WEBVTT
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Let's analyze a
one-dimensional collision,
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where we're in the
laboratory reference frame.
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And we have V1 initial
and V2 initial is 0.
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It's a frictionless surface.
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But there's a collision here.
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And now we're going to make this
collision totally inelastic.
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Now what that means is that the
two particles stick together.
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So if this is our
initial state--
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and let's, again, choose
a direction i hat-- then
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our final state will just stick
those two particles together.
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And they're going to move.
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We only need one velocity
for the final state, i hat.
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Now again, let's assume that
there's no external forces.
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There are clearly
internal forces
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that are acting between the
two particles as they collide.
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But, remember, we know
that internal forces
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cancel in pairs.
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And if there's no
external forces,
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the momentum of the
system is constant.
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So we can write down our
momentum condition very simply.
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And we'll write it again
in terms of components.
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We have some incoming
momentum into the system.
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And all of that incoming
momentum is going out.
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And now, is energy constant?
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Well, a totally
inelastic collision,
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energy cannot be constant.
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And the reason is that
these internal forces
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will cause deformations
that are irreversible.
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So the objects
might be deformed,
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which is a loss of-- some
of that kinetic energy will
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go into deformation.
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There could be noise.
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There could be heat generated in
the collision, lots of sources.
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When objects stick together,
kinetic energy is not constant.
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In fact, we can even figure out
the change in kinetic energy,
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because we can see here that Vx
final is m1 over m1 plus m2 V1
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x initial.
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And now let's ask ourselves, in
a totally inelastic collision
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what is the change in
the kinetic energy?
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So our final kinetic energy
is 1/2 m1 plus m2 V-- here,
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we can, again, right in terms
of components-- Vx final,
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because the component's squared.
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That's no problem.
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And we just have this initial
kinetic energy coming in.
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Now we have a solution
for our problems.
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So let's just put that in.
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m1 plus m2 times m1
over m1 plus m2 times V1
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x initial squared minus 1/2
m1 V1 x initial squared.
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And now let's pull
out the 1/2 m1
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V1 x initial squared--
1/2 one factor of m1 V1 x
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initial squared.
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But what we're left with
here is another factor of m1.
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And downstairs, we
have an m1 plus m2
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outside but an m1 plus m2.
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Now remember, we
didn't square that.
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We have an m1 plus m2 inside.
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So there's that m1
plus m2 minus 1.
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Now this is the
initial kinetic energy.
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And when we subtract
these terms,
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we have negative
m2 over m1 plus m2.
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And so what we see
here is-- I'm just
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going to rewrite this again--
that the kinetic energy,
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the change in
kinetic energy, just
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depends on the initial kinetic
energy and this mass ratio.
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In fact, if we asked
ourselves, what
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is the ratio of that
loss of kinetic energy
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to the initial kinetic
energy, that just
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depends on the mass
ratio of our two objects.
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In particular, it
should be a minus sign,
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because the kinetic
energy is decreasing.
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And so this is the fact that
it's very transparent now
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that in a totally
inelastic collision
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the kinetic energy
is not constant.
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Where did that energy go?
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It went into other
forms of energy.