@JohnRennie Hello

@pi-π hi :-)

I have a bit of work to do at the moment. Let me get on with that then I'll have a look at your problem.

@JohnRennie Okay

@pi-π Are we allowed to take $g=10 \mathrm{m/s^2}$?

Or is that just $9.8$?

@GuruVishnu There's no such restriction. We can either of them.

Ok. I think the result will depend on the value of $g$ we take. But I'm not sure. Let me try and then say.

@GuruVishnu ok

@pi-π I got $a_b=0.753846$ but not exactly $0.75$. Is that an issue? Given this value it would be easy to find other accelerations as well as tension in the strings.

I used $g=9.8$ itself to be on the safer side. If used $10$ it gives a slightly higher value for the above acceleration.

@GuruVishnu If the process is correct then 0.75 or 0.753846 will not bother.

@pi-π I hope I did it correctly. I just wrote $F=ma$ equation for each block after assuming an arbitrary tension in each string. Then computed the acceleration of B by relating it with that of A.

In the beginning of the exercise, were you asked to round off to 2 decimal places or something like that?

@GuruVishnu No.

@JohnRennie Are you free?

I've started a bounty for the above question :)

@GuruVishnu I believe that if that post didn't answered your question then you should have asked a new one. Cause new posts tend to get more attention than old ones.

@JohanLiebert Thanks for your message. I thought of that idea. I didn't do so as I felt my new question might be closed as a duplicate of this one. Further, this one has a lot of views. Now, I don't know whether I made the right decision or not.

I incorporated my question into it through my latest edit.

Does anyone know here why C is the right answer?

I thought electrons only emmit light when they are going down energy levels. And I though they emmit the highest wavelength the shorter the energy level they go down from. So From that I thought it was D but apparently its wrong :/

@amanuel2 For non single electron system we follow aufbau rule

And then check which electron is going down with less difference in energy

@JohnRennie Hello sir:-)

@Jasmine hi :-)

Sir was saying that the answer given to this question is correct

I still think the answer is B

He said in (1) friction will not change but in (2) friction will change

Dont know what exactly that would imply

Suppose in II the friction is high enough for the wedge not to move. We can always make this the case by making the ball speed slow enough. Then the ball will do this:

The only thing moving is the ball, and clearly $v_x$ changes so $p_x$ must change.

I would agree

What does it mean ^

Maybe he meant in (1) friction is non impulsive , but still the collision forces are impulsive

I have no idea what that means.

Friction clearly will change $p_x$ in scenario I

@JohnRennie isnt the value of friction before and after collision same in 1

@JohnRennie yes

All three forces act for the same time, so all three change the momentum by $\Delta p = Ft$ where $t$ is the collision time.

F1 and F2 are equal and opposite by the 3rd law, while F3 will depend on the coefficient of friction and the mass of the wedge.

Okay..

So there will be a net change in $p_x$ of $\Delta p_x = F_3 t$

I suppose if we take the limit of $t \to 0$ then this change will go to zero because $F_3$ is a constant.

Maybe this is what the question means i.e. in a typical collision $t$ is small enough that the change in $p_x$ is negligible.

I'm not sure what that means ...

@JohnRennie Got it :-)

@Jasmine OH Ur right... thanks a ton