WEBVTT
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Let's calculate
the kinetic energy
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in the center of mass frame of
two objects that are colliding.
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I'm going to use prime for the
center of mass frame velocity.
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So we have V1 prime.
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And we have V2 prime.
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And recall two things that
when we calculated V1 prime
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in the center of
mass frame, we found
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that this was equal
to the reduced mass
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divided by m1 times
the relative velocities
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of the two objects, V1
prime minus V2 prime.
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Now in the center of
mass frame, this quantity
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is a reference
frame independent.
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And so we can just
write that as V1, 2.
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It's the relative velocity is
reference frame independent.
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So we also knew
that V2 prime was
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equal to minus mu over m2 V1, 2.
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So now let's calculate
the kinetic energy
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in the center of mass frame.
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So K cm is 1/2 m1
V1 prime squared
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plus 1/2 m2 V2 prime squared.
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Now let's use our results in
terms of relative velocity.
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So we have 1/2 m1 mu
over and m1 times V1,
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2 squared-- because
we're squaring
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that-- plus 1/2 m2 mu-- the
minus sign doesn't matter
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anymore because
we're squaring it--
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V2 squared-- I could put that
in there, it wouldn't matter.
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And now what we
have is we have 1/2.
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One of the m1s cancels so we
have a mu squared over m1 V1, 2
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squared plus 1/2 another
mu square over m2 times
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V1, 2 squared.
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So I pull out the common terms,
mu square, V1, 2 squared.
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I have 1 over m1 plus 1 over m2.
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But recall that the reduced
mass was precisely 1 over m1
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plus 1 over m2.
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And so I get 1/2 mu
times V1, 2 squared
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is the kinetic energy in
the center of mass frame.
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So what that means
is a way of thinking
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if this were just a
simple reduced mass
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problem with a relative
speed of V1, 2 squared,
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I can write down
the kinetic energy.
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But that's the kinetic energy
in the center of mass frame.
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And therefore, the change in
kinetic energy in a collision
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is just 1/2 V1, 2 V
final squared minus--
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and it's the same
mu, so we'll just
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put a parentheses-- minus
V1, 2 initial squared.
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And if the collision
is elastic-- remember,
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our elastic collision
show that V1, 2 initial
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was minus V1, 2 final.
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So an elastic collision
satisfies that condition.
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And you can see directly that
delta Kcm is 0 in that case.
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And a completely
inelastic collision
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has the two objects moving
together at the end.
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So V1, 2 final is 0.
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So completely inelastic is
the condition that V1, 2 final
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is 0 because the
objects stick together.
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And then have a
very simple result
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for the change in kinetic
energy for completely
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inelastic collision.
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It's only negative 1/2 mu V1,
2 initial quantity squared.