From Newsgroup: sci.physics.research

The Wikipedia entry for "Elastic collision" https://en.wikipedia.org/wiki/Elastic_collision
contains the following animation
https://youtu.be/wl0c6NMysY4
where the two bodies collide at point x and instantly reverse direction.

Does this seem correct?

Can the 2m mass body be stopped at point X of the collision and pushed
back by the smaller body?

Luigi Fortunati

[[Mod. note --
The Wikipedia animations assume (1) Newtonian mechanics, (2) 1-D motion
with no other forces acting, and (3) the elastic collisions occur very
quickly (i.e., each body's acceleration is nonzero for only a short time).
And saying that the collisions are *elastic* implies that there's no
permanent deformation of either body after the collision.

Within these assumptions, yes, the Wikipedia animations look correct.

The answer to your question "Can the 2m mass body be stopped at point X
of the collision and pushed back by the smaller body?" is yes, that's how Newtonian mechanics works.

The Wikipedia article includes a section "Derivation of solution" which
nicely explains how to derive the solution from conservation of momentum
(which always holds) and conservation of energy (which holds in an elastic collision).
-- jt]]
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From Newsgroup: sci.physics.research

Il 12/02/2026 07:30, Luigi Fortunati ha scritto:

The Wikipedia entry for "Elastic collision" https://en.wikipedia.org/wiki/Elastic_collision
contains the following animation
https://youtu.be/wl0c6NMysY4
where the two bodies collide at point x and instantly reverse direction.

Does this seem correct?

Can the 2m mass body be stopped at point X of the collision and pushed
back by the smaller body?

Luigi Fortunati

[[Mod. note --
The Wikipedia animations assume (1) Newtonian mechanics, (2) 1-D motion
with no other forces acting, and (3) the elastic collisions occur very quickly (i.e., each body's acceleration is nonzero for only a short time). And saying that the collisions are *elastic* implies that there's no permanent deformation of either body after the collision.

Within these assumptions, yes, the Wikipedia animations look correct.

The answer to your question "Can the 2m mass body be stopped at point X
of the collision and pushed back by the smaller body?" is yes, that's how Newtonian mechanics works.

The Wikipedia article includes a section "Derivation of solution" which nicely explains how to derive the solution from conservation of momentum (which always holds) and conservation of energy (which holds in an elastic collision).
-- jt]]

I dispute what the moderator wrote.

A body of mass 2m cannot bounce back (in place!) when it collides with a
body of mass m, otherwise a body of mass 3m, 10m, or 100m would also
bounce back.

It's obvious that a body of mass 100m, colliding with a body of mass m,
can only slow down but not stop in place and bounce back!

So, there should be a mass limit within which a larger body bounces off
a smaller body and beyond which it slows down but does not stop and does
not come back.

Does this limit exist? I don't think so.

Luigi Fortunati
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From Newsgroup: sci.physics.research

On 13/02/2026 09:26, Luigi Fortunati wrote:

Il 12/02/2026 07:30, Luigi Fortunati ha scritto:

The Wikipedia entry for "Elastic collision"
https://en.wikipedia.org/wiki/Elastic_collision
contains the following animation
https://youtu.be/wl0c6NMysY4
where the two bodies collide at point x and instantly reverse direction.

Does this seem correct?

Can the 2m mass body be stopped at point X of the collision and pushed
back by the smaller body?

Luigi Fortunati

[[Mod. note --
The Wikipedia animations assume (1) Newtonian mechanics, (2) 1-D motion
with no other forces acting, and (3) the elastic collisions occur very
quickly (i.e., each body's acceleration is nonzero for only a short time). >> And saying that the collisions are *elastic* implies that there's no
permanent deformation of either body after the collision.

Within these assumptions, yes, the Wikipedia animations look correct.

The answer to your question "Can the 2m mass body be stopped at point X
of the collision and pushed back by the smaller body?" is yes, that's how
Newtonian mechanics works.

The Wikipedia article includes a section "Derivation of solution" which
nicely explains how to derive the solution from conservation of momentum
(which always holds) and conservation of energy (which holds in an elastic >> collision).
-- jt]]

I dispute what the moderator wrote.

A body of mass 2m cannot bounce back (in place!) when it collides with a
body of mass m, otherwise a body of mass 3m, 10m, or 100m would also
bounce back.

That is not always true. If the smaller body is not stationary before
the collision it may have enough momentum to stop the larger body and
make it bounce back. As long as there are no interaction with any
third body the speed of the center of mass of the pair is unchanged.
--
Mikko
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From Newsgroup: sci.physics.research

Luigi Fortunati <fortunati.luigi@gmail.com> wrote:

A body of mass 2m cannot bounce back (in place!) when it collides with a >body of mass m, otherwise a body of mass 3m, 10m, or 100m would also
bounce back.

3m will stop. Less than 3m will bounce back. More than 3m will only slow down. --- Synchronet 3.21b-Linux NewsLink 1.2

From Newsgroup: sci.physics.research

Luigi Fortunati <fortunati.luigi@gmail.com> wrote:

Il 12/02/2026 07:30, Luigi Fortunati ha scritto:

The Wikipedia entry for "Elastic collision" https://en.wikipedia.org/wiki/Elastic_collision
contains the following animation
https://youtu.be/wl0c6NMysY4
where the two bodies collide at point x and instantly reverse direction.

[ ... ]

I dispute what the moderator wrote.

A body of mass 2m cannot bounce back (in place!) when it collides with a body of mass m,

And yet, it bounces back, though with a reduced velocity.

otherwise a body of mass 3m, 10m, or 100m would also bounce back.

No, a body of mass 3m would stop cold, and bounce the body of mass m with velocity 2*v.
More massive bodies would continue forward with reduced velocity, and bounce the smaller
body with increased velocity.

It's obvious that a body of mass 100m, colliding with a body of mass m,
can only slow down but not stop in place and bounce back!

Slows down a bit and bounces the smaller body with velocity approaching 3*v. The formulas are in the text above the animations. (Mind the signs of the vA1,vB1 and vA2,vB2.)
--
pa at panix dot com
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From Newsgroup: sci.physics.research

On Fri, 13 Feb 2026 18:05:38 PST, kaukasoina3dore73js4@sci.fi (Petri Kaukasoina) wrote:

Luigi Fortunati <fortunati.luigi@gmail.com> wrote:

A body of mass 2m cannot bounce back (in place!) when it collides with a
body of mass m, otherwise a body of mass 3m, 10m, or 100m would also
bounce back.

3m will stop. Less than 3m will bounce back. More than 3m will only

slow down.

On Fri, 13 Feb 2026 18:05:45 PST, pa@see.signature.invalid (Pierre
Asselin) wrote:

Luigi Fortunati <fortunati.luigi@gmail.com> wrote:

Il 12/02/2026 07:30, Luigi Fortunati ha scritto:

The Wikipedia entry for "Elastic collision"
https://en.wikipedia.org/wiki/Elastic_collision
contains the following animation
https://youtu.be/wl0c6NMysY4
where the two bodies collide at point x and instantly reverse

direction.

[ ... ]

I dispute what the moderator wrote.

A body of mass 2m cannot bounce back (in place!) when it collides

with a

body of mass m,

And yet, it bounces back, though with a reduced velocity.

otherwise a body of mass 3m, 10m, or 100m would also bounce back.

No, a body of mass 3m would stop cold, and bounce the body of mass m

with velocity 2*v.

More massive bodies would continue forward with reduced velocity, and

bounce the smaller

body with increased velocity.

It's obvious that a body of mass 100m, colliding with a body of mass m,
can only slow down but not stop in place and bounce back!

Slows down a bit and bounces the smaller body with velocity

approaching 3*v.

The formulas are in the text above the animations. (Mind the signs of

the vA1,vB1 and vA2,vB2.)

Yes, even in the newsgroup free.it.scienza.fisica the discussion has
reached this point: the 3m mass body stops, the smaller mass body moves backward, the larger mass body slows down but continues forward without stopping.

The problem is determining where the body stops, because my objection
doesn't concern the initial and final velocities, but only the location
of the rebound.

In the animation, the two bodies rebound exactly at the point of
contact, and this is only true when the masses of the two bodies are equal.

The reason is simple: if the two bodies have the same mass, the system's center of mass is stationary and, during the compression and subsequent springback, remains stationary (as required by the law of conservation
of momentum), so, in the end, the rebound occurs at the same point of
contact.

However, in our animation, where the masses are different, the system's
center of mass moves continuously to the right (due to the law of
conservation of momentum), even throughout the entire collision.

Consequently, when the rebound ends, the two bodies are no longer where
they began the contact but are further to the right.

All of this is missing in our animation, where body "2m" returns to the
same point of contact and not from a later point, as it should in reality.

Luigi Fortunati


[[Mod. note --
As I pointed out earlier, the animation assumes that the bodys'
accelerations are nonzero for only a short time. This implies that
the bodies move only a very short distances (so short that we can
neglect them) during the time interval that their accelerations are
nonzero.
-- jt]]
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From Newsgroup: sci.physics.research

In article <10mlemh$1pima$1@dont-email.me>, Luigi Fortunati writes:

It's obvious that a body of mass 100m, colliding with a body of mass m,
can only slow down but not stop in place and bounce back!

No, it's not obvious, in fact it's not even always true. Here's a
specific example where the body of mass 100m *does* bounce back:

Suppose we have
body A: mass m_A=100 kg, initial velocity v_A1=+1 m/s (moving right)
body B: mass m_B=1 kg, initial velocity v_B1=-1000 m/s (moving left)
and these bodies have an elastic collision at position x=0 m and t=0 s.

The formulas in the previously-cited Wikipedia article <https://en.wikipedia.org/wiki/Elastic_collision> give the final
velocities after the collision as:
v_A2 = -18.822 m/s (body A recoils to the left)
v_B2 = +982.178 m/s (body B recoils to the right)

But we don't have to trust the Wikipedia article! We can check for
ourselves whether or not these v_A2 and v_B2 are correct by checking
whether or not both linear momentum and kinetic energy are conserved:

Linear momentum:
before the collision: m_A*v_A1 + m_B*v_B1 = -900 kg m/s
after the collision: m_A*v_A2 + m_B*v_B2 = -900 kg m/s
i.e., linear momentum is conserved.

Kinetic energy:
before the collision: 1/2 m_A*v_A1^2 + 1/2 m_B*v_B1^2 = 500050 Joules
after the collision: 1/2 m_A*v_A2^2 + 1/2 m_B*v_B2^2 = 500050 Joules
i.e., kinetic energy is conserved.

Since we find that these values of v_A2 and v_B2 conserve both linear
momentum and kinetic energy, we know that these are in fact the correct
v_A2 and v_B2 for an elastic collision.

From the initial & final velocities, it's easy to calculate the
bodies' positions:
t (s) x_A (m) x_B(m)
-3 -3.0 +3000.0
-2 -2.0 +2000.0
-1 -1.0 +1000.0
0 0.0 0.0 (collision happens here)
+1 -18.822 +982.178
+2 -37.644 +1964.356
+3 -56.465 +2946.535

So in this case, yes, a body of mass 100 kg does "bounce back" (final
velociity is of the opposite sign to initial velocity) after colliding
with a body of mass 1 kg.
--
-- "Jonathan Thornburg [remove -color to reply]" <dr.j.thornburg@gmail-pink.com>
(he/him; on the west coast of Canada)
"All models are wrong, but some are useful" -- George E. P. Box
--- Synchronet 3.21b-Linux NewsLink 1.2

From Newsgroup: sci.physics.research

In article <10mlemh$1pima$1@dont-email.me>, Luigi Fortunati writes:

It's obvious that a body of mass 100m, colliding with a body of mass m,
can only slow down but not stop in place and bounce back!

In article <mt2.1.4-91499-1771143993@gold.bkis-orchard.net>, I gave
an example where the body of mass 100m *does* bounce back.

I realise that although that example is a correct example of an
elastic collision of Newtonian point masses, it would be very hard to
actually do in an engineering sense, because the collision velocities
are very high (~ 1 km/second) and it's very hard to arrange elastic
collisions at such high speeds.

So, to make my example plausible in an engineering sense, let's scale
down all the velocities by a factor of 1000. Now we have:
body A: mass m_A=100 kg, initial velocity v_A1=+0.001 m/s (moving right)
body B: mass m_B=1 kg, initial velocity v_B1=-1.0 m/s (moving left)
and these bodies (both point masses) have an elastic collision at position
x=0 m and t=0 s.

The final velocities (after the collision) are now
v_A2 = -0.0188 m/s (body A recoils to the left)
v_B2 = +0.9822 m/s (body B recoils to the right)

Again, we can check whether total linear momentum and kinetic energy are conserved:

Total Linear momentum:
before the collision: m_A*v_A1 + m_B*v_B1 = -0.9 kg m/s
after the collision: m_A*v_A2 + m_B*v_B2 = -0.9 kg m/s
i.e., total linear momentum is conserved.

Total kinetic energy:
before the collision: 1/2 m_A*v_A1^2 + 1/2 m_B*v_B1^2 = 0.50005 Joules
after the collision: 1/2 m_A*v_A2^2 + 1/2 m_B*v_B2^2 = 0.50005 Joules
i.e., total kinetic energy is conserved.

Since we find that these values of v_A2 and v_B2 conserve both linear
momentum and kinetic energy, we know that these are in fact the correct
v_A2 and v_B2 for an elastic collision.

From the initial & final velocities, it's easy to calculate the
bodies' positions:
t (s) x_A (m) x_B(m)
-3 -0.003 +3.0
-2 -0.002 +2.0
-1 -0.001 +1.0
0 0.000 0.0 (collision happens here)
+1 -0.01882 +0.9822
+2 -0.03764 +1.9644
+3 -0.05647 +2.9465

Once again, we see that body A (mass 100 kg) does indeed bounce back
(final velociity is to the *left*) after colliding with a body of mass
1 kg.
--
-- "Jonathan Thornburg [remove -color to reply]" <dr.j.thornburg@gmail-pink.com>
(he/him; on the west coast of Canada)
"All models are wrong, but some are useful" -- George E. P. Box
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From Newsgroup: sci.physics.research

Il 15/02/2026 09:26, Jonathan Thornburg [remove -color to reply] ha scritto:

In article <10mlemh$1pima$1@dont-email.me>, Luigi Fortunati writes:

It's obvious that a body of mass 100m, colliding with a body of mass m,
can only slow down but not stop in place and bounce back!

No, it's not obvious, in fact it's not even always true. Here's a
specific example where the body of mass 100m *does* bounce back:

Suppose we have
body A: mass m_A=100 kg, initial velocity v_A1=+1 m/s (moving right)
body B: mass m_B=1 kg, initial velocity v_B1=-1000 m/s (moving left)
and these bodies have an elastic collision at position x=0 m and t=0 s.

The formulas in the previously-cited Wikipedia article <https://en.wikipedia.org/wiki/Elastic_collision> give the final
velocities after the collision as:
v_A2 = -18.822 m/s (body A recoils to the left)
v_B2 = +982.178 m/s (body B recoils to the right)

But we don't have to trust the Wikipedia article! We can check for
ourselves whether or not these v_A2 and v_B2 are correct by checking
whether or not both linear momentum and kinetic energy are conserved:

Linear momentum:
before the collision: m_A*v_A1 + m_B*v_B1 = -900 kg m/s
after the collision: m_A*v_A2 + m_B*v_B2 = -900 kg m/s
i.e., linear momentum is conserved.

Kinetic energy:
before the collision: 1/2 m_A*v_A1^2 + 1/2 m_B*v_B1^2 = 500050 Joules
after the collision: 1/2 m_A*v_A2^2 + 1/2 m_B*v_B2^2 = 500050 Joules i.e., kinetic energy is conserved.

Since we find that these values of v_A2 and v_B2 conserve both linear momentum and kinetic energy, we know that these are in fact the correct
v_A2 and v_B2 for an elastic collision.

From the initial & final velocities, it's easy to calculate the
bodies' positions:
t (s) x_A (m) x_B(m)
-3 -3.0 +3000.0
-2 -2.0 +2000.0
-1 -1.0 +1000.0
0 0.0 0.0 (collision happens here)
+1 -18.822 +982.178
+2 -37.644 +1964.356
+3 -56.465 +2946.535

So in this case, yes, a body of mass 100 kg does "bounce back" (final velociity is of the opposite sign to initial velocity) after colliding
with a body of mass 1 kg.

When body A collides with body B at a speed of 1001m/s, body B also
collides with body A at the same speed of 1001m/s.

Why would body A contribute to the collision with a speed of +1m/s and
body B with a speed of -1000m/s?

[[Mod. note -- The pre-collision velocities of the two bodies, with
respect to the laboratory inertial reference frame (IRF), are +1 m/s and
-1000 m/s respectively.
-- jt]]

The collision velocity is unique for both bodies, regardless of the
external observer, who, in your case, is moving to the right at a speed
of +499.5 m/s.

[[Mod. note --
Notice that A has a much larger mass (100 kg) than B (1 kg).
Therefore the average velocity of A and B isn't very interesting;
it's much more informative to consider the velocity of A & B's center
of mass. In the laboratory IRF, the center-of-mass velocity works out
to -8.911 m/s (center of mass is moving to the *left* with respect to
the laboratory IRF).

Equivalently, we could say that in the center-of-mass IRF (i.e., in
the IRF where the center of mass is stationary), the initial velocities
are
COM:v_A1 = +9.911 m/s (A is moving *right* in the COM IRF)
COM:v_B1 = -991.089 m/s (B is moving *left* in the COM IRF)

After the collision, the final velocities in the center-of-mass IRF are
COM:v_A2 = -9.911 m/s (A is moving *left* in the COM IRF)
COM:v_bB = +991.089 m/s (B is moving *right* in the COM IRF)
-- jt]]

It cannot be true that the impact energy of a body decreases when viewed
by one observer and increases when viewed by another. The impact energy
(for example, that of an asteroid crashing into Earth) depends
exclusively on the impact velocity and not on the (variable) position of
the observer.

[[Mod. note --
We need to precisely define what we mean by "impact energy".

In Newtonian mechanics, the total kinetic energy of the system *does*
change when measured in one IRF versus another. For example, for our
100:1 collision example it's easy to work out that the total kinetic
energy as measured in the center-of-mass IRF is 496040.099 Joules,
which is different from that measured in the lab frame (500050 Joules).

If we think of the elastic collision as being implemented by having ideal springs between the bodies, then in the center-of-mass IRF, all of the
kinetic energy goes into compressing the springs during the collision
(and is relased again as kinetic energy as the springs re-expand).

In any other (non-center-of-mass) IRF (e.g., the laboratory IRF), some
of the kinetic energy isn't available for compressing the springs, but
rather corresponds to the overall motion of the system with respect to
the center-of-mass IRF.

So if by the phrase "impact energy" you mean the energy available to
compress the springs, then you're right: that is the same (= that measured
in the COM) regardless of what IRF we measure it in.
-- jt]]

And therefore, it depends on the single velocity common to both bodies,
to which they contribute equally.

Luigi.
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From Newsgroup: sci.physics.research

I wasn't very happy with the way this discussion ended.

The case involved the elastic collision in the animation https://youtu.be/wl0c6NMysY4 where the two bodies A and B collide at
equal and opposite velocities v and -v.

In the end, we concluded the discussion by discussing velocities v_A =
+1 m/s and v_B = -1000 m/s, which have nothing to do with what I had
asked, nor with the animation, where the two velocities are equal and opposite.

I didn't have the presence of mind to clarify this, but now I'm ready to
do so by discussing the interesting clarifications that other comments
have highlighted.

We all agreed that (under the conditions of the animation) with the mass
of body A between "m" and less than "3m" both bodies recoil after the collision, while with the mass of A greater than "3m," body A slows down
but continues to move forward without stopping.

What if the mass of body A is exactly equal to "3m"?

In this case, it doesn't go back or move forward.

So it stops.

But where does it stop?

I say it doesn't stop at the point x=0 where the collision occurs, but
at a point further forward in the previous direction of mass A.

This happens because, during the compression phase, the two bodies don't remain stationary at the point x=0 but advance to the right, and only
then does body A stop, at the end of the elastic recoil.

It's as if body A of three train cars (moving to the right) and body B
of one train car (moving to the left) were to elastically collide on the
same track.

In this case, I expect the three train cars to stop at a point further
forward than the point (x=0) where the collision occurred.

Is this deduction correct?

Luigi Fortunati
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From Newsgroup: sci.physics.research

On 28/02/2026 10:48, Luigi Fortunati wrote:

I wasn't very happy with the way this discussion ended.

The case involved the elastic collision in the animation https://youtu.be/wl0c6NMysY4 where the two bodies A and B collide at
equal and opposite velocities v and -v.

In the end, we concluded the discussion by discussing velocities v_A =
+1 m/s and v_B = -1000 m/s, which have nothing to do with what I had
asked, nor with the animation, where the two velocities are equal and opposite.

I didn't have the presence of mind to clarify this, but now I'm ready to
do so by discussing the interesting clarifications that other comments
have highlighted.

We all agreed that (under the conditions of the animation) with the mass
of body A between "m" and less than "3m" both bodies recoil after the collision, while with the mass of A greater than "3m," body A slows down
but continues to move forward without stopping.

What if the mass of body A is exactly equal to "3m"?

In this case, it doesn't go back or move forward.

So it stops.

But where does it stop?

It stops at the last place where somthing can change its motion.
--
Mikko
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