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We would now like to apply the
momentum principle to analyze

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the motion of the rocket.

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So what is our
momentum principle?

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We know that the
external force at time t

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is equal to the rate of
change of the momentum

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of the system over t.

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Now, recall we're
going to actually use

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the formal definition of
a derivative to write this

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as a limit as delta t
goes to 0 of the momentum

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of the system at
time t plus delta t

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minus the momentum of the system
at time t divided by delta t.

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So in order to
analyse the rocket,

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what we need to do is separately
analyze the momentum at our two

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states.

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So we have a state at time t.

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And we have a state at
time t plus delta t.

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And our goal will be to analyze
the system momentum at time t,

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and separately the
system momentum and time

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t plus delta t.

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And then we can apply
the momentum principle.

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So let's begin by analyzing
the system at time t.

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So recall that at that
time, we had our rocket

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and this is our time t.

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And the rocket had a
velocity V of r or t.

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And what we identified
the mass, m-r of t--

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recall this was the mass--
the dry mass of the rocket

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and the mass of the fuel.

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And so now, it's very simple
to write down the momentum

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of our system at time t.

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That's equal to just the
mass m of r of t times

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the velocity of the
rocket at time t.

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And so we can use
that in our expression

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for the momentum of the system.

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Now our next step is to
consider the system the time

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t plus delta t.

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Now recall, we still
have the rocket

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but the mass of the
rocket has changed.

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That's inside the
rocket, so this

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is dry mass of the rocket
plus mass of the fuel,

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but recall some of the fuel
has been ejected outwards.

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So how do we depict that
in our momentum diagram?

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Well let's just symbolize
that by a certain amount

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and we're going to
call this delta m-fuel.

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Now, what we need to do, again,
is to have our velocities.

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So here, the velocity
of the rocket

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is now the velocity at
time t plus delta t.

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And what about the velocity of
this fuel that's being ejected?

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Well, in our problem the
fuel is ejected at a velocity

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u relative to the rocket.

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So what we have is
u is the velocity

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of the fuel relative
to the rocket.

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But we've been choosing
the ground frame

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as our reference frame.

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And the rocket is
moving at a velocity

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at time t plus delta t with
respect to that ground frame.

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And so recall that the velocity
of the fuel in the ground frame

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is equal to the velocity
relative to the rocket

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plus the speed of the
rocket with respect

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to the ground frame.

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And that's what we're going
to draw on our diagram

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as the fuel.

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And so now we can
finish this analysis--

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and we'll apply that
next-- is to right

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the momentum of the system
at time t plus delta t.

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Now this is going
to have two terms.

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It has the mass of
the rocket at time t

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plus delta t times the
velocity of the rocket.

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And we also have to
add mass of the fuel

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times the velocity of the fuel
with respect to the ground.

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And now we have the momentum
at time t plus delta t.

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So we have both of
our pieces here.

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And we still have a
little bit of analysis

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because we'll use
our mass conservation

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equation and our relative
velocity condition

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to simplify this expression.

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So we'll do that next.