Remember, earlier in the course we measured the average speed
of a bullet which we fired from a rifle. That was because we had
the ability of very fast timing. In the old days,
fast timing was not possible and people measured the speed of
bullets in a very delicate way and all the tools that we have
learned we can apply now to this device which we call
the ballistic pendulum. We have a pendulum
with a very heavy object hanging here at the end--
I call it the block. You see it here. And this pendulum
has length L. Ours is about one meter. I will give you
the exact numbers later. And we have a bullet
of mass little m and the bullet comes in
with velocity v. It gets completely absorbed,
sticks in there. It's a completely
inelastic collision, and the pendulum will then
pick up the velocity v prime with the bullet inside. The bullet is somewhere here. Momentum is conserved,
so we clearly have that, m v equals m plus M
times v prime. So if you could measure v prime, then you could measure
the speed of the bullet, which is v. How do we measure v prime? Well, we wait for the pendulum
to come to a halt, let's say here,
when the speed is zero. When it was here,
it had a speed v prime. And we know that there
was kinetic energy here-- no gravitational
potential energy. I can call this level U = 0, but right here, if this
difference in height is h, then all the kinetic energy
has been converted to gravitational
potential energy. So we apply the theorem,
the work-energy theorem, or you could say... and it's
equally valid, you could say we applied the conservation
of mechanical energy. And so this kinetic energy, which is one-half m plus M
times v prime squared is now converted exclusively to
gravitational potential energy, which equals m plus M
times g times that h. And we lose our m plus M, so v prime would be
the square root of 2gh. And so all you would
have to measure h, and then you know v prime, and if you know v prime, you
know the speed of the bullet. But life is not that simple. It's very difficult
to measure h, and I can make you see that. Suppose this angle-- angle
theta-- when it comes to a halt is only two degrees. Then h, which is L times
1 minus cosine theta, is only 0.6 millimeters for the dimensions
that I have chosen here for a length of one meter. And you can't even see it--
it's invisible-- let alone that you can measure
it to any degree of accuracy. So what are we going to do now? Well, we are going
to not measure h, but we are going to measure x. I call this x = 0. And here, when the pendulum
comes to a halt, I call that x. For a two-degree angle, x is approximately
3½ centimeters. It can easily be checked by you,
of course. So you get a huge displacement
in this direction compared to h. If you use small angle
approximation-- and you better believe
that two degrees is very small-- then you can prove, which is
purely geometrical mathematics-- and I leave you
with that proof-- that this is approximately
x squared divided by 2L. I want you to prove that. You take the expansion
of the cosine, the theories
of the Taylor series, and you cut it off somewhere, and this is not
so difficult to prove. In other words, v prime squared, which is 2gh,
can now be replaced by approximately 2g times
x squared divided by 2L which is g times x squared
divided by L. And so the velocity
of the bullet, v, which is m plus M
divided by m-- I bring the m down there--
times v prime, but v prime is now
the square root of this, so I get an x here times
the square root of g over L. And what you see now,
but in a very clever way, by getting x in here, we can now
do a quite accurate measurement of the speed of this bullet, because we can measure x
with a fair accuracy-- maybe an uncertainty of only
one or two millimeters in x out of the 3½ or four
or five centimeters. We're going to measure
the speed of such a bullet-- we did that before-- to get a number which
is not too different. I think we got something like
200 to 250 meters per second. I will give you the input
for this experiment. The bullet mass is 2
plus or minus 0.2 grams. I apologize that I don't give
you all the numbers in mks units. You will have to convert them,
of course, to mks units. The length of the pendulum is 1.13 meters
plus or minus two centimeters-- we're not certain-- with an accuracy
of about two centimeters. And the mass of that block,
which is huge, I believe is 3,200 grams,
3.2 kilograms, with an uncertainty
of about two grams. So we have a ten percent
uncertainty in the mass, we have a two percent
uncertainty in the length, and we have
a negligible uncertainty in the mass of the block--
that's negligibly small. So if I want to know now
what the velocity is of the speed of the bullet, I can calculate what
m plus M divided by m is and I can calculate
the square root of g over L, and with those numbers I find 4.7 times
ten to the third times x. We have to measure x. Now if you look
at the uncertainties in the whole thing,
we're going to measure x, which may not be 3½ centimeters,
it may be four or five-- but the uncertainty
in our measurement will probably be one
or two millimeters. Let's say it's 0.2 centimeters. So this is not
our real number yet. So that would be an uncertainty
of about two out of 50, is four out of a hundred,
is four percent. This is four percent,
roughly four percent. So I would say we get
the final speed of the bullet to an accuracy of about 15% if we combine
all the uncertainties. So let's give it a shot,
no pun implied. And we'll... we can make you see the pendulum right there
very shortly. And it is very cleverly
designed. When the pendulum
starts to swing, it's going to move
a small object. There you have it. I think I can turn this
on again. I think that's fine. There is a very small wiper,
which is this black wiper, and as the pendulum swings-- this is five centimeters,
this is ten centimeters-- the wiper will stay
at the largest extension when the pendulum swings back. Okay. When we fire bullets,
it's always a little risky. I have here the... bolt. Put the bolt in. The bullet's in my pocket--
there's one. That's right. I take the bullet. And I cock the gun. Everything done right, yes? So you can look there
and then see the swing of this very massive block, and the bullet will get absorbed
in that block. You ready for this? Three, two, one (gun fires), zero. I would say 5.2 centimeters
seems to be about right, so x observed
is about 5.2 centimeters, and we know the uncertainty
of the 15% already, so I have to calculate now
the 4.7 times ten to the third. 4.7 exponent third,
multiplied by... I go mks, of course,
so that this .052, and that is 244 meters
per second. I remember last time we had
something very similar. So the speed of the bullet
is about 244 meters per second. It's a little under
the speed of sound, which is 340 meters per second, and we came to the same
conclusion last time. All right. We have kinetic energy
in the bullet before the bullet hit the block, and you can calculate
how much that is, because you know the speed now
and you know the mass-- one-half mv squared. You can also calculate
how much kinetic energy there is when this bullet
is absorbed in here. That's very easy-- that's one-half times
the total mass, m plus M, times v prime squared,
which you also know now. And so you will see then,
perhaps to your surprise, that if you compare the two, that 99.94% of all
available kinetic energy before the collision
was destroyed, and therefore
was converted to heat. That happened, of course... the
heat was produced in that block. Now I'm going to change
to the concept of impulse. It's not completely unrelated
to what we just did. An impulse is giving someone a
kick, that's what an impulse is. Our bullet gave an impulse to
this block, it gave it a kick. Impulse. Impulse is a vector, and it is defined
as the integral of F dt during a certain
amount of time-- let's say from zero to delta t. Now, F equals ma,
which is also dp/dt-- we have seen this now
several times-- the rate of change of momentum-- and so I can substitute
that in here, and so I find then the integral from zero to delta t
of dp/dt dt, and that makes me move
to the domain of momenta, so I have now simply
the integral over dp from some initial
momentum, pi, to some final momentum, pf. And so that is simply the final momentum
minus the initial momentum. So what an impulse does,
it changes the momentum. There is a force
that acts on something for a short amount of time-- could be a little longer,
as you will see with rockets-- and that gives it
a change in momentum. If we have an object
that we drop on the floor, so we have an object, mass m,
and we drop it on the floor and we let it fall
over a distance h, then it's going to hit the floor
with a certain speed-- we know it's down, the velocity, and that equals
the square root of 2gh. If this were a completely
elastic collision, which depends, of course,
on the quality of the object, and it depends on the quality
of the floor-- maybe a super ball on marble would be almost
completely elastic, then the ball would bounce back
with that same speed. And if that were indeed a
completely elastic collision, then you can see
that the impulse that is given to the ball
as the ball hits the floor-- the floor is giving
an impulse to the ball, and that impulse equals 2mv. The ball changes its momentum. It was first mv
in this direction, and now it's mv
in this direction, so the change is 2mv. So an impulse is given
to the ball. Now, if the collision
were completely inelastic, then the ball would just...
say, like a tomato, I throw a tomato on the floor,
it goes (splat). No speed anymore
when it hits the ground, then, of course,
the impulse would only be mv, because then
it doesn't come back up, so there is no... the change
in momentum is then smaller. We have here two balls
that look alike. They have a mass
of 0.1 kilogram, so m equals 0.1 kilogram, and I will drop them from
a height of about 1½ meters, and that gives them a speed
when they hit the floor of about 5½ meters per second. And so the momentum change
is 2mv, so the impulse equals 2mv,
is about 1.1, and that would be
kilograms-meters per second. And that means that
if the collision time is delta t seconds, that the average force
acting upon this ball during the collision
with the floor equals the impulse
divided by delta t, because remember, that was
our definition of impulse. So if we know the impulse, we get a feeling
for the average force. And for the ball that
I will drop on the floor, we have done fast photography. I will show you some results
of the fast photography with a different ball,
but nevertheless, we did it with the ball that I will drop
on the floor very shortly-- which is this one-- that impact time
is only two milliseconds. It's hard to believe
that in two milliseconds the entire collision occurs. And so if you substitute
in here now two milliseconds, then you get for the average
force 550 newtons. Just imagine, this ball
has a mass of 0.1 kilogram, the weight is one newton, and during the impact,
it weighs 550 times more. What an incredible
weight increase! And the average acceleration that it experiences during
the impact is 550 times g. People play tennis, and they have speeds
of hundreds of miles per hour-- speeds are way higher than we
have here, ten times higher-- and so the weight increase
is even more. Now, if the collision
were completely inelastic, so that if it were a tomato
or an egg, and this one wouldn't come up, the average force would still
be approximately the same, the reason being that
the impulse will be half. But if the impact time
were also half and the impulse were half,
then, of course, the force... the average force
will be the same-- very high, but for
a shorter amount of time. So I want to show... oh, I first want to show you now
the... these two balls. One is almost complete elastic
collision with the floor. Whether this is a complete
elastic collision depends not only on this ball--
whether it is a super ball-- it also depends on the condition
of the floor. This is not a very good floor,
this is not marble. So when I drop this one, it doesn't come up
to this point here. So it's not a completely
elastic collision, but it bounces pretty much. So it is somewhere
in completely inelastic and completely elastic. It's not bad, right? It's not bad. Now this one. Watch it. (splat) Looks alike, but it ain't. This one is completely
inelastic. It goes to the floor
and it goes clunk. You see a small bounce,
but that's it. And so the impact times
are very short-- two milliseconds in the case
of the one that bounces back, one millisecond in the case
of the one that went clunk, and their average weight is about 550 times
their normal weight. I'd like to show you
fast photography on not this very same ball, but on another one that we did. Let me take this... this out. And that is a ball
that comes down with a speed of four meters
per second. And each frame
is one millisecond, so you will see a ruler, and the ruler indicates...
has marks in centimeters, and so you will see it go... in four milliseconds
it will go one centimeter, so it has a speed
of 2½ meters per second. It will hit the floor, and then we can count
the number of milliseconds that it takes contact
and going back up again. It's not going to be
two milliseconds, it's a little longer,
but, again, impressively short. All right, we're going
to make it rather dark in order to get decent quality. I'm going to turn
five of these off and I'm going to set
the TV at two, and now I will start this. There we go. Let's hope that that will go. Okay, there comes the ball down. These marks are in centimeters. So there's a ruler
in centimeters. This is one centimeter. I'll rewind a little, because
we were a little too late. Okay, let's start again. So watch when it passes
this mark-- one, two, three-- you see four milliseconds
for about one centimeter. So that's 2½ meters per second. And now we'll count the number
of milliseconds at impact. One, two, three, four,
five, six, and it's off. About six, maybe
seven milliseconds. And this is no special ball. These impact times
are amazingly short. Now I have something
very special for you-- something really special, something that
has kept me awake-- a lot of things keep me awake
in physics, and not only in physics... (scattered laughter) Um... but this... but this
is very special. This is very special. I have here a basketball. (ball bouncing) Not completely elastic,
but not bad. A tennis ball-- not completely
elastic, but not bad. Now I'm going to drop them
together vertically down, and then this ball
will bounce up somehow. And so the question
that I have for you is, do you think if I drop it
from this height that this tennis ball
will sort of come up at most to this height, or do you think
it will be lower, or do you think
it will be higher? So use your intuition. In the worst case it can be...
(chuckling) it can be wrong-- he is already
pushing his finger up-- so what do you think? Will the tennis ball reach
about the same height? Who is in favor of that? Who is in favor of higher? Wow. Who is in favor of a lot higher? Okay. Okay, I'll try it. Now, I cannot guarantee you that
this ball will go straight up after the impact, because
clearly that's impossible. That has zero chance, so it probably go up
in some direction. But you will see the effect
that I had in mind. So there we go. (balls bounce) And you see that, indeed,
that tennis ball... goes way higher. I'll try it once more to see whether I can get it to go up
a little bit more vertically, but that is very difficult. It goes way higher, and this is something that you
should be able to calculate, and you can, and you will. Believe me, it's part
of assignment number six. You haven't seen it yet. There we go. (ball bounces) Oh, boy, that was better. Well... oh! Do any one of you know approximately how much
higher it goes, if this ball has way higher mass
than this one? Of course, the mass ratio
comes into it. Any idea? Twice as high? You'll be surprised
when you do assignment six. Much higher. Okay, in fact, you could
even see it here all right that it was quite a bit higher
than twice. Great... a great experiment, and it's something you can do
yourself in your dormitory. Now I want to discuss in the
remaining time about rockets. A rocket experiences an impulse
from the engine, and that changes
the momentum of the rocket. But before we go into
the details of the rocket, think about, again, this idea of
throwing objects on the floor. And let's turn to tomatoes. I want tomatoes, because I want
a complete inelastic collision, so tomatoes hit the floor,
and it's not that I... I'm not only going to throw
one tomato on the floor, but I'm very angry today, and I'm going to throw a lot
of these tomatoes on the floor-- n, as in Nancy, tomatoes
on the floor. If one tomato hits the floor, the change of momentum is mv,
if m is the mass of the tomato. But I'm going to throw n
on the floor, so the change of momentum
is n, as in Nancy, times the mass of the tomato
times v. And this is the number
of kilograms per second of tomatoes that I throw
on the floor. So this is a change of momentum. And so this equals delta p
divided by delta t, and that's an average force. So the floor will experience
a force in down direction. Of course that was
also the case here when the ball experienced
a force up, which we calculated here. The floor experienced,
of course, the same force
in down direction-- action equals minus reaction--
Newton's third law. So the force experiences...
the floor experiences a force in down direction. And I can write it down in a
somewhat more civilized form: F equals dm/dt
times the velocity, if the velocity of the tomatoes
that hit the floor is constant. And we're going
to apply this to rockets, whereby the exhaust
out of rockets has, relative to the rocket,
a constant speed. And this is then the number
of kilograms per second that I throw on the floor. This is very real. I can throw these tomatoes
on a bathroom scale, and if I threw four kilograms
per second on a bathroom scale, and they hit the bathroom scale
with five meters per second, you better believe it that you
will see that the bathroom scale will reach an average force of about 20 newtons--
four times five. If it weren't tomatoes, but if they were super balls
which would bounce up, then the momentum change
would be double, and so the bathroom scale
would indicate 40 newtons. So this is a real thing, it's a
real force that you can record. Now I'm going to be
a little bit unpleasant to you. I'm going to throw
rotten tomatoes at you. Here you are,
and I have here a tomato, and this tomato has zero speed
to start with. Let me make you a little bigger, otherwise I won't even...
I won't even hit you. So this is you. And so I give this tomato
a certain velocity, v of x. The tomato hits you, (splat). Maybe it stays there,
it's possible. Maybe (squish),
it will drip down. But in any case, the velocity
in the x direction is gone. So it hits you with a velocity
vx and then v of x equals zero. And it may make a mess. You will experience a force. If I keep throwing these
tomatoes at you all the time... and this is the force
that you will experience-- and that force is, of course,
in this direction. You've got all these tomatoes,
and you feel that as a force. But now look at the symmetry
of the problem. Here the velocity goes
from vx to zero. But I, throwing the tomatoes, have to increase the velocity
from zero to vx. So for obvious reasons, I must then feel a force
in this direction-- think of it as a recoil
when you fire a bullet. So I experience exactly the same
force, but in this direction, and that now is the idea
behind a rocket. A rocket is spewing out
tomatoes-- well, not quite tomatoes-- it's spewing out hot gas
in this direction, and then the rocket
will experience a force in that direction. That is the basic concept
behind a rocket. And the higher the speed
of the gas that it throws out-- the higher that velocity-- the more kilograms per second
it spits out, the higher dm/dt, the higher
will be the force on the rocket, and this force on the rocket is
called the thrust of the rocket. So if we have a rocket
in space-- here's the rocket-- and the rocket is spewing out
gas with a velocity, u, which is fixed relative
to the rocket-- it's a burning
of chemical energy. Chemicals are burned, it comes
out with a certain speed, and the rocket will then
experience a force, which we call the thrust, and that is given
by this equation. If you know how many kilograms
per second are spewed out, and you know what the velocity
is, which I've called u here, this will tell you what
the thrust is of that rocket. If we take the case
of the Saturn rockets that were used for the landing
on the Moon... Saturn. For the Saturn rockets, the speed u was about
2½ kilometers per second. So the gas came out
with 2½ kilometers per second relative to the rocket, and the rate at which this gas
was coming out is phenomenal-- 15 tons of material per second. dm/dt is about 15,000 kilograms
per second-- it's almost unimaginable. And that would give it
then a thrust of about 35 million newtons, which, of course, was higher
than the weight of the rocket, otherwise the rocket
would never go up. An incredible thrust. I have also a nice problem
for you in assignment six with the Saturn rockets. You'll see these numbers again. These are rounded off. So rockets obtain an impulse
from their engines, a force acts upon the rocket
for a certain amount of time-- we call that the burn time-- but as they burn the fuel, the
mass of the rocket goes down, because the fuel leaves. And therefore the acceleration
during the burn goes up, because the mass goes down. And this makes it
somewhat complicated to derive the velocity change that you get during the burn,
during this impulse. It's done in your book--
Ohanian. I find that derivation
a little bit complicated. I looked in other books, and I found one in the book
by Tipler on physics, and his derivation-- and I will go through that
in general terms... I put that on the Web for you. It should be there tonight, so you may want to take
very few notes. The entire derivation, of which I will just highlight
the important issues, is on the Web. And the way that Tipler
is doing it is exclusively
from your frame of reference. You're sitting in 26.100, and you are seeing
this rocket go up. Let's keep that equation, which
is the thrust of the rocket, except that in the case of
a rocket we would call this u, which is the speed of the
exhaust relative to the rocket, and this is what
we call the thrust. So let's now take a rocket,
which at time t as seen from your
frame of reference-- where you are sitting. I use v-- it's your
frame of reference. It's going up with a velocity v. The mass of the rocket is m. And now we're going to look
at time t plus delta t. The rocket has increased
its speed, v plus delta v. The mass is m-- let me do that
in... not in color. The mass is now m minus delta m, and here is
a little bit of exhaust that was spewed out
with velocity u relative to the rocket--
u relative to the rocket. So you in 26.100
will see the velocity of this little piece delta m-- you will see this
to be v minus u. If the velocity of the rocket
is larger than u-- could be-- you will see the exhaust going
up from your frame of reference. If the velocity of the exhaust is larger than the velocity
of the rocket, from your frame of reference,
you will see it go down. That's taken care of
in the signs. Now I'm going to compare
the momentum here with the momentum there, and we take the situation that there are no external
forces acting upon it-- somewhere in outer space,
this rocket burns. Momentum must be conserved. The momentum at time t
equals m times v. That is very simple. That is time t. The momentum at time t plus
delta t of the entire system including the exhaust-- if I'm telling you
that momentum is conserved, you can't ignore the exhaust. It's momentum of the system
that is conserved. The momentum of the rocket
is going to change and the momentum of the exhaust
is going to change, but not of the system. So we're going to get
mass times velocity-- m minus delta m times
v plus delta v plus delta m, which has a velocity v minus u. And how large is this? Well, there is
mv plus m delta v. Here you see the m delta v,
here you see the mv, and then you get
minus u delta m, and the delta mv here
cancels minus delta mv here, and this term, delta m delta v, is the product of two incredibly
small numbers-- I ignore that. So this is the momentum
at t plus delta t and this is the momentum
at time t, and so the change
in momentum, delta p-- which must be zero
because momentum is conserved-- is m delta v minus u delta m. m delta v minus u delta m. I can take the derivative
of this equation, so I get on the left side dp/dt. dp/dt is going to be zero, so I get zero
equals m times dv/dt, but dv/dt is the acceleration
of the rocket minus u dm/dt. And that is the thrust
on the rocket. So what you see here is something that
is very easy to digest-- that ma, which is the... this is
the acceleration of the rocket, this is the mass of the rocket
at time t-- equals the thrust of the rocket,
and that equals u dm/dt. And some people call this
the... the "rocket equation." Now, this is true if there is
no external force on the system. It is interesting to include
a real launch from Earth, and if you have a real launch
from Earth, then the rocket is going up
in this direction, but then gravity is exactly
in the opposite direction. In other words, only when you
launch vertically from Earth would you have
a thrust like this, and you would have mg like this. In that case, this equation
has to be adjusted, and then you get
ma equals m thrust minus mg. Only if you have a launch
from Earth vertically up. Now you have to do
a little bit of massaging, and I will leave you
with that massaging. A couple of integrals are
necessary to convert this into the final velocity
of the rocket after the burn compared to
the initial velocity. And that part
I will leave you with, but you will see that
worked out in all detail on the notes that I left
on the Web. And then you come up
with a very famous equation that the final velocity
of the rocket minus the initial velocity
of the rocket equals minus u
times the logarithm of the final mass of the rocket divided by the initial mass
of the rocket. This is if there were
no gravity at all. Just in case, and only in case
of a vertical launch from Earth, there is also here
a term, minus gt-- only if you have
a vertical launch from Earth. When I watched my lecture,
I noticed that in my enthusiasm I put brackets
around the minus gt. This is quite misleading, because it may give you
the wrong impression that there's a product
at stake here, which is not so. So the brackets really should
not be around the minus gt. It is this term, minus u logarithm of mf
divided by m of i minus gt. Let's now look at this equation
in a little bit more detail, so that we get a little bit
of feeling for it. Suppose we had
a vertical launch from Earth, but we had no rocket. It's possible. So this term does not exist. What do you see? That the velocity
equals v initial-- we have called that before
in 801 v zero-- that's the initial speed--
minus gt. Ha! We had that result
during our first lecture-- completely consistent
with this equation. If you have no rocket, you get
that v equals v zero minus gt, if you throw the object
vertically up and we had a vertical launch. So that looks good. We launch from the Earth, and we now have an initial speed
which is zero. The rocket is standing there,
and we fire the rocket. t, by the way, is the burn time
of the rocket. Let me write that down. t is the burn time. So initial speed is zero. So now, this final velocity...
if we want this final velocity to be anything physically
meaningful-- a positive number-- this thing has
to come out positive. And you will say,
"But how can that be?" Because we have
a minus sign here and we have a minus sign there. How can that ever
become positive? Well, don't forget
that the final mass is smaller always
than the initial mass, because you burn fuel, and so this logarithm
is always going to be negative, and so this term is always
going to be positive, if you burn any fuel. Now, of course for this velocity to be a real value--
physically meaningful-- this term has to kill this one,
has to be larger than this one, otherwise the rocket
won't even go up. You could be spewing out fuel
all the time, and the rocket would just be
sitting there, because the thrust up
effectively is not enough to lift it off. So the first term has to be larger
than the second term in case of a vertical launch. Let's take a example
with some numbers-- it always gives
a little bit of insight. We have a burn time
of about 100 seconds, and the initial speed is zero, and let us take a case whereby u
is 1,000 meters per second. That is less
than the Saturn rocket, but it's still a sizable...
it's a kilometer per second. So the exhaust is coming out
relative to the rocket with one kilometer per second. And the final mass divided by
the initial mass of the rocket is 0.1, so 90% was burned away,
the fuel that is gone. You have only ten percent left
when the burn is over. So now we can calculate-- if this was a launch from Earth,
a vertical launch up-- we can calculate
how large this term is, take the logarithm
of this value, multiply this by minus u, and then we find
that the final velocity... The first term is 2,300, and the second term,
100 seconds, and this is ten,
is minus 1000. So you pay a price
for the gravitational field. And so you have about
1.3 kilometers per second is the final speed
of that rocket. These are meters per second, and that is in
kilometers per second. Now, if there were no gravity, then of course the gain
in speed-- this difference-- would be 2,300 meters
per second. Also, if there
is no vertical launch but if, for instance, the rocket
were in orbit around the Earth-- here we have the rocket and going in orbit
around the Earth-- then, of course,
even though there is gravity, gravity is now
not doing any work. And therefore if you fire this
rocket for 100 seconds, the change in velocity,
the tangential change, will also be 2,300 meters
per second. You cannot say
now there is gravity and therefore we have to take
the gt term into account. It will be the same
for this equation. If you had an object
going in this direction, this minus gt term
wouldn't be there either. It's only, of course, when you
deal with a vertical motion. It is very important to realize,
and very non-intuitive, that when you burn
a certain amount of fuel for a given amount of time you obtain a fixed change
in the velocity. This is fixed
for a given amount of fuel. The change in kinetic energy
is not fixed, and I'll give you some numbers so that you can immediately
check that for yourself. And it is very non-intuitive. It is, in fact, a mistake that
is made by many physicists who think that if you burn
the same amount of fuel of the same rocket
for the same amount of time that the increase
in kinetic energy is a given. That is not true. It's the change in velocity
that is a given. But suppose that v final
minus v initial is 100 meters per second. That's just a given. I have a rocket, I have
a certain amount of fuel, I burn it, and that is
my change in velocity. I start off with v initial zero, so my kinetic energy increase
is one-half m times 100 squared, which is ten to the fourth. That's what I got out
of this burn. Now I use the very same rocket, the same amount of fuel,
the same burn time, so I get again that
the final velocity minus the initial velocity
is still 100. That's exactly the same. But before the burn, this rocket
had an initial velocity of 1,000 meters per second. What is now the gain
of kinetic energy? The final velocity is now 1,100. That's non-negotiable, because
the rocket changes the momentum changes the velocity
by a fixed amount. So now the gain
of kinetic energy, the increase of kinetic energy equals one-half m times 1,100
squared minus 1,000 squared. This is the new velocity
and this was the old one. And this number is
one-half m times 200,000. This could be in joules,
and this is also in joules. This number is 20 times higher. So you see that the change in
velocity was exactly the same. The rocket burned
the same amount of time, the same amount of fuel, but the kinetic energy increase
is way more if the rocket has a higher speed
to start with. We here in 26.100
made our own rocket. It's a very down-to-earth model,
no pun implied. But it is quite powerful, and
I would like to show it to you, because we are
quite proud of it. As a rocket, we use a fire
extinguisher-- carbon dioxide-- and this fantastic invention
I have here. And this powerful rocket is enough to reach
the escape velocity of... (loud whoosing)
... of 26.100. (crash sounds, Lewin and students laugh) I almost reached the escape
velocity, but I crashed. See you next Friday. (laughter and
scattered applause) (more whooshing)
