Problem Solving Strategies - 2021-04-23

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3:30 AM

@Xnero The point of radioactive decay isn’t that. Radioactive decay is a general term. The purpose is to just stabilise the nucleus. What you are saying happens in a category of radioactive decay which is called beta- decay. Sometimes the nucleus may havean excess of protons so beta+ decay may be needed

1 hour later…

4:42 AM

@Xnero The energy states of nuclei are very complicated. So complicated in fact that while we have various approximate ways to describe nuclei we still don't have an exact description of them. In some cases a nucleus can get more stable by turning a neutron into a proton, and in some cases it can get more stable by turning a proton into a neutron.

But as a rule a nucleus with too many neutrons undergoes beta decay and a nucleus with too few neutrons undergoes beta plus or electron capture decay.

@JohnRennie Hello sir :-)

@RobinSingh hi :-)

Sir suppose the bob of a pendulum is drawn aside so that the thread deviated through some angle. Now, we give it a horizontal velocity such that it starts moving in a spiral upwards

So in this situation is the vertical component of tension perpendicular to velocity?

In the situation you describe wouldn't it move in a cone?

i.e. you've just described a conical pendulum.

Ah yes sir a cone

Oh so this is a conical pendulum?

But still I wish to ask the question above

4:54 AM

@RobinSingh we generally use the approximation that the angle is small so cosθ ~ 1.

That means the vertical motion of the bob is ignored.

It that case the motion is in a horizontal plane, so it is always normal to the vertical.

In a real pendulum there is a small up and down motion so the velocity is not quite horizontal.

Ah no sir, in my question the bob is actually moving up

I think you need to draw a diagram for me ...

I’ll post the entire question

user image

Q192

At first, I couldn’t even visualise what they meant here so I had to check the solutions

That's a conical pendulum.

But a conical pendulum moves in a horizontal plane only, right?

5:00 AM

The ball is moving in a horizontal plane ... (approximately)

Then what do they mean by “deviate through the maximum angle pi/2”? Won’t any velocity do the job?

RishiNandha Vanchi

@RobinSingh i meant for the negative work by table

I must admit I don't understand that ...

What I just said?

user image

I don't understand what the underlined text means.

5:04 AM

Morning sir @JohnRennie

The ball will move in an ellipse.

:-)

Here’s a diagram from the solutions-

@Jasmine hi :-)

user image

RishiNandha Vanchi

5:05 AM

I think my max angle pi/2 they mean theta becomes pi/2 in the other side

@JohnRennie I hope you are healthy first of all been long time :-)

Ah, OK, so the small angle approximation is specifially disallowed.

RishiNandha Vanchi

I think we need to look at conservations rather than kinematics, because kinematics is hopeless here lol

@Jasmine covid has settled down a lot here and the lockdown is over. Life is gradually getting back to normal.

@JohnRennie thats nice :-)

5:07 AM

@RishiNandhaVanchi yes

@JohnRennie yes sir

@Jasmine I've been vaccinated now so my chance of getting it is small :-)

@JohnRennie thats nicer :-)

RishiNandha Vanchi

angular mom and energy maybe?

Apparently things arent good here :-(

I’ll have to go for a while, sorry

5:08 AM

@Jasmine No, I saw in the news that India is experiencing a big second wave of infections.

RishiNandha Vanchi

Lsin(theta)v_0 = Lu; v^2/2 + mgL cos(theta) = u^2/2?

Guess it would take time to get better

RishiNandha Vanchi

@JohnRennie Yup its BIG

people grew slack of it

but covid didnt

@JohnRennie Yes sir :-(

RishiNandha Vanchi

Like once vaxx came everyone went "ah yeah covid's over"

5:09 AM

@RobinSingh when you're back I think I know how to do this.

RishiNandha Vanchi

like until a week back I was going to school

@RishiNandhaVanchi :-(

To be fair we run that risk in the UK. The vaccine isn't 100% effective so if people abandon all the precautions there will be more cases even with the vaccine.

RishiNandha Vanchi

Yeah thats what is happening in india

@JohnRennie True :-(

RishiNandha Vanchi

5:11 AM

@RobinSingh v = root(2glsctheta)?

when ur back that is

@JohnRennie I wanted to ask something, can I ask now

RishiNandha Vanchi

shoot

@Jasmine yes, go ahead.

RishiNandha Vanchi

:-D

I wanted to use multisim to complete the observation table for my assignment

But I seem to be clueless how to go about it

5:14 AM

This site multisim.com ?

@JohnRennie Yes :-)

What circuit are you trying to simulate?

user image

So XFG1 is a signal generator and XSC1 is an oscilloscope?

user image

@JohnRennie yes sir

I just set up multisim but I dont know anything more about how it works

5:21 AM

OK, so that's a simple RC circuit and you're measuring the source voltage and the voltage across C. Yes?

That's exactly the example they describe in the introduction here.

@JohnRennie Yes

@JohnRennie oh ! I will go through it!

@RishiNandhaVanchi is that sec(theta)?

@JohnRennie yes sir pls

@JohnRennie I will ping you again if I cant get it :-)

@Jasmine OK :-)

@RobinSingh this is my diagram:

user image

Looking at the side view we have the ball displaced a distance r so the angle will be sin θ= r/l

Then in the three diagrams on the right we are looking vertically down at the system.

OK so far?

Yes sir

5:33 AM

In all cases the ball moves in an ellipse and the eccentricity depends on the magnitude of the velocity v. For v = 0 it is just a regular pendulum that swings along a straight line.

For v > 0 it moves in an ellipse.

If you look at the rightmost diagram where v is large the semimajor axis of the ellipse is y, and I'd guess the question means that y = L i.e. it swings so far that θ = π/2.

@RobinSingh Does this all make sense?

@JohnRennie The rightmost diagram is when the ball is at its highest, right?

The circles are drawn at the starting point i.e. when the ball is at it's lowest. As it rotates θ increases i.e. the ball moves outwards, and it also moves upwards. At the top and bottom of the ellipse it is farthest from the centre and also at the highest point on the vertical axis.

Ah ok sir, everything's good so far

Now at the left and right and top and bottom of the ellipse the velocity is normal to the radius vector, so the angular momentum is just mv times the radial distance. And the angular momentum is conserved.

So
mv r = mv' y
where v' is the speed at the top of the ellipse. Yes?

Yes sir

5:43 AM

user image

And also total energy, KE + PE, is conserved.

Take our zero of PE at the starting point, so at the starting point total energy = ¹⁄₂mv²

Then at the highest point the KE is ¹⁄₂mv'² and the PE is mgh where h is the vertical distance moved upwards.

And you can calculate h from the distance y because at the highest point sinθ = y/L

And h = L cosθ

@JohnRennie But isn't Tension doing work here too? As I asked earlier although the horizontal component of Tension is always perpendicular to velocity, its vertical component isn't.

The tension acts along the string, and the length of the string is constant, so the work done by the tension is ... ?

ZERO

Ok sir thank you :-)

:-)

user image

user image

@JohnRennie this is question from @CHEMOJEE can you explain the situation what's really happening?

5:57 AM

I think the idea is that when the container strikes the ground both the container and air inside it are moving at some velocity given by v² = 2as.

The collision of the container with the ground is elastic so it just bounces back at the same speed.

@JohnRennie yes, so it should go to same height?

However the air inside collides with the inside wall of the container, and this collision is perfectly inelastic.

So the KE of the air gets turned to heat, then on the upwards part of the bounce the container has to accelerate the air back to the same velocity as the container.

Does that sound reasonable?

@JohnRennie Ok right

@JohnRennie yes

Note that it says all the internal fluctuations in the gas damp out

I assume this means the air is viscous and converts motion to heat.

@JohnRennie ok instantaneously

6:02 AM

I guess we assume it's instantaneous for the purposes of this question.

So, at the start velocity of gas turns to be 0 ?

On the way down gravity affects the container and the air equally, so just before the collision both have the same velocity given by v₁² = 2as

At the bottom of the bounce, when the velocity is instantaneously zero, the container has converted the KE to elastic energy but the air has converted its KE to heart so it is lost.

Now the bounce starts and the container converts its elastic energy back to KE.

Okok got it now where energy is lost

However the container has to accelerate the air as well. So we end up with the initial KE of the container being equal to the final KE of the container plus the air.

@JohnRennie yes

6:09 AM

@JohnRennie Hii sir

Free?

@SrijanM.T hi, yes I'm free.

Ohk.

1) if transition for molecules happens from high pressure to low pressure. Then , is it also that it happens form low volume to high , high temp to low temp

2) [![enter image description here][1]][1]

[1]: https://i.stack.imgur.com/p8nsx.jpg

Sir , I told you for this Q that we consider the transition like nL-x and nR+x for the moles. Can we also consider like this for Temp , volume,pressure ?

For this Q , if I make an assumption that n does not change or it remains constant even though transition happens , can it be possible even though pressure changes.

@JohnRennie Sir , these are my Q

If confusion in anyone in understanding,I’m here

@JohnRennie I am still getting 11m as answer?

@Rover I haven't done the calculation. Have you got the answer key?

@JohnRennie answer is 8.1m

At bottom velocity of both container and gas v=$\sqrt{ 2 ×10×12.1}$

6:21 AM

Give me a moment to do the calculation

Now, since K.E of gas is lost , energy =1/2(10)$v^2$

@JohnRennie ok

OK, I agree my method gives 11m

So the question must mean something else

@JohnRennie yes, yesterday also I tried vaguely and got 11m only.

@JohnRennie yes

If h' = 8.1 (i.e. 9²/g) then working backwards the velocity when the box + air bounced up is give by v'² = 2gh' = 2 x 9²

So v' = 9√2

Yes

6:34 AM

So the initial KE is ¹⁄₂ x 11 x 2 x 9² = 11 x 9²

Yes and that means v^2 before hitting is 2.2×$9^2$

No, the initial height is 11²/g

Yes

So the initial KE is (10 + 1) x g x 11²/g = 11 x 11²

So the change in KE is 11 x 11² - 11 x 9²

Yes, right so $11^3-11×9^2$ is lost ?

6:40 AM

Hmm that's 11 x (11² - 9²)

And that difference of squares factors to (11 + 9)(11 - 9)

But I'm not sure this is leading anywhere ...

Yes, so that must mean there was some K.E of gas initially before releasing which damped out on collision?

I guess so ...

I'm sorry but I cannot see the principles behind that question.

@JohnRennie ok, no problem sir maybe something is missing , some information to question..

See you later @JohnRennie . Have a good day :-)

Bye. Sorry I couldn't help :-(

@JohnRennie Hello sir. Can we discuss my Q in half an hour. I have session right now in 5 minutes ?

6:52 AM

@SrijanM.T Yes, I'll be here.

Thank you sir. I’ll come soon

7:18 AM

@JohnRennie Hi :-)

Are you free?

@Wolgwang hi, yes I'm free :-)

yesterday, by Wolgwang

> Two trains A and B of length 400 m each are moving on two parallel tracks with a
uniform speed of 72 km h–1 in the same direction, with A ahead of B. The driver of
B decides to overtake A and accelerates by 1 m s–2. If after 50 s, the guard of B just
brushes past the driver of A, what was the original distance between them ?

yesterday, by Wolgwang

Motion relative to train A

Intial velocity of train B=$0$

Initial distance between the train=$x$
$$800+x=\frac12(1)(50)^2\\x=450m$$

What did I do wrong?

I guess it depends on what exactly we mean by the distance between the trains.

I would take this to be the distance between the fronts of the two trains, so the length of the trains is irrelevant.

I.e. when we say the initial distance is $x$ we mean the distance between the fronts of the trains is $x$, and the moment of overtaking is when $x = 0$.

So you would just use $x = \tfrac12 a t^2$

@JohnRennie I had taken the distance between back of A and front of the train B.

That isn't how I would define the distance

7:30 AM

So $450+800=1250m$ which is the correct answer.

Thanks :-)

@JohnRennie Hello sir

Got free now

@SrijanM.T hi, yes I'm free now.

1) if transition for molecules happens from high pressure to low pressure. Then , is it also that it happens form low volume to high , high temp to low temp

2) [![enter image description here][1]][1]

[1]: i.stack.imgur.com/p8nsx.jpg

Sir , I told you for this Q that we consider the transition like nL-x and nR+x for the moles. Can we also consider like this for Temp , volume,pressure ?

For this Q , if I make an assumption that n does not change or it remains constant even though transition happens , can it be possible even though pressure changes.

@JohnRennie Great.

Yes

Yes, you can solve the question that way using simultaneous equations. But one of the simultaneous equations is that nL + nR = n, where n is the total number of moles.

Ohk. This is for 2nd one I guess

7:34 AM

So by writing nL + x and nL - x you are basically doing the same thing.

@JohnRennie Ohk. I can also write T_L -x

and T_R +x?

Doesn't the question state the temperatures remain constant?

Let’s say they both change

And then come at an equilibrium temperature where both values are same

Then , for a change. We have to do T_L -x
and T_R +x?

Then you need to specify some other condition e.g. is the whole system adiabatic i.e. no heat can enter or leave? In that case the internal energy is conserved.

You also need to specify if heat can travel along the pipe between the two sides. If so then the final temperature will be the same on both sides.

Ohk, let us say adiabatic for now

@JohnRennie This one , is it even Sth like they would to jee students ? Like is it of thermo chapter ?

@JohnRennie This one I know they ask

7:37 AM

Can heat travel along the pipe? i.e. will the temperature end up the same on both sides?

@JohnRennie Yes. Right

Thats how our heat transfer is happening

between the pipes

And I understood that by adiabatic , you meant the temp of surrounding around them(container ) or where the container are right ?

Temp of surrounding, sth like if it container is in a garden of temp 300K.

Then , heat transfer happens there ?

By adiabatic I mean the two spheres are isolated from the environment around them. So the spheres can exhange heat with each other (along the pipe) but not with their surroundings.

@JohnRennie ohk.

So , they exchange with two methods. Pipes and between themselves (exchanging)

Then you can calculate the final temperature because the total internal energy is constant

Ohk.

Sir , about VL-x and Vr+x if I consider ?

@JohnRennie This method which I told you about.

You were right about it being 1/3rd factor of 900K.

Also , I feel when talking about VL-x and Vr+x , That method which I used right

Would be in some way used

7:48 AM

I'm not sure what you are asking. I think this will work better if we consider a particular question.

Ohk

user image

Take this one

So , volume changes , T changes , moles change and so does pressure.

@JohnRennie

Then , Can we consider Vl-x

VR+x

Or there has to be Sth else also

@JohnRennie Sir , pls ask if not understood anything

Are we assuming the spheres can exchange heat with each other?

i.e. they have the same final temperature?

Yes. Same final T

That happens in two ways. Between the pipes and between the sphere

Right ?

You wouldn't do this by assuming some number of moles x is transferred.

That would be quite a hard calculation.

Yes it is.

Right

@JohnRennie That’s why I did this method 2

7:59 AM

The way you do this is to find the number of moles on each side before the tap is opened using n = RT/(PV)

Yes right

Then you find the internal energy on both sides using U = n C_v T

Ohk.

Now in the final state we know that the number of moles can't have changed, and we know the internal energy can't have changed.

No of moles would change right ?

8:01 AM

The total number of moles i.e. n₁ + n₂ is constant.

@JohnRennie Yes right. If we take both cylinder as one system

The number of moles on each side will change, but we don't need to know the final number of moles on each side.

Yeah. Then , we can also write VL+VR

As total volume right ?

Yes. P and T are the same on both sides and V = V₁ + V₂

And n = n₁ + n₂

And the total internal energy is constant.

Yes

8:06 AM

We can go through the calculation if you want.

Just the equation sir

R*n*(VL+VR)/U = to what I didn’t get ? would be the equation be like thissir ?

Start by calculating n₁ and n₂. Are you comfortable with doing that?

Yes. Ohk. I’ll write it

10*8.21/R*U?

N_L

P₁V₁ = n₁RT₁

Yes?

Yes

Sorry. Didn’t get what to write at position of T1

8:10 AM

So n₁ = P₁V₁/(RT₁)

and n₂ = P₂V₂/(RT₂)

OK so far?

Yes

@SrijanM.T n₁ is the number of moles in the left container before the tap is opened.

Yes

V₁ is the volume of the left container, P₁ is the pressure in the left container and T₁ is the temperature of the left container.

yes sir

8:12 AM

So what we are doing is taking the equation of state PV = nRT and applying it to the left sphere before the tap is opened to find n.

Yes sir

I don't get the feeling you are comfortable with this ...

Only with T

@SrijanM.T I don't understand what that means

The problem is teacher has not taught thermo

But went on for advanced Q of gas laws

Which have thermo too

8:15 AM

Then I think you are going to struggle to understand this.

I’ll do one thing

Ill read thermo

And then come back to you maybe after a few days for this Q

2 or 3

Then. It is better

OK :-)

I can also answer then

Thank you sir

@JohnRennie Shall we do 1st and 3rd one

What is the 1st one?

1) if transition for molecules happens from high pressure to low pressure. Then , is it also that it happens form low volume to high , high temp to low temp

8:18 AM

I don't understand what that means.

In the Q above, we consider that moles travel from LHS to RHS?

So , we take nL-x and NR+x. Ok so far ?

OK

Similarly on two buckets connected by a pipe having volume 1L and 100mL

We say exchange happens due to high pressure to low pressure

Ok so far ?

Yes

So , can we also say if PV=nRT

That it happens because of high temp to low temp

Low volume to high volume

8:20 AM

No, because it depends on the ratio T/V not on just T or just V

Since P in directly proportional to 1/V

@JohnRennie Why do we say then that it depends on P?

only

high pressure to low pressure

The force the left sphere exerts on the gas in the tube is F₁ = P₁A where A is the cross sectional area of the tube. Yes?

Yes

And the force the right sphere exerts is F₂ = P₂A

yes

8:24 AM

So the net force is Fnet = F₁ - F₂ = P₁A - P₂A = (P₁ - P₂)A

Yes

So the force on the gas in the tube depends only on the pressure difference and not on the volume or temperature difference.

Ohk. That force , forces the moles to travel ?

Yes

The gas in the tube flows out of one sphere and into the other because that net force is non-zero.

Ohk sir

The 3rd one

We know PV=nRT , So if P changes by a 2/rd factor then , so will T. Does it mean that there is no change in V or n there

Is it necessary for V to change to balance ?

Or only one value is enough

Or T , or n

8:29 AM

What is the 3rd question?

We know PV=nRT , So if P changes by a 2/rd factor then , so will T. So , if a change in P , then a change in T looks necessary. But Does it mean that there is no change in V or n there

@JohnRennie did you get it sir ?

> So if P changes by a 2/rd factor then , so will T

Yes. Is it true ?

No. If P changes by a 2/rd factor then so will T/V not just T.

Ohk, what about n sir ?

8:35 AM

If you allow n to change as well then if P changes by a 2/rd factor then so will nT/V not just T.

Ohk, thanks a lot sir,

I’m done

OK :-)

9:06 AM

wow. exactly one year ago , I stumbled upon this chatroom, it used to be as busy as it is right now, and unfortunately, so was the covid situation in India :(

@JohnRennie why does molecules of solid possess least energy while that of gas possess highest

Energy = 1/mv^2

Then , m of solid is high but not v

Whereas , v of gas is high but not m

So , in terms of potential energy. I think solid has it more

than gas.

What other reason can be the answer

I'm busy at the moment I'm afraid.

Sure.

Np

9:26 AM

@SrijanM.T that's too vague a question to be answered.

For example at < 0°C ice has a lower energy than water vapour. We know that because below 0°C water exists as ice.

But above 100°C ice has a higher energy than water vapour. We know that because above 100°C water exists as the gas.

9:43 AM

user image

@JohnRennie Ohk. So , for Q1

Textbook gave c as right option meaning that it is wrong

so , I should consider a , b and d to be true

So , I can say that even a is false ?

@JohnRennie Your point is right sir.

I think option (a) is too vaguely defined to be able to say. It isn't a good question.

@JohnRennie I hope Q like this don’t come in JEE.

@JohnRennie ohk. I got it sir.

I guess it means that molecules in a gas are free to move around while molecules ina solid are fixed in one average position.

But molecules in a solid have energy associated with their vibratory motion so even though they aren't free to move around they still have energy associated with motion.

@SrijanM.T the JEE questions are generally pretty good and are much clearer than this.

@JohnRennie Ohk.

In that sense

Mostly yes

Thank you for your help sir.

@JohnRennie Do you usually come around 10pm Indian timing

I am sometimes around in the evening from about 10pm until 11pm Indian time, but not always.

9:57 AM

@JohnRennie Ohk.

In morning ?

10am I guess

to 4pm usually as I have noticed

I'm here from about 05:30 until 12:30, then from 16:30 to 18:00 UK time. At the moment India is 4¹⁄₂ hours ahead of the UK so add 4.5 to those times.

Ohk. Almost correct

Thanks a lot sir.

:-)

1 hour later…

11:18 AM

@Rover Thanks for having an interest in it. It is actually a pathfinder question. The author provides following hint.

user image

@Rover @JohnRennie

5 hours later…

4:18 PM

@JohnRennie Hello sir !

Are you here for some time

@Jasmine no, I'm going soon. For some reason I'm feeling tired this afternoon so I'm going to quit early.

@JohnRennie no problem sir take care :-)

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Apr '2123

Problem Solving Strategies

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