*Post by Richard Livingston**Post by Luigi Fortunati*In my animation

https://www.geogebra.org/m/sywszege

...

Regarding my animation, I based myself on the following considerations=

which seem correct to me.

The stone is a set of billions of billions of particles which, in my

animation, for simplicity, I have reduced to 5 (from A to E).

...

Between C and D the forces are of magnitude 3, between B and C of

magnitude 2 and between A and B of magnitude 1.

Always for the third principle.

A more precise way to model this, the way it is done in computer codes

of elastic collisions, is to model the material as a fine array of mass=

es

*Post by Richard Livingston*connected to adjacent masses by short stiff springs. When the point

E contacts F there is actually a short spring that reduces the theoreti=

cal

*Post by Richard Livingston*instantaneous stop of E to a rapid deceleration.

As E comes to a stop the spring to D compresses and causes D to slow

down, then C then B, etc.

The impulse from E colliding with F propagates back through the mass

(D to C to B to A etc) at the speed of sound in that material.

Rich L.

I agree with you: the impulse of the collision goes leftward from

particle E to A but not with the same intensity.

Is it correct to say that the exchange of forces decreases in the

proportion of 5 between F and E, 4 between E and D, 3 between D and C,

2 between C and B and 1 between B and A?

Luigi Fortunati

[[Mod. note --

1. You seem to be using the word "impulse" in a way different from how

it's used in physics ( https://en.wikipedia.org/wiki/Impulse_(physics) ).

2. I don't quite know what you mean by the term "exchange of forces".

But if we interpret your question as asking if the forces

F_FE = the force F exerts on E

F_ED = the force E exerts on D

F_DC = the force D exerts on C

F_CB = the force C exerts on B

F_BA = the force B exerts on A

are in the ratio 5:4:3:2:1, then we can apply Newton's 2nd law to try

to find out. To do this, let's introduce a bit of terminology: Let's

* we'll call the entire ball (the combined system A+B+C+D+E) "ABCDE"

* we'll call the combined system A+B+C+D "ABCD"

* we'll call the combined system A+B+C "ABC"

* we'll call the combined system A+B "AB"

Now we observe that

* the net (horizontal) force acting on ABCDE is F_FE, i.e.,

F_FE is the only (horizontal) external force acting on ABCDE

* the net (horizontal) force acting on ABCD is F_ED, i.e.,

F_ED is the only (horizontal) external force acting on ABCD

* the net (horizontal) force acting on ABC is F_DC, i.e.,

F_DC is the only (horizontal) external force acting on ABC

* the net (horizontal) force acting on AB is F_CB, i.e.,

F_CB is the only (horizontal) external force acting on AB

* the net (horizontal) force acting on A is F_BA, i.e.,

F_BA is the only (horizontal) external force acting on A

* the net (horizontal) force on A is F_BA

If we assume A, B, C, D, and E to all have the same mass m, then

* the mass of ABCDE is m_ABCDE = 5m

* the mass of ABCD is m_ABCD = 4m

* the mass of ABC is m_ABC = 3m

* the mass of AB is m_AB = 2m

* the mass of A is m_A = m

Since the masses m_ABCDE:m_ABCD:m_ABC:m_AB:m_A are in the ratio

5:4:3:2:1, by F_net=ma the net forces will be in the same ratio

if and only if the accelerations of the center-of-masses of

ABCDE, ABCD, ABC, AB, and A are all the same. But in reality these

accelerations *won't* all be the same -- the springs will compress

by different amounts under the different forces, so the balls won't

all accelerate at the same rate, and the center-of-mass accelerations

of ABCDE, ABCD, ABC, AB, and A *won't* all be the same. So, we

conclude that the forces F_FE:F_ED:F_DC:F_CB:F_AB *aren't* in the

ratio 5:4:3:2:1. Figuring out the actual (time-dependent) forces

requires a more careful analysis of the dynamics of the collision.

-- jt]]