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MARKUS KLUTE: Welcome back
to 8.20, Special Relativity.

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In this section we're
going to talk a bit more

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about collisions.

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I've already seen collisions in
study of momentum conservation

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in previous sections.

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So here we can have a collision.

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Then we can describe them
in the center of mass frame,

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for example, where the total
momentum is equal to 0.

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So in the case of the
collision of two particles,

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the momentum of particle
one plus the momentum

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of particle two is equal to 0.

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We can then describe the
energy and the momentum

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of the particles before
and after the collision.

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In the lab frame, the
situation is different.

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Here typically we
have one particle

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with some momentum hitting
another particle, which

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is at rest.

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But we can also have
different types of collisions.

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We describe or characterize
elastic collisions

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where the kinetic energy is
conserved and so is the mass.

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So you can think about two
billiard balls colliding

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without any friction,
in which case

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they don't change their
appearance, their mass.

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Everything is unchanged, so
you must change the direction.

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The total kinetic energy
in these collisions

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are typically conserved.

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But we can also have
inelastic collisions.

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And there's two different kinds.

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There's sticky kinds, where
the mass after the collision

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is greater.

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So you have two
particles, for example,

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maybe they stick together--
they're some, like,

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Play Dough balls--

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and the kinetic energy after
the collision is smaller.

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Or you can have explosive
collisions, where

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the mass afterwards is smaller.

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Maybe you start from
one heavy, big object

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and then which explodes
into many smaller ones.

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But the kinetic energy after
the collisions is much smaller.

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Those are also collisions.

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So here we want
to do an activity

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and study an
inelastic collision.

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So before we have two particles
there, or billiard balls.

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They're exactly the same
and have a velocity u.

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And after the collision their
mass is capital M, big mass.

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And you're going to
describe this collision once

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in the center of mass frame and
one in the laboratory frame.

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And so the question
now is, are the masses

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and is energy conserved
in those collisions?

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And you're going
to just described

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this in both reference forms.

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So again, stop the video here
and try to work this out.

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I already did this,
so I discussed

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before, in those
collision problems

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it's always important
to be really clear.

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The situation
before the collision

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was A. The situation
after the collision was B.

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So I'm describing this here.

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First in the center of
mass frame where the x--

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and I'm just talking
about x component here--

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the x momentum is 0, which
is equal to the mass times u

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times gamma minus the
mass times u terms gamma.

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That's the 0.

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The energy before is 2 times
the mass times gamma times

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c squared.

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After the collision,
the particle is at rest.

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The new one
particular is at rest

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and has an energy,
large M over--

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times c squared.

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In the laboratory frame
situations, different case.

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X momentum 0 minus
m times u prime--

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this is a different velocity--

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times gamma of u prime.

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So here I'm trying to indicate
that this gamma is not

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the same gamma as over here.

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This is a gamma, but it's
the velocity of u prime.

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And the energy is the rest
mass of the particle addressed

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plus the mass times
gamma times c squared off

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the second particle.

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After the collision, the
particle has some velocity u.

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And so the momentum
in x direction

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is minus large M times u
times gamma of u again.

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And the energy is large M
times gamma u times c squared.

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OK, good.

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So now we can use
momentum conservation

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and find this equation here.

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And from which we
can then calculate

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that the large mass is equal
to 2 times the smaller mass.

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So what you find, and this
is the relativistic math,

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you find that at the conclusion
that the rest mass is not

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conserved.

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The mass of this big
ball is not simply

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the mass of the two
rest masses, or 2 times

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the mass of the rest mass.

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You have to consider
this gamma factor here.

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It's 2 times the relativistic
math, if you want.

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But you also find
that the total energy

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is conserved in circulation so
that the sum of m0 gamma times

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c squared is conserved
in the collision,

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irrespectively in how
you actually reference it

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when you discuss the problem.

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I want to close this
part of collisions

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with a small
discussion of units.

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And that will become
interesting and important later

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on when we look at
particle physics examples.

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So in particle
physics, we often talk

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about units of electronvolt
in collision experiments,

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or mega electronvolts,
or kilo electronvolts,

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tera electronvolts.

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So 1 electronvolt is
the kinetic energy

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of the particle with charge
e, which is accelerated

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in a potential of 1 volts.

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So that corresponds--
that's a unit of energy

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and it corresponds
to 1.6 times 10

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to the minus 19
joules or 1.6 times 10

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to the minus 90 kilograms
meter squared over 2nd square.

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But the mass of an electron
is really, really small.

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And those units
here are introduced

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because the mass
is small and you

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want to have reasonable
numbers to work with.

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So the mass of the
electron is 9.11 times

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10 to the minus 31 kilogram.

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So if you just rewrite an m0 as
equal to m0 c squared times 1

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over c squared you
find that, huh,

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now we rewrite this and find
that the masses 8 times 10

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to the minus 14
joules over c squared.

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Or in units of
electronvolts, 5 times 10

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to the 5 electronvolts
over c squared,

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which is 0.511 mega
electronvolts over c squared

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or 511 kilo electrons
over c squared.

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So when we talk about
the mass of an electron,

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we sometimes approach this
with natural units, in which c

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squared is equal to 1.

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And that just simply says
that the mass of an electron

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is 511 kilo electronvolts.

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The math of a neuron
is mega electronvolts,

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and so on, and so on.