inelastic collision (was: Re: Newton e Hooke)

[[I've changed the subject line since this discussion no longer has
anything to do with Hooke's law.]]

In article <vpd0kg$23r6$1@dont-email.me>, Luigi Fortunati writes:

In my animation https://www.geogebra.org/m/rs4cfxzg body A of 5 particles
collides inelastically with body B of 3 particles.

Before the collision, the total momentum p=+2 is entirely owned by
body A

This is incorrect: before the collision, *both* bodies have nonzero
momentum (relative to Luigi's inertial reference frame).  To be precise,
before the collision we have:
mass:        m_A = 5      m_B = 3
velocity:    v_A = +1     v_B = -1
momentum:    p_A = +5     p_B = -3      p_total = p_A+p_B = +2

After the collision, it is true that the total momentum does not change
(it always remains equal to p=+2), [[...]]
>
In fact, at the end of the collision, the positive momentum p=+2 belongs
"only" in part to body A (p=+1.25) and the rest has transferred to body B
(p=+0.75).

This is correct: since momentum is conserved, the total momentum after
the collision is equal to the total momentum before the collision, i.e.,
+2.  Since the total mass is m_A+m_B = 8, the common velocity after the
collision must be v_A = v_B = 0.25, and thus the two bodies' individual
momenta after the collision must be p_A = +1.25 and p_B = +0.75.

How is it possible that, during the collision, there is a transfer of
momentum from body A to body B if the action of body A on body B is
*equal* to the opposite reaction of body B on body A?

The answer is that the momentum transfer during the collision is actually
*two-way*, i.e., *each* body transfers some momentum to the other body.
That is, during the collision A's momentum changes (because B transfers
some momentum to A), AND B's momentum changes (because A transfers some
momentum to B).

Let's work this out in detail:
Before the collision:
   p_A_before = +5
   p_B_before = -3
   p_total_before = p_A_before+p_B_before = +2
After the collision:
  p_A_after = +1.25
  p_B_after = +0.75,
  p_total_after = p_A_after+p_B_after = +2
Thus, *during* the collision, the bodies momenta change by these amounts:
  Delta_p_A = p_A_after-p_A_before = -3.75
  Delta_p_B = p_B_after-p_B_before = +3.75
  Delta_p_total = Delta_p_A+Delta_p_B = 0

In other words, during the collison B transfers momentum -3.75 to A,
so that A's momentum changes by Delta_p_A=-3.75.  AND, during the
collision A transfers momentum +3.75 to B, so that B's momentum changes
by Delta_p_B=+3.75.

And why does this transfer of momentum from A to B not occur in the first
three instants and only occurs in instants 4 and 5?

To answer that we need to delve into the detailed internal structure of
each body and the dynamics of the collision.  I'll write about that in
a following posting.

--
-- "Jonathan Thornburg [remove -color to reply]" <dr.j.thornburg@gmail-red.com>
   (he/him; currently on the west coast of Canada)
   "[I'm] Sick of people calling everything in crypto a Ponzi scheme.
    Some crypto projects are pump and dump schemes, while others are pyramid
    schemes.  Others are just standard issue fraud.  Others are just middlemen
    skimming off the top.  Stop glossing over the diversity in the industry."
                                                 -- Pat Dennis, 2022-04-25