Problem Solving Strategies

Harjot Dhillon

03:26

@KavinIshwaran yes I would like to know that

Kavin Ishwaran

05:08

@HarjotDhillon Hi

Harjot Dhillon

05:32

@KavinIshwaran hello

Kavin Ishwaran

@HarjotDhillon One minute

So this would be the intermediate after aldol reaction. yes ?

Harjot Dhillon

Yes

Kavin Ishwaran

Can you see what is happening ?

Harjot Dhillon

Yes positive charge shift from secondary to tertiary

Kavin Ishwaran

So next step double bond formation

Harjot Dhillon

05:44

Are you talking about acid catalysed or base catalysed aldol condensation?

Kavin Ishwaran

base...

Harjot Dhillon

In base catalysed there is no carbonation formation

Kavin Ishwaran

But in our example there is no two alpha hydrgens...

So carbocation will form during the aqueous work up (consider the OH group in the carbocation position), this will produce carbocation and rearrangement will happen.

Harjot Dhillon

But there are already many hydroxide ions present in solution so I don't think its going to be favoured by equilibrium unless our carbonation react further very quickly.

Kavin Ishwaran

06:04

Aldol addition is kinetic controlled. but the condensation reaction is thermodynamically controlled. so I would say the condition is enough for the rearrangement to happen

Harjot Dhillon

Ok then there would be double bond formation

Kavin Ishwaran

That is what I would answer

Normally it won't happen. I am considering the whole process into two, one is aldol addition in basic condition, and the condensation in aqueous condition at increased temperature

Oh is a bad leaving group in a basic condition. the base catalyzed mechanism you linked there undergoes E1cb elimination. but there is another alpha hydrogen in the place. but in our example it isn't.

Anyhow. It doesn't matter if the reaction happens in base or acid medium. Know that your example can undergo condensation

Harjot Dhillon

Tbh if this question again come in exam I would choose statement 1 to be true. If it was considered true in 2022 then there is no way they can mark it wrong now. I don't want to go from whole process of claiming and challenging.

Kavin Ishwaran

Ok

Harjot Dhillon

Thanks for your help :-)

Kavin Ishwaran

06:14

Like the linked q and a in Chemistry chat, there are exceptions like that. still a five membered ring can undergo condensation considering ring expansion

@HarjotDhillon Welcome :-)

@HarjotDhillon Are you writing NEET this week ?

Harjot Dhillon

Yes

On 5th may

Kavin Ishwaran

All the very best :-)

Harjot Dhillon

Thank :-)

Kavin Ishwaran

How dont know how the exam compares to JEE. But it is truly a challenge considering the amount of immense competition it has.

I*

@JohnRennie Hi !

John Rennie

Hi :-)

Kavin Ishwaran

06:23

John Rennie

Did you find a solution?

Kavin Ishwaran

We were discussing this question before. And I got the answer key

Looks like B and C are right

John Rennie

I can have a go at it using the ideas we discussed, but not right now as I have to go out soon.

Kavin Ishwaran

But I found a solution for this question. and I couldn't able to understand it.

@JohnRennie Ok :-) we will discuss later

John Rennie

Post the link to the solution so I can have a look at it.

Kavin Ishwaran

06:27

It looks vague

sanya

12:18

Hi @JohnRennie

John Rennie

@sanya Hi :-)

You've pinged at a bad time. I'm going out now. I'll be back around 8:30 p.m. Indian time.

sanya

13:24

Okay:) @JohnRennie

tush

14:16

This is a quite basic question, but if anyone can reflect about it, I would be very grateful.
Suppose I use a spectrometer to record the spectrum of the radiation outside.
The data the spectrometer generates is a plot of watts/(meter² nanometer) vs. nanometer.
The fact that the vertical axis is not in watts/meter² but watts per meter² per nm indicates that the instrument uses binning. In other words, the dataset the spectrometer generates is not a scatter plot (also known as point plot) but a histogram.

? Do you see any fualts in it? Otherwise, can you point me to the correct Q&A site to ask this question?

John Rennie

14:40

@tush I suspect this is just a typo, but it should be:

$$ \Phi = \sum_{\lambda_1}^{\lambda_2} \Phi_i \delta\lambda $$

That is Φᵢδλ is the flux per square metre in the range of wavelengths from λᵢ to λᵢ + δλ.
And add up all the bins to get the total.

But you've basically written ∫Φ(λ) dλ so why not just do the numerical integration?

tush

14:57

@JohnRennie Correct, it is a typo

@JohnRennie I want to first understand what I do before letting the computer do the dirty job :)

@JohnRennie What if the binning is not constant?

John Rennie

If the binning is not constant instead of a fixed δλ between each reading you will get a variable δλᵢ

tush

I see. Thanks!

John Rennie

In that case just sum Φ(λ)δλᵢ

Odestheory12

Hello. I have some questions about this problem:

"A platform of M kg moves with a velocity of 2 m/s on straight horizontal rails without friction. On the platform, and at rest relative to it, an elastic spring of negligible mass and a spring constant of 1000 N/m is arranged in the direction of motion, initially compressed by 20 cm, and two bodies, attached to the spring, one at each end, of masses m1 and m2 kg, respectively.

At a certain moment, the spring is released, causing both bodies to start moving on the platform in opposite directions, being shot off. Assuming no frictional forces exist, calculate the velocity of the three objects."

First of all, is not clear if the spring is attached to the platform or if it isnt.

If it isnt, I guess we can solve the problem by doing $m_1v_1 + m_2 v_2 = 0$ and $ 1/2 k x^2 = KE_1 + KE_2$

But if it is attached, the speed of the platform its going to change. Naming it $v_p$ we have: $m_1v_1+m_2v_2 = Mv_p$ and $1/2 k x^2 = KE_1+KE_2 + KE_p$

sanya

@JohnRennie please ping me when you're free :)

John Rennie

15:06

@sanya I'm free now :-)

Odestheory12

But I miss one equation, since i have three variables ($v_1,v_2$ and $v_p$) but two equations

How could I get the third equation?

(I am not sure if it is even solvable)

sanya

Posting the ques....

Position vector of q is î+j(hat)+2k(hat) and position vector of 4q is 2î+j(hat) +3k(hat). Find the position vector of neutral point and find the charge Q that could be placed here so that q is in equilibrium

The image is whats done and I got the positon vector of the neutral point after this I need the magnitude of distance between the neutral point and q for the second apart am I supposed to find the magnitude twice and see which is smaller and then use that or is there a faster way to do this?

John Rennie

@sanya You have the charge q at (1, 1, 2) and 4q at (2, 1, 3) and you find the neutral point. Let's call this (x, y, z).

OK so far?

sanya

Yes

John Rennie

Then the vector from the neutral point to q is (x-1, y-1, z-2). Yes?

sanya

15:18

Yes

John Rennie

And the length of that vector is:
s² = (x-1)² + (y-1)² + (z-2)²
so it's easy to calculate.

sanya

Yes but the position vector of the neutral point has √q1 and √q2 in the equation so wont I have to take out the magnitude twice once with positivel sq root and once with negative

John Rennie

Oh I see.

I think there's only one neutral point ...

The neutral point lies on the line in between q and 4q

And it's one third of the way along that line from q.

Yes?

sanya

Yes because when I tried to find out the magnitude of position vector of Q one of it was √-something so we dont take that..

John Rennie

The vector from q to 4q is (1, 0, 1) so the vector from q to the neutral point is oe third of this i.e. (¹⁄₃, 0, ¹⁄₃)

sanya

15:25

@JohnRennie How'd you see that ?

John Rennie

Call the distance from q to 4q 3s

sanya

Okay

John Rennie

Then on third of the way along this is 𝑠

sanya

Okay

John Rennie

So the distance from q is 𝑠 and the distance from 4q is 2𝑠.

Then the force from q is kq/s² and the force from 4q is k4q/(2s)²

Yes?

sanya

15:30

Force or field?

John Rennie

@sanya Force on unit charge, so yes it's the field since the force on a unit charge is equal to the field.

The point is that k4q/(2s)² = kq/s²

So the forces from the two charges are the same so this is the neutral point.

sanya

Ah ok

@JohnRennie Then I dont have to do this

John Rennie

I'm no sure how you ended up with a √q in the equation. But there is only one neutral point so you only need to do one calculation.

@Odestheory12 If I understand the question correctly the spring pushes the two masses away from each other and they shoot off the ends of the cart. Since there is no friction between the masses and the cart the velocity of the cart is unaffected.

Yes?

Odestheory12

But that is if the spring is not attached, right?

John Rennie

I guess the spring could be attached at its mid point and the outcome would be the same.

But I'd guess the spring is not attached anywhere.

So the two masses fly off the ends of the spring leaving the spring lying on the cart.

Odestheory12

15:40

Yeah, if the spring is not attached I can solve it with conservation of momentum and energy, but I was interested in the case of the spring attached

John Rennie

Do you mean the spring attached to the two masses, or the spring attached to the cart?

Odestheory12

attached to the cart

John Rennie

At it's mid point?

Odestheory12

Yeah

John Rennie

Then that would make no difference since the mid point of the spring doesn't move relative to the cart.

Odestheory12

15:41

You are right, the reaction force is independent of the mass

sanya

@JohnRennie if you do the calculation I get $$ \frac {\vec I (2√q_1 + √q_2) + \vec j ( √q_1 + √q_2) + \vec k (3√-q_1+2√q_2)} {√q_1 + q_2} $$

Not sure if I made a mistake with the tex the formatted version is not coming through :(

John Rennie

Odestheory12

Then its simply $m_1 v_1 + m_2 v_2 = 0$ and $\frac{1}{2} k x^2 = KE_1 + KE_2 $ right?

John Rennie

It's fine here. If you're on a phone I don't think MathJax works.

sanya

@JohnRennie Oh I see why then

John Rennie

15:45

@Odestheory12 If you work in the rest frame of the cart, then in this frame the two masses are initially stationary.

So after they have moved apart their total KE has to be equal to the energy stored in the spring and their momenta have to be equal and opposite.

Odestheory12

Yeah, exactly ^^

John Rennie

@Odestheory12 So yes, this gives the velocities of the masses in the rest frame of the cart.

Odestheory12

Thanks, much appreciated.

John Rennie

Then you need to add the velocity of the cart to get the velocity in the ground frame.

@Odestheory12 You're welcome :-)

@sanya We can go through the calculation if you want ...

sanya

Yes please

Also note that its not -q in the k vector bracket and in the denominator its √q2 not only q2 :( Sorry im using it for the first time should have been careful

John Rennie

15:54

@sanya Is this the method you used?

sanya

Yes

Sorry for the delay in messages my wifi is unstable

John Rennie

So if call the q charge 1 then x₁ = (1, 1, 2) and x₂ = (2, 1, 3)

sanya

Yes

John Rennie

If we call the neutral point n = (nx, ny, nz) then the first equation gives:
q/|x₁ - n|² = 4q/|x₂ - n|²

4|x₁ - n|² = |x₂ - n|²

It feels as though this method is going to get very messy ...

sanya

@JohnRennie yes

@JohnRennie it was fairly ok I didnt substitute though I just used the general equation..so maybe

John Rennie

16:04

How did you get square roots of charges in the final result when there aren't any charges in the equation we've got so far?

I have to say I don't see how this method works ...

I think it's kind of the same as what I did but more complicated.

sanya

@JohnRennie if we go by this we take the square root of the first equation and then divide it with the second equation

John Rennie

Oh I see. But here the charge q cancels out of the first equation doesn't it?

q₁ = q and q₂ = 4q so we can divide both sides by q to get just:

1/|x₁ - x₃|² = 4/|x₂ - x₃|²

sanya

Oh I was trying to get the magntiude with q1 and q2 so I thought of substituting at last but I see that would make it complicated

John Rennie

Then square rooting gives us:
1/|x₁ - x₃| = 2/|x₂ - x₃|

And we only take the positive root because magnitudes are always positive.

sanya

Yes got it

John Rennie

16:10

OK :-)

sanya

Sorry feels like wasted your time , this was simple . Thankyou

John Rennie

You absolutely have not wasted my time!

For you to see how we old timers simplify problems like this is very important!

You are always welcome to ask questions like this :-)

sanya

I see :)) thankyou again

Ronald Becker

16:56

@JohnRennie hello sir

sanya

20:37

@JohnRennie we discussed a problem where we had to prove if a charge was in stable or unstable equilibrium some days back... I just got the answer sheet for it back and it states that it is in stable equilibrium in the line but if we consider all directions (3d space) it is in unstable equilibrium...can you tell me why's that? Its not like theres a charge placed in any other dimension in whose field the charge can move, in any other dimension I would assume df/dx=0,hence neutral eqbm

@JohnRennie it was this conversation

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Problem Solving Strategies