Re: The opposing rockets and the box
- From: "Spaceman" <spaceman@xxxxxxxxxxxxxxxxxxxxxxxxxx>
- Date: Tue, 22 Jul 2008 11:30:49 -0400
Greg Neill wrote:
"Spaceman" <spaceman@xxxxxxxxxxxxxxxxxxxxxxxxxx> wrote in message
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Greg Neill wrote:
"Spaceman" <spaceman@xxxxxxxxxxxxxxxxxxxxxxxxxx> wrote in message
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Greg Neill wrote:
"Spaceman" <spaceman@xxxxxxxxxxxxxxxxxxxxxxxxxx> wrote in message
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Greg Neill wrote:
"Spaceman" <spaceman@xxxxxxxxxxxxxxxxxxxxxxxxxx> wrote in
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You removed the M for no reason at all.
You made it basically dissapear from the problem
without even knowing where you went wrong.
For some stupid ass reason you made m*v
hit a larger mass and still stay as m*v again even though
it combined with a larger mass.
Are you really that dense?
You have m*v = v*m/(M + m)
You are truly lost.
That is not what I wrote. I wrote v2 = v*m/(M + m).
That's the velocity of the combined masses following
the collision.
And then you said that was equal to m*v all over again.
No, I said that the momentum was conserved, and that
the momentum after the collision and the bodies are
stuck together is given by v2*(M + m).
I then plugged in the expression for the velocity v2
into that.
No you did not.
Sorry, but I did.
Also, I corrected myself with regards the above paragraph in
a separate post. I meant to say that the *velocity* after the
collision was given by v2 = v*m/(M + m). The momentum is
thus (M + m) * v2 = (M + m) * v*m/(M + m) = m*v. That is,
momentum is conserved (the same before and after the
collision).
See, you did it again and can't even tell.
You started with the small mass moving (m*v)
And then you played all around and came up yet again
with (m*v) all over again Greg.
Yes, of course! The momentum is conserved so it's numerical
value, after the collision, is equal to the value it had
prior to the collision. That value is m*v. It's also
(M + m)*v2. They are equal. The same.
Wow.
see
again you are stating
(M + m)*v2 = m*v
LOL
Poor Greg.
so stuck in conservation of momentum he can't see the
math of what he is stating is wrong unless of course the
larger mass of M = 0.
Poor Greg "still ignoring the larger mass completely
to come up with his "conservation of momentum" wrong answer.
LOL
--
James M Driscoll Jr
Spaceman
.
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