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Let's look at these
two blocks here.
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Block 1 has a mass of 3m and
Block 2 has a mass of 1 m.
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And they're colliding
and sticking together.
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And then they're going
to move in this direction
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and eventually, slide onto
this rough surface here.
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What we want to
figure out is, what
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was that initial
velocity v0 with which
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the two blocks collided?
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So up to this part here--
before this x equals 0--
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there's no external force
acting on the system.
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So conservation
of momentum holds,
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which means that p
final equals p initial.
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So let's put this together.
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We're going to have 3m v0.
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That block goes in the
positive i hat direction.
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Block 2 goes the opposite
way, so we have m v0.
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And then at the end, when
they're stuck together
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before here, we have to
consider only this portion here
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that pertains to a
frictionless surface.
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We have 4m v final.
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And so the m's fall out.
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This is 2.
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This is 4.
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So we'll see that the
final velocity is v0 half.
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Now, in order to get
to this v0, we also
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have to consider energy.
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And in particular, we have to
use the work energy principle
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here, because this joint
block is going to slide
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onto this rough surface here.
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So we can make use of the
work energy principle,
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as we know what the initial
conditions are at this point
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here, using the
conservation of momentum
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from here in the first place.
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So we want to use work equals
delta K, the work energy
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principle.
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Let's look at the work first.
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Work is the integral
of F dot dS.
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And we're dealing with a
one-dimensional problem
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here, so we can
just write Fx dx.
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And we want to have
this run from 0 to d.
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I should say here that
this block eventually
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comes to rest at x equals d.
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We're going to use
that in a second.
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So let's look at what this F is.
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That's the frictional
force here at work.
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And actually, which
direction does it go?
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It opposes the motion fk.
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So we're going to get a minus
sign here, integral fk dx,
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again going from x prime
equals 0 to x prime equals d.
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Now what is this fk ?We
know that fk equals mu k N.
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And from doing a quick
free body diagram,
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we'll see that we
have 4mg going down,
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and we have a normal
force going up.
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So we'll see that N equals 4mg,
which we can then plug in here.
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And then that goes over there.
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And the coefficient
of kinetic friction
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is actually given to us.
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So that nicely works out.
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So we're going to have minus
integral x prime 0, x prime
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equals d, the coefficient first,
b x squared, and then N 4mg dx.
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If we integrate that, we're
going to get minus 4bmg.
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And we'll integrate this one,
so that's x cubed over 3.
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And we're going to plug
in this value here,
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so we're going to get a d, and
that second term goes away.
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So we are left
with d cubed here.
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So this is our work.
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And now we need to look at
the change in kinetic energy
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between here and there.
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We already know that this
block is at rest at x equals d.
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So if the velocity is 0, our
kinetic energy will be 0,
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and so our K final
minus K initial
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is going to be
just minus initial.
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And that's initial here,
which is the final situation
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of our collision.
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So we're going to look
at the combined block.
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So we'll have minus 1/2 and
we have 4mv final squared.
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We determined here already
that the final velocity is
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the initial 1/2, so have minus.
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Here we're going to have 2m,
and now v0 squared over 4.
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And that actually gives
us minus 1/2 m v0 squared.
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OK, so now we can put
this all together.
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We apply this work
energy principle.
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So we're going to get
minus 4/3 bmg d cubed
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equals minus 1/2 m v0 squared.
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We'll see that this and
this and this and this goes.
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And so we're going
to get v0 equals 8/3,
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and then we have bgd cubed
and the square root of that.
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So this is our initial velocity
with which the two blocks
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collided initially,
then eventually went
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onto this rough surface and
came to rest here at x equals d.