From Newsgroup: sci.physics.research

A stone hits a wall and stops.

In the collision, there are two forces: the force F1 of the stone
against the wall and the force F2 of the wall against the stone.

These are two contact forces, and there are no spring forces because
neither the wall nor the stone have elasticity.

Instead, if a ball with the same mass as the stone hits the wall, there
is both a contact force and a spring force.

In this case, are there always only the contact forces F1 and F2, or are
there other forces besides these?

Luigi Fortunati

[[Mod. note --
Actually, both the stone and the wall are (slightly) compressible
(all solid objects are compressible!), so there are indeed spring
forces acting during the collision.

This is separate from the question of how elastic the collision is.
You wrote that the stone "stops", so I take it that you mean that the
collision is completely inelastic, i.e., the stone *stays stopped*
(at least in terms of perpendicular-to-the-wall motion) after the
collision.
-- jt]]
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From Newsgroup: sci.physics.research

From: SMTP%"fortunati.luigi@gmail.com" 1-APR-2026 22:20:01.22
To: sci-physics-research@moderators.isc.org
CC:
Subj: Re: Contact forces and spring forces.

Elastic collision between two bodies and a spring

I prepared the initial rough diagram for the animation https://www.geogebra.org/classic/p5dmxcae
where bodies A (2m) and B (m) compress the spring during the first part
of the collision, and then the spring reacts by elastically pushing the
bodies back.

The changes in velocity during and after the collision are missing.

To complete the work, I need to know how I should slow down the two
bodies A and B during the compression, before they come to a stop:
- Should I slow them down with the same deceleration?
- Do they stop at the same instant or at different times?

I need the answers to complete the animation.

Luigi Fortunati

[[Mod. note --
Some other assumptions which are sort of implicit here, but are good to
write out explicitly, are
* we're using Newtonian mechanics throught
* the lab frame is an inertial reference frame
* A and B are each rigid with uniform density
-- jt]]
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From Newsgroup: sci.physics.research

In article <10qjun8$hrl8$1@dont-email.me>, Luigi Fortunati asks about
a Newtonian-mechanics elastic collision of two bodies
A: mass 2m, initial velocity +1 (moving to the right in the lab frame)
B: mass 1m, initial velocity -1 (moving to the left in the lab frame)
with a spring in between them.

Let's take the lab frame to be an inertial reference frame, and assume
that A and B are each rigid with uniform density. Luigi's figure
https://www.geogebra.org/classic/p5dmxcae
shows that at the moment A and B first contact the spring (which extends
from x=-1 to x=+1), A extends from x=-5 to x=-1 (so its center of mass
is at x=-3), and B extends from x=+1 to x=+3 (so its center of mass is
at x=+2) [all measurements taken in the lab reference frame].

It will make the analysis simpler if we switch over to the center-of-mass
(COM) inertial reference frame, so let's do that.

And, it's simplest if we assume that the mass of the spring is much less
than the mass of A and B (so that we can neglect the spring's own linear momentum).



For convenience, let's take the mass unit to be 1 kg (so that A has mass
m_A = 2kg and B mass m_B = 1kg). Let's take the length scale to be in
meters (so that A's center of mass is at x_A = -3 m when the front (right
side) of A first contacts the spring, and B's center of mass is at x_B =
+2 m when the front (left side) of B first contacts the spring. Let's
take the initial velocities in the lab frame to be +/- 1 m/s.

Then the total linear momentum in the lab frame is +1 kg m/s. Since
the total mass is m_total = m_A + m_B = 3kg, that means that the COM is
moving to the right at +0.333333 m/s with respect to the lab reference
frame. So, in the COM frame,
A is initially moving to the right: v_A(initial) = +0.666667 m/s
B is initially moving to the left: v_B(initial) = -1.333333 m/s
so that the total linear momentum is zero (as it must be in the COM frame).

The COM position in the lab frame at the time the fronts of A and B
first contact the spring is
x_COM(first contact) = (m_A*x_A + m_B*x_B)/m_total
where x_A and x_B are the positions of A's and B's center of mass at
that time. This works out to
x_COM(first contact) = (2kg*-3m + 1kg*+2m)/3kg = -1.333333 m.

The COM is (by definition) *stationary* in the COM frame, i.e., it is
always at the same position x_COM = -1.333333 m. In order for the COM
to stay stationary in the COM frame as the spring is compressed, A and
B's motions in the COM frame must be in a +1:-2 ratio at all times.
That is, at any time when the spring is compressed, in the COM frame
B must have moved twice as far left since first contacting the spring
as A has moved to the right since first contacting the spring.

Since this is true *throughout* the spring compression, this means that
at any time when the spring is compressed, B's COM velocity must be
-2 * A's COM velocity.

Since this is true *throughout* the spring compression, B's acceleration
must be -2 * A's acceleration. (Note that acceleration doesn't need a
"COM" or "lab" qualifier, because acceleration is the same in any inertial reference frame.)

By Newton's 2nd law, this ratio of accelerations combined with the 2:1
A:B mass ratio means the force the spring applies to B at any time must
be -1 * the force the spring applies to A at that same time. (Note that
this means the sum of these two forces = 0, which is what it must be for
the total linear momentum to remain constant.)



So, the answers to Luigi's two questions are:

- Should I slow them down with the same deceleration?

No, B has twice the deceleration as A.

- Do they stop at the same instant or at different times?

In the COM frame they both stop at the same time.

The COM frame is moving to the right at +0.333333 m/s with respect to
the lab frame, so at the moment when A and B both stop in the COM frame,
A is still moving right at +0.333333 m/s in the lab frame and B has
already stopped and reversed direction and is now moving right at
+0.333333 m/s in the lab frame. So, we conclude that in the lab frame,
B stops before A.

If we assume that the spring obey's Hooke's law, it's actually quite
easy to solve the motion exactly. I'll leave the details for a later
post.
--
-- "Jonathan Thornburg [remove -color to reply]" <dr.j.thornburg@gmail-pink.com>
(he/him; on the west coast of Canada)
"Totalitarianism in power invariably replaces all first-rate talents,
regardless of their sympathies, with those crackpots and fools whose
lack of intelligence and creativity is still the best guarantee of
their loyalty." -- Hannah Arendt, "The Origins of Totalitarianism"
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From Newsgroup: sci.physics.research

[[Mod. note --
I apologise for the dalay in posting this article, which arrived at
my computer on 2026-04-02 but was mistakenly misfiled by an over-agressive
spam filter.
-- jt]]

On 02/04/2026 09:43, Luigi Fortunati wrote:

From: SMTP%"fortunati.luigi@gmail.com" 1-APR-2026 22:20:01.22
To: sci-physics-research@moderators.isc.org
CC:
Subj: Re: Contact forces and spring forces.

Elastic collision between two bodies and a spring

I prepared the initial rough diagram for the animation https://www.geogebra.org/classic/p5dmxcae
where bodies A (2m) and B (m) compress the spring during the first part
of the collision, and then the spring reacts by elastically pushing the bodies back.

The changes in velocity during and after the collision are missing.

To complete the work, I need to know how I should slow down the two
bodies A and B during the compression, before they come to a stop:
- Should I slow them down with the same deceleration?
- Do they stop at the same instant or at different times?

I need the answers to complete the animation.

Luigi Fortunati

[[Mod. note --
Some other assumptions which are sort of implicit here, but are good to
write out explicitly, are
* we're using Newtonian mechanics throught
* the lab frame is an inertial reference frame
* A and B are each rigid with uniform density
-- jt]]

One should also add that the mass of the spring is so small that
assuming it is zero is a good enough approximation.

The spring produces the same force to both bodies. Because the masses
of the bodies are different the same force produces different
accelerations.

Whether they both stop depends on the initial velcities.
--
Mikko
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