| Topic: |
Science > Physics |
| User: |
"nsgi_2004" |
| Date: |
30 Aug 2004 11:23:18 PM |
| Object: |
elastic collision |
Two spheres collide and assume that the collision is perfectly elastic.
I have the relationship:
(va' - vb') dot N = -(va - vb) dot N
Where N is the normal vector at the point of collision.
Now is this relationship true due to the conservation of kinetic energy?
And if so, how do I prove it?
Thanks.
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| User: "Jim Black" |
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| Title: Re: elastic collision |
31 Aug 2004 10:00:22 PM |
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"nsgi_2004" <nospam@nospam.com> wrote in message news:<Y2TYc.31186$bT1.27999@fed1read07>...
Two spheres collide and assume that the collision is perfectly elastic.
I have the relationship:
(va' - vb') dot N = -(va - vb) dot N
Where N is the normal vector at the point of collision.
Now is this relationship true due to the conservation of kinetic energy?
And if so, how do I prove it?
Thanks.
In the center of momentum frame, in which ma va + mb vb = 0:
va' = -va
vb' = -vb.
This follows from the combination of conservation of energy and
conservation of momentum. It is simple to show that this is a
solution of these constraints. In the center of momentum frame, the
spheres are moving directly toward each other, making this a
one-dimensional problem. That leaves us with only two variables to
solve for, va and vb, with two constraints. One would expect the
solution to be unique, or almost unique. Actually solving the
equations yields two solutions to the constraints; the one that
actually happens, and one in which the spheres pass right through each
other.
From that, we know that:
va' - vb' = - (va - vb)
Since this is a statement about the relative velocities, it should be
true in any frame.
Alternatively, one can choose the x-direction to be in the direction
of N, and then use the conservation of energy and the conservation of
momentum.
Given that the force between the spheres during the collision is in
the direction of the N vector, the components of the velocities in the
y and z directions are unchanged and can be subtracted from the
conservation equations, since they are the same on both sides. It is
then possible to rearrange the equations to get:
ma (va_x'^2 - va_x^2) = - mb (vb_x'^2 - vb_x^2)
ma (va_x' - va_x) = - mb (vb_x' - vb_x)
One can then divide the first equation by the second, and the equation
you ask about follows readily.
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| User: "nsgi_2004" |
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| Title: Re: elastic collision |
01 Sep 2004 12:01:17 AM |
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I want to thank everyone for all of the responses. I have worked through
the derivation myself now. And I also learned something new about solving
problems in the best appropriate coordinate system. My lower division
physics book (the only one I have) never mentioned about solving collision
problems in the center of mass frame.
Thanks again.
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| User: "" |
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| Title: Re: elastic collision |
31 Aug 2004 09:20:07 AM |
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"nsgi_2004" <nospam@nospam.com> wrote in message news:<Y2TYc.31186$bT1.27999@fed1read07>...
Two spheres collide and assume that the collision is perfectly elastic.
I have the relationship:
(va' - vb') dot N = -(va - vb) dot N
Where N is the normal vector at the point of collision.
Now is this relationship true due to the conservation of kinetic energy?
And if so, how do I prove it?
Assuming rotation is negligible...
Transform to the centre of momentum frame. In this frame, the two
momentum vectors are "back to back" that is, they have equal
magnitude but are 180 degrees apart. In this frame, the collision
looks pretty simple.
After the collision, transform back to the lab frame.
Socks
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| User: "nsgi_2004" |
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| Title: Re: elastic collision |
31 Aug 2004 03:14:28 PM |
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<puppet_sock@hotmail.com> wrote in message
news:c7976c46.0408310620.724e06ee@posting.google.com...
"nsgi_2004" <nospam@nospam.com> wrote in message
news:<Y2TYc.31186$bT1.27999@fed1read07>...
Two spheres collide and assume that the collision is perfectly elastic.
I have the relationship:
(va' - vb') dot N = -(va - vb) dot N
Where N is the normal vector at the point of collision.
Now is this relationship true due to the conservation of kinetic energy?
And if so, how do I prove it?
Assuming rotation is negligible...
Transform to the centre of momentum frame. In this frame, the two
momentum vectors are "back to back" that is, they have equal
magnitude but are 180 degrees apart. In this frame, the collision
looks pretty simple.
After the collision, transform back to the lab frame.
Socks
Thank you for your reply. Is this what you mean:
Relative to object B, object A has velocity v_ab. Relative to object B,
object B has velocity 0.
m_a*v_a + m_b*v_b = m_a*v_a' + m_b*v_b'
Relative to B:
m_a*v_ab + 0= m_a*v_ab' + 0
v_ab = v_ab'
That would give me this: (va' - vb') dot N = (va - vb) dot N
But I am missing the negative sign, because they should be opposite. Please
advise. Thanks in advance.
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| User: "Bjoern Feuerbacher" |
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| Title: Re: elastic collision |
01 Sep 2004 02:12:26 AM |
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nsgi_2004 wrote:
<puppet_sock@hotmail.com> wrote in message
news:c7976c46.0408310620.724e06ee@posting.google.com...
"nsgi_2004" <nospam@nospam.com> wrote in message
news:<Y2TYc.31186$bT1.27999@fed1read07>...
Two spheres collide and assume that the collision is perfectly elastic.
I have the relationship:
(va' - vb') dot N = -(va - vb) dot N
Where N is the normal vector at the point of collision.
Now is this relationship true due to the conservation of kinetic energy?
And if so, how do I prove it?
Assuming rotation is negligible...
Transform to the centre of momentum frame. In this frame, the two
momentum vectors are "back to back" that is, they have equal
magnitude but are 180 degrees apart. In this frame, the collision
looks pretty simple.
After the collision, transform back to the lab frame.
Socks
Thank you for your reply. Is this what you mean:
No.
Relative to object B, object A has velocity v_ab. Relative to object B,
object B has velocity 0.
m_a*v_a + m_b*v_b = m_a*v_a' + m_b*v_b'
Relative to B:
m_a*v_ab + 0= m_a*v_ab' + 0
Here you go wrong. If you for both sides of the equation a frame in which
before the collision, object B rested, this does in no way imply that it
rests after the collision, too! Alternatively, if you use on the right
hand side of the equation a frame in which object B rests after the
collision, you are using different frames on both sides of the equation,
which makes no sense - and hence the equal sign does *not* hold!
v_ab = v_ab'
That would give me this: (va' - vb') dot N = (va - vb) dot N
But I am missing the negative sign, because they should be opposite. Please
advise. Thanks in advance.
Apparently you are not aware with the term "center of momentum frame"
(comf) (or "center of mass frame"). It is the frame in which the center
of mass of a system is at rest. I.e. in the comf, the following relation
holds:
m_a v_a,comf + m_b v_b,comf = 0,
and hence v_a,comf = - v_b,comf * m_b/m_a. Since the momentum of the com
(=0 in the comf) is conserved in the collision, these equations hold
before and after the collision.
The velocity of the center of momentum in your original frame is
v_com = (m_a v_a + m_b v_b)/(m_a + m_b),
and you get the velocities in the comf from the velocities in your
original frame by
v_a,comf = v_a - v_com
v_b,comf = v_b - v_com
HTH.
Bye,
Bjoern
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| User: "John C. Polasek" |
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| Title: Re: elastic collision |
31 Aug 2004 09:20:57 AM |
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On Mon, 30 Aug 2004 21:23:18 -0700, "nsgi_2004" <nospam@nospam.com>
wrote:
Two spheres collide and assume that the collision is perfectly elastic.
I have the relationship:
(va' - vb') dot N = -(va - vb) dot N
Where N is the normal vector at the point of collision.
Now is this relationship true due to the conservation of kinetic energy?
And if so, how do I prove it?
Thanks.
How about changing the RHS to dot -N?
Mr. Dual Space
(If you have something to say, write an equation.
If you have nothing to say, write an essay).
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| User: "nsgi_2004" |
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| Title: Re: elastic collision |
31 Aug 2004 10:45:40 AM |
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"John C. Polasek" <jpolasek@cfl.rr.com> wrote in message
news:l629j0ls9jiamdakuo5ebgo4nbul899d04@4ax.com...
On Mon, 30 Aug 2004 21:23:18 -0700, "nsgi_2004" <nospam@nospam.com>
wrote:
Two spheres collide and assume that the collision is perfectly elastic.
I have the relationship:
(va' - vb') dot N = -(va - vb) dot N
Where N is the normal vector at the point of collision.
Now is this relationship true due to the conservation of kinetic energy?
And if so, how do I prove it?
Thanks.
How about changing the RHS to dot -N?
Yes, I realize the relative velocities before and after are equal but in
opposite direction.
[1] Is this relationship true due to the conservation of kinetic energy?
[2] How can I derive (va' - vb') dot N = -(va - vb) dot N from the
conservation of KE?
Thanks.
Mr. Dual Space
(If you have something to say, write an equation.
If you have nothing to say, write an essay).
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| User: "" |
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| Title: Re: elastic collision |
31 Aug 2004 06:17:21 PM |
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In article <J21Zc.37217$bT1.2381@fed1read07>, "nsgi_2004" <nospam@nospam.com> writes:
"John C. Polasek" <jpolasek@cfl.rr.com> wrote in message
news:l629j0ls9jiamdakuo5ebgo4nbul899d04@4ax.com...
On Mon, 30 Aug 2004 21:23:18 -0700, "nsgi_2004" <nospam@nospam.com>
wrote:
Two spheres collide and assume that the collision is perfectly elastic.
I have the relationship:
(va' - vb') dot N = -(va - vb) dot N
Where N is the normal vector at the point of collision.
Now is this relationship true due to the conservation of kinetic energy?
And if so, how do I prove it?
Thanks.
How about changing the RHS to dot -N?
Yes, I realize the relative velocities before and after are equal but in
opposite direction.
[1] Is this relationship true due to the conservation of kinetic energy?
Kinetic energy and momentum.
[2] How can I derive (va' - vb') dot N = -(va - vb) dot N from the
conservation of KE?
Do it in the center of mass frame, its a breeze.
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: "Bjoern Feuerbacher" |
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| Title: Re: elastic collision |
31 Aug 2004 11:13:20 AM |
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nsgi_2004 wrote:
"John C. Polasek" <jpolasek@cfl.rr.com> wrote in message
news:l629j0ls9jiamdakuo5ebgo4nbul899d04@4ax.com...
On Mon, 30 Aug 2004 21:23:18 -0700, "nsgi_2004" <nospam@nospam.com>
wrote:
Two spheres collide and assume that the collision is perfectly elastic.
I have the relationship:
(va' - vb') dot N = -(va - vb) dot N
Where N is the normal vector at the point of collision.
Now is this relationship true due to the conservation of kinetic energy?
And if so, how do I prove it?
Thanks.
How about changing the RHS to dot -N?
Yes, I realize the relative velocities before and after are equal but in
opposite direction.
[1] Is this relationship true due to the conservation of kinetic energy?
I would say that it is due to the conservation of momentum,
[2] How can I derive (va' - vb') dot N = -(va - vb) dot N from the
conservation of KE?
I think you can't. For how to derive it from the conservation of momentum,
see the post by puppet_sock.
[snip]
Bye,
Bjoern
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| User: "nsgi_2004" |
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| Title: Re: elastic collision |
31 Aug 2004 03:15:17 PM |
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[snip]
I think you can't. For how to derive it from the conservation of momentum,
see the post by puppet_sock.
Ok thanks for pointing me in the right direction.
[snip]
Bye,
Bjoern
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