Re: The opposing rockets and the box

"Spaceman" <spaceman@xxxxxxxxxxxxxxxxxxxxxxxxxx> wrote in message
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Greg Neill wrote:
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Greg Neill wrote:
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Greg Neill wrote:
"Spaceman" <spaceman@xxxxxxxxxxxxxxxxxxxxxxxxxx> wrote in message
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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. I figured why
carry around the more complicated expression into the
next stage of the analysis when m*v will do nicely.

You might as well ignore the larger mass completely.
You would get the same answer of m*v.

If you were to think about it, James, you'd realize that
when considering momentum the amount of mass is not
critical; It's the product of mass and velocity that
determiness the momentum. A baseball and a cannon ball
can have the same momentum with appropriate velocities
for each, and both will 'deliver' the same momentum to
a target if they become embedded.
.

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