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Let's consider another example
of continuous mass transfer.

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Suppose we have a
truck, and that truck

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has some type of plow.

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And it's plowing snow.

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And there's some type
of external force acting

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on this truck, friction,
pushing the truck forward,

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so let's just assume we
have some type of force, F,

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on the truck.

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And this is our snow.

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And what's happening
in this problem

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is that the truck connects--
picks up the snow.

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And then, which is
at rest initially,

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gets the snow up to
the speed of the truck.

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And then the snow
falls off the plow.

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So how do we model this problem?

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Well, let's look at our
situation at time t.

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And what we're going
to do is, we're

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going to consider a certain
mass of snow, delta ms, that's

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at rest.

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And our truck, it's
a fixed mass truck,

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is moving with a velocity
vt at time t, the truck.

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So now, what happens
at time t plus delta t?

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Well, the truck has picked
up the mass of the snow.

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And the truck has now
changed its speed.

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So that's at time
t plus delta t.

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And now we want to write
down our momentum law.

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So we have our external force.

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Let's call this the
plus i hat direction.

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We have our external force
is equal to the limit

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as delta t goes to 0 of
the momentum at time t.

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So we have t plus
delta t, so what

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we have is the mass of the
truck plus delta ms times

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v of t plus delta t.

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And we have to subtract from it.

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That's divided by delta t.

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And we have to subtract from
this the moment at time t.

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The snow is at
rest, only the truck

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is moving, so we have minus
t of vt divided by delta t.

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Now, as usual, we're going to
say that the truck has changed

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its velocity in this interval.

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And what we want to do now
is write out our equation.

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We have F equals the limit
as delta t goes to 0.

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Now when we write
this out, notice

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that we're going to have
a number of terms here.

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We have mt plus delta
ms times v plus delta v

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of the truce-- that's the first
term-- divided by delta t.

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And the second term is just
minus mass of the truck,

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vt over delta t.

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Well, first off, we
see some cancellations

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between this term,
this, and that.

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And what we're left with is
limit as delta t goes to 0.

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We have the snow term, delta ms
over delta t times the truck.

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Now, here we have
a term, which we're

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going to analyze in a moment.

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It has two infinitesimal
quantities.

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And this term is
of second order,

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which we're going to neglect.

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And finally, we have
the term of mass

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of the truck times
delta vt delta t.

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So, neglecting this
second-order term

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in differentials,
what we get when

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we take the limit is, that
we get the force is the rate

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that the snow is being picked
up times v of the truck,

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plus mass of the truck,
times v truck, dt.

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Now, the only issue that
we have to think about here

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is about the rate that
the truck, the snow,

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is being picked up.

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So that's our last
consideration,

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but this will be our
differential equation

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for adding mass,
continually, to a system.

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So, when we found our equations
for describing the rate

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that the truck changes its
speed-- the snowplow when it's

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pushing snow away, we
wrote down our equation

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in terms of the external force
on the truck, the rate that

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snow is being picked up
by the truck, the velocity

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of the truck, the
mass of the truck,

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the rate of change of
velocity of the truck.

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If we multiply our
equation through by dt,

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we have the following equation.

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And now what we
want to consider is

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we want to focus on how
much dms is picked up

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in our infinitesimal time, dt.

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So one way to think about that
is, let's do a little drawing.

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So here's our truck.

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And let's say at time t--
the plow is right here,

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so here's the picture at time t.

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And let's draw some snow here.

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Here in New England
we have lots of snow.

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And at time t plus
delta t, the truck

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has moved a certain distance.

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And so, this quantity
is the amount of snow

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that's picked up by the
truck and displaced.

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Now, the truck has moved a
distance v truck delta t.

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And so, we can identify the
delta ms, which in the limit

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will be dms, is equal to
the density of snow times

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the cross sectional
area of the plow,

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times the length v
of truck delta t.

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And so now we have an
expression for the rate

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that the mass is being picked
up in this time interval.

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And we can write that
in the following way.

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That we see that our
differential equation

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Fdt equals dms,
which is rho A vt,

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times dt-- This will
be a small interval,

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we're taking a limit now--
times another vt plus mt dvt.

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Now, we can bring this term
over to the other side.

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And so we have F
minus rho A vt squared

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times dt equals the mass
of the truck times dvt.

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And the mass of the truck,
in this case, is fixed.

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Why?

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Because when the snow
gets into the truck,

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at the end of this
delta t time interval,

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it gets displaced to the side.

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So this is an equation that
we can separate and integrate.

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So we have dt is equal to
mt divided by F minus rho

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A vt squared dvt,
and we integrate

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from some initial time
and some final time.

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And here, we're
integrating from v of truck

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from some t initial
time to some final time.

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And that's our integral version
for our-- finding the velocity.

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This interval is an interval
that's not hard to do,

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that you should
try as an exercise

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in elementary calculus.