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.
| Date | Sujet | # | Auteur | |
| 11 Feb 26 |  Elastic Collision | 15 | Luigi Fortunati | |
| 13 Feb 26 |  Re: Elastic Collision | 5 | Luigi Fortunati | |
| 13 Feb 26 |   Re: Elastic Collision | 1 | Mikko | |
| 13 Feb 26 | Re: Elastic Collision | 1 | Petri Kaukasoina | |
| 13 Feb 26 | Re: Elastic Collision | 1 | Pierre Asselin | |
| 15 Feb 26 |  Re: Elastic Collision | 1 | Luigi Fortunati | |
| 15 Feb 26 | Re: Elastic Collision | 1 | Jonathan Thornburg [remove -color to reply] | |
| 15 Feb 26 | Re: Elastic Collision | 1 | Jonathan Thornburg [remove -color to reply] | |
| 16 Feb 26 | Re: Elastic Collision | 7 | Luigi Fortunati | |
| 28 Feb 26 |  Re: Elastic Collision | 1 | Luigi Fortunati | |
| 28 Feb 26 | Re: Elastic Collision | 1 | Mikko | |
| 5 Mar 26 | Re: Elastic Collision | 1 | Luigi Fortunati | |
| 8 Mar 26 | Re: Elastic Collision | 1 | Mikko | |
| 12 Mar 26 | Re: Elastic Collision | 1 | Luigi Fortunati | |
| 16 Mar 26 | Re: Elastic Collision | 1 | Mikko |
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