English subtitles for clip: File:Prof Walter Lewin Elastic Collisions.ogv

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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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[…]

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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.