Good morning sir :-) @JohnRennie

@user8718165 morning :-)

@JohnRennie hi morning

@yuvrajsingh morning :-)

@JohnRennie i.stack.imgur.com/eRqES.jpg

@JohnRennie when k1 is cut

I think tension between string is zero

You mean K1 is cut so the three blocks A, B and C are falling freely?

Because all the block are in free fall. Condition

Before K1 is cut the weight of the bottom block has stretched the spring K2, so K2 is in a stretched state. If you now cut K1 then the blocks are falling freely so it's like they were floating in space.

That means the spring K2 is now pulling the blocks B and C together.

@yuvrajsingh OK so far?

@JohnRennie ok

And that means block B is being pulled inwards, so block A is also being pulled inwards by the string attached to block B.

So there will be a tension in the string equal to F = ma where m is the mass of block A and a is the acceleration of block A.

@JohnRennie have a question we assume free fall, when there Is No Other Force Accept Gravity But For kx Is The Other Force

And Secondly Block Earlier K1 Supprted Three Mass A, b, c so It Ks Strecheched Postion , when I Cut K1 ShOuld N, t It Apply Upward Force

You mean because when you cut K1 half the spring is still attached to block A?

I Did N, t Get What yu Mean Inward It Mean upward

@yuvrajsingh I meant inwards towards the centre of mass of the three blocks.

@JohnRennie @ yes

We assume that the springs are massless

So the half of K1 still attached to the top of block A hasn't got anything to pull on. It instantly relaxes to its equilibrium length.

Where a=g

T=mg yes @JohnRennie

@yuvrajsingh No.

Imagine the three blocks just floating in space, or maybe it's easier to imagine them lying on a frictionless table top.

Why

Start with the spring K2 at its equilibrium length i.e. no tension in the spring. So the three blocks just remain motionless. OK so far?

Ok

Now grab hold of blocks B and C and pull them apart, so now there is a tension in the spring K2.

Ok

If you now let go of the blocks the spring starts to pull the blocks B and C together, and because block A is tied to block B it pulls A towards block C as well.

The centre of mass of the three blocks cannot move, so what happens is that Cand A/B accelerate towards the centre of mass.

I'm still thinking of the blocks floating in space. So in that frame the COM doesn't move.

OK

But yes, if you're in the lab frame watching the blocks fall then the COM accelerates downards at g.

Tension will be zero

@yuvrajsingh tension in what part of the system?

@JohnRennie A and b

No. because the acceleration of B in the lab frame will be greater than g, because the spring K2 is pulling B downwards as well as gravity pulling B downwards.

@JohnRennie at b forces are kx+mg-t=ma

I would work in the centre of mass frame as it will be a lot easier.

@JohnRennie sir what will be the final value of tension between block A and B

@yuvrajsingh What do mean by the final value? The AB tension will start off at some value greater than zero (which we have to calculate) then it will fall as the blocks B and C approach their equilibrium distance. The AB tension will fall to zero then the string will go slack.

I need to work now for a bit. Back later.

@JohnRennie one more question two point charges are placed at distance R from each other in vacuum, the force between them is f1, one if charges is spread uniformly over the surface of a hemispherical bowl of radius R while the other is spread r and (r<R/2) and both have common Centre o, force on sphere is f1/n then n is what I did I know potential kq/8r at Centre f=kq^2/8r^2

Distance between charges is R

@JohnRennie hi

@JohnRennie Sir, are you working now?

@user8718165 @yuvrajsingh I have a fair bit of work to do this morning I'm afraid. I'm writing some code.

@JohnRennie okay sir....do peacefully :-)

@JohnRennie hi

@Aladdin morning :-)

Morning!

If you are free, can u ping me

For python

I'm free now. Codeclub room?

Yep

@JohnRennie hi

@yuvrajsingh hi

@JohnRennie i have two ball one big and other is small with mass m1 and m2 both ball are toching each other and in contact with each with other like a sonwmean head and stmoach where the radius of bigger ball of mass m1 is R and bigger ball icentre of mass is 3R above the ground ,ball droped from this height ,if the height of small ball m2 is 4nr then value of n will be assume all collisions are elastic

@yuvrajsingh I can't remember how this is done, but it's a standard problem and should be easy to Google ...

@JohnRennie sorry sir to distrub you ,and didn,t ask you before asking question

@yuvrajsingh it's OK, I've finished work now.

Have a look at this web page

@JohnRennie it is not on google

It explains how to do the calculation.

@JohnRennie i almost done it since collosions are elastic so firs ball will get momentum of 2mv after collosion to ground

@JohnRennie can you make diagram

@JohnRennie i got my answer from your link

@yuvrajsingh cool :-)

@JohnRennie hi.

@Nobodyrecognizeable hi

@JohnRennie ^^

@Nobodyrecognizeable Presumably you do this by writing an equation for the interaction energy between the two dipoles and then minimising the energy with respect to the angle of the dipole P.

@JohnRennie yes. I used to. I just forgot now.

I can't remember the equation for the interaction energy between two dipoles, but presumably it's easy to Google ...

@JohnRennie ^^

That should be straightforward to minimise because a.b = |a| |b| cos\theta and only the angle is changing.

@JohnRennie ie theta should be zero.

You need to consider p2.r as well. i.e. the angle between P and the vector (a,a).

Suppose we take theta to be the angle measured anticlockwise from the x axis then p.P = |p| |P| cos theta and P.r = |P| |r| (cos(theta - 45°)

@JohnRennie OK.

@JohnRennie so theta =18.4 degrees.

@JohnRennie are you here?

@Nobodyrecognizeable hi

@JohnRennie whatll happen if both are allowed to rotate?

They will align antiparallel and at right angles to the line joining them.

@JohnRennie yep.

@JohnRennie ^^^

Suppose you have a population that follows a normal distribution with a standard deviation $\sigma$ and you are trying to determine the value of the peak of that distribution.

@JohnRennie OK.

There is a standard equation for the error in the peak position based on the standard deviation and the number of measurements. Can you remember it?

@JohnRennie to be honest have heard very little about it.

If the standard deviation is $\sigma$ and you make $N$ measurements and calculate the average then the error in your calculated average is $\sigma/\sqrt{N}$.

@JohnRennie all right.

So I think this question is expecting you to calculate the standard deviation from the data you have, then work out how many samples you need to get the error down to 1% of the mean.

@JohnRennie so the standard deviation remains same?

@JohnRennie really you are a champion.