@JohnRennie

Hi :-)

Hi:)

Can you help me understanding a question

Yes of course. Post the question.

How do I post a picture of a question

In here there is no option

Can you post the picture to Imgur and post the link here?

Ok I'll try

g.co/about/k9jfpf

Wait I'll send a clear link

This question?

Yes

My doubt is not solving it but something else

OK ... ?

In a similar question in the previous chapter of my book this was the same type of question and

I solved it using constraints

It had no friction

By assuming acceleration in X and in both y direction

When I am applying the same concept here the solution says acceleration in y direction is zero

And they didn't use constraints to solve it

Do you mean in part (b) the y acceleration is zero?

How come in y acceleration is zero in this question while the previous one had no friction but it had assumed the acceleration in y direction

No I mean a question from the previous chapter not this one

It essentially is the same question but had no friction

The surface was smooth

I'd have to see the previous solution to understand what it was doing.

Does it have to do something with friction

Ok wail I'll send the link

images.app.goo.gl/odXeG8oz1AryqBDr5

I solved it by using the constraint relation of acceleration by measuring distance to x till block of mass m and X till wedge the in y direction I took total height of were h and the height till block is kept y

Then I subtracted x-X/h-y = tan45

x-X=h-y

Then I differentiated it twice and a-x(double dot)=y double dot

But when I apply the same logic to friction question it's wrong and in solution it has assumed no acceleration in y direction

I got it has got acceleration in y direction but it has been asked to find maximum acceleration and minimum applied on wedge in x direction specifically

@JohnRennie

So I just solely focus on x direction only

I'm sorry but I'm busy with another question for a while.

I'll ping you as soon as I'm free.

Ok no problem I solved it so no worries

@JohnRennie Hey! How are you? :-)

H :-)

I'm good thanks. How are you?

I guess stressed about end semester exams!

@JohnRennie Damnn How do you know hehe?

@JohnRennie I am alive :-)

How's your life going on? Same daily schedule like before?

I have several students preparing for the end sem exams :-)

:P

@Wolgwang Yes, though it's gone a bit quiet. It always goes quiet immediately after the Mains exams. Then it ramps up again as the Advanced exam approaches.

I see.

How's your health?

I'm fine. Trying to shed a bit of weight after staying with my mother over the Easter holiday. Mother's never think you eat enough :-)

How are you? Has the first year at college been OK?

@JohnRennie Yes it has been "ok".

Not as enjoyable as you had hoped then? :-(

Yeah :'-) Students are not even free to study . Exam oriented studies and syllabus

@JohnRennie Free for 2 minutes. I have a real quick physics question :/

To be fair the first year is always a bit like that. You'll get more freedom in later years.

@Wolgwang Yes, what's the question?

@JohnRennie I hope. Though I enjoyed studying data structures.

Why are we not dividing by $\int_0^1 a^2x^2$ while finding the expectation value?

I would guess it is assumed that ψ is normalised so that integral is one.

And while checking if the $\Psi$ is normalised or not . We calculate $\int_{-\infty}^{\infty} |\Psi|^2 dx$

But here from 0 to 1 it is not 1

How come then it is normalised?

Whether it's normalised or not depends on the value of the constant 𝑎.

I would guess it is assumed that 𝑎 is chosen to make ψ normalised. You could easily calculate what value 𝑎 must have if you felt the urge ...

Ah

@JohnRennie I did hehe $a=\sqrt3$

If you look at equation 5.19 above that makes the point about normalisation.

Yes..?

Hmm they have assumed psi to be normalised.

That's my guess. It doesn't explicitly say so, but it seems likely.