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- [Instructor] What does momentum mean?

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The definition of momentum is
the mass times the velocity.

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So the formula is simple,
it's just m times v.

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And why do we care about momentum?

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We care about momentum
because if there's no net

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external force on a system,

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the momentum of that
system will be conserved.

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In other words, the total
initial momentum of that system

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would equal the total final
momentum of that system.

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So momentum will be conserved

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if there's no net external force.

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And momentum is a vector,
that means it has components.

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The total momentum will
point in the direction

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of the total velocity, and
the momentum in each direction

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can be conserved independently.

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In other words, if there's no
net force in the y direction,

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then the momentum in the y
direction will be conserved,

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and if there's no net
force in the x direction,

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the momentum in the x
direction will be conserved.

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Since the momentum is m
times v, the units are

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kilograms times meters per second.

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So what's an example problem
involving momentum look like?

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Let's say two blocks of mass 3M and M

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head toward each other, sliding
over a frictionless surface

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with speeds 2V and 5V, respectively,

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and after the collision
they stick together.

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Which direction will the two masses head

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after the collision?

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To figure this out, we can
just ask what direction is the

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total momentum of the system initially.

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Since momentum's gonna be
conserved, that'll have to be the

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direction of the momentum finally.

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So the momentum of the 3M mass

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is going to be the mass, which is 3M,

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times the velocity, which is 2v,

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so we get a momentum of 6Mv.

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And the momentum of the mass
M is gonna be the mass M,

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times the velocity, which is negative 5v.

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Momentum is a vector, so you can't

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forget the negative signs.

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Which gives a momentum of negative 5Mv.

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So the total initial momentum
of the system would be

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6Mv plus -5Mv, which is one
Mv, and that's positive,

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which means the total momentum,
initially, is to the right.

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That means, after the collision,

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the total momentum will also
have to be to the right,

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and the only way that could be the case

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if these two masses joined
together, is for the total

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combined mass to also move to the right.

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What does impulse mean?

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The impulse is the amount of
force exerted on an object

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or system multiplied by
the time during which

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that force was acting.

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So in equation form, that means
J, the impulse, is equal to

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the force multiplied by how
long that force was acting.

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And the net impulse is gonna
be equal to the net force times

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the time during which
that net force was acting.

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And this is also going
to be equal to the change

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in momentum of that system or object.

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In other words, if a mass
had some initial momentum

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and ends with some final momentum,

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the change in momentum of that mass,

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p final minus p initial, is
gonna equal the net impulse,

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and that net impulse is
gonna equal the net force

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on that object multiplied by

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the time during which
that force was acting.

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And since impulse is a change in momentum,

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and momentum is a vector,
that means impulse is also

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a vector, so it can be
positive and negative.

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And the units are the same as momentum,

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which is kilograms
times meters per second.

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Or, since it's also force
times time, you could write

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the units as Newtons times seconds.

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So what's an example problem
involving impulse look like?

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Let's say a bouncy ball
of mass M is initially

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moving to the right with a speed 2v.

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And it recoils off a wall with a speed v.

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We want to know, what's the
magnitude of the impulse

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on the ball from the wall?

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So the impulse, J, is going to be equal

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to the change in momentum.

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The change in momentum is
p final minus p initial,

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so the final momentum is gonna
be the mass times the final

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velocity, but this
velocity's heading leftwards,

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so you can't forget the negative sign,

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minus the initial momentum,
which would be M times 2v,

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which gives a net impulse of -3Mv.

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This makes sense.

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The net impulse has to point in the

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same direction as the net force.

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This wall exerted a force to the left,

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that means the impulse also points left,

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and has a magnitude of 3Mv.

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If you get a force versus time graph,

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the first thing you
should think about is that

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the area under that
graph is going to equal

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the impulse on the object.

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So if you graph the force on
some object as a function of

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time, the area under that
curve is equal to the impulse.

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Just be careful, since
area above this time axis

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is going to count as positive
impulse, and area underneath

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the time axis would count
as negative impulse,

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since those forces would be negative.

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Why do we care that the
area's equal to the impulse?

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Well, if we can find the area,
that would equal the impulse,

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and if that's the net
impulse on an object,

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that would also equal the change
in momentum of that object.

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Which means we could figure out

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the change in velocity of an object.

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So what's an example
problem involving impulse

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as the area under a graph look like?

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Let's say a toy rocket
of mass two kilograms

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was initially heading to the right

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with a speed of 10 meters per second,

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and a force in the horizontal direction

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is exerted on the rocket,
as shown in this graph,

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and we want to know, what's
the velocity of the rocket

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at the time t equals 10 seconds?

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To figure that out, we'll figure out

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the area under the curve.

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This triangle would
count as positive area.

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This triangle would
count as negative area.

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And since this triangle
is just as positive

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as this triangle is negative,
these areas cancel completely.

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And the only area we'd have
to worry about is the area

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between eight seconds and 10 seconds.

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This is going to end up
being a negative area,

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since the height of the rectangle is -30,

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and the width of the rectangle
is going to be two seconds.

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This gives an impulse
of -60 Newton seconds.

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So if the impulse on this
object is -60 Newton seconds,

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that's going to equal the change
in momentum of that object.

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How much momentum did
this object start with?

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The initial momentum of this object

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is going to be two kilograms
times the initial velocity,

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which was 10 meters per
second to the right,

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which is positive 20
kilogram meters per second.

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So if the initial momentum
of the rocket is positive 20

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and there was a change in momentum of -60,

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the final momentum just has to be -40.

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Or in other words, since the
change in momentum would have

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to be the final momentum
minus the initial momentum,

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which was positive 20, we could
find the final momentum by

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adding 20 to both sides,
which would give us -60

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plus 20, which is -40.

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What's the difference between an elastic

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and an inelastic collision?

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What we mean by an
elastic collision is that

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the total kinetic energy of that system

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is conserved during the collision.

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In other words, if a
sphere and a cube collide,

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for that collision to be elastic,

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the total kinetic energy
of the sphere plus

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the kinetic energy of the
cube before the collision

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would have to equal the
kinetic energy of the sphere

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plus the kinetic energy of
the cube after the collision.

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If the total kinetic
energy before the collision

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is equal to total kinetic
energy after the collision,

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then that collision is elastic.

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It's not enough for the system
to just bounce of each other.

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If two objects bounce,
the total kinetic energy

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might not be conserved.

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Only when the total
kinetic energy is conserved

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can you say the collision is elastic.

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For an inelastic collision,
the kinetic energy is not

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conserved during the collision.

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In other words, the total
initial kinetic energy

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of the sphere plus cube would not equal

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the total final kinetic energy
of the sphere plus cube.

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Where does this kinetic energy go?

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Typically, in an inelastic collision,

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some of that kinetic energy is transformed

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into thermal energy during the collision.

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While masses could bounce
during an inelastic collision,

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if they stick together, the
collision is typically called

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a perfectly inelastic collision,

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since in this collision
you'll transform the most

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kinetic energy into thermal energy.

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And when two objects stick
together, it's a surefire sign

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that that collision is
definitely inelastic.

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So what's an example problem
that involves elastic

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and inelastic collisions look like?

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Let's say two blocks of mass
2M and M head toward each other

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with speeds 4v and 6v, respectively.

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After they collide,
the 2M mass is at rest,

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and the mass M has a
velocity of 2v to the right.

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And we want to know,

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was this collision elastic or inelastic?

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Now you might want to say that,

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since these objects
bounced off of each other,

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the collision has to be
elastic, but that's not true.

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If the collision is elastic,
then the objects must bounce,

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but just because the objects bounce

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does not mean the collision is elastic.

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In other words, bouncing
is a necessary condition

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for the collision to be elastic,
but it isn't sufficient.

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If you really want to know
whether a collision was elastic,

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you have to determine whether
the total kinetic energy

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was conserved or not.

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And we can figure that
out for this collision

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without even calculating anything.

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Since the speed of the 2M mass decreased,

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the kinetic energy of
the 2M mass went down.

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And since the speed of
the M mass also decreased

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after the collision,

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the kinetic energy of the
mass M went down, as well.

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So if the kinetic energy
of both masses go down,

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then the final kinetic
energy after the collision

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has to be less than the
initial kinetic energy.

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Which means kinetic
energy was not conserved,

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and this collision had to be inelastic.

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One final note, even though kinetic energy

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wasn't conserved during this process,

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the momentum was conserved.

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The momentum will be conserved

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for both elastic and inelastic collisions.

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It's just kinetic energy
that's not conserved

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for an inelastic collision.

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How do you deal with
collisions in two dimensions?

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Well the momentum will be
conserved for each direction

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in which there's no net impulse.

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If there's no net impulse
in both directions,

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then the momentum in both directions

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will be conserved independently.

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In other words, if there's no
net force in the x direction,

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the total x momentum has to be constant,

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and if there's no net
force in the y direction,

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the total momentum in the y
direction has to be constant.

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So in other words, if two spheres collide

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in a glancing collision,
the total momentum

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in the x direction initially should equal

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the total momentum in
the x direction finally,

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if there's no net impulse
in that x direction.

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And the total momentum in
the y direction initially,

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of which there is none in this case,

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would have to equal the total
momentum in the y direction

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finally, if there's no net
impulse in the y direction.

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So what's an example involving collisions

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in two dimensions look like?

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Let's say a metal sphere
of mass M is traveling

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horizontally with five meters per second

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when it collides with an
identical sphere of mass M

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that was at rest.

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After the collision,
the original sphere has

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velocity components of
four meters per second and

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three meters per second
in the x and y directions.

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And we want to know, what
are the velocity components

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of the other sphere right
after the collision?

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00:09:41,716 --> 00:09:43,046
So assuming there were no net forces

250
00:09:43,046 --> 00:09:45,249
in the x or y direction in this case,

251
00:09:45,249 --> 00:09:47,487
then the momentum will be
conserved for each direction,

252
00:09:47,487 --> 00:09:49,824
and since the mass of
each sphere's the same,

253
00:09:49,824 --> 00:09:52,083
we can simply look at
the velocity components.

254
00:09:52,083 --> 00:09:53,914
So if we started with
five units of momentum

255
00:09:53,914 --> 00:09:55,724
in the x direction, we have to end with

256
00:09:55,724 --> 00:09:57,680
five units of momentum in the x direction.

257
00:09:57,680 --> 00:09:59,726
So the x component of the second sphere

258
00:09:59,726 --> 00:10:01,661
has to be one meter per second.

259
00:10:01,661 --> 00:10:03,323
And since we started
with no momentum in the

260
00:10:03,323 --> 00:10:04,641
vertical direction initially,

261
00:10:04,641 --> 00:10:06,741
we have to end with no
momentum vertically.

262
00:10:06,741 --> 00:10:08,738
So if the first sphere has
three units of momentum

263
00:10:08,738 --> 00:10:11,042
vertically after the collision,
then the second sphere

264
00:10:11,042 --> 00:10:13,801
has to have three units of
momentum vertically downward

265
00:10:13,801 --> 00:10:16,694
after the collision, which
gives us an answer of D.

266
00:10:16,694 --> 00:10:18,426
What's the center of mass mean?

267
00:10:18,426 --> 00:10:21,019
The center of mass of an
object or a system is the point

268
00:10:21,019 --> 00:10:23,360
where that object or system would balance.

269
00:10:23,360 --> 00:10:24,917
And the center of mass is also the point

270
00:10:24,917 --> 00:10:28,159
where you can treat the entire
force of gravity as acting.

271
00:10:28,159 --> 00:10:29,599
The way you can solve
for the center of mass

272
00:10:29,599 --> 00:10:31,011
is by using this formula.

273
00:10:31,011 --> 00:10:33,720
You multiply each mass
by how far that mass is

274
00:10:33,720 --> 00:10:35,252
from the reference point.

275
00:10:35,252 --> 00:10:36,894
If there's no reference point specified,

276
00:10:36,894 --> 00:10:39,312
you get to choose the
arbitrary reference point,

277
00:10:39,312 --> 00:10:41,813
which would designate where x equals zero.

278
00:10:41,813 --> 00:10:44,838
You continue adding each
mass times its position.

279
00:10:44,838 --> 00:10:47,065
For positions to the left
of the reference point,

280
00:10:47,065 --> 00:10:49,295
those would count as negative positions.

281
00:10:49,295 --> 00:10:50,872
And when you're done
accounting for every mass

282
00:10:50,872 --> 00:10:53,580
in your system, you
divide by the total mass,

283
00:10:53,580 --> 00:10:55,916
which would be all the masses
added up, and the number

284
00:10:55,916 --> 00:10:59,312
you get would be the position
of the center of mass.

285
00:10:59,312 --> 00:11:01,168
The center of mass is going
to have units of meters,

286
00:11:01,168 --> 00:11:02,710
since it's a location.

287
00:11:02,710 --> 00:11:05,172
The location where the system
or object would balance,

288
00:11:05,172 --> 00:11:06,513
and the location where you can treat

289
00:11:06,513 --> 00:11:08,750
the entire force of gravity as acting.

290
00:11:08,750 --> 00:11:10,721
Something else that's
extremely important to remember

291
00:11:10,721 --> 00:11:13,942
is that the center of mass of
a system will not accelerate

292
00:11:13,942 --> 00:11:16,858
unless there's an external
force on that system.

293
00:11:16,858 --> 00:11:19,126
In other words, the
center of mass of a system

294
00:11:19,126 --> 00:11:20,810
follows Newton's first law.

295
00:11:20,810 --> 00:11:23,233
If the center of mass
of a system is at rest,

296
00:11:23,233 --> 00:11:25,082
then even if the masses in that system

297
00:11:25,082 --> 00:11:27,559
exert forces on each
other and move around,

298
00:11:27,559 --> 00:11:30,030
the center of mass of
that system will stay put

299
00:11:30,030 --> 00:11:33,037
until there's a net external
force on the system.

300
00:11:33,037 --> 00:11:34,419
And if the center of mass was initially

301
00:11:34,419 --> 00:11:36,322
moving to the right at some speed,

302
00:11:36,322 --> 00:11:37,590
that center of mass will continue

303
00:11:37,590 --> 00:11:39,266
moving to the right at that speed,

304
00:11:39,266 --> 00:11:41,931
even if the masses are moving
in different directions,

305
00:11:41,931 --> 00:11:44,617
until there's a net force on that system.

306
00:11:44,617 --> 00:11:45,961
So what's an example problem involving

307
00:11:45,961 --> 00:11:47,540
center of mass look like?

308
00:11:47,540 --> 00:11:50,001
Lets say a remote control car of mass M

309
00:11:50,001 --> 00:11:53,130
is sitting at rest on a
wooden plank, also of mass M,

310
00:11:53,130 --> 00:11:54,925
in the position seen here.

311
00:11:54,925 --> 00:11:56,994
There is friction between
the wheels of the car

312
00:11:56,994 --> 00:11:59,392
and the plank, but there's
no friction between the plank

313
00:11:59,392 --> 00:12:01,930
and the ice upon which
the plank is sitting.

314
00:12:01,930 --> 00:12:04,288
The remote control car
is turned on and off.

315
00:12:04,288 --> 00:12:06,218
What would be a possible final position

316
00:12:06,218 --> 00:12:07,908
of the car and the plank?

317
00:12:07,908 --> 00:12:10,913
Now because the car's at rest
and the plank is at rest,

318
00:12:10,913 --> 00:12:12,838
that means the center
of mass of this system

319
00:12:12,838 --> 00:12:14,444
is also at rest.

320
00:12:14,444 --> 00:12:17,066
And since there's no net
force on this system,

321
00:12:17,066 --> 00:12:19,616
the center of mass is going
to have to remain at rest.

322
00:12:19,616 --> 00:12:21,029
Where is the center of mass?

323
00:12:21,029 --> 00:12:22,856
Well the car's mass is at three,

324
00:12:22,856 --> 00:12:25,105
the plank's center is at five,

325
00:12:25,105 --> 00:12:27,567
so the center of mass
between the car and the plank

326
00:12:27,567 --> 00:12:29,753
would be at the location of four.

327
00:12:29,753 --> 00:12:31,095
So to find the correct solution,

328
00:12:31,095 --> 00:12:32,746
we just need to figure
out which one of these

329
00:12:32,746 --> 00:12:35,116
also has the center of mass at four.

330
00:12:35,116 --> 00:12:37,040
Option A has the car at three

331
00:12:37,040 --> 00:12:38,964
and the center of the plank at three.

332
00:12:38,964 --> 00:12:41,315
That'd put the center
of mass at three meters,

333
00:12:41,315 --> 00:12:43,690
but that can't be right, the
center of mass can't move,

334
00:12:43,690 --> 00:12:46,449
there were no external
forces on our system,

335
00:12:46,449 --> 00:12:48,013
and the center of mass started at rest,

336
00:12:48,013 --> 00:12:49,628
so it's got to remain at rest.

337
00:12:49,628 --> 00:12:52,232
For option B, the center
of the car is at four,

338
00:12:52,232 --> 00:12:54,070
the center of the plank is at three,

339
00:12:54,070 --> 00:12:56,169
this would put the center
of mass somewhere between

340
00:12:56,169 --> 00:12:58,380
three and four, but again,
that can't be right.

341
00:12:58,380 --> 00:13:01,263
We need our center of mass
to be at the location four.

342
00:13:01,263 --> 00:13:03,392
Option C has the car at six

343
00:13:03,392 --> 00:13:04,916
and the center of the plank at four.

344
00:13:04,916 --> 00:13:07,536
This would put the center of
mass of the system at five,

345
00:13:07,536 --> 00:13:08,369
that can't be right.

346
00:13:08,369 --> 00:13:09,906
We need our center of mass at four.

347
00:13:09,906 --> 00:13:11,822
Option D has the car at five,

348
00:13:11,822 --> 00:13:13,612
and the center of the plank at three.

349
00:13:13,612 --> 00:13:15,824
That puts the center of
mass at location four,

350
00:13:15,824 --> 00:13:17,078
just like it was before.

351
00:13:17,078 --> 00:00:00,000
So D is a possible solution.