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Let's consider a
collision in a lab frame
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in which one
particle is coming in
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and the other
particle is at rest.
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So here's particle 2.
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V2 initial is 0.
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And we could say
that m2 is twice m1.
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And this is our lab frame.
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And now let's consider
the same collision
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in the center of mass frame.
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And in that reference
frame, we have
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particle 1 moving with
velocity V1 prime initial
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and particle 2 moving with
velocity V2 initial prime.
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Now, the way we're going
to analyze this problem
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is that we know the speeds
in the center of mass frame
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relative to the lab frame.
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So we actually know
these initial speeds.
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And we'll write that
result down in a moment.
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Because we know
the initial speeds,
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we also know that V1 final
in the center of mass frame
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is just minus the
initial velocity.
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That's the beauty of the
center of mass frame.
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The velocities just
change direction.
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They don't change magnitude.
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Because we know this
we can get V1 final.
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And then it's just
a simple exercise
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to go back to the lab
frame to calculate
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the quantity, the velocity of
the object 1 in the lab frame.
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So that will be our
sequence of ideas.
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And the key fact that
we know is that V1 prime
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is equal initially to
the reduced mass divided
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by m1 times V1, 2,
where V1, 2 initial
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is just equal to
V1 initial minus 0.
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So that's V1 initial.
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And the ratio, mu over m1, it's
very simple to calculate that.
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That's just m2 over m1 plus
m2, or our case, that's 2/3.
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And so from our result,
we now have very simply
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that V1 final prime is
minus V1 initial prime.
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So it's minus 2/3 V1 initial.
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And that's a very
straightforward calculation.
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We can do exactly the same
thing with V2 final prime.
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V2 final prime is
minus V2 initial prime.
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And V2 initial prime is equal
to minus mu over m2 times V1, 2
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initial.
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This ratio instead of being
2/3, it's a simple exercise.
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It will be m1 over m1
plus m2 is 1/3 V1 initial.
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And so we have solved for the
final velocities in the center
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of mass reference frame.
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Now let's just double
check our results.
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We have V2 final is
minus V2 initial.
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So that's minus.
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But there's another
minus sign here.
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So we have 2 minus signs.
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So that's a plus.
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And that's why we have
a plus 1/3 V1 initial.
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And finally, if we want
to ask the question, what
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are the velocities
in the lab frame,
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it's now a very simple exercise
to do reference frame change.
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We do need to know what
V center of mass is.
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That's V1 V initial over m1
plus m2, because remember
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that V2 initial is 0.
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So that's another
factor, V1 initial.
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And now for conclusion, we
have that the final velocity
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in the lab frame is
equal to the velocity
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in the center of mass frame
plus V center of mass.
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And we just collect our
results, minus 2/3 V1
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initial plus 1/3 V1 initial.
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So that's minus 1/3 V1 initial.
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The outgoing velocity of
particle 1 in the lab frame
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and the outgoing velocity of
particle 2 in the lab frame,
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again, we have 1/3 V1 initial
plus another 1/3 V1 initial
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00:05:26,370 --> 00:05:31,080
is equal to 2/3 V1 initial.
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So we were able to solve this
problem by switching reference
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frame using our basic fact
in the center of mass frame
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that the speeds remain the
same but the direction changes
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and able to solve this
problem without any
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of the traditional ways of
applying the energy momentum
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relationship and kinetic energy
or having quadratic equations.