| Topic: |
Science > Physics |
| User: |
"Gunnar G" |
| Date: |
10 May 2006 12:31:36 PM |
| Object: |
collision |
If two 2D circular shaped objects (with different velocities and masses)
collides total elastically I find that there are four unknown variables,
the two vellocities and their direction in x- and y-direction.
With conservation of momentum (mass times velocity) I get two equations, and
conservation of energy gives a third equation.
Then there is one degree of freedom left. How can I resolve the situation
and find the velocities after the collision?
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| User: "Thomas Smid" |
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| Title: Re: collision |
11 May 2006 10:55:15 AM |
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Gunnar G wrote:
If two 2D circular shaped objects (with different velocities and masses)
collides total elastically I find that there are four unknown variables,
the two vellocities and their direction in x- and y-direction.
With conservation of momentum (mass times velocity) I get two equations, and
conservation of energy gives a third equation.
Then there is one degree of freedom left. How can I resolve the situation
and find the velocities after the collision?
See my webpage http://www.plasmaphysics.org.uk/collision2d.htm which
gives the exact and general solution for this (including links to
corresponding C++ and Fortran codes). If you want this in 3D, then have
a look at http://www.plasmaphysics.org.uk/collision3d.htm .
Thomas
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| User: "" |
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| Title: Re: collision |
10 May 2006 01:32:09 PM |
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Gunnar G wrote:
If two 2D circular shaped objects (with different velocities and masses)
collides total elastically I find that there are four unknown variables,
the two vellocities and their direction in x- and y-direction.
With conservation of momentum (mass times velocity) I get two equations, and
conservation of energy gives a third equation.
Then there is one degree of freedom left. How can I resolve the situation
and find the velocities after the collision?
What PD tells you is completely correct. He does not tell you how
to do it though.
The trick is to transfer to the centre of momentum frame for the
impact. This is a frame in which the total momentum is zero.
It is zero before the impact, and zero after. And since you've
got total energy conserved, the two masses have to be going
the same speed after the collision, just at different angles.
That is, you've taken care of nearly everything just by transferring
to the COM frame.
So all you need to do is find the relative angle of in and out for
each mass. Then transfer back to the lab frame when you are
done all that.
How do you find the change in angle? Well, that's where what
PD says comes in. When the two particles collide, if you
assume there is no friction going on (so the masses don't
change their rotation, for example) then the change all has
to be impact. And it has to be perpendicular to the surface
of each object.
So consider the first mass, call it A. You find where it contacts
the other mass. Then you draw the line from the centre of
the mass to that point. The component of momentum perpendicular
to that line can't change, as the force is all parallel to that line.
But the total length of the momentum can't change.
We decided that previously. So the component parallel to that
line must just change sign. It's like a reflection.
So, the only hard part is deciding where the disks contact.
That's the impact parameter that PD talked about. Think about
the two masses in COM frame. Their momentums have to be
equal and opposite, and they have to be approaching eachother.
But if they were not on *exactly* the same line, then they
could miss. Projecting their paths forward, you will find that
the paths are some distance apart. If that distance is zero,
they hit head on. If it is more than zero, but less than the
sum of the radii of the two disks, they contact. So, you want
to work out the location of the impact based on the impact
parameter. You then have a keen little problem in geometry.
Socks
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| User: "PD" |
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| Title: Re: collision |
10 May 2006 12:52:54 PM |
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Gunnar G wrote:
If two 2D circular shaped objects (with different velocities and masses)
collides total elastically I find that there are four unknown variables,
the two vellocities and their direction in x- and y-direction.
With conservation of momentum (mass times velocity) I get two equations, and
conservation of energy gives a third equation.
Then there is one degree of freedom left. How can I resolve the situation
and find the velocities after the collision?
What you need is the direction of the momentum-transfer, which must lie
along the line connecting their centers on contact. This is equivalent
to what's called an "impact parameter" in scattering theory.
PD
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| User: "" |
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| Title: Re: collision |
10 May 2006 05:41:13 PM |
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In article <Yrp8g.56600$d5.210721@newsb.telia.net>, Gunnar G <debian@comhem.se> writes:
If two 2D circular shaped objects (with different velocities and masses)
collides total elastically I find that there are four unknown variables,
the two vellocities and their direction in x- and y-direction.
With conservation of momentum (mass times velocity) I get two equations, and
conservation of energy gives a third equation.
Then there is one degree of freedom left. How can I resolve the situation
and find the velocities after the collision?
You can't with the information provided. To see this, consider the
collision in the CM reference frame of the two objects. What you've
before the collision is both objects approaching the CM, from opposite
sides, with the ratio of the magnitudes of the velocities being
inversely proportional to the ratio of masses (follows from the
defintion of CM). After the collision you've them receding from the
CM (which didn't change) due to conservation of momentum), with the
ratio of the magnitudes of the velocities still being the same. And,
after you add the requirement of elasticity, you find that not only
the ratio but, in fact, the actual magnitudes remain the same. So,
the picture after the collision simply looks like the one before, just
run backwards. Only ... the straight line along which the two objects
move after the collision, doesn't have to be the same as the one
before, since rotation of the picture doesn't change the physics. All
such lines are equally consistent with the conservation laws.
So, you need some additional information about the direction of the
momentum exchange during the collision, in order to fix this last
degree of freedom.
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
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| User: "" |
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| Title: Re: collision |
11 May 2006 12:02:07 PM |
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In article <d_t8g.25$25.2115@news.uchicago.edu>, writes:
In article <Yrp8g.56600$d5.210721@newsb.telia.net>, Gunnar G <debian@comhem.se> writes:
If two 2D circular shaped objects (with different velocities and masses)
collides total elastically I find that there are four unknown variables,
the two vellocities and their direction in x- and y-direction.
With conservation of momentum (mass times velocity) I get two equations, and
conservation of energy gives a third equation.
Then there is one degree of freedom left. How can I resolve the situation
and find the velocities after the collision?
You can't with the information provided. To see this, consider the
collision in the CM reference frame of the two objects. What you've
before the collision is both objects approaching the CM, from opposite
sides, with the ratio of the magnitudes of the velocities being
inversely proportional to the ratio of masses (follows from the
defintion of CM). After the collision you've them receding from the
CM (which didn't change) due to conservation of momentum), with the
ratio of the magnitudes of the velocities still being the same. And,
after you add the requirement of elasticity, you find that not only
the ratio but, in fact, the actual magnitudes remain the same. So,
the picture after the collision simply looks like the one before, just
run backwards. Only ... the straight line along which the two objects
move after the collision, doesn't have to be the same as the one
before, since rotation of the picture doesn't change the physics. All
such lines are equally consistent with the conservation laws.
So, you need some additional information about the direction of the
momentum exchange during the collision, in order to fix this last
degree of freedom.
If you've ever played pool, the required information is conveyed by
where on the one ball the other ball strikes. To a first approximation,
the impulse that is exchanged is along the line that goes from the
center of the one ball to the center of the other at the instant of
contact.
For perfectly elastic, perfectly rigid and perfectly slippery
[so that we may ignore angular momentum] balls, this first approximation
is exact.
You control the direction taken by the target ball by controlling
where the cue ball hits it. That's your degree of freedom.
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