Problem Solving Strategies

tatan

3:44 AM

@JohnRennie What do you think will be the solution to this one?

Some of my friends are having doubt over this one

John Rennie

4:06 AM

@tatan If no rotational or vibrational modes are active the energy of the molecule will be $\tfrac{3}{2}k_bT$ so we get $\tfrac{1}{2}mv^2 = \tfrac{3}{2}k_bT$

tatan

So its (3) ?

John Rennie

$$ T = \frac{mv^2}{3k_b} $$

user8718165

Good morning :-)@JohnRennie

tatan

Yes

Thanks

But why do you say " If no rotational or vibrational modes are active the energy of the molecule"

I mean on what basis?

John Rennie

I was thinking aloud.

tatan

4:08 AM

If we include them then it should be (1) right?

John Rennie

The KE is always just $\tfrac{3}{2}k_bT$ and rotations or vibrations won't change that.

So I need not have made the restriction that there is no rotation or vibration.

tatan

isn't it the case that each Degree of freedom contributes 1/2KT?

So, 6 DoF including rotational and vibrational should contribute 6/2 KT=1/2 M (V_{rms} )^2

why is this not correct?

John Rennie

@tatan yes, but we are relating the temperature to the RMS velocity

tatan

yes

John Rennie

Total energy is KE + rotational energy + vibrational energy, but only the KE is related to the rms velocity

tatan

4:14 AM

I didn't know this

if so, then it should be 3/2

John Rennie

Consider two gases, one monatomic and the other diatomic with the same atomic/molecular mass at the same temperature. Both will have the same rms velocity and both have a KE of $\tfrac{3}{2}kT$.

tatan

ok... that solves the doubt

John Rennie

The diatomic also has rotational energy, so it has a higher specific heat, but the rms velocity is still just $\tfrac{3}{2}kT$.

tatan

ok

user8718165

Hii @JohnRennie are you free?

John Rennie

4:22 AM

@tatan the speed of gas molecules is related to the temperature by the Maxwell-Boltzmann equation:

Maxwell–Boltzmann distribution

In physics (in particular in statistical mechanics), the Maxwell–Boltzmann distribution is a particular probability distribution named after James Clerk Maxwell and Ludwig Boltzmann. It was first defined and used for describing particle speeds in idealized gases, where the particles move freely inside a stationary container without interacting with one another, except for very brief collisions in which they exchange energy and momentum with each other or with their thermal environment. The term "particle" in this context refers to gaseous particles only (atoms or molecules), and the system of...

This depends only on the temperature and molecular mass, not on whether the molecule is monatomic, diatomic or polyatomic.

tatan

Ok... i have understood it :-)

John Rennie

@user8718165 yes, I think we're done with molecules and their velocities

user8718165

@JohnRennie Ok sir...I have a doubt

John Rennie

@user8718165 Yes ... ?

user8718165

@JohnRennie imgur.com/hvu8Kxx

@JohnRennie I don't understand how they write $$I=\frac q{dt}$$ shouldn't it be $$I=\frac {dq}{dt}$$

John Rennie

4:33 AM

I'm not sure what they mean by $q$ there.

I guess it's either a treatment for a single particle of charge $q$, or it's a charge element $q$ in which case I think it should be $dq$.

Note that the equation is for $dB$, so there should be some infinitesimal quantity on the right side as well.

user8718165

can you tell some other way to prove this which is correct?

@JohnRennie yes sir

John Rennie

I'd need to see the whole section of the book to judge what exactly the book means.

user8718165

whole page?

John Rennie

Since they are talking about the velocity of the charge I guess it's the field due to a single charge with velocity $v$.

user8718165

@JohnRennie here is the page imgur.com/2560IYB

@JohnRennie Hello

John Rennie

4:52 AM

Aha, the book refers to the current density $\vec J$ and a volume element $dV$.

user8718165

@JohnRennie I got that current density part:-)

John Rennie

So it is writing $\vec J dV = q \vec v$ where $q$ is the charge within the volume element $dV$. It should really be written as $dq$ because it is an infinitesimal quantity.

user8718165

@JohnRennie so the book is wrong...

@JohnRennie moreover won't the current be infinitely large if this equation were used? $$I=\frac q{dt}$$

John Rennie

@user8718165 yes, it doesn't make sense to divide a non-infinitesimal quantity by an infinitesimal quantity.

It's unfortunate notation by your book, but the point it's making is still fine.

user8718165

@JohnRennie because its not HC Verma. Those books never use bad notations or hardly have any errors. Anyways, thanks a lot for helping me:-)

Scáthach

5:07 AM

@JohnRennie hello. Good morning. Are you free for some time

John Rennie

@Scáthach hi :-) I'm around for a few hours. I have to go out around 11 a.m. UK time i.e. in about 5 hours.

Though I'll be dropping out to work from time to time.

Scáthach

Okay

@JohnRennie imgur.com/a/SHe7DVr

I didn't get how to calculate Fy

You said adding centripetal acceleration to mg

Why

John Rennie

Suppose it was a point mass on the end of a string rather than a bar. If you whirl a mass on a string then obviously you feel an outwards force due to the motion of the mass. This force is equal to the centripetal force $mr\omega^2$.

So in this case if you were holding the string as the mass falls downwards you'd feel the weight of the mass plus the centrifugal force due to the motion of the mass.

@Scáthach Is this OK so far?

Scáthach

Yes

John Rennie

So you have to calculate the centrifugal force exerted on the hinge by the rotating bar due to its angular velocity $\omega$. Offhand I don't know the equation for the centrifugal force from a solid bar pivoting about one end, but we can easily derive it.

OK, a bit of scribbling later and I get the same result as if we just considered the mass of the rod at the centre of mass i.e. at a distance $L/2$ from the pivot i.e. the centrifugal force is:

$$ F = M \omega^2 (L/2) $$

So the total vertical force on the hinge would by $F_y = Mg + M\omega^2 L/2$

@Scáthach shall I explain how I worked out the centrifugal force or shall we just go with this?

Scáthach

5:24 AM

hmm i got it

John Rennie

So you just need to work out the angular velocity, and that comes from energy conservation i.e. change in gravitational PE = change in rotational KE.

Scáthach

5:39 AM

@JohnRennie hi

John Rennie

@Scáthach hi

Scáthach

@JohnRennie imgur.com/a/2Y8YsS5

About the block question , I get time to be 4sec

I worked in com frame

The Intial velocity of the rear block is 5m/s

So at the time of. Losing contact, the distance it travels is 10/π

Will this be shm?

John Rennie

I get the time to be $\sqrt{2}$ i.e. option (1)

Scáthach

How

John Rennie

It's simpler than you think.

Remember that the period of a simple harmonic oscillator does not depend on the amplitude of the oscillation.

It is simply given by:

$$ T = 2\pi \sqrt{\frac{\mu}{k}} $$

where $\mu$ is the reduced mass of the oscillating system.

Scáthach

5:45 AM

Okay

John Rennie

If we work in the centre of mass then we start with the two masses moving inwards and compressing the spring. The masses come to a halt at the maximum compression then they start moving outwards again.

Scáthach

Yes

Abcd

@JohnRennie What's the answer? Which option?

John Rennie

But as soon as the spring has returned to its original, uncompressed, length the masses come apart from the spring and no longer perform SHM. So we get just half a cycle of SHM.

Scáthach

@Abcd 1

John Rennie

5:48 AM

@Scáthach Make sense so far?

5 mins ago, by John Rennie

I get the time to be $\sqrt{2}$ i.e. option (1)

Abcd

@JohnRennie I was asking wrt Tatan's question

John Rennie

@Abcd (3)

Scáthach

@JohnRennie Yes

John Rennie

@Scáthach so the time is just T/2, which is:

$$ \pi \sqrt{\frac{\mu}{k}} $$

Scáthach

got it

i get 1 option too now

John Rennie

5:54 AM

@Scáthach cool :-)

PolarBear

A carom striker and a carom coin have masses 25 gm and 15 gm. When the striker, hits a
stationary coin elastically, the angle that it (striker) can be deflected from original
direction is :-
(A) between zero to 53° (B) between zero to 37°
(C) between 37° to 53° (D) All angles are possible

Why are all angles not possible?

John Rennie

@PolarBear A collision has to simultaneously conserve momentum and energy. There will be angles for which this is impossible.

PolarBear

@JohnRennie So I have to write both those expressions and then compare?

John Rennie

Yes.

I'm wondering if there is an easy way to do this by working in the centre of mass frame ...

In the centre of mass frame the maximum deflection angle is 90°. You can work out what the angle of this trajectory is in the lab frame. That should give the answer.

Hmm, I get the maximum as 31° not 37°.

On that grounds I'd choose option B

PolarBear

6:14 AM

@JohnRennie I think they do something similar in the solution

They get an exact answer

I don't really understand the solution though.

John Rennie

Yes, I got $3/5$, but it's $\tan\theta = 3/5$ not $\sin\theta = 3/5$

@PolarBear I can draw a diagram to explain if you want ...

PolarBear

So, I should read up about oblique collision in COM frame?

@JohnRennie I would be grateful!

John Rennie

OK, give me a moment ...

PolarBear

Sure

John Rennie

This is the initial state in the lab and COM frame:

PolarBear

6:21 AM

Okay

John Rennie

OK, now consider the collision in the COM frame:

tatan

@JohnRennie please ping once you are free:-)

John Rennie

In the COM frame the big striker is deflected by some angle $\theta$

PolarBear

Following

John Rennie

The collision is elastic, so the velocity of the big striker is still $15/40 V$, so the vertical component of the velocity is $V_y = 15/40 V \sin\theta$. OK so far?

PolarBear

6:27 AM

@JohnRennie Why will velocity of big striker remain same?

John Rennie

@PolarBear we are working in the centre of mass frame and it's an elastic collision. That means the velocities of the two objects are unchanged in magnitude, though they can change in direction.

That has to be the case to conserve momentum and energy.

PolarBear

I see

Got that

John Rennie

OK. So the maximum vertical velocity the striker can have is when the scattering angle is 90° in which case the vertical velocity is $15/40 V$ and the horizontal velocity is zero.

PolarBear

Yes

John Rennie

Like so.

PolarBear

6:33 AM

I get it

John Rennie

Now, to transform back into the lab frame we need to add back the COM velocity of $25/40 V$

PolarBear

@JohnRennie add back meaning?

Oh oh, I get it

Oh, no I don't.

The COM has no velocity?

John Rennie

PolarBear

Oh, I see. The horizontal one!

John Rennie

Yes.

PolarBear

6:38 AM

And now, we will add the components for each disc

And we can find the angle between those?

John Rennie

Yes, though actually the angle is just the arctan of the ratio

PolarBear

Ahh, I see. This makes the problem so much easier.

Thank you for explaining it to me!

John Rennie

@PolarBear Actually I've just realise my answer isn't quite correct ...

PolarBear

@JohnRennie What's the error?

John Rennie

@PolarBear I've assumed that a 90° scattering angle in the COM frame gives the maximum scattering angle in the lab frame, and I suspect that while that's almost right it's not exactly right.

PolarBear

6:45 AM

@JohnRennie I see, but the maximum angle can be 90° only, right?

John Rennie

If the scattering angle in the COM frame is $\theta$ then the components of the velocity are $F_y = 15/40V\sin\theta$ and $F_x = 15/40V\cos\theta$

We transform into the lab frame by adding a horizontal velocity of $25/40V$, so in the lab frame we get $F_y = 15/40V\sin\theta$ and $F_x = (25/40 + 15/40\cos\theta)V$

In the lab frame the scattering angle $\phi$ is $\tan\phi = F_y/F_x$ so it's:

$$ \tan\phi = \frac{15/40 V \sin\theta}{(25/40 + 15/40\cos\theta)V} $$

I suspect the maximum value of the right hand side is for $\theta$ slightly greater than 90°

So my answer is close but not exactly correct.

@PolarBear anyhow, I need to work now for a bit. Back later.

PolarBear

Okay, sure. Thanks for your time!

tatan

Is this discussion complete?

Dante

7:07 AM

brb

John Rennie

@tatan yes, I think we're done with the scattering question

Scáthach

@JohnRennie Hi.Are you done with @tatan

John Rennie

@Scáthach he doesn't seem to be around, so ask now

tatan

No... i would start now@Scáthach

John Rennie

Ah :-)

Scáthach

7:15 AM

okay @tatan

tatan

I am here

John Rennie

@tatan your turn

tatan

I have 2 questions

I have seen q no 12 in the main site... however i cant find it now... can you share the link if you find it?

John Rennie

17

Q: "Find the net force the southern hemisphere of a uniformly charged sphere exerts on the northern hemisphere"

This is Griffiths, Introduction to Electrodynamics, 2.43, if you have the book. The problem states Find the net force that the southern hemisphere of a uniformly charged sphere exerts on the northern hemisphere. Express your answer in terms of the radius $R$ and the total charge $Q$. Note: I...

tatan

Q no 9

And q no 11 in the previous image... isn't the unit vector of the area =\frac{1}{\sqrt2}(i-j). Doesn't that mean the flux \frac{E_0a^2}{\sqrt2}?

John Rennie

7:23 AM

The first step is to combine the 2 and 3 uF capacitors into a single 5 uF capacitor to simplify the calculation to a 4 uF and 5 uF capacitor in series. Then when we've worked out the charge on the 5 uF capacitor we can split it between the 2 and 3 uF capacitors.

@tatan let's do one question at a time.

tatan

Yes sure... i am just posting all in one go

Ok... done with q no9

John Rennie

@tatan you mean you've solved q9?

Yes

(C) ?

Yes

7:28 AM

OK, do you want to go back to Q11?

tatan

Yes

John Rennie

@tatan I'd probably consider the strip of the surface between $x$ and $x+dx$

tatan

Then?

John Rennie

The area of the strip normal to the $x$ axis is just $dA=adx$

So the flux is $d\phi = EdA = E_0 x a dx$

Oh wait, I misread the question

tatan

Can you tell me what is the unit vector of the area of the region?

Ok

Scáthach

7:36 AM

@JohnRennie Why not jut find area using vecors

John Rennie

I thought the field was $E = E_0 x$ i.e. dependent on $x$, but it's just a constant $E_0 \hat x$

tatan

Yee

John Rennie

It's just $\phi = E_0 a^2$ isn't it? Option (C).

tatan

Yes it is but I fail to understand

Scáthach

the angle is 45 or 135 i guess

@Dante Hi .Are you still here

John Rennie

7:40 AM

@tatan You can do it using the vector area, or just take a quick shortcut. I can explain both if you want.

tatan

I think the area is $a^2\frac{1}{\sqrt2}(i-j)$ Isn't it? @JohnRennie

Yes please

Dante

@Scáthach Hi, I'm having my lunch.

John Rennie

@tatan The magnitude of the area is the area of the shaded sheet. Yes?

tatan

Yes

a^2 by magnitude

Scáthach

@Dante Sorry to disturb.I just wanted to say I will ask after u

John Rennie

7:42 AM

@tatan The edge that lies along the y axis has length $a$, and the length of the other edge is $a\sqrt{2}$, so the area is $|A| = a^2\sqrt{2}$

Dante

Np, you can ask. I won't be back soon.

okay.Thanks

Oh

@tatan :-)

I thought it was a square

John Rennie

7:44 AM

Is the answer obvious now?

tatan

No... please show it

John Rennie

That's how we get the area of the shaded sheet

@tatan OK so far?

tatan

1sec

yes its ok

John Rennie

@tatan OK. The flux is $\phi = \vec E \cdot \vec A = |E| |A| cos \theta$ where $\theta$ is the angle between $E$ and $A$. Yes?

tatan

yes

John Rennie

7:51 AM

And $\theta=45°$ so $\cos\theta = 1/\sqrt{2}$

tatan

yes

John Rennie

So $\phi = E_0 \times a^2\sqrt{2} \times 1/\sqrt{2} = E_0 a^2$

tatan

oh yes

30 mins ago, by John Rennie

The first step is to combine the 2 and 3 uF capacitors into a single 5 uF capacitor to simplify the calculation to a 4 uF and 5 uF capacitor in series. Then when we've worked out the charge on the 5 uF capacitor we can split it between the 2 and 3 uF capacitors.

John Rennie

@tatan I thought you'd done that?

tatan

Can you explain what the earthing does here?

@JohnRennie yes.. but I am not too convinced myself

John Rennie

7:55 AM

Earthing allows a charge of -80uC to flow from earth onto the lower plates of the two capacitors.

tatan

ok so that becomes 0 C?

John Rennie

There's a class of JEE problems involving parallel plates with charges on them. The sort of thing I describe in this question:

5

Q: Why are the two outer charge densities on a system of parallel charged plates identical?

One of the ways examiners torture students is by asking them to calculate charge distributions and potentials for systems of charged parallel plates like this: the ellipsis is meant to indicate any number of additional plates could be inserted where I've placed the ellipsis. The plates are ass...

electrostatics electric-fields symmetry capacitance gauss-law

tatan

Ok thanks

a lot!

I will go through it... as of now I think I have no more questions :-)

@Scáthach Go ahead

John Rennie

Cool :-)

Scáthach

@JohnRennie One face of an equiconvex lens is silvered.then after silvering the lens will have more covering power or less

how to see this

John Rennie

8:00 AM

@Scáthach If you have a lens with no mirroring then the light goes through the lens once i.e. it goes in one side and out the other side. Yes?

Scáthach

yes

John Rennie

When you mirror one face this is equivalent to placing a concave mirror behind the lens.

I can draw a diagram to show this if it isn't clear.

Scáthach

yes a diagram please

John Rennie

The red line shows where we've created the mirror.

Ok so far?

Scáthach

yes

John Rennie

8:07 AM

So when you shine light in from left it:
1. goes through the lens and is converged
2. reflects off the concave mirror and is converged again
3. goes back through the lens and is converged a third time

While a lens with no mirror only converges the light once.

Scáthach

1st surface-2nd surface (convergence)

mirror(relection)

2nd surface-st suraace(converging)

so this ?

John Rennie

Yes

Scáthach

so I just add their focal length now

reciprocal of focal length

John Rennie

Yes. If the lens focal length is $f_l$ and the mirror focal length is $f_m$ then the total focal length is:

Scáthach

2Plens+P mirror

P(Power)

John Rennie

8:11 AM

$$ \frac{1}{f} = \frac{1}{f_l} + \frac{1}{f_m} + \frac{1}{f_l} $$

Scáthach

so positive convverging power

sorry mirror

it will act as a mirror

John Rennie

Yes

Scáthach

@JohnRennie The answer is mirror with more converging power

what is converging power

John Rennie

The power is just $1/f$

Scáthach

yes

John Rennie

8:16 AM

The lens with a mirrored face obviously acts as a mirror, and we've just worked out that its power is greater than a single lens, so the power is increased.

And it's obviously converging because all the elements in the combination are converging.

Scáthach

yes power gets increased

John Rennie

I need to work now. I'll be one to two hours - I'm not sure exactly how long.

Scáthach

okay