Re: inelastic collision (was: Re: Newton e Hooke)

Jonathan Thornburg [remove -color to reply] il 26/02/2025 09:02:15 ha
scritto:

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.

...

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.

Body B has a momentum -3 and, therefore, cannot transfer -3.75 to body
A because it does not have it.

What happens is something else.

In the first 3 instants, body B experiences a counterforce +3 from body
A and stops because it no longer has any momentum (pB=-3+3=0).

In the following instants, a further +0.75 hits it that it can no
longer counteract and that +0.75 becomes a "net" force that accelerates
it to the right (remember that it had just stopped after the first 3
instants!).

Therefore, body B experiences the force FA_B=+3.75, reacts with the
counterforce FB_A=-3 and accelerates due to the residual uncontested
force +0.75.

Ciao, Luigi.