All right, last time
inelastic collisions. Today I will talk about collisions
in more general terms. Let’s take
a one-dimensional case. We have here m1
and we have here m2, and to make life a little easy,
we’ll make v2 zero and this particle has
velocity v1. After the collision,
m2 has a velocity v2 prime, and m1, let it have
a velocity v1 prime. I don’t even know whether
it’s in this direction or whether it is
in that direction. You will see
that either one is possible. To find v1 prime
and to find v2 prime, it’s clear that you
now need two equations. And if there is no net external
force on the system as a whole during the collisions,
then momentum is conserved. And so you can write down that m1 v1 must be
m1 v1 prime plus m2 v2 prime. Now, you may want to put
arrows over there to indicate
that these are vectors, but since it’s
a one-dimensional case, you can leave the arrows off and the signs will then
automatically take care of the direction. If you call this plus,
then if you get a minus sign, you know that the velocity is
in the opposite direction. So now we need
a second equation. Now, in physics we do believe very strongly
in the conservation of energy, not necessarily in the
conservation of kinetic energy. As you have seen last time,
you can destroy kinetic energy. But somehow we believe
that if you destroy energy, it must come out
in some other form, and you cannot create energy
out of nothing. And in the case of the
completely inelastic collisions that we have seen last time, we lost kinetic energy,
which was converted to heat. There was internal friction. When the car wreck plowed
into each other, there was internal friction–
no external friction– and that took out
kinetic energy. And so, in its most general
form, you can write down that the kinetic energy
before the collision plus some number Q equals the kinetic energy
after the collision. And if you know Q,
then you have a second equation, and then you can solve
for v1 prime and for v2 prime. If Q is larger than zero, then
you have gained kinetic energy. That is possible;
we did that last time. We had two cars which
were connected by a spring, and we burned the wire and each went off
in the opposite direction. There was no kinetic energy before… if you want
to call it the collision, but there was
kinetic energy afterwards. That was the potential energy
of the spring that was converted
into kinetic energy. So Q can be larger than zero. We call that
a superelastic collision. It could be an explosion. That’s a
superelastic collision. And then there is the
possibility that Q equals zero, a very special case. We will deal with that today, and we call that
an elastic collision. I will often call it
a completely elastic collision, which is really not necessary. “Elastic” itself already means
Q is zero. And then there is a case– of which we have seen
several examples last time– of inelastic collisions,
when you lose kinetic energy, so this is
an inelastic collision. And so, if you know what Q is, then you can solve
these equations. Whenever Q is less than zero, whenever you lose
kinetic energy, the loss, in general,
goes into heat. Now I want to continue a case whereby I have
a completely elastic collision. So Q is zero. Momentum is conserved, because
there was no net external force, so now kinetic energy is
also conserved. And so I can write down now
one-half m1 v1 squared– that was the kinetic energy
before the collision– must be the kinetic energy
after the collision one-half m1 v1 prime squared plus one-half
m2 v2 prime squared. This is my equation number one, and this is
my equation number two. And they can be solved;
you can solve them. They are solved in your book. I will simply give you
the results, because the results are
very interesting to play with. That’s what
we will be doing today. v1 prime will be m1 minus m2
divided by m1 plus m2 times v1 and v2 prime will be 2 m1
divided by m1 plus m2 times v1. The first thing
that you already see right away is that v2 prime is always
in the same direction as v1. That’s completely obvious, because the second object
was standing still, remember? So if you plow something
into the second object, they obviously continue
in that direction. That’s clear. So you see you can never have
a sign reversal here. Here, however, you can have
a sign reversal. If you bounce a ping-pong ball
off a billiard ball, the ping-pong ball
will come back and this one becomes negative, whereas if you plow a billiard
ball into a ping-pong ball, it will go forward. And so this can both be
negative and can be positive depending upon whether
the upstairs is negative or positive. So this is the result which
holds under three conditions: that the kinetic energy
is conserved, so Q is zero; that momentum is conserved; and that v2 before
the collision equals zero. Let’s look at three interesting
cases whereby we go to extremes. And let’s first take the case that m1 is much,
much larger than m2. m1 is much, much larger than m2. Another way of thinking about
that is that let m2 go to zero. Extreme case, the limiting case. So it’s like having
a bowling ball that you collide
with a ping-pong ball. If you look at that equation
when m2 goes to zero– this is zero, this is zero–
notice that v1 prime equals v1. That is completely intuitive. If a bowling ball collides
with a ping-pong ball the bowling ball doesn’t
even see the ping-pong ball. It continues its route
as if nothing happened. That’s exactly what you see. After the collision, the bowling
ball continues unaltered. What is v2 prime? That is not so intuitive. If you substitute in there
m2 equals zero, then you get plus 2 v1–
not obvious at all, plus 2 v1. It’s not something
I even want you to see; I can’t see it either. I’ll do a demonstration. You can see
that it really happens. So, now you take a bowling ball and you collide the bowling ball
with the ping-pong ball and the ping-pong ball will get
a velocity 2 v1– not more, not less– and the bowling ball continues
at the same speed. Now let’s take a case whereby m1
equals much, much less than m2; in other words, in the limiting
case, m1 goes to zero. And we substitute that in here. So m1 goes to zero,
so this is zero and so you see
v1 prime equals minus v1. v1 prime equals minus v1,
completely obvious. The ping-pong ball bounces
off the bowling ball and it just bounces back. And this is what you see. And the bowling ball
doesn’t do anything, because m1 goes to zero,
so v2 prime goes to zero. So that’s very intuitive. And now we have a very cute case
that m1 equals m2. And when you substitute
that in here– when m1 equals m2–
v1 prime becomes zero. So the first one stops
with v2 prime becomes v1. If m1 equals m2, you have two downstairs here
and two upstairs and you see
that v2 prime equals v1. And that is a remarkable case– you’ve all seen that, you’ve all
played with Newton’s cradle. You have two billiard balls. One is still
and the other one bangs on it. The first one stops and the second one takes off
with the speed of the first. An amazing thing. We’ve all seen it. I presume you have all seen it. Most people do this
with pendulums where they bounce these balls
against each other. I will do it here with a model that you can see
a little easier. I have here billiard balls, and if I bounce this one
on this one, then we have case number three. Then you see
this one stands still and this one takes over
the speed– quite amazing. Every time I see this,
I love it. It is a wonderful thing to see. Nature… just imagine
you are nature, and this ball comes on,
and in no time at all you have to solve these
two equations very quickly. There’s only one solution,
and nature knows how to do that. This one stops
and this one goes on. It’s an amazing result. And I’m sure that you have seen these pendulums
that you can play with. Here we have not a bowling ball
onto a ping-pong ball but we have a billiard ball. The mass ratio is
not infinity to one but it’s 500 to one,
which is quite large. And so what I will first do is
very intuitive. I will first bounce
the ping-pong ball off the bowling ball…
the billiard ball. The ping-pong ball comes back
almost with the same speed– not quite, because the ratio
is not infinity to one but it is 500 to one, and the billiard ball will do
practically nothing. It’s exactly what you see
there in case two. So, there we go. You see, the ping-pong ball
comes back almost as far as I let it go. It tells you that the speed… Oh! That the speed has not changed. It just bounces back and the billiard ball
almost does nothing. Now comes case number one. That’s the nonintuitive case. It is intuitive that if the billiard ball hits
the ping-pong ball that it will continue. As you will see, v1 prime is v1,
with the same speed. It’s not at all intuitive that the ping-pong will get
twice the speed of the billiard ball, and, of course, you cannot see
that quantitatively because we don’t do
a quantitative measurement of the speed
of the ping-pong ball, but you will see
that it bounces up quite high. So, there we go. The billiard ball
onto the ping-pong ball. Look at
the billiard ball alone– forget the ping-pong ball– and try to see that the speed
of the billiard ball is practically unaffected
by the collision. That’s what you see there–
v1 prime is v1. You see, it’s
practically unaffected. Now, of course,
it is way harder to see that the ping-pong ball gets twice the speed
of the billiard ball, because we don’t do… because we don’t do
a quantitative measurement. All right. So, those were examples, then,
of those three possibilities that you all see
on the blackboard there. Now I want to do more
completely elastic collisions, and I’m going to do that
with the air track. I’m going to try to make
completely elastic collisions. That’s not so easy. I will have one object stand
still, so always v2 equals zero. And they are completely elastic. Kinetic energy is conserved. This word “completely”
is not necessary. I always add it in my mind. I’m going to have one object m1
which I bang against object m2, and object m2 will be standing
here, will have no speed. And object m1 comes in
from this side and I’ll try to make
the collision elastic. And the way I will do that
is by using springs which are attached to each mass. And springs are
conservative forces so there is almost no heat
that is generated in the springs during the collision. And so to a reasonable
approximation, you will get
an elastic collision, but, of course,
there’s always air drag. I can never take air drag out. So there’s always some
external force on the system. So momentum is never
exactly conserved, and kinetic energy is never
exactly conserved either, so it’s only an approximation. So I have one mass of unit 1– I will tell you what that is–
and I bang that into number 2. And the masses that I have… One mass is 241
plus or minus one gram, and the other mass that I have
is 482 plus or minus one gram. And I have two of these. And in my first experiment
I will use these two, so this is the ratio
that you see, one to one. I will give it
a certain velocity, which I cannot tell
you what that is. It depends upon
how happy I feel. If I push hard, then
the velocity will be high. If I push softly,
the velocity will be low. But it is something. And then we are going
to get here v1 prime, and there is a prediction
that this velocity will be zero. Look all the way
on the blackboard there. You will see
if the masses are the same and there is
an elastic collision, this one will stand still and this one, v2 prime,
will have a velocity v1. So that’s a prediction. The second case,
I bounce one car onto another one
which is twice the mass– certain velocity
which I don’t know what it is– and now
what are we going to get? So mass m1 is half
the mass of m2. So this is a 1 and
this is a 2 and this is 3. 1 minus 2 is minus 1
divided by 3 is minus 1/3. So number one comes back–
no surprise. If you bounce a lighter object
off a more massive object, no surprise
that it bounces back. So that’s minus one-third,
and when we look here, we have 2 times 1 divided
by 1 plus 2 is 3. That is plus two-thirds, so number one will come back
with a speed one-third– which, of course,
since this is sign-sensitive, we get one-third v1 and
this one is plus two-thirds v1. That’s the prediction. And the way we’re going
to measure it is by measuring the time
for the objects to move over a distance
of ten centimeters. No matter how long
you see these cars to be, there is here a piece of metal
which is ten centimeters long. It goes through a slot
with a diode. We have one system here
and one there. And during this ten-centimeter
movement, we measure the time. We begin when this piece of
metal enters the diode system and we stop the time when it
leaves the diode system. And each one of those cars has
this ten-centimeter metal plate. So, the way
we’re going to do this is by measuring timing, of course. This velocity will give me
a certain time, and this velocity will give me
a certain time. Whatever comes out,
you will see that on this timer. Then t1 prime… is… in this case… it stands still,
so it must be zero. And t2 prime must be
the same as t1. So these two numbers, you should
be able to compare directly. What kind of uncertainties
do we have? It’s hard to tell. But I would say, as I argued
last time, that you should allow for at least about 2½ percent
uncertainty in each time. If it comes out better,
then you are lucky. If it comes out worse,
then it is a bad day. 2½ percent,
and I argued last time how I reasoned
about the 2½ percent. Now we do this experiment. We get a certain t1. And so this one goes back
with one-third of the speed, so it will take three times
longer to go ten centimeters. So I’m going to multiply
this time by one-third, and whatever comes out
should be the same as this. So, let me move this up
a little… t2 prime. The speed here, the forward
speed, is two-thirds, so it will go slower. So if I multiply this time
by two-thirds, then I should be able
to compare it with t1. And all of that,
all these times, I think, will not be any better
than roughly 2½ percent, except if we are a little lucky. (system powers up) Okay, the system is up. The timers are up,
the timers are up. This will be the time t1. This will be the time t2 prime, and this is the one
when the object bounces back, so this is t1 prime. You won’t see
that it is negative. It will go back
through the same slot, and the electronics
is arranged in such a way that when it goes back
through the same slot that it will initiate this time. I will zero them. First you have to tell me
whether they are working. I will just let it go
through these slots. Is this one working? Tell me
that this one is working. If I send this one back,
is this one working? Okay. Okay, there we go. This is the one
that will have no speed, and this is the one that
we’re going to give a velocity. You are ready? I hope you know
what you are going to see. This one will come to a halt,
and this will go on. They have the same mass. There we go. This one is indeed
coming to a halt and this one took over
the speed. What are the numbers? 194 and 196– only
two milliseconds’ difference. That is an incredible result. 0.194 and 0.196. There’s only one percent
difference between them– way within
my wildest expectations. Because my expectations were that they could be off each
by 2½ percent. Now we go to the one-to-two. So here is the car
which has twice the mass. It’s important that I zero it. And now this one
is going to come back. So you’re going to see this one
come in, give you the time here. This one goes through here,
it gives you the time here. This one will come back,
it gives you the time here. Are they zero? Okay? You’re ready? Yeah? There we go. Okay. Now, now comes
the real acid test. 123… 186. 123… And what is the last one? 375. Okay, let me be lazy
and let me use my calculator. 375 divided by three– I could have done that
by heart, of course– this is 0.125,
an amazing agreement! Amazing! Only two percent off,
less than two percent. Now here, 0.186 times 2
divided by 3– 0.124. I can’t believe it–
less than a percent off. So, you see
in front of your eyes that we were able
to create something that was indeed extremely close
to elastic collisions in spite of all the problems: that we have air drag,
and that, of course there is always
some loss of kinetic energy. But it’s so little that it doesn’t show up
in these measurements. Now I want you to have
a sleepless night. I want you to think
about the following and if you can’t solve it
before the night is over, then I really think
you should lie awake. Here is the wall. Here is a tennis ball that
comes in with a certain mass m, and it has a certain velocity v,
whatever that is. It’s a nearly elastic collision and it bounces back
from the wall. And we all know that if it is
a nearly elastic collision that it comes back
with the same velocity. Kinetic energy is conserved. All the kinetic energy is
in the tennis ball; nothing is in the wall. The wall has
an infinitely large mass, but the momentum
of this tennis ball has changed by an amount 2 mv. That momentum must be
in the wall– it’s nonnegotiable, because
momentum must be conserved. So now here you see
in front of your eyes a case that the wall has momentum,
but it has no kinetic energy. Can you understand that?
Can you reconcile that? Can you show me mathematically
that that is completely kosher? That the wall has momentum 2 mv,
it’s nonnegotiable. It must have momentum, and
yet it has no kinetic energy. How is that possible? Think about it,
and if you can’t solve it, call me at 3:00 a.m. and
I’ll tell you the solution. Okay, now let’s look at this from the center-of-mass
frame of reference. The center of mass
is very special, and physicists love to work
in the center of mass for reasons
that you will understand. In the absence
of any net external forces on a system as a whole–
as we discussed last time– the center of mass will
always have the same velocity. We did a demonstration there with two vibrating objects
with a spring, but yet the center of mass was
moving with a constant velocity. The beauty now is if you jump into the frame
of the center of mass– that means you move with the same velocity
of the center of mass– the center of mass stands still
in your frame of reference. And if the center of mass
stands still, the momentum of the particles
in your frame of reference– in the center-of-mass frame
of reference– is zero. It is zero before the collisions and it is zero
after the collisions. And this gives the center
of mass some amazing properties that I will discuss
with you now. First, we have a particle m1
and we have a particle m2. And let this one
in the center-of-mass frame have a velocity u1, and this in the center-of-mass
frame has a velocity u2. I give it specifically u’s so that you can separate
the u’s from the v’s. The v’s are always
in your frame of reference; the u’s are in the frame of
reference of the center of mass. And I take a case whereby I have
a completely elastic collision– that means Q is zero– the kind
that we have just discussed. Momentum is not only conserved
but it is also zero at all moments in time
before and after the collision. After the collision let’s say m2 goes back
with velocity u2 prime, and let’s say
m1 has a velocity u1 prime. That’s the situation
after the collision. Now, I know that momentum is
zero, so I only can write down for this situation
after the collision that m1 times u1 prime plus m2
times u2 prime must be zero. I don’t write them down
as vectors. That’s not necessary, because it’s
a one-dimensional collision and the signs will automatically
take care of the directions. I told you I chose the case of
a completely elastic collision– Q is zero– and so kinetic
energy must be conserved. So I have…
before the collision, I have one-half m1 u1 squared
plus one-half m2 u2 squared, and after the collision, I have
one-half m1 u1 prime squared plus one-half
m2 u2 prime squared. Kinetic energy before;
kinetic energy afterwards. Equation one; equation two. Nature can solve that
quicker than we can, and the result is amazing. The result
in the center of mass is that the velocities reverse directions
but the speeds remain the same. And that is a remarkable,
a remarkable result. If you ever want to move yourself
to the center-of-mass you will have to know what
the center-of-mass velocity is. How do we calculate the velocity
of the center of mass? So we’re dealing here
with the center-of-mass frame, and now I’m going back
to the laboratory frame. And we know that M total times the position vector
of the center of mass– this is the way
we defined it last time– equals m1 times the position
vector of particle 1 plus m2 times the position
vector of particle 2. And so if you take
the derivative of this equation, then the positions
become velocities, so the velocity
of the center of mass equals l divided by m1 plus m2, because that’s the M total
which I bring under here, and upstairs I get
m1 v1 plus m2 v2. And notice I left the arrows off because, since it’s
a one-dimensional problem, signs will take care
of the directions. So this is the velocity
of the center of mass. I wrote it also there
on the blackboard because later on
in this lecture I will need it, and it is possible that
by that time I have erased it, and so that’s
why I wrote it down there, too. So if now you want to know
what u1 is, so we are… Now you want to know what the velocity is
in the center-of-mass frame, that, of course, equals v1 minus the velocity
of the center of mass and u2 equals v2 minus the
velocity of the center of mass. So this is a way
that you can transfer, if you want to, into
the center-of-mass frame. And it sometimes pays off,
for reasons that I mentioned, that the momentum in the
center-of-mass frame is zero– always zero before the collision
and after the collision, independent of whether
it is an elastic collision, whether it’s
an inelastic collision or whether it’s
a superelastic collision. Now, if you later wanted
to transfer back to your laboratory frame, then,
of course, you will have to add the velocity of the center
of mass again to the u1 prime, and you have to add the velocity
of the center of mass to the u2 prime. The velocity of the center
of mass has not changed as seen from your frame
of reference, because the velocity
of center of mass is always the same, remember,
because momentum is conserved. So to get into
the center-of-mass frame, you must subtract the velocity
of the center of mass from the initial velocities. To get out of it,
you must add them. Now, the kinetic energy
and the momentum depend on your reference frame. In general, the total momentum
as seen from your seats is not zero. That’s only
in a very special case. In the case
of the center of mass, the total momentum is
always zero. The kinetic energy
as seen from the lab frame is certainly, in general,
not the same as the kinetic energy
from the center-of-mass frame. And now comes another unique
property of the center of mass. If I have a completely
inelastic collision, then all energy in the
center-of-mass frame is lost. That’s obvious– in the center-of-mass frame,
remember, momentum is zero. So you’re in the center of mass. One particle comes to you
and the other comes to you. You’re not moving,
you’re in the center of mass. They get stuck together because it’s a completely
inelastic collision. If they get stuck together after the collision
they stand still. That means all kinetic energy
that was there before is all destroyed,
and this kinetic energy– as observed in
the center-of-mass frame– we call the internal energy, and that is the maximum energy
in a collision that can ever be converted
into heat. And I will show that
to you partially. I will do some of the work and I will let you do
some of the work as well. So I will first calculate– in your frame of reference,
where you are sitting– how much energy is lost when we have a completely
inelastic collision. I will then transfer
to the center-of-mass frame, and I will show you, then,
this quite amazing property. So now we are back in 26.100, and we’re going to make a
completely inelastic collision. That means
they stick together, remember? And m2, we will have again,
to make life a little simple, no speed, v2 equals zero,
and m1 has a velocity v1. You’ve seen this now
a zillion times. They get together,
they stick together, and so I have here a velocity
which I call v prime, and the mass is m1 plus m2. That’s after the collision. Momentum is conserved if there’s no external… net
external force on the system, and so I can write down that m1 v1 must be
m1 plus m2 times v prime. And so v prime equals m1 v1
divided by m1 plus m2. That’s a very
simple calculation. This, by the way–
it’s not so obvious– is also the velocity
of the center of mass. And how can I see that
so quickly? Well, what was the velocity
of the center of mass? Here you have it. This was in general. This was not for the case
that v2 was zero; this was more general. Make v2 zero,
and you see exactly that you see here
the same result, so this must be the velocity
of the center of mass. Now we can calculate
what the difference is between the kinetic energy
after the collision and the kinetic energy
before the collision. That is, of course, something
that is rather trivial. You know what the kinetic energy
is before the collision– it’s one-half m1 v1 squared– and you know what it is
after the collision. I calculated v prime, and so
you have take half this mass, multiply it
by this velocity squared. You can do that,
I am sure you can do that. And you will be able to see
that this equals minus… and you have to massage
the algebra a little bit– minus one-half m1 m2 divided
by m1 plus m2 times v1 squared. That’s what you will find. The minus sign is predictable. We lose kinetic energy when there is a completely
inelastic collision. We’ve done many last lecture. You lose kinetic energy– we
saw it in every single case. That’s what
the minus sign means. This is Q. You lose kinetic energy
and that goes into heat. So you’ve done your homework now
in the laboratory frame and you are home free–
very well. Now I’m going to do
the same calculation in the center-of-mass frame. And I will show you, now–
that’s the purpose– that this amount of energy,
which is what is lost, that that is all there was
to start with in the center-of-mass frame, and that’s a unique property
of the center of mass. And so I’m going to convert now, to transfer you
to the center-of-mass frame and then we will calculate
how much energy there was in the center-of-mass frame
before the collision, because after the collision
there is nothing. There is zero. In the case of a completely
inelastic collision, there is no kinetic energy left
in the center-of-mass frame. So we go to the
center-of-mass frame. So we first have to calculate
what u1 is. Well, u1 equals v1 minus
v center of mass. And we know what v center
of mass is– it’s right there. That’s where it is. And if you do the subtraction,
which is by no means difficult, you will find m2 divided by m1
plus m2 times v1. And you checked that, I hope. And now we go to calculate u2. We want to know
what the velocity is of the center of mass
of the other one. That, of course, is
v2 minus v center of mass, but this was zero. This m1 divided
by m1 plus m2 times v1, so the difference is only
the m1 upstairs and the m2. Now we are going to calculate the kinetic energy
in the center-of-mass frame. Well, that equals
one-half of m1 times u1 squared plus one-half m2
times u2 squared. That’s all we have
before the collision occurs. Oh, by the way,
this is not a minus… This is a plus sign
and this is a minus sign. This one comes this way
and this one goes in that way. Now, I can…
I can calculate that for you. You know u1 and you know u2. If that’s a plus
or a minus sign, it makes no difference
because they cancel anyhow. What are you going to find? One-half m1 m2 divided
by m1 plus m2 times v1 squared. And this is exactly the same
that we had there. And so what you see here– if you allow me for having skipped
some steps in the algebra; you will have to do
a little massaging to get from here to here–
you see here this is the kinetic energy
before the collision, and all that kinetic energy
is removed, went to heat. This is the maximum
you can ever lose, and this is what we call the internal kinetic energy
of the system. And so going
to the center-of-mass system, you can always
immediately calculate what the maximum heat is that
you can expect from a collision. We can take a very special case, and we can take m2 going
to infinity. It’s like having a piece
of putty I slam on the wall. It gets stuck, and what is the maximum heat
that you can produce that’s all the energy there is? If m2 becomes infinitely high,
then m1 can be ignored. m2 cancels m2, you get
one-half m1 v1 squared. And that’s obvious. That’s completely trivial. I have a piece of putty,
I slam it against the wall. It has a certain amount
of kinetic energy. Whether you stay
in your reference frame or in the reference frame
of the center of mass, it’s immediately obvious that
all the kinetic energy is lost. And that’s exactly what you see
comes out of these equations whether you go
to the center of mass or whether you do it
from 26.100. I now would like to return
to the air track and do several completely
inelastic collisions with you. Again, we have to assume
that momentum is conserved. It never is completely,
but we can come close. And we will have
two cars that… Now we’re going
completely inelastic. Completely inelastic. So they hit each other
and they get stuck. I get a certain velocity,
which I put into the first car. It hits the second car
and it gets stuck. And I see there
what my v prime is. I see there on the blackboard.
that if the two are the same, I have a one here,
a one and a one here. If that’s the ratio, so the outcome is that
v prime must be one-half v1. So this must be one-half v1. Now I have a mass which is
half the other one. I plow it into the other,
they get stuck together and now I get one divided by
one plus two– I get one-third. Notice in both cases
I get a plus sign. That’s, of course, obvious. If I plow into something
and they stick together, they continue
in the same direction. And so now I will do the timing in exactly the same way
that I did before, except that now I have a
completely inelastic collision. I will have a timing t1 and
I will have a timing t prime. The cars have
a slightly different mass: 237 plus or minus one gram and I have one that is
474 plus or minus one gram– not too different from this. I have two of these cars
and I have one of these. And first I’m going to slam
these two onto each other, and so when they collide,
I expect the speed to be half. So I get
a certain amount of t1– so t prime will be twice as long because the speed goes down
by a factor of two– so when I multiply
this number by one-half, I would like to get
that number back. Here I get the time
for this car. The first one to come in
gives me a certain time. When they continue together,
the speed is three times lower so this time will be
three times higher. So I multiply this by one-third and I would like to get
that time back. And those are the two experiments
that I would like to do now. So we have to get
the noise back on. Air track one. (air track humming) Air track two. Here we have the two cars. They have Velcro, so when they
hit each other, they get stuck. Sometimes they bounce, actually. You have to do
the experiment again. This one goes here;
this one goes here. Are these two timers on? We don’t need
this timer anymore. Are they both on? Zero them. Zero them. Are they zero? Okay, are you ready? This one is going to plow
into this one, they merge and they continue, and then you’ll see
the time t prime here. There we go. Ready… All right. What do we see? 138, 288. 138… 288… 138… 288, divide this by 2 is 0.144. That is…
the difference is only six, and that is only
four percent off. That’s completely
within my prediction that each time could be off
by 2½ percent. Now I will have a car
which is twice the mass and this is once the mass. I will set them again to zero and now we get one
onto double the mass. Ready? There we go. Oh! You see? They bounced. They didn’t stick,
so we have to start all over. If they don’t stick,
of course, then it’s… it’s not a completely
inelastic collision. There we go. They made a slight bounce,
which I didn’t like. Let’s see what we get. We can always do it again. 168, 545. 168… Multiply this by three… 504, 545– eight-percent
difference is a little high. But these things can happen. I told you they bounced,
and I didn’t like that, and then they got stuck again. I will do it once more. Maybe I’m a little bit luckier. Again, they bounced a little
before they finally merged. 603, 187. 187… 603. If I divide this
by three, I get 0.201, and that is off,
so it’s a little bit more than the five percent
that I allowed for. Now, to make it worse
for you tonight, I would like you to think
against the wall– whereby the wall has momentum
but no kinetic energy– but now I would like you to
probably also have eight balls hanging
from pendulums. This is easier to demonstrate
in 26.100. Now I let two balls bang
onto the other six. And I know that you predict
what’s going to happen and your prediction is correct. If you slam two balls
onto the six nature is going
to calculate like mad to conserve momentum,
to conserve kinetic energy closely enough to
a completely elastic collision, and out comes that the only way
that nature can do it is the following. They all stand still
and these two take off. Look again. They all stand still
and these two take off. That’s not an easy calculation. But now look at the following. Hold it– what happens if I take three and I bang three
on the other five? What do you predict
is going to happen? How many will take off? Three, you think. You sure? You were right. Okay. Now… five. Five on three. This is a tough problem
even for nature. What do you think will happen? (students respond) LEWIN: Five will go again and
three will stay– you’re good. If any one of you can show me
analytically that this is what has to happen, I would love to see
those results. Okay, see you Wednesday.

## Only registered users can comment.

1. Durbadal chakraborty says:

If kinetic energy after collision in inelastic collision in centre of mass frame is zero then What is kinetic energy after collision in elastic collision in centre of mass frame???

2. Zonnymaka says:

My take on the tennis ball problem 🙂
The total momentum of the system before and after the collision is zero.
Indeed if we calculate the velocity of the center of mass:
(mv+M0)/(m+M) =mv/(m+M) = v(CM)
we notice that if the mass of the wall is infinite/big, then v(CM)=0
Hence, because of the wall/enormous mass our frame reference is incredibly near to the center of mass of the system.
Hence it doesn't matter if the collision is elastic, inelastic or superelastic…the total momentum is always zero!
By not recognizing that we are in the center of the mass can lead to the paradoxical (yet correct) result that the wall has a momentum = 2mv.

3. Eric Su says:

A wonderful video about elastic collision for everybody. The more you learn, the smarter you are.

4. Rishabh jain says:

The answer to the tennis ball question could be given easily as follows:
(First things to be taken into consideration mass of tennis is negligible in comparison with the ball so as per the equations you had given v1'=v1 and v2'=0)
Since v2 and v2' are going to be zero before and after collision we can say that KE of wall will be zero since KE=1/2 mv squared and velocity is zero so it is clear that kinetic energy will be zero I don't know this is right or wrong but the result Is indeed correct.

5. John Walker says:

Hello professor, to make it worse for you tonight sir, I have watched this lesson, and at 46:29 you show us a very interesting experiment. You show a few Newtonian billiard ball collissions, and I learn a lot here. Thanks.
However, at 48 minutes you state that this is a tough problem, even for nature. I don't agree with you sir. I want to show you, analytically, that seen from natures point of view, being the center of mass, there would not be any difference between three balls comming from the right and five balls comming from the left sir, and three balls comming from the left and five balls comming from the right. In all cases the result after the collission will be three balls going in one direction and five balls going in the opposite direction. Do you love this result professor ?

John

6. Nerd says:

Suppose a ball is thrown upwards in a room and it hits the ceiling and comes back down. We observe that the velocity of the ball increases. Conservation of momentum does not take place in this case, does it? External force is indeed applied initially.

7. Srijan Guha says:

Hello Sir! What are glancing collisions? Is any collision in 2-D or 3-D called glancing collision? Or is there something special about glancing collisions. Im not clear with Google's answers. Help please!

8. sachin B S says:

Momentum is scaled velocity vector. If m is very high then P will be high (velocity vector gets scaled by mass even if it is very low).
Kinetic energy is a number. If velocity vector is low , then it's magnitude is also low . Then |v|^2 is also low. So KE~=0.
Thus wall has p=2mv but KE~=0.
Is this correct?

9. Chetan Krishna Palicherla says:

professor this is my answer on the momentum problem:-
m – mass of the tennis ball ; M – mass of the wall ; u – initial velocity of ball ; u' – final velocity of ball ; V – final velocity of the wall
1. Change is momentum must be zero, [ mu = mu' + MV ——- (1)]
2. From (1) we can get [ V={m(u-u')}/M——(2) ]
3. In (2) we know the mass of wall(M) is infinity so the velocity of the wall (V) is zero.
4. So, |u| = |u'| but the directions are opposite.
5. So, the KE of the wall is zero(1/2M(0) but the momentum is 2mu not 0.
Is it correct professor???

10. Lord Naver says:

Tennis ball on wall problem, my humble perspective:

Momentum balance:

mv + P(wall before collision) = -mv + P(wall after collision)

ΔP (wall) = P(wall before collision) – P(wall after collision) = -2mv

ΔP (wall) = -2mv

KE balance:

1/2 mv^2 + KE(wall before collision) = 1/2 m(-v)^2 + KE(wall after collision)

Δ KE (wall) = KE(wall before collision) – KE(wall after collision) = 1/2 mv^2 – 1/2 mv^2 = 0

Δ KE (wall) = 0

to me the difference lies between the natures of odd and even powers, i.e., the difference between v^1 and v^2.

one could say it lies in the difference between scalar and vector quantities or in the difference between constructive and destructive combinations.

or as I'd like to put it in a slightly embarrassing couple of lines, as written below,

"oh momenta! you fled away from my insurmountable mass.
you may think you made no impact on me, but I still have your scars.
and yet you think there is nothing you have left behind, but that is only because I have no energy"

11. vaishnavi vaishu says:

thank you so much sir… your videos help me a lot in understanding physics.

12. Prakhar Bhalla says:

since velocity of center of mass is constant as told at 30:25 then on collision of a ball with the wall the velocity of center of mass was towards the wall before collision then after collision it should again be towards the wall. but it doesn't seem so as center of mass move away from wall after collision. so please help me out. thanks in advance.

13. Krokodyle Sama says:

from momentum conservation v(wall)=2*m(ball)*v(ball)/m(wall) and since m(wall)->infinity v(wall)->0 and his KE->0 and a big thank for you legend

14. John Cena says:

How can one find the center of mass of a moving system?

15. sawan kumawat says:

The wall has gained momentum with respect to the ball (i.e. we are in the frame of wall), Now the velocity of wall is zero and since energy is absolute (K.E. is scalar) the K.E. after the collision is zero. Is it correct?

16. S5DEB says:

@Professor Lewin….. From the equations v'1=(m1-m2)*v1/(m1+m2)….&……v'2=2m1*v1/(m1+m2)………as m2>>m1….we have v'1=(-v1);.
… and…v'2=0…… Then the tennis ball's momentum after bouncing back is (-m1v1)…and the Wall's momentum is 0……… isn't it?…. I'm not clear how wall has momentum of 2m1v1……I know total momentum must be same before and after collision…but from the equations I'm unable to prove that.

17. Bullfrog FPV says:

Thank you

18. Corbu Lucian says:

The ball is pushing twice the wall! The first time, the ball is hitting the wall, and it is compressed, and the two bodies are accelerating in the same direction. The momentum mv= (m+M) v’ so the wall gets a first impulse Mv’= mv- mv’ with v’ almost 0, that mean Mv’=mv. Then the ball is decompressed by the elastic forces to his normal shape, the ball is pushing again the wall, and the two bodies are accelerating in opposite directions! Which mean the wall gets a second impulse Mv’’= mv. The ball is pushing twice the wall, first time because is coming with the speed v, the second time because is departing with the speed –v, so therefore wall gets a double impulse from the ball, in the same direction p=MV=2mv. The impulse of wall is p=2mv, but the kinetic energy of the wall is E=pV/2, with V almost zero, therefore E=0.
The speed of the ball is the same, before and after hitting the wall, and this can be proved, but we can not prove the momentum of the wall, is based just on the theory. Today, based on this theory, the momentum of the wall is p=2mv. But I'm not so sure about tomorrow…

19. berke kocadere says:

Can someone clearly explain the ball-wall question? I couldn't understand it perfectly.

20. Luis Breva says:

For the problem of the wall and the ball. Consider instead of a wall, a huge cement block sitting on the ground (acts like wall). When the ball hits the block, it exerts a momentaneous force on the block. This force does transmit kinetic energy to the block, but at the same time, the ground exerts a static friction force on the block, taking away that energy, hence, the block doesn't move. Momentum isnt conserved as there is a net external force acting on the system (the friction force) as the collision takes place. The ball bounces back due to action-reaction principle (the block exerts an equal but opposite force on the ball, that accelerates it, changing the direction of its velocity). In other words, there is a net impulse over tge system that changes its momentum. We can replace the block with a wall and the friction f. with forces due to the configuration of the building. If the block was sitting on a frictionless surface or the wall was inserted in a peculiar system of frictionless rails attached to the ceiling and to the ground, then there wouldnt be a net external force and momentum would be conserved. In this case we would see indeed that both the wall and the block would move (with the corresponding velocities arising from the equations).Would this reasoning be correct sir?

21. Losat of the Lord says:

I have a really stupid question I guess; Why use V1, V2, etc.; why not use V, V2, etc. (isn't the first V assumed to be V1, so using V1 would be redundant?)? Same for Time, Mass, etc. Also, what is Prime and its value? I'm not a math guy, but I love watching these lectures; for some reason this seems to make more sense to me than algebra – algebra makes me hate math so I never bothered trying to move past it (is it really necessary to understand algebra in order to under stand stuff like this?). Thanks for your time!

22. Joey Demiane says:

What is the definition of momentum professor ?? It's mv I know that but what is it in real life ? It's not energy it's not a force … What is it ?? And how can the wall have a change in momentum if v is always 0 for the wall ! I am really confused !

23. Mayur Patel says:

He is the only person which can make us love physics and proceed physics as our carrier

24. Bry WHsbg says:

problem in halfway through the lecture, the wall has a momentum because it has infinite mass but its velocity is zero (because, infinity*0 = some real number)

25. David Viper says:

27:44 when did you write the solutions for equation 1 and equation 2. And you didn't talk anything about the result as well. I guess there was some problem with the video and you had to edit the video or something.

26. Vijay Prasanna says:

I have a doubt in a collision question
A block A of mass 2m is placed on another block of mass 4m which in turn is placed on a fixed table. The two blocks have the same length (not height) 4d. The coefficient of friction (both static and kinetic) between block B and table is μ. There is no friction between the two blocks. A small object of mass m moving horizontally along a line through the center of mass of the block B (parallel to the ground) and perpendicular to the face with a speed v collides elastically with the block B at a height d above the table. What is the minimum value of v required to make the block A to topple?

27. Shree Ayilognag says:

Love

28. Pot Head says:

Relating to the balls on a string problem… could it be because the number of balls create a pulse (like a wave) with a wavelength equal to 2n (where n is length of the number of balls pushed), and since the balls were only given a positive velocity (not oscillating), only the positive amplitude of the wavelength is represented. The positive pulse then travels through the balls when they strike each other until it reaches the end, where it knows it has to create momentum as the preexisting momentum is gone, and that this momentum must affect the number of balls which could not get rid of the pulse, equal the length of the pulse (n) which is also the total length of the balls?

29. HAIDAR HAYEK says:

its a shame to our world that a great and amazing professor like that haven't gain an award for teaching or a status for his great achievement and he make physics loved by people THANKS PROFESSOR : )

30. Kate Tsynaeva says:

Wow!!! I love your lectures so much!!! It's really amazing and wonderful

31. Thet Aung says:

I call what is happening to billiard balls when they collide "Momentum Transfer".
Here is why. If the balls have the same mass and there is no friction or external force between them, the moment and kinetic energy is conserved. Now, if a ball with mass 'm'–b1–collide with 2 balls at rest, its momentum is transferred to the first ball in contact–b2–and again, b2's moment is transferred to the third ball–b3–. So, mv1=mv3 and b3 takes off.
If we use mathematics, v1´=0 and v2´=v1(before it transfers momentum to b3). When v2 transfers, v2´=0 and v3´=v1. As an example, if 3 balls collide with 2 balls, the two balls in the 3 balls system push 2 balls at rest with 2mv and momentum is transferred ,so, the two balls takes off and 2 balls momentarily at rest and the remaining ball collide with them and as in previous argument, another 1 takes off. This all happened in a flash so, we see 3 balls take off. Professor Lewin,please correct me if I'm wrong.

32. Prakhar Bhalla says:

in completely inelastic collision you told that energy lost is in the form of heat but the sound produced during collision is also a form of energy. where does that come from?

33. Jarka Metelka says:

Thank you very much Mr. Lewin. I studied twenty years ago and I watch your show some evenings. It is fantastic and MIT is fantastic too. I am studied in the Czech republic unfortinetely.
Your last question is collision balls. My answer: Balls do not collision in the same time, but first ball move last ball, second ball move second last ball, third ball move thirt last ball.

34. Alfio Marrone says:

Regarding the last problem of the Newton's cradle, I guess the balls are not really touching, so at each collision we simply apply the case of a ball colliding with a stationary ball of the same weight.
So if we start with one ball moving then this ball collides with the second (and stops), the second travels a very little distance and collides with the third and so on to the last.
If we have more than one moving ball at the start (for example, as you showed, three balls) then the third ball (the central one) collides with the fourth and stops, in a supershort time the second arrives from behind and the central starts again, now the central hits again the fourth that has just hit the fifth and stopped almost instantaneously, etc…

35. Daniel Scimé says:

I always liked to picture the extreme case as a train hitting a golf ball. A change of reference frame /almost/ makes it intuitive!

From the perspective of someone on the train, the train isn't moving, the golf ball is approaching them at velocity V, and it bounces off the front of the train with velocity -V. That's a delta of 2V.

From the perspective of someone on the ground, watching the train barrel toward the golf ball at velocity V, they must observe the same delta 2V! So they should see the golf ball at rest, fly away from the train, at 2V.

36. Divyanshu Kumar says:

Thanks sir

37. Efe Güleroğlu says:

For the last question.
Let's say we have any number of balls. I pulled and dropped from one side "x" number of balls with velocity v. After the elastic collision "y" number of balls moves with Φv.
(1/2)xmv^2=(1/2)ym(Φ^2)(v^2) so x=yΦ^2 (1)
xmv=ymΦv so x=yΦ (2)
combining (1) and (2) yΦ=yΦ^2 and Φ=1.
Which means x=y.

38. JR 2020 says:

I've read most of the comments regarding the wall/tennis ball scenario could someone please explain this to me like I'm a 4 year Old please. SOS

39. Oishika Chaudhury says:

I cannot understand why professor Lewin said that since spring force is a conservative force, kinetic energy remains conserved in this case? The total mechanical energy has to remain conserved right? Not just the kinetic energy?

40. tortue de la nuit says:

One of the funniest things about these videos is watching the purple circles under the students' eyes get deeper as the semester wears on.