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

The animation https://www.geogebra.org/classic/hxvcaphh shows how two
rigid bodies interact when they collide head-on.

When body A comes into contact with body B, an often imperceptible
contraction occurs (first phase) and a subsequent elastic return to the
initial form (second phase).

The quantitative model is the following, before the collision the
masses are both equal to 1 (m_A=1 and m_B=1), the initial velocities
are vi_A=+1 and vi_B=-1, and the momentum quantities are pi_A=+1 and
p_Bi=-1.

In the first phase of the collision, a compression zone and a pair of
action and reaction forces F1 and F2 are activated simultaneously in
the contact area (the compression is the cause, the opposite forces are
the effects).

The force F1 slows the motion of body B to the left until it stops and
the force F2 does the same, slowing the motion of body A to the right
until it stops (the forces are the cause, the decelerations are the
effects).

During this phase, the changes in the value of the two opposing blue
and red action and reaction forces are displayed.

When the two bodies have stopped, the contraction is at maximum (1),
the velocities have become zero and the opposing forces measure F1=+1
and F2=-1.

You can stop the motion at this moment with the "Max compression"
button.

In the second phase of the impact, the opposing forces F1 and F2
accelerate the two bodies A and B in the opposite direction until the
initial position of contact where both the compression, the forces and
the acceleration cease to exist.

From this point on, the motion of the two bodies returns to being
inertial, the velocities have been inverted:
body A from vi_A=+1 to vf_A=-1 and body B from vi_B=-1 to vf_B=+1
and the quantities of motion as well:
body A from pi_A=+1 to pf_A=-1 and body B from pi_B=-1 to pf_B=+1.


All this is very simple because the 2 bodies have the same mass but
things get complicated when we choose to increase (with the appropriate
button) the mass of body A from m_A=1 to m_A=2 because, to the main
contraction between the particles A1 and B1, the secondary compression
between A2 and A1 is added whose action on the left concerns
exclusively body A, while that on the right concerns both body A and
body B.

But I will talk about this in the next post because here I have gone on
too long.

Luigi Fortunati