have fallen backwards.

And so, it means that the force he exerted on you is *greater* (and not
equal) to the one you exerted on him: it is the greater force of him
that threw you to the ground!

With your force F=10 you only slowed down the other person, while he
only used half of his strength F=-20 to stop you and had the other half
left to push you backwards and make you fall.

The third principle applies to the compressive force (the action and
reaction compressed you *exactly* as much as it compressed him) but
does not apply to the force that accelerated you backwards (the residue
of his preponderant force, to which no you were able to object!).

In my animation https://www.geogebra.org/m/snjpnvsu there is a
photograph of a schematized collision between two generic bodies A and
B at the instant of maximum compression and there is the elastic ring
whose deformation measures the intensity of the action and reaction
force, which depends exclusively on the smaller body and remains
unchanged when the larger body varies.

Luigi Fortunati

[[Mod. note --

There are several mistaken statements here, including
(a) "the force that you exerted on him is the same as yesterday",
(b) "the force he exerted on you is *greater* (and not equal) to the one
you exerted on him", and
(c) "the action and reaction compressed you *exactly* as much as it
compressed him".

To analyze a 2-body collision, it's useful to start by considering two
extreme cases.

First, let's consider the case where the collision is *perfectly elastic*,
i.e., there is no energy dissipation in the collision. This means that,
in any inertial reference frame (IRF), the final kinetic energy of the
two bodies is equal to the initial kinetic energy of the two bodies.
Another way to think of this case is that the bodies "bounce back" after
the collision.

It's particularly convenient to analyze collisions in the center-of-mass
(COM) IRF, i.e., the IRF where the COM of the two bodies is at rest
(equivalently, this is the IRF where the total linear momentum is zero).
Unless stated otherwise, EVERYTHING I write below is in the COM IRF.

Before the collision:
MA has some velocity, say VA_initial (assume this is nonzero)
MB has some velocity, say VB_initial (assume this is nonzero)
the condition that the total linear momentum is zero says
MA*VA_initial + MB*VB_initial = 0 (1)
i.e., the initial velocities must be in the ratio
VA_initial/VB_initial = - MB/MA (2)

Since the total linear momentum is conserved, *after* the collision
we also have
MA*VA_final + MB*VB_final = 0 (3)
i.e., the final velocities must be in the ratio
VA_final/VB_final = - MB/MA (4)

In order to also conserve kinetic energy, we must have
VA_final = - VA_initial (5a)
VB_final = - VB_initial (5b)

For example, if the two masses are equal, we might have
VA_initial = +10, VB_initial = -10 (6a)
VA_final = -10, VB_final = +10 (6b)
so that the net velocity *changes* between before & after the collision
are
VA_final - VA_initial = -20 (7a)
VB_final - VB_initial = +20 (7b)
i.e., the net velocity changes are equal.

In contrast, if B has twice the mass of A, we might have
VA_initial = +10, VB_initial = -5 (8a)
VA_final = -10, VB_final = +5 (8b)
so that the net velocity *changes* between before & after the collision
are
VA_final - VA_initial = -20 (9a)
VB_final - VB_initial = +10 (9b)
i.e., the net velocity change of A is twice that of B.


So far we've only considered a perfectly elastic collision, one where
the bodies recoil after the collision. Now let's consider the opposite
extreme, a perfectly *inelastic* collision, where after the collision
the bodies don't bounce back at all, but instead stick to each other
like lumps of clay.

Everything I wrote above about the situation before the collision still
applies here. Equation (3) also still holds. After the collision, the
two bodies stick to each other, i.e.,
VA_final = VB_final (10)
Combining this with (3), we see that we must have
VA_final = VB_final = 0 (11)

So, returning to our earlier two examples, if the two masses are equal,
we might have (6a) before the collision, but (11) after the collision,
so that the net velocity *changes* between before & after the collision
are
VA_final - VA_initial = -10 (12a)
VB_final - VB_initial = +10 (12b)
i.e., the net velocity changes are equal (but both are smaller than the
net velocity changes (7a) and (7b) for an elastic collision with the
same initial velocities).

If B has twice the mass of A, we might have (8a) before the collision,
but (11) after the collision, so that the net velocity *changes*
between before & after the collision are
VA_final - VA_initial = -10 (11a)
VB_final - VB_initial = +5 (11b)
i.e., the net velocity change of A is twice that of B (but both are
smaller than the net velocity changes (9a) and (9b) for an elastic
collision with the same initial velocities).


Real collisions fall somewhere between the perfectly elastic and perfectly
inelastic cases; in practice they are probably closer to the perfectly
inelastic case.


A few more final points:

1. At *every* time during the collision, the force A exerts on B is equal
to precisely -1 times the force B exerts on A (this is just Newton's 3rd
law). But if the masses are different, these same-magnitude forces result
in different accelerations, and thus different net velocity changes.

2. As I noted above, if we compare elastic and inelastic collisions with
the same initial velocities, the net velocity *changes* are smaller for
the inelastic case. This is why cars are engineered with "crumple zones"
which crush in accidents -- this absorbs kinetic energy, reducing the
velocity changes (and hence accelerations & forces) of the occupants.
(As the article I cited earlier in this thread noted, having crumple
zones also spreads out the velocity changes over longer times, further
reducing the instantaneous accelerations).

If we compare collisions with different ratios of the two masses, the
smaller mass always has a larger velocity change, and in fact a larger
instantaneous acceleration at each time during the collision. Thus,
if you're in a car accident, the very worst case for you (in terms of
how severe your injuries are likely to be) is to collide with another
vehicle that's both very rigid (it doesn't crumple much) and much more
massive than your vehicle.

3. Suppose that instead of the COM IRF, we were to use some other IRF,
say one moving at a relative velocity V_IRF with respect to the COM IRF.
This would mean that every velocity we've worked out above would have
V_IRF subtracted from it, but the net velocity *changes* would be the
same as they are in the COM IRF.
-- jt]]

[Luigi Fortunati]

Post by Luigi Fortunati
[[Mod. note --
1. At *every* time during the collision, the force A exerts on B is equal
to precisely -1 times the force B exerts on A (this is just Newton's 3rd
law).

In this way, you (without proof) assume that Newton's 3rd law is correct.

Instead, I will *prove* that it is incorrect.

First of all, I'll start with the fundamental error in your calculations, which is to base yourself on the variations between the final and initial speeds, as if they depended only on the action and reaction forces.

If this were really the case, it is obvious that the third principle would be right in all cases but this is not the case and I demonstrate it to you visually and mathematically with my animation https://www.geogebra.org/m/tmwxph9z where there is the body A with mass mA=2 and body B with mass mB=1.

At the moment of the inelastic collision, the momentum of half the mass of A (mAv/2=+1) and the momentum of the full mass of B (mBv=-1) cancel each other's effects and half the mass of A and the entire mass of B come to rest.

So far, the forces of the two impulses are effectively equal and opposite, and are, without a doubt, action and reaction forces between A and B.

Instead, the force of the impulse of the remaining half mass of A (mAv/2=+1) not only acts on body B (now stationary) but also on the entire mass of body A!

In fact, this +1 impulse is divided proportionally into a part (+1/3) which acts on body B accelerating it from zero to 1/3 of v to the right and a part (+2/3) on the mass of the same body A.

The first part (+1/3) is an action and reaction force between A and B, the second part (+2/3) is not.

The second part (which you use unduly to make balance the accounts) is an *internal* force in body A that has nothing to do with the action and reaction between bodies A and B!

Luigi Fortunati

[Luigi Fortunati]

I had the audacity to point out a case in which Newton's 3rd law is wrong, citing numbers, formulas and drawings.

It is something extraordinary: a law considered perfect is not always so.

Yet, there was no one who found a single mistake in what I wrote.

So, I'll go ahead and propose my latest animation
https://www.geogebra.org/m/azp3jkja
in which body A of mass m=15 (15 particles of mass m=1) collides with body B of mass m=9.

The impact lasts only a very small fraction of time but it is still sufficient to be able to divide it into 5 successive instants.

In the first instant the 3 blue surface particles of body A and the 3 red surface particles of body B collide.

The action of the blue particles and the reaction of the red ones occurs according to Newton's 3rd law: the 3 pulses of the blue particles push to the right, the 3 pulses of the red particles push to the left in the same way.

It is the equality of the 3 opposing blue and red forces that guarantees the stopping of the contact particles on the vertical of the y axis, which would not happen if these opposing action and reaction forces were not equal.

In this first phase, the remaining particles of the 2 bodies are not yet affected by the collision.

In the second instant, the 3 particles that are just behind the first ones collide with the particles in front (which have stopped) and push them, adding the force of their impulse.

And since the forces of the 2 second rows are also equal and opposite (always according to Newton's 3rd law), they cancel each other out and, even at the second instant, the first particles stop on the vertical of the y axis.

The same thing happens at time 3.

At instant 4 things change, because the fourth row of red particles (in body B) is not there and the lack of their strength prevents any reaction to the push of the fourth row of blue particles of body A.

These blue particles of the fourth row of B act on the other blue particles in front and their impulse propagates until it reaches the blue particles of the first row, where the respective red counterforce does not reach!

In this way, this blue force becomes a preponderant net force that accelerates the (now) unified body AB towards the right.

And the same happens at instant 5 when the impulsive force of the fifth row also arrives on the surface of body A without encountering any reaction and, thus, contributes to increasing the speed towards the right even further.

As you have noticed, the equality provided by Newton's 3rd law is valid only as long as the reaction can resist it and, obviously, it ceases to be valid when the reaction can no longer resist it.

It is the point of impact that we must observe to understand if the action is equal to the reaction: if it stays still, the equality is there, if it moves it is not.

And here, I want to remind you that Newton, speaking of the third principle, says that the horse pulls a stone tied to a rope with the same force with which the stone pulls the horse.

But this is only true while the horse pulls the stone at a constant speed and not when, at the start, the horse has to accelerate the stone from zero speed to speed v!

Luigi Fortunati

[Luigi Fortunati]

In my animation https://www.geogebra.org/m/qterew9m the momentum of
body A before the collision is p=mv=15v.

After the collision, the momentum of body A is reduced to
p=m*(1/4)v=15*(1/4)v=(15/4)v=3.75v

During the collision, body A lost a momentum equal to 15v-3.75v=11.25v

Where did this momentum lost by body A during the collision go?

Luigi Fortunati

[[Mod. note --
Hint: What is body B's mass? What was its velocity before the collision?
What is its velocity after the collision?
-- jt]]

[Luigi Fortunati]

Post by Luigi Fortunati
In my animation https://www.geogebra.org/m/qterew9m the momentum of
body A before the collision is p=mv=15v.
After the collision, the momentum of body A is reduced to
p=m*(1/4)v=15*(1/4)v=(15/4)v=3.75v
During the collision, body A lost a momentum equal to 15v-3.75v=11.25v
Where did this momentum lost by body A during the collision go?
Luigi Fortunati
[[Mod. note --
Hint: What is body B's mass? What was its velocity before the collision?
What is its velocity after the collision?
-- jt]]

Thank you for your suggestion.

It's true (and it's also obvious): there are only two bodies and what
one loses can only be gained by the other and vice versa.

However, Newton's 3rd law dictates that there are no gains or losses,
otherwise the equality between action and reaction would fail.

If body A passes some momentum to body B, body B must politely return
to body A the *same* momentum it received, no more and no less.

[[Mod. note -- Newton's laws say that (assuming that there are no external
forces acting) the *total* momentum of the system remains constant. In
this case, that's saying that
p_A + p_B = p_total = constant .
In other words, if we define
p_total_before = p_A_before + p_B_before
p_total_after = p_A_after + p_B_after
then we have
p_total_after = p_total_before .
This statement holds in any inertial reference frame.

This statement implies that
(p_A_after - p_A_before) = - (p_B_after - p_B_before) ,
i.e., the *change* in A's momentum during the collision is precisely
the opposite of the *change* in B's momentum during the collision.
-- jt]]

And instead, in the case of my animation this is precisely what doesn't
happen and I demonstrate it.

Before the collision, body A had 6 more momentum than body B (+15 for
body A and -9 for body B), after the collision it has only one and a
half points more momentum than body B ( body A +3.75, body B +2.25).

Body A gave (and received nothing), body B received (and gave nothing).

[[Mod. note -- I don't see any problem here.
Before the collision,
p_A_before = +15
p_B_before = -9 ,
hence
p_total_before = p_A_before + p_B_before = +6 ,
and after the collision
p_A_after = +3.75
p_B_after = +2.25 ,
hence
p_total_after = p_A_after + p_B_after = +6 ,
so we do indeed have
p_total_after = p_total_before .

If we look at the momentum *changes* during the collision, we have
p_A_after - p_A_before = 3.75 - 15 = -11.25 ,
while B's momentum change during the collision is
p_B_after - p_A_after = 2.25 - -9 = +11.25
so the momentum *changes* are indeed precisely opposite.
-- jt]]

The action was greater than the reaction and not equal: body A won and
moved forward, body B lost and went back.

[[Mod. note -- What you're describing is consistent with the mass of A
being larger than the mass of B. -- jt]]

My animation https://www.geogebra.org/m/tgtrjjuw clearly demonstrates
that if the action and reaction were *always* equal and opposite, the
rods would never tilt.

Luigi Fortunati

[Luigi Fortunati]

In my animation https://www.geogebra.org/m/qterew9m ...

If body A passes some momentum to body B, body B must politely return to body A the *same* momentum it received, no more and no less.
[[Mod. note -- Newton's laws say that (assuming that there are no external forces acting) the *total* momentum of the system remains constant.

Sure! The *total* momentum of the system remains (undoubtedly) constant, this is established by the law of conservation of momentum which I have never disputed.

If we look at the momentum *changes* during the collision, we have p_A_after - p_A_before = 3.75 - 15 = -11.25 , while B's momentum change during the collision is p_B_after - p_B_before = 2.25 - -9 = +11.25 so the momentum *changes* are indeed precisely opposite. -- jt]]

Yes, even the *changes* in momentum during the collision are equal and opposite, otherwise the law of conservation of momentum which I have never disputed would not be correct.

My objection concerns something else, it concerns the equality between action and reaction (the third principle).

In the case of my animation, the action of body A is worth +15 because there are 15 particles and their impulses directed towards the right are worth +15 (during the collision the momentum p=mv becomes impulsive force F=ma because the velocity changes and the change in speed is acceleration).

Instead, the reaction of body B is worth -13.75 because it is formed by the impulses of the 9 particles (directed towards the left) which are worth -9 and by the inertial reaction of body B (which opposes the external force of body A) which is worth -3.75.

The residue +2.25 (+15-13.75) is the net force that accelerates body B to the right.

Therefore, the +15 action of body A is greater (and not equal) than the -13.75 reaction of body B.

Luigi Fortunati

[Mikko]

Post by Luigi Fortunati

Post by Luigi Fortunati

In my animation https://www.geogebra.org/m/qterew9m ...

If body A passes some momentum to body B, body B must politely return
to body A the *same* momentum it received, no more and no less.
[[Mod. note -- Newton's laws say that (assuming that there are no
external forces acting) the *total* momentum of the system remains
constant.

Sure! The *total* momentum of the system remains (undoubtedly)
constant, this is established by the law of conservation of momentum
which I have never disputed.

Post by Luigi Fortunati
If we look at the momentum *changes* during the collision, we have
p_A_after - p_A_before = 3.75 - 15 = -11.25 , while B's momentum change
during the collision is p_B_after - p_B_before = 2.25 - -9 = +11.25
so the momentum *changes* are indeed precisely opposite. -- jt]]

Yes, even the *changes* in momentum during the collision are equal and
opposite, otherwise the law of conservation of momentum which I have
never disputed would not be correct.
My objection concerns something else, it concerns the equality between
action and reaction (the third principle).

The law of conservation of momentum is equivalent to lawes of inertia
and reaction together. If conseervation of momentum is violated without
interaction the law of inertia is violated. If conservation of momentum
is violated in an interaction the law of reaction is violated.

[Luigi Fortunati]

Post by Mikko
The law of conservation of momentum is equivalent to lawes of inertia
and reaction together. If conseervation of momentum is violated without
interaction the law of inertia is violated. If conservation of momentum
is violated in an interaction the law of reaction is violated.

This is not the case and my animation
https://www.geogebra.org/m/qterew9m proves it.

Is it true or not that the momentum of the 15 particles is transformed
(in the collision) into 15 blue impulsive forces directed towards the
right which (inside body A) accumulate in the 3 particles of the first
row and discharge their action on the B body?

Is it true or not that body B does the same thing with its 9 particles
and its 9 red impulsive forces directed towards the left?

It is true or not that the two bodies react (inertially) to external
forces with their internal forces.

Is it true or not that the action and reaction forces would not be
activated if there were not the opposite inertia of the two bodies?

Is it true or not that *mathematically* the action of body A against
body B is +15 and the reaction of body B against body A is -12.75? If
not, how much are they worth?

Luigi Fortunati

[[Mod. note --
Your questions are in some sense ill-posed: they use words which don't
have well-defined meanings in physics. Notably, it's not clear what
it means to say that impulsive forces do or don't "accumulate"; it's
not clear what it means to say that action-reaction forces are or are
not "activated"; and it's certainly not clear what it means to ask
how much an action-reaction force "is worth".

To analyze a collision at the level of granularity you're trying to
achieve (i.e., forces acting on the individual mass particles which
make up bodies A and B, you need to include the inter-particle forces
between the particles which make up each body. That is, if we label the
columns of particles in your animations from left to right as A5, A4,
A3, A2, A1 (comprising body A) and B1, B2, B3 (comprising body B),
then we need to take into account that the actual structure is
effectively
A5 <--> A4 <--> A3 <--> A2 <--> A1 B1 <--> B2 <--> B3
where each '<-->' denotes a spring with some energy dissipation
(we need the dissipation in order for the collision to be inelastic).
If, for example, we push B1 to the right relative to B2 and B3), that
will compress the B1 <--> B2 spring, but B3 won't be (immediately)
affected. However, the compressed B1 <--> B2 spring will then
accelerate B2 to the right, compressing the B2 <--> B3 spring,
which will then accelerate B3 to the right.

To analyze the collision, we really need to write out Newton's 2nd
law for all the particles, taking into account all the forces. It's
a bit of a mess, but it will give a precise answer.
-- jt]]

[Luigi Fortunati]

Post by Luigi Fortunati
[[Mod. note --
To analyze a collision at the level of granularity you're trying to
achieve (i.e., forces acting on the individual mass particles which
make up bodies A and B, you need to include the inter-particle forces
between the particles which make up each body. That is, if we label the
columns of particles in your animations from left to right as A5, A4,
A3, A2, A1 (comprising body A) and B1, B2, B3 (comprising body B),
then we need to take into account that the actual structure is
effectively
A5 <--> A4 <--> A3 <--> A2 <--> A1 B1 <--> B2 <--> B3
where each '<-->' denotes a spring...

Exactly, between the particles there are elastic forces that act like springs.

And as in all springs, if there is compression there are forces, if there is no compression there are no forces.

Before the collision there are no forces because there is no compression, during the collision suddenly the forces are there because the compression is there: this is why I talk about forces that are activated.

Post by Luigi Fortunati
To analyze the collision, we really need to write out Newton's 2nd
law for all the particles, taking into account all the forces. It's
a bit of a mess, but it will give a precise answer.
-- jt]]

Great, I use Newton's 2nd law F=ma.

In my animation https://www.geogebra.org/m/qterew9m the force that brakes body A is F=ma=15*-3/4=-11.25.

The one that stops body B and then accelerates it backwards is F=9*(1/4--1)=9*+5/4=+11.25.

Therefore, the forces acting on bodies A and B during the collision are equal and opposite.

This result was tormenting for me because I was certain (and I am certain!) that the action of body A on body B cannot be equal to the reaction of body B on body A.

But my certainty was shattered on the rock of the mathematical result mentioned above.

With these doubts, I spent days and days trying to clarify every little thing, I consumed an entire notepad filling it with notes, numbers, formulas and concepts.

And finally, I understood where the solution was.

The forces acting on bodies A and B are (as I have already said) certainly equal to +11.25 and -11.25.

But how much of that +11.25 is action force exerted by A against B and how much of that -11.25 is reaction force exerted by B against A?

Certainly, the force +11.25 on body B is exerted entirely by body A.

However, not all of the -11.25 force on body A is exerted by body B.

In fact, body B can only exert a force equal to -9 on body A, which is all the force it has and with which it manages to stop body A in the first 3 instants of my animation.

After the first 3 moments, body B no longer reacts against body A and, therefore, retreats and goes away!

So, the action and reaction ends here, with the action being worth +11.25 and exceeding the -9 reaction by +2.25 (+11.25-9).

Obviously, the matter does not end here because the answer to the question is missing: if body A, during the collision, receives a force equal to -11.25, who provides the remaining force -2.25 that body B does not provide (since it limits to -9)?

It's an answer that I will have to give but I can't give it here because the post would become too long and I will do it in the next post (if the moderators allow me to do this double intervention).

Luigi Fortunati.

[Luigi Fortunati]

This is a continuation of my previous post.

In my animation https://www.geogebra.org/m/qterew9m the action of body A and the reaction of body B occur exclusively through the two first rows, because the actions and reactions that occur behind the first row of body A are *internal* to body A and the same happens to the actions and reactions behind the front row of body B, which are internal to body B.

At instant 1, the action is +3, the reaction -3 and the two first rows stop and compress.

At instant 2, the action is +6, the reaction -6 and the first two rows of each body stop and compress more.

At instant 3, same thing: the action is +9, the reaction -9 and the first three rows of each body (9+9 particles) stop, compressing to the maximum.

Why did the compression progressively increase? Because the *internal* forces of the internal particles have *accumulated* on the first row of each of the two bodies (the place where the action and reaction take place).

Why were there these actions and these compressions? They occurred because the B corps was able to resist without retreating.

Instead, at instant 4, when the first row of body A receives a further increase in forces directed towards the right from its internal particles and increases its action from +9 to +12, on the other side there is the first row of body B which has not received any further help from its internal particles, can no longer increase its reaction from -9 to -12, must suffer the net external force +3 (+12-9) and must retreat accordingly, without being able to to oppose.

So, already here, the action of body A (+12) is greater than the reaction of body B (-9).

At instant 5, the action of body A increases even more (+15), while the reaction of body B always remains equal to -9.

Obviously, the actions and reactions don't end here because the dark area of the animation remained compressed like a spring.

And like all compressed springs, it will have to release its internal force from one side to the other.

So, there is still more to write and I cannot do it in this post which has become quite long and there is no need to make it even longer.

Furthermore, these other considerations are more than enough to reflect on while waiting for my next post, always assuming that the moderators allow me to divide my writings as I am doing.

Luigi Fortunati

[Mikko]

Post by Luigi Fortunati

Post by Luigi Fortunati
[[Mod. note --
To analyze a collision at the level of granularity you're trying to
achieve (i.e., forces acting on the individual mass particles which
make up bodies A and B, you need to include the inter-particle forces
between the particles which make up each body. That is, if we label the
columns of particles in your animations from left to right as A5, A4,
A3, A2, A1 (comprising body A) and B1, B2, B3 (comprising body B),
then we need to take into account that the actual structure is
effectively
A5 <--> A4 <--> A3 <--> A2 <--> A1 B1 <--> B2 <--> B3
where each '<-->' denotes a spring...

Exactly, between the particles there are elastic forces that act like springs.
And as in all springs, if there is compression there are forces, if there is no compression there are no forces.
Before the collision there are no forces because there is no compression, during the collision suddenly the forces are there because the compression is there: this is why I talk about forces that are activated.

Post by Luigi Fortunati
To analyze the collision, we really need to write out Newton's 2nd
law for all the particles, taking into account all the forces. It's
a bit of a mess, but it will give a precise answer.
-- jt]]

Great, I use Newton's 2nd law F=ma.
In my animation https://www.geogebra.org/m/qterew9m the force that brakes body A is F=ma=15*-3/4=-11.25.
The one that stops body B and then accelerates it backwards is F=9*(1/4--1)=9*+5/4=+11.25.
Therefore, the forces acting on bodies A and B during the collision are equal and opposite.
This result was tormenting for me because I was certain (and I am certain!) that the action of body A on body B cannot be equal to the reaction of body B on body A.
But my certainty was shattered on the rock of the mathematical result mentioned above.
With these doubts, I spent days and days trying to clarify every little thing, I consumed an entire notepad filling it with notes, numbers, formulas and concepts.
And finally, I understood where the solution was.
The forces acting on bodies A and B are (as I have already said) certainly equal to +11.25 and -11.25.
But how much of that +11.25 is action force exerted by A against B and how much of that -11.25 is reaction force exerted by B against A?

The two forces are equal. If they are not equal to the number shown then
there is no reason to show that numbmer.

Post by Luigi Fortunati
Certainly, the force +11.25 on body B is exerted entirely by body A.
However, not all of the -11.25 force on body A is exerted by body B.

Yes, it is. You can prove that from the third law or from the conservation
of momentum. Each momentum is changed by the product of the force and
the duration of the interaction.

[Luigi Fortunati]

Post by Mikko

Post by Luigi Fortunati
But how much of that +11.25 is action force exerted by A against B and how
much of that -11.25 is reaction force exerted by B against A?

The two forces are equal.

Obviously: 11.25 is equal to 11.25!

You didn't realize that my question is about something else.

Post by Mikko

Post by Luigi Fortunati
Certainly, the force +11.25 on body B is exerted entirely by body A.
However, not all of the -11.25 force on body A is exerted by body B.

Yes, it is.

And yet it isn't.

Follow my reasoning to the end (I haven't completed it yet) and you
will find out why it isn't.

Post by Mikko
You can prove that from the third law...

Maybe you haven't noticed but it's the third law I'm contesting.

Luigi Fortunati

[Luigi Fortunati]

Force compresses and accelerates.

Compression depends on the reaction because there can be no compression if there is no reaction.

Instead, the acceleration depends precisely on the lack of the reaction that makes the force "net", i.e. not counteracted by an opposite force (it is precisely the net force F=ma of the 2nd law).

The force, therefore, is divided into force that compresses and force that accelerates (which is the excess, i.e. what remains of the total force diminished by the reaction).

If I push against the wall I only generate compression because there is no excess of the action over the reaction.

If I push a cart, with part of my force I generate compression (at the point of contact against the reaction of the cart) and with the remaining part I generate acceleration of the cart (without reaction).

At instant 3 of my animation https://www.geogebra.org/m/qterew9m the impulsive force of body A against body B (+9) and that of body B against body A (-9) generate compression and no acceleration.

At instant 5, the force +15 of body A generates an opposite reaction -13.75 of body B which compresses the two bodies (+13.75 and -13.75) and with the remaining part +2.25 (+15-13.75) accelerates body B , without reaction.

At the end of the collision, the +6 sum of the momentum of body A (+3.75) and body B (+2.25) is perfectly equal to the total momentum before the +6 collision: the conservation of momentum hi is safe.

But the total +15 impulsive force exerted by body A on body B is not at all equal to the -13.75 impulsive reaction force of body B on body A.

Therefore, the equality of only the opposite compressive forces (+13.75 and -13.75) is true and is correct but the equality of all the forces exchanged between body A and body B is not true (+15 and -13.75) .

Luigi Fortunati

[Luigi Fortunati]

There is some error in my previous post, which I have corrected here.

The force compresses if it encounters a reaction and accelerates if it does not.

If the reaction exists but is insufficient, the force compresses until the reaction is canceled out and then (with what remains, i.e. with the "net" force F=ma of Newton's 2nd law) accelerates.

If I push against the wall I only generate compression because my force does not overcome the wall's reaction and there is no net residual force left to accelerate.

Instead, if I push a cart, I obtain both compression (at the point of contact where the cart reacts) and acceleration with what remains of my force net of what was used for the compression.

The sum of my force that compresses plus the residual "net" force that accelerates is exactly equal to the overall force I exerted on the cart.

This division between compression and acceleration can be clearly seen in the animation https://www.geogebra.org/m/qterew9m where there are two bodies A (mass=15) and B (mass=9) colliding at speed v= 1 and v=-1.

At instant 3, the action of body A (+9) and the reaction of body B (-9) are exactly equal and opposite and, therefore, compress the contact zone without accelerating.

In fact, the 18 particles involved were globally still before (+9v and -9v) and are still after (+0 and -0).

Instead, at instant 5, the impulsive force +15 (in some unit of measurement) of body A on body B generates a further opposite reaction which increases from -9 (what it was at instant 3) to -12.75, so that the remaining force +2.25 (+15-12.75) which finds no reaction, instead of compressing, accelerates body B.

At the end of the collision, the +6 sum of the momentum of body A (+3.75) and body B (+2.25) is perfectly equal to what it was before the collision (+15-9=+6), in accordance with the law of conservation of momentum.

The equality of the opposite action and reaction forces exerted by body A on body B and vice versa is valid if it is limited to only the two opposite compressive forces (+12.75 and -12.75) but not if we consider the entire impulsive force (action ) of body A (+15) and the entire reaction of body B (-12.75).

The forces that compress are equal and opposite, those that accelerate are not.

Luigi Fortunati

[Mikko]

Post by Luigi Fortunati
There is some error in my previous post, which I have corrected here.
The force compresses if it encounters a reaction and accelerates if it does not.

In a typical collision the interaction first compresses both bodies and
then acclerates them. The amount of compression and acceleration of the
bodies need not be same. In come cases the direction of acceleration can
be opposite to the direction of motion so the speed is reduced.

Sometimes one of the bodies is so massive that the acceleration is too
small to be detected, e.g. when someting falls to the ground.

[Luigi Fortunati]

Post by Mikko

Post by Luigi Fortunati
There is some error in my previous post, which I have corrected here.
The force compresses if it encounters a reaction and accelerates if it does not.

In a typical collision the interaction first compresses both bodies and
then acclerates them. The amount of compression and acceleration of the
bodies need not be same. In come cases the direction of acceleration can
be opposite to the direction of motion so the speed is reduced.
Sometimes one of the bodies is so massive that the acceleration is too
small to be detected, e.g. when someting falls to the ground.

These are generic considerations that demonstrate neither equality nor inequality between the action and reaction of the two bodies.

Instead, my animation https://www.geogebra.org/m/qterew9m (together with the numbers and correct reasoning), explains well what is the right relationship between the force that body A exerts on body B and what body B returns to body A.

First of all, this exchange of forces takes place in a limited and very specific area: it is the one at the point x=0 where the collision between the first rows of the two bodies occurs.

All the exchanges of forces that take place behind these two first rows do not take place between one body and another but within each individual body.

At instant 1, the forces are certainly equal and opposite (+3 and -3): body A exerts a +3 impulsive force on body B and body B exerts a -3 impulsive force on body A.

At instant 2, the second row of body A slams into the first, doubling the force it exerts on body B from +3 to +6 but the first row of body B also receives the "help" of its second row and can also oppose a doubled force -6, maintaining the equality between the action and the reaction unchanged.

Even at instant 3, the action remains the same as the reaction because both the first two rows receive help from the third rows and simultaneously increase each other's forces to +9 and -9.

At instant 4 the equilibrium breaks down.

The first two rows do not know what happened behind them, but the first row of body A receives the push of its fourth row and increases the force of its action towards the right from +9 to +12, while the first row of body B cannot do the same because he receives no help from his fourth row (which is not there) and cannot increase his reaction which remains unchanged at -9.

Thus, they face the first row of body A which acts to the right against body B with force +12 and the first row of body B which reacts to the left against body A with unchanged force -9 and is accelerated backwards .

Where is the equality between action and reaction predicted by Newton's 3rd law?

Luigi Fortunati.

[Mikko]

Post by Luigi Fortunati

Post by Mikko

Post by Luigi Fortunati
There is some error in my previous post, which I have corrected here.
The force compresses if it encounters a reaction and accelerates if it does not.

In a typical collision the interaction first compresses both bodies and
then acclerates them. The amount of compression and acceleration of the
bodies need not be same. In come cases the direction of acceleration can
be opposite to the direction of motion so the speed is reduced.
Sometimes one of the bodies is so massive that the acceleration is too
small to be detected, e.g. when someting falls to the ground.

These are generic considerations that demonstrate neither equality nor
inequality between the action and reaction of the two bodies.

My intent was not to demonstrate what needn't be demostrated, only that
your reasoning involves false assumptions. Those whose opöinion matters
believe in Newtons's third law until otherwise is proven and, so it more
than enough to point out that you have not proven.