Problem Solving Strategies

12:15 AM

Can anyone tell me if suspend a body with a spring, can I take spring potential energy to be zero at any point. If not why?

sammy gerbil

12:31 AM

@LoopBack Yes. Potential energy is always relative to some reference point. It is only differences in PE which matter. You can choose whichever reference point is convenient and assign that point as zero or any other figure. It will make no difference to the result of your calculation. The only time it matters is when you are asked for the absolute potential (or potential energy).

Loop Back

@sammygerbill but this ain't true in my case, wait I'll write what I did

Let the spring potential energy as well as gravitational potential energy to be zero at equilibrium position. If a sudden blow is given to the block it's total mechanical energy just after the blow will be ½mv². When the block reach the extreme position the total mechanical energy will be ½kx²-mgx.

As no external force is acting total mechanical energy is conserved (here the spring and earth are considered as the system therefore gravitational force is not external force) equating both I'll get a quadratic in x. ½mv²=½kx²-mgx.

Next case: if I take potential energy of the spring to be zero in it's natural length and gravitational potential energy to be zero at equilibrium position. In equilibrium position mg=ky. Or y=mg/k. Just after a sharp blow total mechanical energy will be ½mv²+½ky²=½mv²+m²g²/2k. At the extreme position total mechanical energy will be ½k(mg/k+x)²-mgx. Equating both I will get ½mv²=½kx². So where have I gone wrong

Nobody recognizeable

1:07 AM

@sammygerbil are you here?

sammy gerbil

@Nobodyrecognizeable hello

Nobody recognizeable

@sammygerbil I'm not getting a big derivation from collisions. Will you help me?

sammy gerbil

@Nobodyrecognizeable A big derivation? What help do you need?

Nobody recognizeable

@sammygerbil just the paperwork and big calculations. The question is how much kinetic energy of the particles is lost after collision. Assuming both had initial masses $m_1$ and $m_2$ and non zero velocities $v_1$ and $v_2$. And coefficient of restitution is $\epsilon$ .

I am doing mistakes in calculation.

The answer they are giving is $\frac{m_1 m_2(1-\epsilon^2)(v_1-v_2)^2}{2(m_1+ m_2)}$

@sammygerbil are you trying this?

sammy gerbil

@Nobodyrecognizeable And what answer are you getting?

Nobody recognizeable

1:20 AM

@sammygerbil i'm coming back after taking that picture please wait a while.

sammy gerbil

@Nobodyrecognizeable ok that looks horrible. I will need several minutes to think about this.

Nobody recognizeable

@sammygerbil well thats surely horrible question to do! But i gotta say i used $u_1$ and $u_2$ instead of $v_1 $ and $v_2$

@sammygerbil I'm doing some different stuff. Please ping me whenever you come back.

sammy gerbil

@Nobodyrecognizeable Yes I noticed, but it makes no difference. Yes I will ping you.

Nobody recognizeable

1:35 AM

@sammygerbil ok thanks bye for now.

Nobody recognizeable

1:50 AM

@sammygerbil have you got that ?

sammy gerbil

@Nobodyrecognizeable Yes but I haven't found a solution yet.

Nobody recognizeable

2:05 AM

@sammygerbil ok ping me when you solve that.

Loop Back

@Nobodyrecognizeable if you don't mind I have moved this question for you. I can't type it here but I'll a snapshot of the solution. I have take initial speeds to be u1 and u2

And if you are thinking what is k in the first paper see this

And I meant to say I have solved this question not moved it(damn autocorrect)

Nobody recognizeable

2:27 AM

@LoopBack many many thanks man.....

sammy gerbil

@LoopBack Your mistake is the calculation of elastic PE.

Loop Back

@Nobodyrecognizeable np

@sammygerbil can you elaborate where is my mistake

sammy gerbil

@LoopBack Although elastic PE can be given the value of zero at the equilibrium position, it does not mean that the difference in elastic PE can be calculated using $\frac12 k x^2$ with $x$ as difference in position.

In this formula $x$ refers to extension from natural length.

Loop Back

So I cannot take elastic pe 0 at equilibrium

sammy gerbil

You can but it makes the calculation far more difficult.

Loop Back

2:38 AM

Can you tell me the correct calculation

Taking elastic pe and gravitational pe 0 at equilibrium

sammy gerbil

The correct calculation is to use $x$ as defined by the formula : it is the extension from natural length.

Loop Back

@sammygerbil Ohh now I get thanks

sammy gerbil

This sounds confusing, but what you were intending to ask was not "Can I define elastic energy as zero at any point?" which is true, but "Can I use the formula $E=\frac12 kx^2$ using $x$ relative to any point?" which is not true.

Nobody recognizeable

@LoopBack your solving method is great man. This resolved the problem in very short. Thanks man

@sammygerbil anyway thanks for trying to help.

Loop Back

@sammygerbil exactly

Nobody recognizeable

2:44 AM

Goodbye everyone.

Goodbye

goodbye

@sammygerbil wait

ok

I have to ask another thing

sammy gerbil

2:50 AM

ok

Loop Back

Let's suppose I chose elastic pe at equilibrium position 0. So the elastic potential energy at the extreme should be working where x is extension from natural length

Which will be ½k(mg/y+x)²

So it will be same no matter what 0 reference I chose for elastic pe

Correction in the first statement --> should be ½kx²

Abcd

@sammygerbil ???

sammy gerbil

@LoopBack Sorry but I don't see what problem you have here. Elastic PE at extreme is ½k(mg/y+x)² which is correct. So?

@Abcd Sorry I thought you had gone to sleep.

Abcd

@sammygerbil Yes slept and back. Were you able to solve that problem??

sammy gerbil

@Abcd No there have been several interruptions.

Abcd

2:59 AM

@sammygerbil $:($

sammy gerbil

I have a question though : is the separation of the slits 1mm? That does not seem right.

Abcd

@sammygerbil please see the diagram. Its clearly given as 1 mm.

Loop Back

Wait, you know what taking elastic pe at equilibrium simply means that I have to subtract a term m²g²/2k nothing else

Abcd

@sammygerbil Can you please solve it now??

sammy gerbil

@Abcd Also yes but this is much larger than the wavelength, which means that the $d\sin\alpha$ is very much bigger than $t(n-1)$.

I think maybe $d=1\mu m$ was intended. Then the 2 terms are comparable.

Abcd

3:03 AM

@sammygerbil NO

@sammygerbil if you do calculations then you'll see it is not.

@sammygerbil Is it possible for you to solve it right now?

sammy gerbil

Also I did not understand how you got your value for $\Delta x$.

No I cannot solve it right now. Sorry.

@Abcd ok I suppose if you used $d=1mm$ and got the other parts correct then it cannot be $1\mu m$.

Abcd

@sammygerbil yes :(

sammy gerbil

@Abcd How did you get $d\sin\alpha=10^{-5}$m?

Abcd

@sammygerbil $d = 10^{-3}$ ;$\sin \alpha \approx \alpha = 10^{-2}$

sammy gerbil

@Abcd But the question states $\alpha=\frac{1.8}{\pi}=0.573$.

Abcd

3:13 AM

@sammygerbil question says $\alpha = \left({\dfrac{1.8}{\pi}}\right)^\circ = 10^{-2} \pu{rad}$

sammy gerbil

Ah. I didn't understand the handwritten symbol outside the brackets.

Why didn't they just say $10^{-2}$ rad?!

Abcd

@sammygerbil dont know, they are crazy

sammy gerbil

@Abcd The formula you need to use for resultant intensity when waves of unequal intensity interfere is $I=I_1+I_2+2\sqrt{I_1}{I_2}\cos{\phi}$ where $\phi$ is phase difference.

Abcd

@sammygerbil Are you getting right answer using that

sammy gerbil

@Abcd Not yet. I'm getting tired. I may have to leave this for John Rennie to look at. I guess he might be starting soon. I think I need to sleep a while, and I have something else I need to do when I get up.

Abcd

3:26 AM

@sammygerbil Okay bye ...

sammy gerbil

Sorry. :(

Goodbye.

sammy gerbil

4:23 AM

@Abcd I couldn't sleep so I had another try and got it using the formula I quoted along with your value of $\phi=\frac23\pi$. If the incident intensity is $I_s$ then the intensities on the screen from each slit are $I_1=0.5I_s$ and $I_2=2I_s$ (slit 2 is twice as wide as slit 1). The maximum intensity on the screen is $I=4.5I_s$ while the intensity at O is $I_O=1.5I_s$. Therefore $I_O=\frac{1.5}{4.5}I=\frac13I$.

Abcd

7:04 AM

@sammygerbil I will see this after some time.

@sammygerbil Are you retiring to sleep? Because I have a question ...(with my attempt)

sammy gerbil

@Abcd I am not able to sleep!

Abcd

@sammygerbil Why?

sammy gerbil

The long telephone call yesterday evening gave me things to think about.

Abcd

@sammygerbil Oh I hope everything is fine!!

sammy gerbil

Yes I think it will be.

Abcd

7:08 AM

@sammygerbil Can I ask my question?

sammy gerbil

Yes.

You said you got all the answers correct. But I don't understand how you got minimum intensity not equal to zero using your formula.

Abcd

@sammygerbil I will think about that question later and will send you attempt of all parts.

sammy gerbil

ok

Abcd

@sammygerbil Question and attempt^

sammy gerbil

@Abcd What do you think is wrong here?

Abcd

7:15 AM

@sammygerbil it looks perfect to me

sammy gerbil

@Abcd But ... ?

Abcd

@sammygerbil but answer given is $-6.8 eV$

sammy gerbil

@Abcd Your answer of $-10.2eV$ is the difference between the ground state and 1st state energy which is $-3.4eV$. (I recall we had this value in a question before I went on my vacation.)

Abcd

@sammygerbil i dont get you.

sammy gerbil

@Abcd I mean

that I recognise that your answer is probably wrong anyway.

Abcd

7:28 AM

@sammygerbil Whats my mistake but??

sammy gerbil

@Abcd I think your formula for $x$ is difference in total energy between ground and 1st excited state.

Abcd

@sammygerbil questions asks for total energy only

sammy gerbil

@Abcd It says potential energy in 1st excited state is zero.

Abcd

@sammygerbil Pleas see I Have clearly written $TE = PE/2$

sammy gerbil

@LoopBack Maybe you can look at this question? ... My brain isn't focusing on this.

Loop Back

7:38 AM

@sammygerbil Ok I'll give it a try

@Abcd My answer is -6.8eV

Abcd

@LoopBack I want to know the mistake in my attempt.

@LoopBack How did you solve it? Let me see your attempt to figure out my mistake.

Loop Back

@Abcd I don't know why are you wasting you time writing such a big derivation. Total energy of hydrogen atom is -13.6/n² eV. And the potential energy by -27.2/n² eV. The question says potential energy is take as zero and not total energy.

Potential energy in the second excited state is -6.8eV so if this is zero means you will have to add 6.8eV to total energy at the ground state. So energy will become -6.8eV

Abcd

@LoopBack Oh I see I have got my error!

sammy gerbil

@LoopBack well done

2

Abcd

@LoopBack Are you a JEE Aspirant?

@LoopBack thanks.

Loop Back

7:46 AM

Don't mind but if ever you don't get your answer in the first attempt you should read the question thoroughly

Yup

Abcd

@LoopBack Class 12?

Loop Back

@Abcd thanks

Dropper

Self study

WBU

Abcd

FIITJEE class 12

Loop Back

Ok

I don't think Sammy is Indian

sammy gerbil

namaste!

Abcd

7:49 AM

@LoopBack he is from UK

Loop Back

Yes I saw his stackexchange profile

Hema

8:16 AM

Would someone please help me with this question? The answer is A-p, B-q, C-t, D-s. The only way I can think of is using the result pressure at any point in the box at a depth h fromupper surface and distance l from left surface is hpg + lpa. But how do I integrate this? For instance I can't divide the upper face into strips since pressure will vary throughout each strip. This is confusing me. Or is there another way to do this?

Loop Back

8:35 AM

I if you want to solve the question within a minute you can follow my steps.

Now the answer is clear R3 should be ma+something so it will be t, which means R4 should be q. Similarly R1 should be mg+something which means R1 is s, so R2 should be p. These are the reaction forces the liquid. Liquid will exert equal and opposite force on the container

@Hema This ain't the proper solution but we have to solve questions quickly.

Hema

9:32 AM

@LoopBack thank you!

@LoopBack I have a small doubt, why did you take direction of R2 as down?

And R1 as up? Couldn't it be the opposite, since they are reaction forces of the liquid on the container walls?

Loop Back

10:19 AM

@Hema a simple answer reaction is a pushing force. You cannot lift a block just by placing your hand over it. You can can lift it only when you place your hand beneath it

Avnish Kabaj

10:42 AM

Solving questions like this is the word thing you could ever do

Right now you aren't attempting a paper

The purpose of solving that question was to understand the concept behind it

Not reaching at the answer

user237650

11:58 AM

Hello! Is this chat room meant for asking doubts. Can I also participate.

Hema

@AvnishKabaj how do I do it then?

Would you please help?

Avnish Kabaj

12:25 PM

@user237650 sure

@Hema if the box wasn't accelerated that is only in the prescence of gravity would you be able to solve it then?

Hema

1:09 PM

@AvnishKabaj according to my understanding the net force on upper face would be P0A and on lower face would be (P0 + lpg) A where l is side or P0A+mg,is that right?

Loop Back

1:49 PM

@Hema solution to your question

Hema

2:07 PM

@LoopBack I understood now :)

@LoopBack just a minute

I got it

For a moment I wondered why we weren't taking atmospheric pressure but then realized we are dealing with force due to fluid pressure

Loop Back

2:56 PM

@Hema exactly

Btw atmospheric pressure should not act on water as the container is rigid and will nullify atmospheric pressure by applying equal reactionary force

Anonymous

@LoopBack Are you a high school student?

Abcd

7 hours ago, by Loop Back

Dropper

7 hours ago, by Abcd

@LoopBack Are you a JEE Aspirant?

7 hours ago, by Loop Back

Yup

Anonymous

Yeah, I read that now.

Abhas Kumar Sinha

3:29 PM

hello

Are there any resources to read more about the projectiles in inclined plane?

any book?

Loop Back

@AbhasKumarSinha Basic idea is to resolve gravity into it's components along the plane and perpendicular

Abhas Kumar Sinha

@LoopBack Find the range of the Projectile on the inclined plane which is projected perpendicular to the inclined plane with the velocity 20m/s Angle of incline = 37 deg

@blue what is quantum computing?

Anonymous

3:45 PM

Quantum computing

Quantum computing is computing using quantum-mechanical phenomena, such as superposition and entanglement. A quantum computer is a device that performs quantum computing. Such a computer is different from binary digital electronic computers based on transistors. Whereas common digital computing requires that the data be encoded into binary digits (bits), each of which is always in one of two definite states (0 or 1), quantum computation uses quantum bits or qubits, which can be in superpositions of states. A quantum Turing machine is a theoretical model of such a computer, and is also known as...

Abhas Kumar Sinha

is that related to quantum mechanics/field mechanics?

Anonymous

@AbhasKumarSinha Yes

Abhas Kumar Sinha

what is has to do with computing?

programming?

or it sounds like astrophysics and stuff like stars and celestial bodies

Anonymous

In short, you use quantum mechanical phenomena to make computers faster in some specific tasks.

Abhas Kumar Sinha

to make computer faster, we've to work at quantum level?

Anonymous

3:48 PM

Yes. The computer you're using right now, already utilizes phenomena at the quantum level.

Abhas Kumar Sinha

how?

Anonymous

Google "working of transistors"

Abhas Kumar Sinha

silicon chips and transistors hardly seems related to any quantum mechanics stuff!!

Anonymous

It's called semiconductor physics. It's quantum mechanics all the way down

Anonymous

Although, you probably wouldn't study too much of the details at an undergraduate level

Abhas Kumar Sinha

3:50 PM

hmmmm.....

So, is there any role of uncertainty principle here? XD :P

Anonymous

Of course.

Abhas Kumar Sinha

seems quite complicated

Loop Back

@AbhasKumarSinha solution to your question

Abhas Kumar Sinha

yes or hints work too

Loop Back

Abhas Kumar Sinha

4:00 PM

@LoopBack That's correct, thanks :)

Loop Back

Np

Abhas Kumar Sinha

how's the gravity there is g cos theta?

and at the next moment it is sin theta

I've not understood this part, rest is okay

Loop Back

R u in class 11th

Abhas Kumar Sinha

yes

oh okay gotcha

but the path of projectiles in horizontal gravity and the gravity at an angles will the different?

or will they be same?

if paths are different, then how can they have the same time period/range?

Loop Back

Why don't you start with the chapter vectors before doing such questions

Abhas Kumar Sinha

4:05 PM

I've ended

XD

is there something I've missed?

eh....?

Loop Back

Yes your basics

Abhas Kumar Sinha

what?

Loop Back

Maybe you should read it thoroughly and ask every single doubt to your teacher

Abhas Kumar Sinha

I don't have joined any coaching

Loop Back

School teacher

Abhas Kumar Sinha

4:07 PM

They don't have the answer

Loop Back

What's your problem

I'll clear it

Abhas Kumar Sinha

how can you split the gravity in such that manner, I think that the place where you've mentioned $\cos \theta$ should be $\sin \theta$

Loop Back

remember that angle between two lines is equal to the angle between there perpendiculars

Abhas Kumar Sinha

oh sorry, I was writing the angle in the wrong place every time, I've figured the problem, yes yours is correct, sorry :P

what is the best way to revise the physics for JEE?

revise everyday or revise a week before AITS?

Abcd

5:14 PM

@AbhasKumarSinha Hey, AITS starting this week. Are you in FIITJEE?

@AbhasKumarSinha AITS JEE Advanced Physics Questions.......Diagrams are enough to scare you away from the question...We had mock tests of previous year papers and I experienced that ... At least thats the case in Mechanics.

Jasmine

6:03 PM

I seem to have forgot this
Consider Rutherford's experiment and alpha particle at the other end and some already very close to nucleus,, in which case will P.E be maximum and in which case K.E maximum?
I am confused because of sign

I need to urgently know this please someone help..

Nevermind

Abcd

11:21 PM

@Jasmine where has your syllabus reached in Physics.

Transcript for

Problem Solving Strategies