All right, last time

we talked exclusively about completely

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,

again, by about 13 units, about seven-percent difference,

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

about something else, not only about the tennis ball

against the wall– whereby the wall has momentum

but no kinetic energy– but now I would like you to

think about this amazing thing. I have eight

billiard balls here, and your Newton cradles

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.

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???

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.

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

https://www.youtube.com/watch?v=8urHLQ68f7M

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.

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

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.

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!

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?

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???

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"

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

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.

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

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

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?

@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.

.. could you please help….

Thank you

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…

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

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?

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!

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 !

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

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)

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.

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?

Love

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?

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 : )

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

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.

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?

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.

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…

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.

Thanks sir

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.

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

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?

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