Re: The opposing rockets and the box

Greg Neill wrote:
"Spaceman" <spaceman@xxxxxxxxxxxxxxxxxxxxxxxxxx> wrote in message
news:jJmdndeTStPRRBnVnZ2dnUVZ_ofinZ2d@xxxxxxxxxxx
Greg Neill wrote:
"Spaceman" <spaceman@xxxxxxxxxxxxxxxxxxxxxxxxxx> wrote in message
news:TaudnbOIq_1SSBnVnZ2dnUVZ_j-dnZ2d@xxxxxxxxxxx
Greg Neill wrote:
"Spaceman" <spaceman@xxxxxxxxxxxxxxxxxxxxxxxxxx> wrote in message
news:DdmdnQ0H-4-fVBnVnZ2dnUVZ_h7inZ2d@xxxxxxxxxxx
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".
:)


.

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