Problem Solving Strategies

Nobody recognizeable

05:19

@JohnRennie hi

John Rennie

@Nobodyrecognizeable hi :-)

Nobody recognizeable

I had a question

John Rennie

@Nobodyrecognizeable yes ...?

Nobody recognizeable

@JohnRennie is it readable?

John Rennie

@Nobodyrecognizeable yes. How far have you got with this?

Nobody recognizeable

05:28

@JohnRennie I don't understand why the magnetic field would be zero. Are talking about motional emf?

user102532

Hi everyone

I have question.pls check

Nobody recognizeable

@JohnRennie isn't B=$\mu_0 I/2\pi r$ ?

user102532

T is is train moving towards - x axis and C is car moving towards + y axis
So , my teacher made a resultant velocity with -5i and -20j
Then he extended that black

Line which is resultant velocity.Then , my sir said that the car sees the motion of train along that black line.I am confused with this since the black line does not even touch the car
so how is that possible ?

John Rennie

@Nobodyrecognizeable You nee to use the Biot-Savart law here. This tells you that:

$$ dB = \frac{\mu_o}{4\pi} \frac{I d\ell \times \mathbf r}{|r|^3} $$

Nobody recognizeable

Yes

dl and r are perpendicular

John Rennie

05:34

The important thing here is $Id\ell \times \mathbf r$

This is a cross product so it is zero when $d\ell$ and $\mathbf r$ are in the same direction. Yes?

Nobody recognizeable

@JohnRennie yes

John Rennie

So the magnetic field at A is zero when the charged particle is moving directly towards A or directly away from A. Yes?

Shall I draw a diagram?

Nobody recognizeable

So we need to find the time period that's all.

@JohnRennie I get it professor

John Rennie

OK :-)

Nobody recognizeable

@JohnRennie so $t'=t+2\pi/\omega$

So a is right

user102532

05:40

@JohnRennie can you check my question as well sir

Nobody recognizeable

@JohnRennie a confusion

John Rennie

@Nobodyrecognizeable no

Nobody recognizeable

@JohnRennie later?

John Rennie

@Nobodyrecognizeable there

@Nobodyrecognizeable those are the two moments when the particle is moving directly towards or away from A. Yes?

Nobody recognizeable

@JohnRennie yes

John Rennie

05:51

So the time is the time it takes the particle to move between those two points.

Nobody recognizeable

Yes

@JohnRennie yes

John Rennie

So the time is the angle between the two red lines divided by $\omega$

Nobody recognizeable

Angle is $2\pi /3$

John Rennie

Yes, so if the time is $t$ then $\omega = angle/time = 2\pi/3t$. Yes?

Nobody recognizeable

So B

Thanks professor

John Rennie

05:56

Yes :-)

Nobody recognizeable

Will come back with more questions.

John Rennie

@user102532 hi, are you around?

Prateek Mourya

Hello @JohnRennie sir

John Rennie

@PrateekMourya hi :-)

Prateek Mourya

Sir doesn't the derivation suggest that the definition of coefficient of restitution isn't applicable everywhere

I mean they assume the momentum to be conserved for the system

But I have seen so many question where the momentum is not conserved

user102532

06:05

@JohnRennie could you also please check my question

John Rennie

@user102532 hi.

We talked about relative velocities yesterday didn't we?

In this case we need the velocity of the train relative to the care. Yes?

Prateek Mourya

Sir please can you just answer yes or no here

John Rennie

user102532 was first, let me finish answering his question.

@user102532 hello?

Prateek Mourya

Ok please ping me after that so that i can receive notification

John Rennie

@PrateekMourya will do :-)

@user102532 you need to answer promptly. It isn't fair to keep other people waiting while you take a long time to answer.

user102532

06:10

@JohnRennie hello

John Rennie

user102532

@JohnRennie Yes sir

@JohnRennie also sorry to keep you waiting

John Rennie

@user102532 on the left I've drawn the positions and velocities in the ground frame.

user102532

@JohnRennie ok

John Rennie

And on the right I have subtracted (20,0) from all the velocities to switch to the rest frame of the car.

OK so far?

user102532

06:12

@JohnRennie yes

John Rennie

So in the car frame the train is at the point (200,0) and moving with velocity (-20,05).

user102532

@JohnRennie ok . Let me see 1sec

John Rennie

That's what your teacher means.

user102532

@JohnRennie yes

He does

John Rennie

So in the car frame the train moves along the black line.

user102532

06:14

So , when we extent the black line touching the y axis , but car is not present there

John Rennie

You are correct that the line does not pass through the car.

user102532

When it touches the y axis point

John Rennie

If it did pass through the car that would mean the car and train would collide. Yes?

user102532

@JohnRennie i think no

that would mean the way the car is looking at train

the direction with which it is looking at train

the train doesn’t move on black line right

it just appears for car it is along the black line

That’s what I think

John Rennie

That's the line extended to meet the y axis.

user102532

06:16

How is it even possible that when car looks at train , it is along that meets y axis but not car

John Rennie

> the train doesn’t move on black line right

user102532

Yes

John Rennie

In the car frame the train does move along the black line.

user102532

one more thing , I think it will help.My sir Has also told this.one sec@JohnRennie

John Rennie

In the car frame the train starts at the point (200, 0) and moves with velocity (-200, -5), so it travels along the black line.

user102532

06:18

If here this is C car looking at train T

This is an example my sir gave also

John Rennie

You mean the line CT gives the angle that the driver in the car has to turn to look at the train?

user102532

@JohnRennie just thinking . If it travels along the black lime , it would just move out of the road it is initially on right

I am very sorry that.It is creating more confusion then@JohnRennie

John Rennie

In the ground frame the track runs west (to negative x) and the train is moving west along the track. Yes?

user102532

@JohnRennie yes sir

agreed

John Rennie

But in the car frame the track is moving south (negative y) at 20m/s

user102532

06:21

@JohnRennie how is that possible

John Rennie

The train is still moving along the track, but now the track is moving south.

user102532

Train is moving south sir @JohnRennie

?

How is that possible still @JohnRennie

John Rennie

In the ground frame the track is stationary and the car is moving north. Yes?

user102532

Yes

John Rennie

In the car frame the car is stationary and the track is moving south.

user102532

06:23

@JohnRennie yes

John Rennie

What my right diagram shows is the car frame. To get to that frame we subtracted the car velocity (+20,0) from all velocities. This makes the car have zero velocity i.e. we are now in the rest frame of the car.

15 mins ago, by John Rennie

user102532

@JohnRennie one sec,I reading

@JohnRennie yes m we are in rest frame of car

John Rennie

So in the rest frame of the car:
- the car is motionless
- the track now has velocity (-200,0)
- the train has velocity (-200,-5)

user102532

3rd velocity that you wrote is relative velocity of train with respect to car right

John Rennie

Yes

user102532

06:27

2nd,that you wrote we will not consider since we are talking about train when it motionless.Right ?

John Rennie

"the track now has velocity (-200,0)"

user102532

You wrote it has -200,-5 right ?

Train

I meant

John Rennie

Oops, I've just realised that should be (-20,-5) not (-200,-5)

But yes, that's the velocity of the train relative to the car.

user102532

What is the difference in track and train have different velocities.The train has to move along track . The car sees both train and track right

John Rennie

In the car frame the velocity of the train is (-20, -5) and the velocity of the track is (-20, 0). Yes?

user102532

06:31

@JohnRennie this one I didn’t get. The train has to move along track . The car sees both train and track

John Rennie

It's because the train and the track are moving south at the same speed, so the train stays on the track.

i.e. the y positions of the track and the train are always the same.

user102532

yes.Right

You have made x position same sir

not y position

y position is different I think

John Rennie

Oops, you're right :-(

I switched the coordinates round!

user102532

It’s ok sir.@JohnRennie

John Rennie

The track velocity is (0, -20) and the train velocity is (-5, -20)

Sorry, I can see why you were confused!

user102532

06:34

ok.Aait let me see it 1sec

0,-20

ok,it means the track is moving along negative y ax

John Rennie

Yes

user102532

when car is at rest

Ok.

John Rennie

Yes.

user102532

Train is moving slightly bended

ok.

So , my question is

The train has to go to west right.

but it reaches south

how is that possible

Prateek Mourya

Sir my coaching timing is now

John Rennie

06:37

In the ground frame the train goes west, and since the track goes west as well the train stays on te track.

Prateek Mourya

I will talk tomorrow

John Rennie

@PrateekMourya OK :-)

user102532

@JohnRennie ok

You know what I am doing is

I am imagining in real that I am sitting in a car

and track is going south and train in south east

John Rennie

@user102532 in the car frame the track still runs east-west, but in the car frame the whole track is moving south at -20 m/s.

user102532

Is this where I am wrong

John Rennie

06:39

@user102532 the whole track is moving south, and the train is moving south-west.

user102532

@JohnRennie is the way I am thinking wrong ?

John Rennie

The end result is that the train still moves west along the track.

user102532

Ok.So we make the question in such a way that we still get train and track in west

just the way we write is different

or solve

Is it that train is on y axis , behind car is train

John Rennie

Well in the car frame, the train starts out to the east of the car, and the train is moving south-west.

user102532

so , in reality with ground frame , what does that mean

Ok.South west

John Rennie

06:42

In the ground frame the train travels west and reaches x = 0, but at this time the car has travelled north up the y axis.

user102532

yes

It has reached some +y axis point

John Rennie

So in the ground frame the car is to the north of the train at the moment the train reaches x=0.

user102532

we don’t know that

John Rennie

Yes we do.

user102532

ok.When it reaches origin

John Rennie

06:43

The car is driving north up the y axis. Yes?

user102532

we just shifted the car ahead and train at 0,0

John Rennie

So when the train reaches the origin the car will be north of the train i.e. up the y axis.

user102532

Ao ofc in real life we can’t see car coming along south west and track south .It is just for calculation.In real , our answers will be same

Isnt it

Track and train is along east only when car is also moving

John Rennie

Yes, really the track is motionless. Railway tracks don't go speeding around the place :-)

When we switch to the car rest frame this is just a convenient way of doing the calculation.

user102532

Yeah.That’s the point

@JohnRennie thanks professor

John Rennie

06:49

:-)

user102532

They don’t tell that.That’s where I get confused

I meant my teachers

napstablook

12:43

@JohnRennie Hi :-)

John Rennie

@napstablook hi :-)

napstablook

can you look at a problem right now?

it might take some time

prnt.sc/wkz888

John Rennie

I'm about to go I'm afraid.

napstablook

Ok i will send. Just look when you find it convinient.

thanks

actually nevermind I got my error. . .

John Rennie

That was easy :-)

I like questions like this :-)

napstablook

12:54

Don't we all :-)

I think my errors always land in my blind spots :-))

anyways have a good evening

John Rennie

:-)

pi-π

16:12

@JohnRennie Hello. Are you there?

John Rennie

@pi-π hi :-)

pi-π

@JohnRennie Can we obtain bright central spot in Newton's Ring in reflected system?

John Rennie

Isn't the central spot dark in Newton's rings?

It's dark because the light gets a phase shift of pi when it reflects.

pi-π

@JohnRennie Yes the central spot is dark. But this question asks if it is possible to obtain bright spot at centre.

John Rennie

Well you could lift the lens a quarter of a wavelength off the mirror, so the gap at the centre was λ/4 instead of zero.

I can't think of any other way to get a bright spot at the centre.

pi-π

16:20

Okay. Can I ask another?

John Rennie

Yes

pi-π

A soap film $5 \times 10^{-5}$cm thick is viewed at an angle of 35 degrees to the normal. Calculate the wavelength of the visible light which will be absent from the reflected light.

John Rennie

Are you told the refractive index of the soap film?

pi-π

Its not given in the question. But we may use 1.34.

John Rennie

Let me draw a quick diagram ...

pi-π

16:28

Okay.

John Rennie

@pi-π It's going to look like this.

You get interference between the rays reflected off the two surfaces.

Not that the top ray phase shifts by π when it reflects, but the bottom ray does not.

pi-π

@JohnRennie Why no phase shift for bottom ray?

John Rennie

You need to calculate the phase difference between the two reflected rays as a function of wavelength, and find the wavelength where the phase difference is π.

@pi-π you get the phase shift when a ray reflects off a higher refractive index, but not when it reflects off a lower refractive index.

pi-π

@JohnRennie Are we going to use $2\mu t + \frac {\lambda }{2}$ as path difference?

John Rennie

No, because the light is not travelling normal to film but at an angle to the film.

We are told the angle of incidence is 35°, and the angle inside the film will be less than this due to refraction.

pi-π

16:37

@JohnRennie Is it $2\mu t \cos r + \frac {\lambda }{2}$?

John Rennie

The length of the ray in the film is $t/\cos r$ not $t \cos r$

Apart from that, yes, that's the correct approach.

pi-π

@JohnRennie So it is $ \frac {2\mu t }{\cos r} + \frac {\lambda }{2}$?

John Rennie

Yes

No, wait, it asks for the wavelength at which the light will be absent.

So you want the path length to be an integral number of wavelengths.

pi-π

@JohnRennie Path length wil be integral multiple of wavelength for constructive interference?

John Rennie

You want $\frac {2\mu t }{\cos r}$ to be an integer number of wavelengths because the top ray gets phase shifted by $\pi$.

That means the two rays will be out of phase by $\pi$ and will destructively interfere.

The question asks what wavelength there is no reflection i.e. destructive interferenece.

pi-π

16:45

@JohnRennie You mean $ \frac {2\mu t}{\cos r} = n \lambda$?

John Rennie

Yes

pi-π

@JohnRennie We calculate $\lambda $ for $ n= 1$ only or for other values of n as well?

John Rennie

Try it and see. You might find there is only one wavelength that gives destructive interference in the visible light range.

pi-π

@JohnRennie Okay.

Thank you!

John Rennie

:-)

Transcript for

Problem Solving Strategies