WEBVTT
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Let's look at some
examples of one
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dimensional elastic collisions
with no external forces
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between two particles.
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So suppose I have
particle 1 and particle 2,
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and we have them moving
on a frictionless surface.
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And let's choose
a reference frame
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in which we'll
call-- the laboratory
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frame-- in which the initial
velocity of particle 2 0.
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So our frame is called
the laboratory frame.
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And in this frame we're thinking
of particle 2 as our target.
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And the target is at rest.
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And particle 1 is moving in
with some initial velocity.
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And so here is our
initial picture.
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And then in our final
state, both the target
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and the particles can be moving.
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So we'll just indicate v1 final.
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And we'll have target was
also moving with v2 final.
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And we have i hat.
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So let's now look
at some examples
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where we're analyzing
this type of collision
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in which both energy and
momentum are constant.
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And for a particular example
that we want to look at,
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we'll have m2 is twice m1.
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Now, for our two equations that
we're going to write about,
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we'll choose components, and
we'll write our energy equation
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as 1/2 m1 v1 x initial
squared plus 1/2 m2--
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so now we're going
to write 2 m1.
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But object 2 is at
rest, so we don't have
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to worry about that one yet.
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And so, our initial
kinetic energy
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is just equal to the
final kinetic energy--
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1/2 m1 v1 x final squared.
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And now I'm going to substitute
in 2m1 for the mass of m2
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v2 x final squared.
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And that's our energy condition.
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And then our
momentum condition is
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that the incoming
momentum, 1/2 x initial
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is equal to the outgoing
momentum, m1 v1 x final.
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And again, I'm going
to substitute for m2.
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That's 2m1 v2 x final.
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And these two
equations represent
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a system of two equations
with two unknowns.
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We'll treat this as givens
along with the initial velocity
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of particle 1.
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And our target here is to
solve for the 1x final and v2 x
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final.
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So we have two equations
and two unknowns.
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Now it's a quadratic equation.
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So we have to identify-- from
a problem solving strategy,
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we have to identify
which quantity
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we're going to solve first.
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And so, let's solve
for v1 x final, which
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means we need to
eliminate v2 x final.
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And the way I'll
eliminate v2 x final
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is I'll use the
momentum equation.
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And notice that
the m's will cancel
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in the momentum equation.
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And so, if I divide through
by 2m in the momentum equation
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and bring this term
over to the other side,
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equation 2a becomes
v2 x final equals--
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I'm dividing through by 2--
1/2 v1 x initial minus 1/2 v1 x
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final.
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So this gives me a
target-- v2 x final--
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I can substitute into there.
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This equation, again,
can be cleaned up.
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Let's clean it up before.
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We can get rid of the halves.
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We can get rid of the m1's.
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And so the quadratic
equation that we're
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going to be working with
is v1 x initial squared
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equals v1 x final squared.
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And I'm dividing through by 2.
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So I have the factor 2
plus 2 v2 x final squared.
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Now that's our energy equation.
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And here's our
momentum equation.
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So if we substitute
this in and square it,
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we can now substitute and we
get v1 x initial squared equals
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v1 x final squared plus 2.
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Now we make our substitution.
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So that's 1/2 v1 x initial
minus 1/2 v1 x final,
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quantity squared.
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So let's now expand this.
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And we have to be careful
not to make any mistakes.
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v1 x initial squared
equals v1 x final squared.
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Now if I pull the 2 out,
I got a quarter in front,
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and divide that by 1/2.
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So I get 1/2 v1 x initial
squared plus another 1
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plus a 1/2 v1 x final squared.
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Those represent the--
squaring out those two terms.
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Now the cross-term will
have a factor of 1/2
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in the front but a factor of 2.
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So we have a simple cross-term
of v1 x initial v1 x final.
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Now, when we--
let's collect terms.
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Let's bring everything
over to this side
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so we have a 0 equals 3/2 v1
x final squared minus 1/2 v1
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x initial squared and minus
v1 x initial v1 x final.
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Now I always like to just
write this up in a simple way
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to use the quadratic formula.
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So I'm going to
divide through by 2/3.
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And I get minus 1/3 v1 x
initial squared minus 2/3 v1
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x initial v1 x final.
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And this is now a
simple application
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of the quadratic formula.
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Negative b is plus 2/3 v1
x initial plus or minus.
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And we're going to interpret
those two roots in a moment.
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We have to factor 2/3 v1 x
initial squared minus 4ac.
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So that's another
minus sign with a plus.
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So that's plus 4/3
v1 x initial squared,
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all to the square root, and
everything divided by 2.
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Now here, let's just
look at this factor.
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This is 4/9 plus 4/3.
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4/3 is 12/9, so
that's 16/9, which
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is very convenient, because when
you take the square root that's
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4/3.
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So we get equal to 2/3 v1 x
initial plus or minus 4/3 v1 x
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initial divided by 2.
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Now we see that there's
two different roots.
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So when you add them, you get
2 v1 x initial divided by 2.
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So there's one solution.
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And when you subtract them,
you're getting 2/3 minus 4/3.
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That's negative
2/3 divided by 2.
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So that is another solution--
v1 x final equals negative 1/3
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v1 x initial.
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Now let's think about
the meaning of these two
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possible solutions.
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This solution has v1 x
final equals v1 x initial.
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So that's the
initial conditions.
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Just repeat it.
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And that will
always be the case.
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One solution will describe
the initial state,
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and the other solution will
define the final state.
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You can check for
yourself that if you just
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put vx1 final into this--
equal to vx1 in the initial--
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into this momentum
equation, then v2 x final
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is 0, which just repeats
the initial conditions.
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So this solution is
the initial state.
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And this solution, here,
is the final state.
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Now just to complete
the picture,
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v2 x final-- well,
that's equal to 1/2 v1
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x initial minus 1/2
times v1 x final.
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But v1 x final is negative 1/3.
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So we have a 1/2 minus
1/2 half times minus 1/3.
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So that's 1/2 plus 1/6, which
is 4/6 or 2/3 v x initial.
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So that's 2/3 the v1 x initial.
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And that represents the solution
to this particular problem.
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Now, of course what
you want to do is,
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you want to check-- we know
the momentum condition is
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already satisfied.
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But just as a check,
you would like
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to put it into the
energy condition
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just to make sure that when
you square these things out
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you get the right terms.
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However, we're very
confident of our result,
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because we've already reproduced
the initial conditions.
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And that wouldn't happen if we
made some algebraic mistake.
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So that, by itself,
is a sufficient check
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that this is a correct solution.
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One thing we should sit
back and think about
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is that when we use
the energy and momentum
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that we're inevitably
dealing with quadratics.
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And so we expect to use-- to
solve a quadratic equation
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at one point.
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And here it was right here.
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Now we're going to
find another way
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to do this by linearizing
the system, which
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will be a lot easier.