Re: The opposing rockets and the box

Greg Neill wrote:
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Greg Neill wrote:
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Greg Neill wrote:
"Spaceman" <spaceman@xxxxxxxxxxxxxxxxxxxxxxxxxx> wrote in
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A box is placed in orbit,
On two opposite sides of the box there
are two equal rockets placed that can fire
in opposite directions.
Rocket (A) and rocket (B).

If you fire rocket (A) for 10 seconds)
and then shut it down.
Will it now take only 10 seconds for rocket
(B) to slow the box to an "at rest" state again?

No, the box will be lighter than before since the
first rocket threw away some mass. It should take
less time if the rocket thrust is the same.

Ok,
You found a flaw, I will admit.
Lets change it to this problem instead.

2 cannons are placed 200 ft away from a mass
1 cannon is fired at the mass and the collision causes
the mass to move towards the other cannon,
Will the other cannon be able to stop the mass
with 1 shot?

You need to define what happens to the cannon balls.
Do they bounce off or become embedded in the mass?

Try the "embedded in the mass" since that is what
occurs in the propulsion posts I made.

Then if you wish, also try the bouncing off.
:)


Also, is the mass on a frictionless surface? Or
perhaps the whole lot is simply floating in deep
space?

They are all floating in space and we can ignore
the motion of the cannons after they fire since both
will move the same away from the firing direction.

First cannon fires ball with mass m at velocity v
(for momentum m*v) at object with mass M. Mass
M is initially at rest in our chosen frame of reference,
so its momentum is initially zero.

Ball strikes object and is embedded. Total momentum
of the the ball and object must remain the same before
and after collision:

m*v = (M + m)*v2 v2 is the velocity of the object+ball

v2 = v*m/(M + m)

So v2 is slower than the original speed of the cannon
ball by the ratio m/(M + m). But the total momentum
of the pair is still m*v in our frame of reference.

Hang on Greg,
You just made
v*m/(M + m) = m*v
It seems you have a slight problem with your "totalling".

v*m/(M + m) is the velocity of the first ball and object
after their collision.

So it is v2

Their momentum is then mass times velocity, the mass being
(M + m) because they're stuck together, so:

(M + m) * v*m/(M + m) = m*v

If v*m/(M+m) truly equals = v2
you have
(M + m) * v2 = m*v

Yes. Plug in the expression for v2 in terms of v and
you see that the (M + m) bits cancel, as I showed.

No,
you multiplied an extra (M+m) to do such and because you did such
you completely removed the larger mass and showed you are lost
and could never figure out how the opposing forces could not
be the same once a relative mass increase occurs to the M.
You basically left the 2 masses moving (M+m) with the same
momentum the single mass had to begin with.
That is a freakin joke.
Why even bother doing anything.
Just ignore the large mass completely and you get the
same result.
You are ignorant to your own "wrong" methods.


No, momentum is a combination of both mass and velocity.
If the mass is larger, a smaller velocity is required to
have the same momentum. That's why after the collision
of the first ball with the object the ensemble travels
slower than the ball did alone. In fact, it travelled
slower by the ratio m/(M + m).

You are "removing" that larger mass completely and don't even
see why such would be wrong.
And of course, removing the larger mass would make both
smaller masses equal.
Sorry you are just wrong by removing the larger mass like
you are doing.

James, at least learn a little about conservation of momentum
and calculating the results of collisions. It's high school
Newtonian physics.

Greg,
You just proved to me that you are clueless.
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.

If m = 30 and M = 100
You have 10* 30 = 30*10/(100 + 30)
Why you can just multiply the (M +m) for no reason to the v2 is
amazing that you would try to pull such *** off at all..

--
James M Driscoll Jr
Spaceman






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