Re: Inertia and third principle

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]]