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All right, last time
we talked exclusively

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about completely
inelastic collisions.

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Today I will talk

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about collisions
in more general terms.

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Let's take
a one-dimensional case.

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We have here m1
and we have here m2,

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and to make life a little easy,
we'll make v2 zero

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and this particle has
velocity v1.

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After the collision,
m2 has a velocity v2 prime,

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and m1, let it have
a velocity v1 prime.

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I don't even know whether
it's in this direction

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or whether it is
in that direction.

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You will see
that either one is possible.

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To find v1 prime
and to find v2 prime,

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it's clear that you
now need two equations.

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And if there is no net external
force on the system as a whole

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during the collisions,
then momentum is conserved.

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And so you can write down

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that m1 v1 must be
m1 v1 prime plus m2 v2 prime.

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Now, you may want to put
arrows over there

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to indicate
that these are vectors,

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but since it's
a one-dimensional case,

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you can leave the arrows off

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and the signs will then
automatically take care

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of the direction.

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If you call this plus,
then if you get a minus sign,

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you know that the velocity is
in the opposite direction.

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So now we need
a second equation.

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Now, in physics

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we do believe very strongly
in the conservation of energy,

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not necessarily in the
conservation of kinetic energy.

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As you have seen last time,
you can destroy kinetic energy.

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But somehow we believe
that if you destroy energy,

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it must come out
in some other form,

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and you cannot create energy
out of nothing.

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And in the case of the
completely inelastic collisions

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that we have seen last time,

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we lost kinetic energy,
which was converted to heat.

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There was internal friction.

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When the car wreck plowed
into each other,

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there was internal friction--
no external friction--

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and that took out
kinetic energy.

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And so, in its most general
form, you can write down

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that the kinetic energy
before the collision

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plus some number Q

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equals the kinetic energy
after the collision.

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And if you know Q,
then you have a second equation,

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and then you can solve
for v1 prime and for v2 prime.

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If Q is larger than zero, then
you have gained kinetic energy.

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That is possible;
we did that last time.

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We had two cars which
were connected by a spring,

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and we burned the wire

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and each went off
in the opposite direction.

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There was no kinetic energy

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before... if you want
to call it the collision,

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but there was
kinetic energy afterwards.

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That was the potential energy
of the spring

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that was converted
into kinetic energy.

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So Q can be larger than zero.

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We call that
a superelastic collision.

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It could be an explosion.

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That's a
superelastic collision.

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And then there is the
possibility that Q equals zero,

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a very special case.

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We will deal with that today,

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and we call that
an elastic collision.

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I will often call it
a completely elastic collision,

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which is really not necessary.

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"Elastic" itself already means
Q is zero.

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And then there is a case--

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of which we have seen
several examples last time--

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of inelastic collisions,
when you lose kinetic energy,

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so this is
an inelastic collision.

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And so, if you know what Q is,

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then you can solve
these equations.

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Whenever Q is less than zero,

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whenever you lose
kinetic energy,

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the loss, in general,
goes into heat.

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Now I want to continue a case

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whereby I have
a completely elastic collision.

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So Q is zero.

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Momentum is conserved, because
there was no net external force,

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so now kinetic energy is
also conserved.

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And so I can write down now
one-half m1 v1 squared--

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that was the kinetic energy
before the collision--

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must be the kinetic energy
after the collision

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one-half m1 v1 prime squared

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plus one-half
m2 v2 prime squared.

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This is my equation number one,

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and this is
my equation number two.

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And they can be solved;
you can solve them.

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They are solved in your book.

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I will simply give you
the results,

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because the results are
very interesting to play with.

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That's what
we will be doing today.

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v1 prime will be m1 minus m2
divided by m1 plus m2 times v1

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and v2 prime will be 2 m1
divided by m1 plus m2 times v1.

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The first thing
that you already see right away

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is that v2 prime is always
in the same direction as v1.

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That's completely obvious,

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because the second object
was standing still, remember?

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So if you plow something
into the second object,

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they obviously continue
in that direction.

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That's clear.

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So you see you can never have
a sign reversal here.

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Here, however, you can have
a sign reversal.

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If you bounce a ping-pong ball
off a billiard ball,

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the ping-pong ball
will come back

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and this one becomes negative,

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whereas if you plow a billiard
ball into a ping-pong ball,

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it will go forward.

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And so this can both be
negative and can be positive

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depending upon whether
the upstairs is negative

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or positive.

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So this is the result which
holds under three conditions:

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that the kinetic energy
is conserved, so Q is zero;

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that momentum is conserved;

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and that v2 before
the collision equals zero.

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Let's look at three interesting
cases whereby we go to extremes.

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And let's first take the case

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that m1 is much,
much larger than m2.

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m1 is much, much larger than m2.

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Another way of thinking about
that is that let m2 go to zero.

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Extreme case, the limiting case.

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So it's like having
a bowling ball

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that you collide
with a ping-pong ball.

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If you look at that equation
when m2 goes to zero--

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this is zero, this is zero--
notice that v1 prime equals v1.

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That is completely intuitive.

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If a bowling ball collides
with a ping-pong ball

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the bowling ball doesn't
even see the ping-pong ball.

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It continues its route
as if nothing happened.

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That's exactly what you see.

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After the collision, the bowling
ball continues unaltered.

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What is v2 prime?

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That is not so intuitive.

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If you substitute in there
m2 equals zero,

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then you get plus 2 v1--
not obvious at all, plus 2 v1.

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It's not something
I even want you to see;

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I can't see it either.

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I'll do a demonstration.

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You can see
that it really happens.

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So, now you take a bowling ball

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and you collide the bowling ball
with the ping-pong ball

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and the ping-pong ball will get
a velocity 2 v1--

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not more, not less--

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and the bowling ball continues
at the same speed.

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Now let's take a case whereby m1
equals much, much less than m2;

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in other words, in the limiting
case, m1 goes to zero.

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And we substitute that in here.

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So m1 goes to zero,
so this is zero

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and so you see
v1 prime equals minus v1.

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v1 prime equals minus v1,
completely obvious.

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The ping-pong ball bounces
off the bowling ball

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and it just bounces back.

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And this is what you see.

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And the bowling ball
doesn't do anything,

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because m1 goes to zero,
so v2 prime goes to zero.

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So that's very intuitive.

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And now we have a very cute case
that m1 equals m2.

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And when you substitute
that in here--

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when m1 equals m2--
v1 prime becomes zero.

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So the first one stops
with v2 prime becomes v1.

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If m1 equals m2,

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you have two downstairs here
and two upstairs

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and you see
that v2 prime equals v1.

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And that is a remarkable case--

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you've all seen that, you've all
played with Newton's cradle.

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You have two billiard balls.

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One is still
and the other one bangs on it.

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The first one stops

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and the second one takes off
with the speed of the first.

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An amazing thing.

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We've all seen it.

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I presume you have all seen it.

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Most people do this
with pendulums

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where they bounce these balls
against each other.

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I will do it here with a model

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that you can see
a little easier.

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I have here billiard balls,

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and if I bounce this one
on this one,

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then we have case number three.

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Then you see
this one stands still

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and this one takes over
the speed-- quite amazing.

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Every time I see this,
I love it.

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It is a wonderful thing to see.

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Nature... just imagine
you are nature,

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and this ball comes on,
and in no time at all

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you have to solve these
two equations very quickly.

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There's only one solution,
and nature knows how to do that.

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This one stops
and this one goes on.

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It's an amazing result.

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And I'm sure that you have seen

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these pendulums
that you can play with.

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Here we have not a bowling ball
onto a ping-pong ball

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but we have a billiard ball.

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The mass ratio is
not infinity to one

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but it's 500 to one,
which is quite large.

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And so what I will first do is
very intuitive.

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I will first bounce
the ping-pong ball

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off the bowling ball...
the billiard ball.

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The ping-pong ball comes back
almost with the same speed--

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not quite, because the ratio
is not infinity to one

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but it is 500 to one,

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and the billiard ball will do
practically nothing.

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It's exactly what you see
there in case two.

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So, there we go.

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You see, the ping-pong ball
comes back

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almost as far as I let it go.

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It tells you that the speed...

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Oh!

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That the speed has not changed.

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It just bounces back

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and the billiard ball
almost does nothing.

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Now comes case number one.

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That's the nonintuitive case.

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It is intuitive

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that if the billiard ball hits
the ping-pong ball

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that it will continue.

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As you will see, v1 prime is v1,
with the same speed.

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It's not at all intuitive

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that the ping-pong will get
twice the speed

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of the billiard ball,

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and, of course, you cannot see
that quantitatively

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because we don't do
a quantitative measurement

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of the speed
of the ping-pong ball,

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but you will see
that it bounces up quite high.

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So, there we go.

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The billiard ball
onto the ping-pong ball.

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Look at
the billiard ball alone--

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forget the ping-pong ball--

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and try to see that the speed
of the billiard ball

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is practically unaffected
by the collision.

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That's what you see there--
v1 prime is v1.

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You see, it's
practically unaffected.

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Now, of course,
it is way harder to see

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that the ping-pong ball

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gets twice the speed
of the billiard ball,

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because we don't do...

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because we don't do
a quantitative measurement.

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All right.

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So, those were examples, then,
of those three possibilities

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that you all see
on the blackboard there.

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Now I want to do more
completely elastic collisions,

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and I'm going to do that
with the air track.

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I'm going to try to make
completely elastic collisions.

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That's not so easy.

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I will have one object stand
still, so always v2 equals zero.

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And they are completely elastic.

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Kinetic energy is conserved.

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This word "completely"
is not necessary.

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I always add it in my mind.

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I'm going to have one object m1
which I bang against object m2,

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and object m2 will be standing
here, will have no speed.

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And object m1 comes in
from this side

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and I'll try to make
the collision elastic.

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And the way I will do that
is by using springs

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which are attached to each mass.

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And springs are
conservative forces

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so there is almost no heat
that is generated in the springs

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during the collision.

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And so to a reasonable
approximation,

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you will get
an elastic collision,

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but, of course,
there's always air drag.

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I can never take air drag out.

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So there's always some
external force on the system.

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So momentum is never
exactly conserved,

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and kinetic energy is never
exactly conserved either,

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so it's only an approximation.

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So I have one mass of unit 1--

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I will tell you what that is--
and I bang that into number 2.

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And the masses that I have...

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One mass is 241
plus or minus one gram,

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and the other mass that I have
is 482 plus or minus one gram.

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And I have two of these.

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And in my first experiment
I will use these two,

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so this is the ratio
that you see, one to one.

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I will give it
a certain velocity,

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which I cannot tell
you what that is.

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It depends upon
how happy I feel.

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If I push hard, then
the velocity will be high.

285
00:14:28 --> 00:14:34
If I push softly,
the velocity will be low.

286
00:14:30 --> 00:14:36
But it is something.

287
00:14:32 --> 00:14:38
And then we are going
to get here v1 prime,

288
00:14:36 --> 00:14:42
and there is a prediction
that this velocity will be zero.

289
00:14:40 --> 00:14:46
Look all the way
on the blackboard there.

290
00:14:42 --> 00:14:48
You will see
if the masses are the same

291
00:14:45 --> 00:14:51
and there is
an elastic collision,

292
00:14:46 --> 00:14:52
this one will stand still

293
00:14:48 --> 00:14:54
and this one, v2 prime,
will have a velocity v1.

294
00:14:53 --> 00:14:59
So that's a prediction.

295
00:14:55 --> 00:15:01
The second case,
I bounce one car

296
00:14:58 --> 00:15:04
onto another one
which is twice the mass--

297
00:15:01 --> 00:15:07
certain velocity
which I don't know what it is--

298
00:15:05 --> 00:15:11
and now
what are we going to get?

299
00:15:07 --> 00:15:13
So mass m1 is half
the mass of m2.

300
00:15:12 --> 00:15:18
So this is a 1 and
this is a 2 and this is 3.

301
00:15:18 --> 00:15:24
1 minus 2 is minus 1
divided by 3 is minus 1/3.

302
00:15:24 --> 00:15:30
So number one comes back--
no surprise.

303
00:15:28 --> 00:15:34
If you bounce a lighter object
off a more massive object,

304
00:15:32 --> 00:15:38
no surprise
that it bounces back.

305
00:15:34 --> 00:15:40
So that's minus one-third,
and when we look here,

306
00:15:38 --> 00:15:44
we have 2 times 1 divided
by 1 plus 2 is 3.

307
00:15:41 --> 00:15:47
That is plus two-thirds,

308
00:15:43 --> 00:15:49
so number one will come back
with a speed one-third--

309
00:15:48 --> 00:15:54
which, of course,
since this is sign-sensitive,

310
00:15:51 --> 00:15:57
we get one-third v1 and
this one is plus two-thirds v1.

311
00:15:58 --> 00:16:04
That's the prediction.

312
00:15:59 --> 00:16:05
And the way we're going
to measure it

313
00:16:01 --> 00:16:07
is by measuring the time
for the objects to move

314
00:16:05 --> 00:16:11
over a distance
of ten centimeters.

315
00:16:07 --> 00:16:13
No matter how long
you see these cars to be,

316
00:16:10 --> 00:16:16
there is here a piece of metal
which is ten centimeters long.

317
00:16:15 --> 00:16:21
It goes through a slot
with a diode.

318
00:16:17 --> 00:16:23
We have one system here
and one there.

319
00:16:19 --> 00:16:25
And during this ten-centimeter
movement, we measure the time.

320
00:16:23 --> 00:16:29
We begin when this piece of
metal enters the diode system

321
00:16:27 --> 00:16:33
and we stop the time when it
leaves the diode system.

322
00:16:30 --> 00:16:36
And each one of those cars has
this ten-centimeter metal plate.

323
00:16:37 --> 00:16:43
So, the way
we're going to do this is

324
00:16:40 --> 00:16:46
by measuring timing, of course.

325
00:16:42 --> 00:16:48
This velocity will give me
a certain time,

326
00:16:45 --> 00:16:51
and this velocity will give me
a certain time.

327
00:16:49 --> 00:16:55
Whatever comes out,
you will see that on this timer.

328
00:16:53 --> 00:16:59
Then t1 prime...

329
00:17:01 --> 00:17:07
is... in this case...

330
00:17:06 --> 00:17:12
it stands still,
so it must be zero.

331
00:17:09 --> 00:17:15
And t2 prime must be
the same as t1.

332
00:17:13 --> 00:17:19
So these two numbers, you should
be able to compare directly.

333
00:17:17 --> 00:17:23
What kind of uncertainties
do we have?

334
00:17:20 --> 00:17:26
It's hard to tell.

335
00:17:21 --> 00:17:27
But I would say, as I argued
last time, that you should allow

336
00:17:25 --> 00:17:31
for at least about 2½ percent
uncertainty in each time.

337
00:17:29 --> 00:17:35
If it comes out better,
then you are lucky.

338
00:17:31 --> 00:17:37
If it comes out worse,
then it is a bad day.

339
00:17:34 --> 00:17:40
2½ percent,
and I argued last time

340
00:17:36 --> 00:17:42
how I reasoned
about the 2½ percent.

341
00:17:39 --> 00:17:45
Now we do this experiment.

342
00:17:41 --> 00:17:47
We get a certain t1.

343
00:17:43 --> 00:17:49
And so this one goes back
with one-third of the speed,

344
00:17:47 --> 00:17:53
so it will take three times
longer to go ten centimeters.

345
00:17:52 --> 00:17:58
So I'm going to multiply
this time by one-third,

346
00:17:57 --> 00:18:03
and whatever comes out
should be the same as this.

347
00:18:01 --> 00:18:07
So, let me move this up
a little... t2 prime.

348
00:18:05 --> 00:18:11

349
00:18:08 --> 00:18:14
The speed here, the forward
speed, is two-thirds,

350
00:18:12 --> 00:18:18
so it will go slower.

351
00:18:14 --> 00:18:20
So if I multiply this time
by two-thirds,

352
00:18:17 --> 00:18:23
then I should be able
to compare it with t1.

353
00:18:21 --> 00:18:27
And all of that,
all these times, I think,

354
00:18:23 --> 00:18:29
will not be any better
than roughly 2½ percent,

355
00:18:28 --> 00:18:34
except if we are a little lucky.

356
00:18:31 --> 00:18:37
(system powers up )

357
00:18:34 --> 00:18:40
Okay, the system is up.

358
00:18:37 --> 00:18:43
The timers are up,
the timers are up.

359
00:18:41 --> 00:18:47
This will be the time t1.

360
00:18:44 --> 00:18:50
This will be the time t2 prime,

361
00:18:50 --> 00:18:56
and this is the one
when the object bounces back,

362
00:18:54 --> 00:19:00
so this is t1 prime.

363
00:18:58 --> 00:19:04
You won't see
that it is negative.

364
00:19:00 --> 00:19:06
It will go back
through the same slot,

365
00:19:02 --> 00:19:08
and the electronics
is arranged in such a way

366
00:19:04 --> 00:19:10
that when it goes back
through the same slot

367
00:19:07 --> 00:19:13
that it will initiate this time.

368
00:19:10 --> 00:19:16
I will zero them.

369
00:19:13 --> 00:19:19
First you have to tell me
whether they are working.

370
00:19:15 --> 00:19:21
I will just let it go
through these slots.

371
00:19:18 --> 00:19:24
Is this one working?

372
00:19:20 --> 00:19:26
Tell me
that this one is working.

373
00:19:23 --> 00:19:29
If I send this one back,
is this one working? Okay.

374
00:19:28 --> 00:19:34

375
00:19:30 --> 00:19:36
Okay, there we go.

376
00:19:35 --> 00:19:41
This is the one
that will have no speed,

377
00:19:38 --> 00:19:44
and this is the one that
we're going to give a velocity.

378
00:19:44 --> 00:19:50
You are ready? I hope you know
what you are going to see.

379
00:19:48 --> 00:19:54
This one will come to a halt,
and this will go on.

380
00:19:51 --> 00:19:57
They have the same mass.

381
00:19:52 --> 00:19:58
There we go.

382
00:19:54 --> 00:20:00

383
00:19:56 --> 00:20:02
This one is indeed
coming to a halt

384
00:19:58 --> 00:20:04
and this one took over
the speed.

385
00:20:01 --> 00:20:07
What are the numbers?

386
00:20:04 --> 00:20:10
194 and 196-- only
two milliseconds' difference.

387
00:20:08 --> 00:20:14
That is an incredible result.

388
00:20:12 --> 00:20:18
0.194 and 0.196.

389
00:20:18 --> 00:20:24
There's only one percent
difference between them--

390
00:20:21 --> 00:20:27
way within
my wildest expectations.

391
00:20:26 --> 00:20:32
Because my expectations were

392
00:20:28 --> 00:20:34
that they could be off each
by 2½ percent.

393
00:20:30 --> 00:20:36
Now we go to the one-to-two.

394
00:20:35 --> 00:20:41
So here is the car
which has twice the mass.

395
00:20:40 --> 00:20:46
It's important that I zero it.

396
00:20:45 --> 00:20:51
And now this one
is going to come back.

397
00:20:50 --> 00:20:56
So you're going to see this one
come in, give you the time here.

398
00:20:54 --> 00:21:00
This one goes through here,
it gives you the time here.

399
00:20:58 --> 00:21:04
This one will come back,
it gives you the time here.

400
00:21:01 --> 00:21:07
Are they zero?

401
00:21:03 --> 00:21:09

402
00:21:05 --> 00:21:11
Okay?

403
00:21:07 --> 00:21:13
You're ready?

404
00:21:10 --> 00:21:16
Yeah?

405
00:21:12 --> 00:21:18
There we go.

406
00:21:14 --> 00:21:20

407
00:21:18 --> 00:21:24
Okay. Now, now comes
the real acid test.

408
00:21:24 --> 00:21:30
123... 186.

409
00:21:29 --> 00:21:35
123...

410
00:21:30 --> 00:21:36

411
00:21:37 --> 00:21:43
And what is the last one?

412
00:21:40 --> 00:21:46
375.

413
00:21:42 --> 00:21:48

414
00:21:48 --> 00:21:54
Okay, let me be lazy
and let me use my calculator.

415
00:21:54 --> 00:22:00

416
00:21:55 --> 00:22:01
375 divided by three--

417
00:21:57 --> 00:22:03
I could have done that
by heart, of course--

418
00:22:00 --> 00:22:06
this is 0.125,
an amazing agreement!

419
00:22:06 --> 00:22:12
Amazing!

420
00:22:07 --> 00:22:13
Only two percent off,
less than two percent.

421
00:22:10 --> 00:22:16
Now here, 0.186 times 2
divided by 3-- 0.124.

422
00:22:21 --> 00:22:27
I can't believe it--
less than a percent off.

423
00:22:25 --> 00:22:31
So, you see
in front of your eyes

424
00:22:27 --> 00:22:33
that we were able
to create something

425
00:22:29 --> 00:22:35
that was indeed extremely close
to elastic collisions

426
00:22:34 --> 00:22:40
in spite of all the problems:

427
00:22:37 --> 00:22:43
that we have air drag,
and that, of course

428
00:22:41 --> 00:22:47
there is always
some loss of kinetic energy.

429
00:22:46 --> 00:22:52
But it's so little

430
00:22:47 --> 00:22:53
that it doesn't show up
in these measurements.

431
00:22:51 --> 00:22:57

432
00:22:54 --> 00:23:00
Now I want you to have
a sleepless night.

433
00:22:58 --> 00:23:04
I want you to think
about the following

434
00:23:01 --> 00:23:07
and if you can't solve it
before the night is over,

435
00:23:03 --> 00:23:09
then I really think
you should lie awake.

436
00:23:07 --> 00:23:13

437
00:23:09 --> 00:23:15
Here is the wall.

438
00:23:11 --> 00:23:17
Here is a tennis ball that
comes in with a certain mass m,

439
00:23:15 --> 00:23:21
and it has a certain velocity v,
whatever that is.

440
00:23:18 --> 00:23:24
It's a nearly elastic collision

441
00:23:20 --> 00:23:26
and it bounces back
from the wall.

442
00:23:21 --> 00:23:27
And we all know that if it is
a nearly elastic collision

443
00:23:25 --> 00:23:31
that it comes back
with the same velocity.

444
00:23:27 --> 00:23:33
Kinetic energy is conserved.

445
00:23:29 --> 00:23:35
All the kinetic energy is
in the tennis ball;

446
00:23:32 --> 00:23:38
nothing is in the wall.

447
00:23:33 --> 00:23:39
The wall has
an infinitely large mass,

448
00:23:36 --> 00:23:42
but the momentum
of this tennis ball

449
00:23:39 --> 00:23:45
has changed by an amount 2 mv.

450
00:23:42 --> 00:23:48
That momentum must be
in the wall--

451
00:23:45 --> 00:23:51
it's nonnegotiable, because
momentum must be conserved.

452
00:23:48 --> 00:23:54
So now here you see
in front of your eyes a case

453
00:23:51 --> 00:23:57
that the wall has momentum,
but it has no kinetic energy.

454
00:23:55 --> 00:24:01
Can you understand that?
Can you reconcile that?

455
00:23:58 --> 00:24:04
Can you show me mathematically
that that is completely kosher?

456
00:24:02 --> 00:24:08
That the wall has momentum 2 mv,
it's nonnegotiable.

457
00:24:06 --> 00:24:12
It must have momentum, and
yet it has no kinetic energy.

458
00:24:10 --> 00:24:16
How is that possible?

459
00:24:12 --> 00:24:18
Think about it,
and if you can't solve it,

460
00:24:15 --> 00:24:21
call me at 3:00 a.m. and
I'll tell you the solution.

461
00:24:18 --> 00:24:24
Okay, now let's look at this

462
00:24:20 --> 00:24:26
from the center-of-mass
frame of reference.

463
00:24:24 --> 00:24:30
The center of mass
is very special,

464
00:24:26 --> 00:24:32
and physicists love to work
in the center of mass

465
00:24:30 --> 00:24:36
for reasons
that you will understand.

466
00:24:33 --> 00:24:39
In the absence
of any net external forces

467
00:24:35 --> 00:24:41
on a system as a whole--
as we discussed last time--

468
00:24:39 --> 00:24:45
the center of mass will
always have the same velocity.

469
00:24:43 --> 00:24:49
We did a demonstration there

470
00:24:44 --> 00:24:50
with two vibrating objects
with a spring,

471
00:24:47 --> 00:24:53
but yet the center of mass was
moving with a constant velocity.

472
00:24:52 --> 00:24:58
The beauty now is

473
00:24:53 --> 00:24:59
if you jump into the frame
of the center of mass--

474
00:24:56 --> 00:25:02
that means you move

475
00:24:57 --> 00:25:03
with the same velocity
of the center of mass--

476
00:24:59 --> 00:25:05
the center of mass stands still
in your frame of reference.

477
00:25:03 --> 00:25:09
And if the center of mass
stands still,

478
00:25:05 --> 00:25:11
the momentum of the particles
in your frame of reference--

479
00:25:08 --> 00:25:14
in the center-of-mass frame
of reference-- is zero.

480
00:25:11 --> 00:25:17
It is zero before the collisions

481
00:25:13 --> 00:25:19
and it is zero
after the collisions.

482
00:25:15 --> 00:25:21
And this gives the center
of mass some amazing properties

483
00:25:20 --> 00:25:26
that I will discuss
with you now.

484
00:25:22 --> 00:25:28
First, we have a particle m1
and we have a particle m2.

485
00:25:28 --> 00:25:34
And let this one
in the center-of-mass frame

486
00:25:33 --> 00:25:39
have a velocity u1,

487
00:25:35 --> 00:25:41
and this in the center-of-mass
frame has a velocity u2.

488
00:25:39 --> 00:25:45
I give it specifically u's

489
00:25:41 --> 00:25:47
so that you can separate
the u's from the v's.

490
00:25:44 --> 00:25:50
The v's are always
in your frame of reference;

491
00:25:46 --> 00:25:52
the u's are in the frame of
reference of the center of mass.

492
00:25:51 --> 00:25:57
And I take a case whereby I have
a completely elastic collision--

493
00:25:55 --> 00:26:01
that means Q is zero-- the kind
that we have just discussed.

494
00:26:00 --> 00:26:06
Momentum is not only conserved
but it is also zero

495
00:26:03 --> 00:26:09
at all moments in time
before and after the collision.

496
00:26:08 --> 00:26:14
After the collision

497
00:26:10 --> 00:26:16
let's say m2 goes back
with velocity u2 prime,

498
00:26:17 --> 00:26:23
and let's say
m1 has a velocity u1 prime.

499
00:26:24 --> 00:26:30
That's the situation
after the collision.

500
00:26:27 --> 00:26:33
Now, I know that momentum is
zero, so I only can write down

501
00:26:32 --> 00:26:38
for this situation
after the collision

502
00:26:34 --> 00:26:40
that m1 times u1 prime plus m2
times u2 prime must be zero.

503
00:26:42 --> 00:26:48
I don't write them down
as vectors.

504
00:26:44 --> 00:26:50
That's not necessary,

505
00:26:45 --> 00:26:51
because it's
a one-dimensional collision

506
00:26:47 --> 00:26:53
and the signs will automatically
take care of the directions.

507
00:26:52 --> 00:26:58
I told you I chose the case of
a completely elastic collision--

508
00:26:56 --> 00:27:02
Q is zero-- and so kinetic
energy must be conserved.

509
00:27:01 --> 00:27:07
So I have...
before the collision,

510
00:27:03 --> 00:27:09
I have one-half m1 u1 squared
plus one-half m2 u2 squared,

511
00:27:12 --> 00:27:18
and after the collision, I have
one-half m1 u1 prime squared

512
00:27:19 --> 00:27:25
plus one-half
m2 u2 prime squared.

513
00:27:23 --> 00:27:29
Kinetic energy before;
kinetic energy afterwards.

514
00:27:27 --> 00:27:33
Equation one; equation two.

515
00:27:30 --> 00:27:36
Nature can solve that
quicker than we can,

516
00:27:33 --> 00:27:39
and the result is amazing.

517
00:27:36 --> 00:27:42
The result
in the center of mass

518
00:27:39 --> 00:27:45
is that u1 prime equals minus u1
and u2 prime equals minus u2.

519
00:27:47 --> 00:27:53
And that is an amazing result
when you think about it.

520
00:27:50 --> 00:27:56
It means that in the center
of mass, all that happens is

521
00:27:53 --> 00:27:59
that the speeds reverse
directions, but the speed...

522
00:27:57 --> 00:28:03
The velocities
reverse directions,

523
00:27:59 --> 00:28:05
but the speeds remain the same.

524
00:28:01 --> 00:28:07
And that is a remarkable,
remarkable result.

525
00:28:05 --> 00:28:11

526
00:28:07 --> 00:28:13
If you ever want to move
yourself to the center of mass,

527
00:28:11 --> 00:28:17
you will have to know what
the center-of-mass velocity is.

528
00:28:15 --> 00:28:21
How do we calculate the velocity
of the center of mass?

529
00:28:19 --> 00:28:25
So we're dealing here
with the center-of-mass frame,

530
00:28:23 --> 00:28:29
and now I'm going back
to the laboratory frame.

531
00:28:26 --> 00:28:32
And we know that M total

532
00:28:29 --> 00:28:35
times the position vector
of the center of mass--

533
00:28:31 --> 00:28:37
this is the way
we defined it last time--

534
00:28:34 --> 00:28:40
equals m1 times the position
vector of particle 1

535
00:28:38 --> 00:28:44
plus m2 times the position
vector of particle 2.

536
00:28:43 --> 00:28:49
And so if you take
the derivative of this equation,

537
00:28:46 --> 00:28:52
then the positions
become velocities,

538
00:28:49 --> 00:28:55
so the velocity
of the center of mass equals

539
00:28:52 --> 00:28:58
l divided by m1 plus m2,

540
00:28:55 --> 00:29:01
because that's the M total
which I bring under here,

541
00:28:59 --> 00:29:05
and upstairs I get
m1 v1 plus m2 v2.

542
00:29:07 --> 00:29:13
And notice I left the arrows off

543
00:29:09 --> 00:29:15
because, since it's
a one-dimensional problem,

544
00:29:12 --> 00:29:18
signs will take care
of the directions.

545
00:29:15 --> 00:29:21
So this is the velocity
of the center of mass.

546
00:29:18 --> 00:29:24
I wrote it also there
on the blackboard

547
00:29:20 --> 00:29:26
because later on
in this lecture I will need it,

548
00:29:23 --> 00:29:29
and it is possible that
by that time I have erased it,

549
00:29:26 --> 00:29:32
and so that's
why I wrote it down there, too.

550
00:29:30 --> 00:29:36
So if now you want to know
what u1 is, so we are...

551
00:29:35 --> 00:29:41
Now you want to know

552
00:29:37 --> 00:29:43
what the velocity is
in the center-of-mass frame,

553
00:29:41 --> 00:29:47
that, of course, equals

554
00:29:43 --> 00:29:49
v1 minus the velocity
of the center of mass

555
00:29:47 --> 00:29:53
and u2 equals v2 minus the
velocity of the center of mass.

556
00:29:56 --> 00:30:02
So this is a way
that you can transfer,

557
00:29:58 --> 00:30:04
if you want to, into
the center-of-mass frame.

558
00:30:02 --> 00:30:08
And it sometimes pays off,
for reasons that I mentioned,

559
00:30:06 --> 00:30:12
that the momentum in the
center-of-mass frame is zero--

560
00:30:09 --> 00:30:15
always zero before the collision
and after the collision,

561
00:30:13 --> 00:30:19
independent of whether
it is an elastic collision,

562
00:30:16 --> 00:30:22
whether it's
an inelastic collision

563
00:30:18 --> 00:30:24
or whether it's
a superelastic collision.

564
00:30:20 --> 00:30:26
Now, if you later wanted
to transfer back

565
00:30:23 --> 00:30:29
to your laboratory frame, then,
of course, you will have to add

566
00:30:28 --> 00:30:34
the velocity of the center
of mass again to the u1 prime,

567
00:30:32 --> 00:30:38
and you have to add the velocity
of the center of mass

568
00:30:36 --> 00:30:42
to the u2 prime.

569
00:30:37 --> 00:30:43
The velocity of the center
of mass has not changed

570
00:30:40 --> 00:30:46
as seen from your frame
of reference,

571
00:30:42 --> 00:30:48
because the velocity
of center of mass

572
00:30:44 --> 00:30:50
is always the same, remember,
because momentum is conserved.

573
00:30:47 --> 00:30:53
So to get into
the center-of-mass frame,

574
00:30:50 --> 00:30:56
you must subtract the velocity
of the center of mass

575
00:30:53 --> 00:30:59
from the initial velocities.

576
00:30:56 --> 00:31:02
To get out of it,
you must add them.

577
00:30:59 --> 00:31:05

578
00:31:00 --> 00:31:06
Now, the kinetic energy
and the momentum depend

579
00:31:04 --> 00:31:10
on your reference frame.

580
00:31:05 --> 00:31:11
In general, the total momentum
as seen from your seats

581
00:31:09 --> 00:31:15
is not zero.

582
00:31:11 --> 00:31:17
That's only
in a very special case.

583
00:31:13 --> 00:31:19
In the case
of the center of mass,

584
00:31:15 --> 00:31:21
the total momentum is
always zero.

585
00:31:17 --> 00:31:23
The kinetic energy
as seen from the lab frame

586
00:31:20 --> 00:31:26
is certainly, in general,
not the same

587
00:31:23 --> 00:31:29
as the kinetic energy
from the center-of-mass frame.

588
00:31:27 --> 00:31:33
And now comes another unique
property of the center of mass.

589
00:31:32 --> 00:31:38
If I have a completely
inelastic collision,

590
00:31:36 --> 00:31:42
then all energy in the
center-of-mass frame is lost.

591
00:31:41 --> 00:31:47
That's obvious--

592
00:31:42 --> 00:31:48
in the center-of-mass frame,
remember, momentum is zero.

593
00:31:46 --> 00:31:52
So you're in the center of mass.

594
00:31:48 --> 00:31:54
One particle comes to you
and the other comes to you.

595
00:31:51 --> 00:31:57
You're not moving,
you're in the center of mass.

596
00:31:54 --> 00:32:00
They get stuck together

597
00:31:55 --> 00:32:01
because it's a completely
inelastic collision.

598
00:31:58 --> 00:32:04
If they get stuck together

599
00:31:59 --> 00:32:05
after the collision
they stand still.

600
00:32:02 --> 00:32:08
That meansall kinetic energy
that was there before

601
00:32:04 --> 00:32:10
is all destroyed,
and this kinetic energy--

602
00:32:08 --> 00:32:14
as observed in
the center-of-mass frame--

603
00:32:11 --> 00:32:17
we call the internal energy,

604
00:32:13 --> 00:32:19
and that is the maximum energy
in a collision

605
00:32:17 --> 00:32:23
that can ever be converted
into heat.

606
00:32:21 --> 00:32:27
And I will show that
to you partially.

607
00:32:24 --> 00:32:30
I will do some of the work

608
00:32:26 --> 00:32:32
and I will let you do
some of the work as well.

609
00:32:29 --> 00:32:35
So I will first calculate--

610
00:32:31 --> 00:32:37
in your frame of reference,
where you are sitting--

611
00:32:35 --> 00:32:41
how much energy is lost

612
00:32:36 --> 00:32:42
when we have a completely
inelastic collision.

613
00:32:40 --> 00:32:46
I will then transfer
to the center-of-mass frame,

614
00:32:43 --> 00:32:49
and I will show you, then,
this quite amazing property.

615
00:32:47 --> 00:32:53
So now we are back in 26.100,

616
00:32:50 --> 00:32:56
and we're going to make a
completely inelastic collision.

617
00:32:57 --> 00:33:03
That means
they stick together, remember?

618
00:33:00 --> 00:33:06
And m2, we will have again,
to make life a little simple,

619
00:33:06 --> 00:33:12
no speed, v2 equals zero,
and m1 has a velocity v1.

620
00:33:14 --> 00:33:20
You've seen this now
a zillion times.

621
00:33:17 --> 00:33:23
They get together,
they stick together,

622
00:33:19 --> 00:33:25
and so I have here a velocity
which I call v prime,

623
00:33:25 --> 00:33:31
and the mass is m1 plus m2.

624
00:33:28 --> 00:33:34
That's after the collision.

625
00:33:31 --> 00:33:37
Momentum is conserved

626
00:33:32 --> 00:33:38
if there's no external... net
external force on the system,

627
00:33:35 --> 00:33:41
and so I can write down

628
00:33:37 --> 00:33:43
that m1 v1 must be
m1 plus m2 times v prime.

629
00:33:45 --> 00:33:51
And so v prime equals m1 v1
divided by m1 plus m2.

630
00:33:54 --> 00:34:00
That's a very
simple calculation.

631
00:33:57 --> 00:34:03
This, by the way--
it's not so obvious--

632
00:34:01 --> 00:34:07
is also the velocity
of the center of mass.

633
00:34:04 --> 00:34:10
And how can I see that
so quickly?

634
00:34:07 --> 00:34:13
Well, what was the velocity
of the center of mass?

635
00:34:11 --> 00:34:17
Here you have it.

636
00:34:12 --> 00:34:18
This was in general.

637
00:34:14 --> 00:34:20
This was not for the case
that v2 was zero;

638
00:34:17 --> 00:34:23
this was more general.

639
00:34:18 --> 00:34:24
Make v2 zero,
and you see exactly

640
00:34:21 --> 00:34:27
that you see here
the same result,

641
00:34:24 --> 00:34:30
so this must be the velocity
of the center of mass.

642
00:34:29 --> 00:34:35
Now we can calculate
what the difference is

643
00:34:33 --> 00:34:39
between the potential energy...

644
00:34:35 --> 00:34:41
the kinetic energy
after the collision

645
00:34:38 --> 00:34:44
and the kinetic energy
before the collision.

646
00:34:40 --> 00:34:46
That is, of course, something
that is rather trivial.

647
00:34:44 --> 00:34:50
You know what the kinetic energy
is before the collision--

648
00:34:47 --> 00:34:53
it's one-half m1 v1 squared--

649
00:34:49 --> 00:34:55
and you know what it is
after the collision.

650
00:34:52 --> 00:34:58
I calculated v prime, and so
you have take half this mass,

651
00:34:56 --> 00:35:02
multiply it
by this velocity squared.

652
00:34:58 --> 00:35:04
You can do that,
I am sure you can do that.

653
00:35:01 --> 00:35:07
And you will be able to see
that this equals minus...

654
00:35:04 --> 00:35:10
and you have to massage
the algebra a little bit--

655
00:35:06 --> 00:35:12
minus one-half m1 m2 divided
by m1 plus m2 times v1 squared.

656
00:35:15 --> 00:35:21
That's what you will find.

657
00:35:17 --> 00:35:23
The minus sign is predictable.

658
00:35:20 --> 00:35:26
We lose kinetic energy

659
00:35:21 --> 00:35:27
when there is a completely
inelastic collision.

660
00:35:25 --> 00:35:31
We've done many last lecture.

661
00:35:26 --> 00:35:32
You lose kinetic energy-- we
saw it in every single case.

662
00:35:30 --> 00:35:36
That's what
the minus sign means.

663
00:35:32 --> 00:35:38
This is Q.

664
00:35:33 --> 00:35:39

665
00:35:35 --> 00:35:41
You lose kinetic energy
and that goes into heat.

666
00:35:39 --> 00:35:45
So you've done your homework now
in the laboratory frame

667
00:35:42 --> 00:35:48
and you are home free--
very well.

668
00:35:46 --> 00:35:52
Now I'm going to do
the same calculation

669
00:35:49 --> 00:35:55
in the center-of-mass frame.

670
00:35:51 --> 00:35:57
And I will show you, now--
that's the purpose--

671
00:35:54 --> 00:36:00
that this amount of energy,
which is what is lost,

672
00:35:59 --> 00:36:05
that that is all there was
to start with

673
00:36:01 --> 00:36:07
in the center-of-mass frame,

674
00:36:03 --> 00:36:09
and that's a unique property
of the center of mass.

675
00:36:07 --> 00:36:13
And so I'm going to convert now,

676
00:36:09 --> 00:36:15
to transfer you
to the center-of-mass frame

677
00:36:13 --> 00:36:19
and then we will calculate
how much energy there was

678
00:36:18 --> 00:36:24
in the center-of-mass frame
before the collision,

679
00:36:22 --> 00:36:28
because after the collision
there is nothing.

680
00:36:24 --> 00:36:30
There is zero.

681
00:36:26 --> 00:36:32
In the case of a completely
inelastic collision,

682
00:36:29 --> 00:36:35
there is no kinetic energy left
in the center-of-mass frame.

683
00:36:32 --> 00:36:38
So we go to the
center-of-mass frame.

684
00:36:35 --> 00:36:41
So we first have to calculate
what u1 is.

685
00:36:38 --> 00:36:44
Well, u1 equals v1 minus
v center of mass.

686
00:36:45 --> 00:36:51
And we know what v center
of mass is-- it's right there.

687
00:36:50 --> 00:36:56
That's where it is.

688
00:36:52 --> 00:36:58
And if you do the subtraction,
which is by no means difficult,

689
00:36:57 --> 00:37:03
you will find m2 divided by m1
plus m2 times v1.

690
00:37:06 --> 00:37:12
And you checked that, I hope.

691
00:37:08 --> 00:37:14
And now we go to calculate u2.

692
00:37:12 --> 00:37:18
We want to know
what the velocity is

693
00:37:14 --> 00:37:20
of the center of mass
of the other one.

694
00:37:17 --> 00:37:23
That, of course, is
v2 minus v center of mass,

695
00:37:20 --> 00:37:26
but this was zero.

696
00:37:21 --> 00:37:27
This m1 divided
by m1 plus m2 times v1,

697
00:37:26 --> 00:37:32
so the difference is only
the m1 upstairs and the m2.

698
00:37:33 --> 00:37:39
Now we are going to calculate

699
00:37:37 --> 00:37:43
the kinetic energy
in the center-of-mass frame.

700
00:37:43 --> 00:37:49
Well, that equals
one-half of m1 times u1 squared

701
00:37:52 --> 00:37:58
plus one-half m2
times u2 squared.

702
00:37:56 --> 00:38:02
That's all we have
before the collision occurs.

703
00:38:01 --> 00:38:07

704
00:38:03 --> 00:38:09
Oh, by the way,
this is not a minus...

705
00:38:05 --> 00:38:11
This is a plus sign
and this is a minus sign.

706
00:38:09 --> 00:38:15
This one comes this way
and this one goes in that way.

707
00:38:12 --> 00:38:18
Now, I can...
I can calculate that for you.

708
00:38:14 --> 00:38:20
You know u1 and you know u2.

709
00:38:16 --> 00:38:22
If that's a plus
or a minus sign,

710
00:38:18 --> 00:38:24
it makes no difference
because they cancel anyhow.

711
00:38:20 --> 00:38:26
What are you going to find?

712
00:38:22 --> 00:38:28
One-half m1 m2 divided
by m1 plus m2 times v1 squared.

713
00:38:31 --> 00:38:37
And this is exactly the same
that we had there.

714
00:38:34 --> 00:38:40
And so what you see here--

715
00:38:36 --> 00:38:42
if you allow me

716
00:38:37 --> 00:38:43
for having skipped
some steps in the algebra;

717
00:38:40 --> 00:38:46
you will have to do
a little massaging

718
00:38:42 --> 00:38:48
to get from here to here--
you see here

719
00:38:44 --> 00:38:50
this is the kinetic energy
before the collision,

720
00:38:48 --> 00:38:54
and all that kinetic energy
is removed, went to heat.

721
00:38:51 --> 00:38:57
This is the maximum
you can ever lose,

722
00:38:54 --> 00:39:00
and this is what we call

723
00:38:56 --> 00:39:02
the internal kinetic energy
of the system.

724
00:38:59 --> 00:39:05

725
00:39:02 --> 00:39:08
And so going
to the center-of-mass system,

726
00:39:05 --> 00:39:11
you can always
immediately calculate

727
00:39:07 --> 00:39:13
what the maximum heat is that
you can expect from a collision.

728
00:39:11 --> 00:39:17
We can take a very special case,

729
00:39:13 --> 00:39:19
and we can take m2 going
to infinity.

730
00:39:16 --> 00:39:22
It's like having a piece
of putty I slam on the wall.

731
00:39:20 --> 00:39:26
It gets stuck,

732
00:39:22 --> 00:39:28
and what is the maximum heat
that you can produce

733
00:39:26 --> 00:39:32
that's all the energy there is?

734
00:39:28 --> 00:39:34
If m2 becomes infinitely high,
then m1 can be ignored.

735
00:39:34 --> 00:39:40
m2 cancels m2, you get
one-half m1 v1 squared.

736
00:39:41 --> 00:39:47
And that's obvious.

737
00:39:43 --> 00:39:49
That's completely trivial.

738
00:39:44 --> 00:39:50
I have a piece of putty,
I slam it against the wall.

739
00:39:47 --> 00:39:53
It has a certain amount
of kinetic energy.

740
00:39:49 --> 00:39:55
Whether you stay
in your reference frame

741
00:39:51 --> 00:39:57
or in the reference frame
of the center of mass,

742
00:39:53 --> 00:39:59
it's immediately obvious that
all the kinetic energy is lost.

743
00:39:56 --> 00:40:02
And that's exactly what you see
comes out of these equations

744
00:40:00 --> 00:40:06
whether you go
to the center of mass

745
00:40:02 --> 00:40:08
or whether you do it
from 26.100.

746
00:40:05 --> 00:40:11
I now would like to return
to the air track

747
00:40:09 --> 00:40:15
and do several completely
inelastic collisions with you.

748
00:40:14 --> 00:40:20
Again, we have to assume
that momentum is conserved.

749
00:40:18 --> 00:40:24
It never is completely,
but we can come close.

750
00:40:22 --> 00:40:28
And we will have
two cars that...

751
00:40:28 --> 00:40:34

752
00:40:30 --> 00:40:36
Now we're going
completely inelastic.

753
00:40:33 --> 00:40:39

754
00:40:38 --> 00:40:44
Completely inelastic.

755
00:40:41 --> 00:40:47

756
00:40:42 --> 00:40:48
So they hit each other
and they get stuck.

757
00:40:46 --> 00:40:52
I get a certain velocity,
which I put into the first car.

758
00:40:52 --> 00:40:58
It hits the second car
and it gets stuck.

759
00:40:56 --> 00:41:02
And I see there
what my v prime is.

760
00:41:00 --> 00:41:06
I see there on the blackboard.
that if the two are the same,

761
00:41:05 --> 00:41:11
I have a one here,
a one and a one here.

762
00:41:07 --> 00:41:13
If that's the ratio,

763
00:41:09 --> 00:41:15
so the outcome is that
v prime must be one-half v1.

764
00:41:13 --> 00:41:19
So this must be one-half v1.

765
00:41:18 --> 00:41:24
Now I have a mass which is
half the other one.

766
00:41:21 --> 00:41:27
I plow it into the other,
they get stuck together

767
00:41:25 --> 00:41:31
and now I get one divided by
one plus two-- I get one-third.

768
00:41:31 --> 00:41:37
Notice in both cases
I get a plus sign.

769
00:41:34 --> 00:41:40
That's, of course, obvious.

770
00:41:36 --> 00:41:42
If I plow into something
and they stick together,

771
00:41:39 --> 00:41:45
they continue
in the same direction.

772
00:41:42 --> 00:41:48
And so now I will do the timing

773
00:41:44 --> 00:41:50
in exactly the same way
that I did before,

774
00:41:47 --> 00:41:53
except that now I have a
completely inelastic collision.

775
00:41:52 --> 00:41:58
I will have a timing t1 and
I will have a timing t prime.

776
00:41:58 --> 00:42:04
The cars have
a slightly different mass:

777
00:42:02 --> 00:42:08
237 plus or minus one gram

778
00:42:10 --> 00:42:16
and I have one that is
474 plus or minus one gram--

779
00:42:16 --> 00:42:22
not too different from this.

780
00:42:20 --> 00:42:26
I have two of these cars
and I have one of these.

781
00:42:24 --> 00:42:30
And first I'm going to slam
these two onto each other,

782
00:42:30 --> 00:42:36
and so when they collide,
I expect the speed to be half.

783
00:42:35 --> 00:42:41
So I get
a certain amount of t1--

784
00:42:39 --> 00:42:45
so t prime will be twice as long

785
00:42:42 --> 00:42:48
because the speed goes down
by a factor of two--

786
00:42:46 --> 00:42:52
so when I multiply
this number by one-half,

787
00:42:49 --> 00:42:55
I would like to get
that number back.

788
00:42:53 --> 00:42:59
Here I get the time
for this car.

789
00:42:56 --> 00:43:02
The first one to come in
gives me a certain time.

790
00:43:00 --> 00:43:06
When they continue together,
the speed is three times lower

791
00:43:04 --> 00:43:10
so this time will be
three times higher.

792
00:43:06 --> 00:43:12
So I multiply this by one-third

793
00:43:08 --> 00:43:14
and I would like to get
that time back.

794
00:43:11 --> 00:43:17
And those are

795
00:43:12 --> 00:43:18
the two experiments
that I would like to do now.

796
00:43:14 --> 00:43:20
So we have to get
the noise back on.

797
00:43:17 --> 00:43:23
Air track one.

798
00:43:18 --> 00:43:24
(air track humming )

799
00:43:21 --> 00:43:27
Air track two.

800
00:43:23 --> 00:43:29

801
00:43:28 --> 00:43:34
Here we have the two cars.

802
00:43:30 --> 00:43:36
They have Velcro, so when they
hit each other, they get stuck.

803
00:43:35 --> 00:43:41
Sometimes they bounce, actually.

804
00:43:37 --> 00:43:43
You have to do
the experiment again.

805
00:43:39 --> 00:43:45
This one goes here;
this one goes here.

806
00:43:43 --> 00:43:49
Are these two timers on?

807
00:43:45 --> 00:43:51
We don't need
this timer anymore.

808
00:43:47 --> 00:43:53
Are they both on?

809
00:43:49 --> 00:43:55
Zero them.

810
00:43:51 --> 00:43:57

811
00:43:52 --> 00:43:58
Zero them.

812
00:43:53 --> 00:43:59
Are they zero?

813
00:43:55 --> 00:44:01
Okay, are you ready?

814
00:43:56 --> 00:44:02
This one is going to plow
into this one,

815
00:43:59 --> 00:44:05
they merge and they continue,

816
00:44:01 --> 00:44:07
and then you'll see
the time t prime here.

817
00:44:06 --> 00:44:12
There we go.

818
00:44:08 --> 00:44:14
Ready...

819
00:44:09 --> 00:44:15

820
00:44:12 --> 00:44:18
All right.

821
00:44:14 --> 00:44:20
What do we see?

822
00:44:16 --> 00:44:22

823
00:44:18 --> 00:44:24
138, 288.

824
00:44:22 --> 00:44:28
138... 288...

825
00:44:26 --> 00:44:32
138...

826
00:44:31 --> 00:44:37
288, divide this by 2 is 0.144.

827
00:44:38 --> 00:44:44
That is...
the difference is only six,

828
00:44:41 --> 00:44:47
and that is only
four percent off.

829
00:44:44 --> 00:44:50
That's completely
within my prediction

830
00:44:47 --> 00:44:53
that each time could be off
by 2½ percent.

831
00:44:50 --> 00:44:56
Now I will have a car
which is twice the mass

832
00:44:57 --> 00:45:03
and this is once the mass.

833
00:45:01 --> 00:45:07
I will set them again to zero

834
00:45:04 --> 00:45:10
and now we get one
onto double the mass.

835
00:45:08 --> 00:45:14
Ready?

836
00:45:10 --> 00:45:16
There we go.

837
00:45:11 --> 00:45:17

838
00:45:12 --> 00:45:18
Oh!

839
00:45:13 --> 00:45:19
You see? They bounced.

840
00:45:15 --> 00:45:21
They didn't stick,
so we have to start all over.

841
00:45:20 --> 00:45:26
If they don't stick,
of course, then it's...

842
00:45:23 --> 00:45:29
it's not a completely
inelastic collision.

843
00:45:25 --> 00:45:31
There we go.

844
00:45:27 --> 00:45:33
They made a slight bounce,
which I didn't like.

845
00:45:30 --> 00:45:36
Let's see what we get.

846
00:45:32 --> 00:45:38
We can always do it again.

847
00:45:34 --> 00:45:40
168, 545.

848
00:45:38 --> 00:45:44

849
00:45:40 --> 00:45:46
168...

850
00:45:42 --> 00:45:48

851
00:45:48 --> 00:45:54
Multiply this by three...

852
00:45:51 --> 00:45:57

853
00:45:54 --> 00:46:00
504, 545-- eight-percent
difference is a little high.

854
00:46:00 --> 00:46:06
But these things can happen.

855
00:46:02 --> 00:46:08
I told you they bounced,
and I didn't like that,

856
00:46:05 --> 00:46:11
and then they got stuck again.

857
00:46:07 --> 00:46:13
I will do it once more.

858
00:46:09 --> 00:46:15
Maybe I'm a little bit luckier.

859
00:46:11 --> 00:46:17

860
00:46:13 --> 00:46:19
Again, they bounced a little
before they finally merged.

861
00:46:17 --> 00:46:23

862
00:46:21 --> 00:46:27
603, 187.

863
00:46:23 --> 00:46:29

864
00:46:30 --> 00:46:36
187... 603.

865
00:46:34 --> 00:46:40
If I divide this
by three, I get 0.201,

866
00:46:38 --> 00:46:44
and that is off,
again, by about 13 units,

867
00:46:40 --> 00:46:46
about seven-percent difference,
so it's a little bit more

868
00:46:43 --> 00:46:49
than the five percent
that I allowed for.

869
00:46:48 --> 00:46:54
Now, to make it worse
for you tonight,

870
00:46:52 --> 00:46:58
I would like you to think
about something else,

871
00:46:56 --> 00:47:02
not only about the tennis ball
against the wall--

872
00:47:00 --> 00:47:06
whereby the wall has momentum
but no kinetic energy--

873
00:47:03 --> 00:47:09
but now I would like you to
think about this amazing thing.

874
00:47:07 --> 00:47:13
I have eight
billiard balls here,

875
00:47:09 --> 00:47:15
and your Newton cradles
probably also have

876
00:47:12 --> 00:47:18
eight balls hanging
from pendulums.

877
00:47:15 --> 00:47:21
This is easier to demonstrate
in 26.100.

878
00:47:18 --> 00:47:24
Now I let two balls bang
onto the other six.

879
00:47:23 --> 00:47:29
And I know that you predict
what's going to happen

880
00:47:25 --> 00:47:31
and your prediction is correct.

881
00:47:27 --> 00:47:33
If you slam two balls
onto the six

882
00:47:29 --> 00:47:35
nature is going
to calculate like mad

883
00:47:32 --> 00:47:38
to conserve momentum,
to conserve kinetic energy

884
00:47:36 --> 00:47:42
closely enough to
a completely elastic collision,

885
00:47:39 --> 00:47:45
and out comes that the only way
that nature can do it

886
00:47:42 --> 00:47:48
is the following.

887
00:47:44 --> 00:47:50
They all stand still
and these two take off.

888
00:47:47 --> 00:47:53
Look again.

889
00:47:49 --> 00:47:55
They all stand still
and these two take off.

890
00:47:52 --> 00:47:58
That's not an easy calculation.

891
00:47:53 --> 00:47:59
But now look at the following.

892
00:47:56 --> 00:48:02
Hold it-- what happens

893
00:47:57 --> 00:48:03
if I take three and I bang three
on the other five?

894
00:48:01 --> 00:48:07
What do you predict
is going to happen?

895
00:48:04 --> 00:48:10
How many will take off?

896
00:48:06 --> 00:48:12
Three, you think.

897
00:48:07 --> 00:48:13
You sure?

898
00:48:08 --> 00:48:14

899
00:48:09 --> 00:48:15
You were right.

900
00:48:11 --> 00:48:17
Okay.

901
00:48:12 --> 00:48:18
Now, five.

902
00:48:16 --> 00:48:22
Five on three.

903
00:48:18 --> 00:48:24
This is a tough problem
even for nature.

904
00:48:22 --> 00:48:28
What do you think will happen?

905
00:48:25 --> 00:48:31
(students respond )

906
00:48:27 --> 00:48:33
LEWIN: Five will go again and
three will stay-- you're good.

907
00:48:31 --> 00:48:37

908
00:48:33 --> 00:48:39
If any one of you can show me
analytically

909
00:48:35 --> 00:48:41
that this is what has to happen,

910
00:48:37 --> 00:48:43
I would love to see
those results.

911
00:48:39 --> 00:48:45
Okay, see you Wednesday.

912
00:48:41 --> 00:48:47

913
00:48:51 --> 00:48:57.000