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Welcome back.

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We will now do a momentum
problem in two dimensions.

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So let's see what
we have here.

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So we have this ball A and we
could maybe even think of it

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as maybe what's going on on the
surface of a pool table.

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We have ball A and it's moving
with its 10 kilograms. So

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these numbers are the
mass of the balls.

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This is a 10 kilogram ball and
it's moving to the right at 3

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meters per second.

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And then it hits this ball, B,
which is a 5 kilogram ball.

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And then we know that ball A,
ball A kind of ricochets off

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of ball B and gets set onto
this new trajectory.

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Now, instead of going due right,
it's going at a 30

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degree angle to, I guess we
could say, horizontal.

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It's going in a 30 degree angle
at 2 meters per second.

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And the question is, what is
the velocity of ball B?

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So velocity is both magnitude
and direction.

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So we need to figure
out essentially,

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what is ball B doing?

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Ball B is going to be going--
We can just think about it.

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If you ever played pool, we
could guess that ball B is

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going to go roughly
in that direction.

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But we need to figure out
exactly what the angle is and

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exactly what its velocity is.

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So let's do this problem.

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So at first you're saying, oh,
Sal, this looks confusing.

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You know, I know momentum
should be

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conserved and all that.

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But now we have these vectors
and there's two dimensions and

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how do I do that?

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And the key here is that there's
just really one more

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step when you're working on it
in two dimensions or really

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three dimensions or an
arbitrary number of

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dimensions.

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When we did one dimension, you
made sure that momentum was

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conserved in that
one dimension.

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So when you do two dimensions,
what you do is you figure out

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the initial momentum in each
of the dimensions.

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So you break it up into the
x and y components.

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And then you say the final
momentum of both objects are

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going to equal the initial x
momentum and are going to

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equal the initial y momentum.

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So let's figure out the
initial x momentum.

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So P for momentum.

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Because the m is for mass.

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So let's say the initial
momentum in the x direction--

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And we don't have to write
initial or final because

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really, the total momentum in
the x direction is always

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going to be the same.

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So let's say what the initial--
Actually, let me

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write initial just so it hits
the point home that initial

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and final don't change.

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So the initial momentum
in the x direction.

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So i for initial, x-- I should
do something better than keep

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writing these subscripts
--is equal to what?

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Well ball B has no
initial velocity,

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so it has no momentum.

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Ball A is 10 kilograms.

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And what is its velocity
in the x direction?

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Well all of its velocity
is in the x direction.

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So it's 3.

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I mean, this is only moving
in the x direction.

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So the momentum in the x
direction is 30 kilogram

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meters per second.

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Mass times velocity, kilogram
meters per second.

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And what's the initial momentum
in the y direction?

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Well B isn't moving at
all, so it has no

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momentum in any direction.

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And A, all of A's movement
is in the x direction.

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It's not moving at an angle
or up at all, so it has no

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momentum in the y direction.

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So we immediately know that
after the collision, the

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combined momentum of both of
these balls in the x direction

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has to be 30, and the combined
momentum of both of these

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balls in the y direction
has to be 0.

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So let's figure out what A's
momentum in the x and y

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directions are.

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So to figure out what A's
momentum is, we just have to

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figure out what A's velocity in
the x and y directions are

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and then multiply that
times the mass.

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Because mass doesn't
have any direction.

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So let's figure out the x and y
components of this velocity.

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Let's do the x component
of the vector first.

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So the x is just this vector.

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Change colors to keep
things interesting.

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The y is this vector.

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That is the y component.

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And so, what are those?

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And this hopefully, is going to
be almost second nature to

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you if you've been watching
all of the other videos on

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Newton's laws.

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This is just our trigonometry
and we can write out our

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SOH-CAH-TOA again.

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And I reassure you, this is
the hardest part of any of

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these multi-dimensional trig
problems-- Multi-dimensional

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physics problems, which really
are just trig problems.

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So if we want to figure out
the x component, so the

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velocity of A in the
x direction,

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what is it equal to?

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Well this is adjacent
to the angle.

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We know the hypotenuse.

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So we know VA sub x or the
velocity of A in the x

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direction over the hypotenuse,
over 2 meters per second, is

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equal to what?

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Adjacent over hypotenuse.

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Cosine.

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Is equal to cosine
of 30 degrees.

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Or the velocity of A in the x
direction is equal to 2 cosine

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of 30 degrees.

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What's cosine of 30 degrees?

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It's square root of 3 over 2.

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This is square root
of 3 over 2.

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And square root of 3 over
2 times 2 is equal to

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square root of 3.

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So this is equal to the square
root of 3 meters per second.

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And what is the velocity of
A in the y direction?

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Well hopefully, this second
nature to you as well.

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But since opposite over
hypotenuse is equal to the

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sine of 30.

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So VA in the y direction is
equal to 2 times the sine of

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30 degrees.

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The sine of 30 degrees is 1/2.

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So this is 1/2.

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1/2 times 2 is equal to
1 meter per second.

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So after the collision,
A is moving at 1

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meter per second up.

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One meter per second in
the upwards direction.

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And it's moving at square root
of 3 meters per second in the

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rightwards direction.

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So what is going to be A's
momentum in each of the

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directions?

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Well, we figured out its
velocity, so we just multiply

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each of the velocities
times the mass.

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So A has a mass of 10 kilograms.
And this is going

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to be the final momentum.

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Momentum of A in the x direction
is going to equal

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square root of 3 times 10.

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Square root of 3 is the
velocity, 10 is the mass.

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So it's 10 square roots of 3
kilogram meters per second.

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And the momentum of A in the y
direction is going to be-- and

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since it's going up, we'll say
its positive --it's 1 meters

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per second is the velocity
times the mass.

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So 10 times 1 is 10 kilogram
meter per second.

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So now let's figure out B.

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Let's do the y direction first
because they add up to 0.

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I'm going to switch colors.

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We know that the momentum
of-- and this

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is after the collision.

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The momentum of A in the y
direction plus momentum of B

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in the y direction have
to equal what?

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What was the initial momentum
in the y direction?

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Right, it was 0.

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There was no movement in the
y direction initially.

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We know the momentum of
A in the y direction.

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It's 10.

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10 kilogram meters per second
plus the momentum of B in the

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y direction is equal to 0.

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So solving for this, just
subtract 10 from both sides.

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So the momentum of B in the
y direction is equal to 10

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kilogram meters per second.

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You know the units.

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So if its momentum is 10 in the
y direction, what is its

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velocity in the y direction?

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Well, momentum is equal to
mass times velocity.

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So we know that 5 times the
velocity in the y direction--

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that's its mass --is
equal to 10.

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10 is its momentum.

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So the velocity of the y
direction of B must be 2

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meters per second.

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So there we go.

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We figured out B's velocity.

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And so let's say this is B's
velocity vector in the y

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direction is-- And this is
a minus because this is

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equal to minus 10.

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So it's going down.

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It was a velocity of positive
1 going up and then the

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minuses carry through and this
is a velocity of minus 2

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meters per second for B
in the y direction.

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So now let's figure
out the velocity

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of B in the x direction.

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And I'm running out of space
and it's getting messy.

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But we just have to remember
that the momentum of B in

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the-- The momentum of A in the
x direction, which is 10

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square root of 3, plus momentum
of B in the x

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direction has to equal what?

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It has to equal the initial
momentum in the x direction,

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which is 30.

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So to figure out the momentum
of B in the x direction, we

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just subtract 10 square
root of 3 from 30.

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And let's do that.

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So let's figure out 3 square
root times 10 equals.

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And then subtract
that from 30.

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And we get let's
just say 12.7.

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So we know that the momentum
of B in the x direction is

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equal to 12.7, 12.7 kilogram
meters per second.

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And we know the momentum, so we
just divide by the mass and

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we get its velocity in
the x direction.

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So 12.7 divided by 5.

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So velocity of B in
the x direction is

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12.7 divided by 5.

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12.7 divided by 5 is equal to
2.54 meters per second.

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So its velocity in the
x direction is

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2.54 meters per second.

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So it's going faster
in both directions.

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I'm not going to figure out the
angle here because I've

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actually run out of time.

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But if you were to add these
two vectors, you'd get an

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angle something like this.

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And you could figure out the
angle by taking the arc tan.

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Well, I won't go into the--
that's a complexity right now.

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Actually, I'll do that in the
next video just so I won't

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leave you hanging.

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But we know what the x and y
components of B's velocity is.

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See you in the next video.