The bad news today is that there
will be quite a bit of math.
But the good news
is that we will only do it once
and it will only take
something like half-hour.
There are quantities in physics
which are determined uniquely
by one number.
Mass is one of them.
Temperature is one of them.
Speed is one of them.
We call those scalars.
There are others where
you need more than one number
for instance, on a one-
dimensional motion, velocity
it has a certain magnitude--
that's the speed--
but you also have to know
whether it goes this way
or that way.
So there has to be a direction.
Velocity is a vector and
acceleration is a vector
and today we're going to learn
how to work with these vectors.
A vector has a length
and a vector has a direction
and that's why we actually
represent it by an arrow.
We all have seen...
this is a vector.
Remember this--
this is a vector.
If you look at the vector
head-on, you see a dot.
If you look at the vector
from behind, you see a cross.
 
This is a vector
and that will be
our representation of vectors.
Imagine that I am standing
on the table in 26.100.
This is the table and I am
standing, say, at point O
and I move along a straight line
from O to point P
so I move like so.
That's why I am on the table
and that's where you will see me
when you look from 26.100.
It just so happens
that someone is also going
to move the table--
in that same amount of time--
from here to there.
So that means that the table
will have moved down
and so my point P will have
moved down exactly the same way
and so you will see me
now at point S.
You will see me
at point S in 26.100
although I am still standing
at the same location
on the table.
The table has moved.
This is now the position
of the table.
See, the whole table
has shifted.
Now, if these two motions
take place simultaneously
then what you will see
from where you are sitting...
you will see me move in 26.100
from O straight line to S
and this holds the secret
behind the adding of vectors.
We say here that the vector OS--
we'll put an arrow over it--
is the vector OP, with an arrow
over it, plus PS.
This defines how we add vectors.
There are various ways
that you can add vectors.
Suppose I have here vector A
and I have here vector B.
Then you can do it this way
which I call
the "head-tail" technique.
I take B and I bring it
to the head of A.
So this is B, this is a vector
and then the net result
is A plus B.
This vector C equals A plus B.
That's one way of doing it.
It doesn't matter
whether you take B...
the tail of B to the head of A
or whether you take the tail of
A and bring it to the head of B.
You will get the same result.
There's another way
you can do it
and I call that
"the parallelogram method."
Here you have A.
You bring the two tails
together, so here is B now
so the tails are touching
and now you complete
this parallelogram.
And now this vector C
is the same sum vector
that you have here,
whichever way you prefer.
You see immediately
that A plus B is the same
as B plus A.
There is no difference.
What is the meaning
of a negative vector?
Well, A minus A equals zero--
vector A subtract
from vector A equals zero.
So here is vector A.
So which vector do I have
to add to get zero?
I have to add minus A.
Well, if you use
the head-tail technique...
This is A.
You have to add this vector
to have zero
so this is minus A
and so minus A is nothing
but the same as A
but flipped over 180 degrees.
We'll use that very often.
And that brings us to the point
of subtraction of vectors.
How do we subtract vectors?
So A minus B equals C.
Here we have vector A
and here we have--
let me write this down here--
and here we have vector B.
One way to look at this
is the following.
You can say A minus B
is A plus minus B
and we know how to add vectors
and we know what minus B is.
Minus B is the same vector
but flipped over
so we put here minus B
and so this vector now
here equals A minus B.
Here's vector C,
here's A minus B.
And, of course, you can do it
in different ways.
You can also think of it
as A plus... as C plus B is A.
Right? You can say you can
bring this to the other side.
You can say C plus B is A,
C plus B is A.
In other words, which vector
do I have to add to B to get A?
And then you have the
parallelogram technique again.
There are many ways
you can do it.
The head-tail technique
is perhaps the easiest
and the safest.
So you can add
a countless number of vectors
one plus the other,
and the next one
and you finally have the sum
of five or six or seven vectors
which, then, can be
represented by only one.
When you add scalars,
for instance, five and four
then there is only one answer,
that is nine.
Five plus four is nine.
Suppose you have two vectors.
You have no information
on their direction
but you do know that the
magnitude of one is four
and the magnitude
of the other is five.
That's all you know.
Then the magnitude of
the sum vector could be nine
if they are both in the same
direction-- that's the maximum--
or it could be one, if they
are in opposite directions.
So then you have
a whole range of possibilities
because you do not know
the direction.
So the adding and the
subtraction of vectors
is way more complicated
than just scalars.
As we have seen,
that the sum of vectors
can be represented by one vector
equally can we take one vector
and we can replace it
by the sum of others.
And we call that
"decomposition" of a vector.
And that's going to be
very important in 801
and I want you to follow this,
therefore, quite closely.
I have a vector which is
in three-dimensional space.
This is my z axis...
this is my x axis,
y axis and z axis.
This is the origin O
and here is a point P
and I have a vector OP--
that's the vector.
And what I do now,
I project this vector
onto the three axes,
x, y and z.
So there we go.
Each one has her or his
own method of doing this.
There we are.
I call this vector
vector A.
Now, this angle will be theta,
and this angle will be phi.
Notice that the projection of A
on the y axis has here
a number which I call A of y.
This number is A of x and
this number here is A of z--
simply a projection of
that vector onto the three axes.
We now introduce
what we call "unit vectors."
Unit vectors are always pointing
in the direction
of the positive axis
and the unit vector in
the x direction is this one.
It has a length one,
and we write for it "x roof."
"Roof" always means unit vector.
And this is the unit vector
in the y direction
and this is the unit vector
in the z direction.
And now I'm going
to rewrite vector A
in terms of the three components
that we have here.
So the vector A,
I'm going to write
as "A of x times x roof,
plus A of y times y roof
plus A of z times z roof."
And this A of x times x
is really a vector
that runs from the origin
to this point.
So we could put in that
as a vector, if you want to.
This makes it a vector.
This is that vector.
A of y times... oh, sorry,
it is A of x, this one.
A of y times y roof is this one
and A of z times z roof
is this one.
And so these three green vectors
added together
are exactly identical
to the vector OP
so we have decomposed one vector
into three directions.
And we will see that very often,
this is of great use in 801.
The magnitude of the vector is
the square root of Ax squared
plus Ay squared plus Az squared
and so we can take
a simple example.
For instance,
I take a vector A--
this is just an example,
to see this in action--
and we call A three X roof,
so A of axis is three
minus five y roof plus 6 Z roof
so that means that it's
three units in this direction
it is five units
in this direction--
in the minus y direction--
and six in the plus z direction.
That makes up a vector
and I call that vector A.
What is the magnitude
of that vector--
which I always write down
with vertical bars--
if I put two bars on one side,
that's always the magnitude
or sometimes
I simply leave the arrow off,
but to be always on
the safe side, I like this idea
that you know it's
really the magnitude
becomes the scalar
when you do that.
So that would be the square root
of three squared is nine
five squared is 25,
six squared is 36
so that's the square root of 70.
And suppose I asked you,
"What is theta?"
It's uniquely determined,
of course.
This vector
is uniquely determined
in three-dimensional space
so you should be able to find
phi and theta.
Well, the cosine of theta...
See, this angle here...
90 degrees projection.
So the cosine of theta
is A of z divided by A itself.
So the cosine of theta equals
A of z divided by A itself
which in our case
would be six divided
by the square root of 70.
And you can do fine.
It's just simply a matter
of manipulating some numbers.
We now come to a much more
difficult part of vectors
and that is multiplication
of vectors.
We're not going to need this
until October, but I decided
we might as well
get it over with now.
Now that we introduced vectors,
you can add and subtract
you might as well learn
about multiplication.
It's sort of, the job is done,
it's like going to the dentist.
It's a little painful,
but it's good for you
and when it's behind you,
the pain disappears.
So we're going to talk about
multiplication of vectors
something that will not
come back until October
and later in the course.
There are two ways
that we multiply vectors
and one is called
the "dot product"
often also called
the scalar product.
A dot B, a fat dot, and that
is defined as it is a scalar.
A of x times B of x,
just a number
plus A of y times B of y--
that's another number--
plus A of z times B of z--
that's another number.
It is a scalar.
It has no longer a direction.
That is the dot product.
So that's method number one.
That's completely legitimate
and you can always use that.
There is another way
to find the dot product
depending upon
what you're being given--
how the problem is
presented to you.
If someone gives you the vector
A and you have the vector B
and you happen to know
this angle between them,
this angle theta--
which has nothing to do
with that angle theta;
it's the angle between the two--
then the dot product
is also the following
and you may make an attempt
to prove that.
You project the vector B on A.
This is that projection.
The length of this vector
is B cosine theta.
And then the dot product
is the magnitude of A
times the magnitude of B
times the cosine
of the angle theta.
The two are
completely identical.
Now, you may ask me,
you may say,
"Gee, how do I know
what theta is?
"How do I know
I should take theta this angle
"or maybe I should take
theta this angle?
I mean, what angle
is A making with B?"
It makes no difference
because the cosine
of this angle here
is the same as the cosine
of 360 degrees minus theta
so that makes no difference.
Sometimes this is faster
depending upon how the problem
is presented to you;
sometimes the other is faster.
You can immediately see
by looking at this--
it's easier to see
than looking here--
that the dot product can be
larger than zero
it can be equal to zero
and it can be smaller than zero.
A and B are, by definition,
always positive.
They are a magnitude.
That's always determined
by the cosine of theta.
If the cosine of theta
is larger than zero,
well, then
it's larger than zero.
The cosine of theta can be zero.
If the angle for theta
is pi over two--
in other words,
if the two vectors
are perpendicular
to each other--
then the dot product is zero,
and if this angle theta
is between 90 degrees
and 180 degrees
then the cosine is negative.
We will see that at work,
no pun implied
when we're going to deal
with work in physics.
You will see
that we can do positive work
and we can do negative work
and that has to do
with this dot product.
Work and energy are
dot products.
I could do an extremely
simple example with you;
the simplest
that I can think of.
Perhaps it's almost an insult--
it's not meant that way.
Suppose we have A dot B
and A is the one that you really
have on the blackboard there.
Right here, that's A.
But B is just two y roof.
Two y roof, that's all it is.
Well, what is A dot B?
A dot B... there's
no x component of B
so that becomes zero,
this term becomes zero.
There is only
a y component of B
so it is minus five
times plus two
so I get minus ten, because
there was no z component.
Simple as that,
so it's minus ten.
I can give you another example,
example two.
Suppose A itself is the unit
vector in the y direction
and B is the unit vector
in the z direction.
Then A dot B is what?
I want to hear it
loud and clear.
CLASS:
Zero.
LEWIN:
Yeah! Zero.
It is zero-- you don't even
have to think about anything.
You know that these two
are at 90 degrees.
If you want to waste your time
and want to substitute it
in here
you will see
that it comes out to be zero.
It should work, because clearly
A of y means
that this... this is one.
That's what it means.
And B is z, that means
that B of z... this is one
and all the others do not exist.
Well, I wish you luck
with that and we now go
to a way more difficult
part of multiplication
and that is
vector multiplication
which is called
"the vector product."
Or also called...
most of the time
I refer to it
as "the cross product."
The cross product is written
like so: A cross B equals C.
It's a cross, very clear cross.
And I will tell you
how I remember...
that is, method number one.
I'm going to teach you--
just like with the dot product--
two methods.
I will tell you
method number one
which is the one
that always works.
It's time-consuming,
but it always works.
You write down here
a matrix with three rows.
The first row is x roof,
y roof, z roof.
The second one is
A of x, A of y, A of z.
It's important,
if A is here first
that that second row must be A
and the third row is then B.
B of x, B of y, B of z.
So these six are numbers
and these are the unit vectors.
I repeat this here verbatim--
you will see in a minute
why I need that--
and I will do the same here.
Okay, and now comes the recipe.
You take... you go from
the upper left-hand corner
to the one in this direction.
You multiply them, all three,
and that's a plus sign.
So you get Ay... so C
which is A cross B equals Ay,
times Bz, times the x roof--
but I'm not going to put
the x roof in yet--
because I have to subtract
this one... minus sign
which has Az By
so it is minus Az By, and
that is in the direction x.
The next one is this one.
Az Bx...
minus this one
Ax Bz
in the direction y.
And last but not least
Ax By...
minus Ay Bx...
in the direction
of the unit vector z.
So this part here is
what we call "C of x".
It's the x component
of this vector
and this we can call "C of y"
and this we can call "C of z."
So we can also write
that vector, then,
that C equals C of x, x roof,
plus C of y, y roof
plus C of z, z roof.
Cross product of A and B.
We will have lots of exercises,
lots of chances you will have
on assignment, too
to play with this a little bit.
Now comes my method number two
and method number two is, again
as we had with the dot product,
is a geometrical method.
Let me try to work
on this board in between.
If you know vector A
and you know vector B
and you know
that the angle is theta
then the cross product, C,
equals A cross B
is the magnitude of A
times the magnitude of B
times the sine of theta
not the cosine of theta as we
had before with the dot product.
It is the sine of theta.
So you can really immediately
see that this will be zero
if theta is either zero degrees
or 180 degrees
whereas the dot product was zero
when the angle between them
was 90 degrees.
This number can be larger
than zero
if the sine theta
is larger than zero.
It can also be smaller
than zero.
Now we only have
the magnitude of the vector
and now comes the hardest part.
What is the direction
of the vector?
And that is something
that you have to engrave
in your mind and not forget.
The direction is found
as follows.
You take A,
because it's first mentioned
and you rotate A over the
shortest possible angle to B.
If you had in your hand
a corkscrew--
and I will show that
in a minute--
then you turn the corkscrew as
seen from your seats clockwise
and the corkscrew would go
into the blackboard.
And if the corkscrew goes
into the blackboard
you will see the tail
of the vector
and you will see a cross,
little plus sign
and therefore we put that
like so.
A cross product is always
perpendicular to both A and B
but it leaves you
with two choices:
It can either come
out of the blackboard
or it can go
in the blackboard
and I just told you
which convention to use.
And I want to show that to you
in a way that may appeal
to you more.
This is what I have used before
on my television help sessions
that I have given at MIT.
I have an apple--
not an apple...
This is a tomato--
not a tomato...
It's a potato.
(class laughs )
I have a potato here
and here is a corkscrew.
Here is a corkscrew.
I'm going to turn the corkscrew
as seen from your side,
clockwise.
And you'll see that the
corkscrew goes into the potato
in -- that's the direction,
then, of the vector.
If we had B cross A,
then you take B in your hands
and you rotate it over
the shortest angle to A.
Now you have to rotate
counterclockwise
and when you rotate
counterclockwise
the corkscrew comes to you--
there you go--
and so the vector is now
pointing in this direction.
And if the vector
is pointing towards you
then we would indicate that
with a circle and a dot.
In other words, for this vector
B cross A would have
exactly the same magnitude--
no difference-- but it would
be coming out of the blackboard.
In other words, A cross B
equals minus B cross A
whereas A dot B
is the same as B dot A.
We will encounter cross products
when we deal with torques
and when we deal
with angular momentum
which is not
the easiest part of 801.
Let's take
an extremely simple example.
Again, I don't mean to insult
you with such a simple example
but you will get chances,
more advanced chances
on your assignment.
Suppose I gave
to vector A this x roof.
It's a unit vector
in the x direction.
That means A of x is one
and A of y is zero
and A of z is zero.
And suppose B is y roof.
That means B of y is one
and B of x is zero
and B of z is zero.
What, now, is the dot product,
the cross product, A cross B?
Well, you can apply that recipe
but it's much easier to go
to the x, y, z axes
that we have here.
A was in the x direction,
the unit vector
and B in the y direction.
I take A in my hand, I rotate
over the smallest angle
which is 90 degrees to y,
and my corkscrew will go up.
 
So I know
the whole thing already.
I know that this cross product
must be z roof.
The magnitude must be one.
That's immediately clear.
But I immediately have
the direction
by using the corkscrew rule.
Now if you're very smart
you may say,
"Aha! You find plus z
"only because you have used
this coordinate system.
"If this axis had been x,
and this one had been y
"then the cross product
of x and y would be
in the minus z direction."
Yeah, you're right.
But if you ever do that,
I willkill you!
(class laughs )
You will always,
always have to work
with what we call "a right-
handed coordinate system."
And a right-handed coordinate
system, by definition
is one whereby the cross product
of x with y is z
and not y minus z.
So whenever you get,
in the future, involved
with cross products and torques
and angular momentum
always make yourself
an xyz diagram
for which x cross y is z.
Never, ever make it such
that x cross y is minus z.
You're going to hang yourself.
For one thing,
that wouldn't work anymore.
So be very, very careful.
You must work... if you use
the right-hand corkscrew rule
make sure you work with the
right-handed coordinate system.
All right,
now the worst part is over.
And now I would like
to write down for you...
We pick up some
of the fruits now
although it will
penetrate slowly.
I want to write down for you
equations for a moving particle
a moving object
in three-dimensional space--
very complicated motion
which I can hardly imagine
what it's like.
It is a point that is going
to move around in space
and it is this point P
this point P is going
to move around in space
and I call this vector OP,
I call that now vector r
and I give it a sub-index t
which indicates
it's changing with time.
I call this location A of y,
I am going to call that y of t.
It's changing with time.
I call this x of t--
it's going to change with time--
and I call this point z of t
which is going
to change with time
because point P
is going to move.
And so I'm going to write down
the vector r
in its most general form
that I can do that.
R, which changes with time
is now x of t-- which is the
same as a over x there, before--
times x roof plus y of t,
y roof plus z of t, z roof.
I have decomposed my vector r
into three independent vectors.
Each one of those change
with time.
What is the velocity
of this particle?
Well, the velocity is the first
derivative of the position
so that it is dr dt.
So there we go-- first
the derivative of this one
which is dx dt, x roof.
I am going to write for dx dt
"x dot," because I am lazy
and I am going to write for
d2x dt squared, "x double dots."
It's often done,
but not in your book.
But it is a notation
that I will often use
because otherwise the equations
look so clumsy.
Plus y dot times y roof
plus z dot times z roof.
So z dot is the dz/dt.
What is the acceleration
as a function of time?
Well, the acceleration
as a function of time
equals dv/dt.
So that's the section derivative
of x versus time
and so that becomes
x double dot times x roof
plus y double dot times y roof
plus z double dot times z roof.
And look
what we have now accomplished.
It looks like minor, but
it's going to be big later on.
We have a point P going
in three-dimensional space
and here we have the entire
behavior of the object
as it moves its projection
along the x axis.
This is the position,
this is its velocity
and this is its acceleration.
And here you can see the
entire behavior on the z axis.
This is the position
on the z axis
this is the velocity component
in the z direction
and this is the acceleration
on the z axis.
And here you have the y.
In other words, we have now...
the three-dimensional motion
we have cut into three
one-dimensional motions.
This is
a one-dimensional motion.
This is behavior
only along the x axis
and this is a behavior
only along the y axis
and this is a behavior
only along the z axis
and the three together make up
the actual motion
of that particle.
What have we gained now?
It looks like... this looks
like a mathematical zoo.
You would say, "Well, if this
is what it is going to be like
it's going to be hell."
Well, not quite--
in fact, it's going
to help you a great deal.
First of all, if I throw up
a tennis ball in class
like this, then
the whole trajectory is...
the whole trajectory is
in one plane
in the vertical plane.
So even though it is
in three dimensions
we can always represent it by
two axes, by two dimensionally
a y axis and an x axis
so already
the three-dimensional problem
often becomes
a two-dimensional problem.
We will, with great success,
analyze these trajectories
by decomposing
this very complicated motion.
Imagine what an incredibly
complicated arc that is
and yet we are going
to decompose it
into a motion
in the x direction
which lives a life of its own
independent of the motion
in the y direction
which lives a life ofits own
and, of course, you always have
to combine the two
to know what the particle
is doing.
We know the equations so well
from our last lecture
from one-dimensional motion
with constant acceleration.
The first line tells you
what the x position is
as a function of time.
The index t tells you
that it is changing with time.
It is the position
at t equals zero
plus the velocity
at t equals zero
times t plus
one-half ax t squared
if there is an acceleration
in the x direction.
The velocity immediately comes
from taking the derivative
of this function
and the acceleration comes
from taking the derivative
of this function.
Now, if we have a motion
which is more complicated--
which reaches out to two
or three dimensions--
we can decompose the motion
in three perpendicular axes
and you can replace
every x here by a y
which gives you the entire
behavior in the y direction
and if you want to know
the behavior in the z direction
you replace every x here by z
and then you have decomposed
the motion in three directions.
Each of them are linear.
And that's
what I want to do now.
I'm going to throw up an object,
golf ball or an apple in 26.100
and we know that it's in the
vertical plane, so we have...
we only deal
with a two-dimensional problem
this being...
I call this my x axis and I'm
going to call this my y axis.
I call this
increasing value of x
and I call this
increasing value of y.
I could have calledthis
increasing value of y.
Today I have decided to call
this increasing value of y.
I am free in that choice.
I throw up an object
at a certain angle
and I see a motion
like this-- boing!--
and it comes back to the ground.
My initial speed
when I threw it was v zero
and the angle here is alpha.
The x component
of that initial velocity
is v zero cosine alpha
and the y component
equals v zero sine alpha.
So that's the "begins" velocity
of the x direction.
This is the "begins" velocity
in the y direction.
A little later in time,
that object is here at point P
and this is now
the position vector
which we have called r of t,
it's this vector.
That's the vector
that is moving through space.
At this moment in time,
x of t is here
and at this moment in time,
y of t is here.
And now you're going to see,
for the first time
the big gain by the way that we
have divided the two axes
which live an independent life.
First x.
I want to know everything about
x that there has to be known.
I want to know where it is
at any moment in time
velocity and the acceleration,
only in x.
First I want to know
that at t = 0.
Well, at t = 0, I look there
X zero-- that's the,
I can choose that to be zero.
So I can say x zero is zero,
that's my free choice.
Now I need v zero x--
what is the velocity?
The velocity at t = 0,
which we have called v zero x
is this velocity--
v zero cosine alpha.
And it's not going to change.
Why is it not going to change?
Because there is no a of x,
so this term here is zero
we only have this one.
So at all moments in time
the velocity in the x direction
is v zero cosine alpha
and the a of x equals zero.
Now I want to do the same
in the x direction for time t.
Well, at time t, I look there
at the first equation.
There it is-- x zero is zero.
I know v zero x,
that is v zero cosine alpha
so x of t is
v zero cosine alpha times t
but there is no acceleration,
so that's it.
What is vx of t?
The velocity in the x direction
at any moment in time.
That is that equation,
that is simply v zero x.
It is not changing in time
because there is
no acceleration.
So the initial velocity
at t zero is the same
as t seconds later
and the acceleration is zero.
Now we're going to do this
for the y direction.
And now you begin to see
the gain for the decomposition.
In the y direction,
we change the x by y
and so we do it first
at t = 0.
So look there.
This becomes y zero--
I call that zero.
I can always call
my origin zero.
I get v zero y times t.
Well, v zero y is this quantity
is v zero sine alpha,
v zero sine alpha.
This is v zero sine alpha.
That is the velocity at
time zero, and this is zero.
At time zero...
this is zero at time zero.
What is the acceleration
in the y direction at time zero?
What is the acceleration?
That has to do with gravity.
There is no acceleration
in the x direction
but you better believe that
there is one in the y direction.
So only when we deal
with the y equations
does this acceleration come in--
not at all when we deal
with the x direction.
Well, if we call the
acceleration due to gravity
g equals plus 9.80,
and I always call it g
what would be the acceleration
in the y direction
given the fact that I call this
increasing value of y?
CLASS:
Minus 9.8.
LEWIN:
Minus 9.8,
which I will also say
always call minus g
because my g is always positive.
So it is minus g.
So that tells the story of t
equals zero in the y direction
and now we have to complete it
at time t equals t.
At time t equals t,
we have the first line there.
Y zero is zero.
So we have y as a function
of time, y zero is zero
so we don't have to work
with that.
Where is my... so this is zero,
so I get v zero y times t
so I get
v zero sine alpha times t
plus one-half, but it
is minus one-half g t squared
and now I get the velocity
in the y direction at time t--
that is my second line.
That is going to be
v zero sine alpha minus g t
and the acceleration in the y
direction at any moment in time
equals minus g.
And now I have done all I can
to completely decompose
this complicated motion
into two entirely independent
one-dimensional motions.
And the next lecture
we're going to use this again
and again and again and again.
This lecture is not over yet
but I want you to know
that this is
what we're going to apply
for many lectures to come--
the decomposition
of a complicated trajectory
into two simple ones.
Now, when you look at this
there is something
quite remarkable
and the remarkable thing
is that the velocity
in the x direction
throughout
this whole trajectory--
if there is no air draft,
if there is no friction--
is not changing.
It's only the velocity in the
y direction that is changing.
It means
if I throw up this golf ball--
I throw it up like this--
and it has a certain component
in x direction
a certain velocity
if I move myself with
exactly that same velocity--
with exactly the same
horizontal velocity--
I could catch the ball here.
It would have to come back
exactly in my hands.
That is because there is only an
acceleration in the y direction
but the motion
in the y direction
is completely independent
of the x direction.
The x direction
doesn't even know
what's going on
with the y direction.
In the x direction,
if I throw an object like this
the x direction simply,
very boringly,
moves with a constant velocity.
There is no time dependence.
And the y direction,
on its own, does its own thing.
It goes up, comes to a halt
and it stops.
And, of course, the
actual motion is the sum
the superposition of the two.
We have tried to find a way
to demonstrate
this quite bizarre behavior
which is not so intuitive.
That the x direction
really lives a life of its own.
And the way we want
to do that is as follows.
We have here a golf ball
a gun we can shoot
up the golf ball
and we do that in such a way
that the golf ball,
if we do it correctly
exactly comes back here.
That's not easy-- that takes
hours and hours of adjustments.
The golf ball goes up
and comes back here.
Not here, not here,
not there-- that's easy.
You can shoot it up
a little at an angle
and the golf ball
will come back here.
Once we have achieved that--
that the golf ball
will come back there--
then I'm going to give
this cart a push
and the moment that it passes
through this switch
the golf ball will fire
so that the golf ball
will go straight up
as seen from the cart
but it has a horizontal velocity
which is exactly the same
horizontal velocity as the cart
so the cart are like my hands.
As the golf ball goes like this
the cart stays always exactly
under the golf ball
always exactly
under the golf ball
and if all works well
the ball ends up exactly
on the cart again.
Let me first show you--
otherwise, if that doesn't work,
of course, it's all over--
that if we shoot the ball
straight up
that it comes back here.
If it doesn't do that
I don't even have to try this
more complicated experiment.
So here's the golf ball.
I'm going to fire the gun now.
Close... close.
Reasonably close.
Well, since it's only
reasonably close, perhaps...
(class laughs )
Perhaps it would help if we
give it a little bit of leeway.
There goes the gun.
Here comes the ball.
And this is just in case.
Tape it down.
So as I'm going to push
this now, give it a push
the gun will be triggered
when the middle
of the car is here.
You've seen how high
that ball goes
so that ball will go...
(makes whooshing sound )
And depending upon
how hard I push it
they may meet here
or they may meet there.
You ready for this?
You ready?
CLASS:
Ready.
LEWIN:
I'm ready.
Physics works.
(class applauds )
LEWIN:
See you Wednesday.
