hi

@GaurangTandon By interaction energy, do you mean potential Energy of the shell - point charge pair?

@MadhuchhandaMandal yes I think I meant that

@GaurangTandon In that case there exists two Shell -Point Pairs and a Shell-Shell pair.. we need to consider them too !!!

I just used the formula for resultant of two vectors , and since $\theta$ is small , I took $\cos \theta=1$ , but no options match this reasoning.

@Tanuj Does that help?

no , I can see its the triangle law , but ..

A and B are both radii of a circle with radius $|A|$ so $A-B$ is approximately an arc of the circle.

@JohnRennie I get that.And answer according to that should be option 3 , but then , couldn't it be $|\vec{B}|\Delta \theta$

@Tanuj Yes

@Tanuj $A$ and $B$ are the same vector, just rotated. So $|A| = |B|$

@MadhuchhandaMandal no, they won't. note that the solid part of the shell, which is between surfaces A and B, will have no electrical lines of force by property of conductor; so, the field lines of point charge and inner surface won't interact with those of the outer surface; they are electrically insulated from each other...

@JohnRennie okay , thanks.

@JohnRennie Have a look

Okay , since the balls / beads are identical , I know that after elastic collision , the velocities of the two bodies get exchanged.

The question says after a long time, and after a long time we expect the energy to be equally distributed between all the beads i.e. each bead has an energy $\tfrac{1}{2}mv^2/n$

Also , there would be $n$ collisions , so the net change in momentum should be $nmv$ , where $m$ is the mass of each bead and $v$ is the initial velocity given to one bead.

@JohnRennie oh , so is it average kinetic energy ?

@Tanuj there is no potential energy in this system. All the energy is kinetic.

@JohnRennie ok

$F=-dU/dx$ , but how would kinetic energy relate to force ?

And from the (average) kinetic energy we get the average momentum using $p^2 = 2mE$.

@JohnRennie yeah

In a collision with the support the momentum change is twice the average momentum, because the particle reverses direction when it hits the support.

So now you know the momentum change per collision. Next you need to work out the number of collisions per second to get the momentum change per second. The momentum change per second is of course just the impulse and is equal to the force.

@JohnRennie okay

@Tanuj answer is C?

idk , man . It might be , they have given it to be (B)

@JohnRennie I get change in average momentum as $2mv/\sqrt{n}$

i was thinking like net kinetic energy of all beads = average force by each support * distance travelled by a bead * 2 (for 2 supports) * number of beads; so, 1/2*mn*v^2=F*(L-2nr)*2*n

@JohnRennie can you tell why my approach is wrong? ^^

If the collisions are ideal then the distance travelled during the collision is zero.

So you can't use work done by the support in the calculation.

In calculations like this you almost always need to use the fact that force is the rate of change of momentum.

@JohnRennie How do I calculate the time ?

This is the way I would do it ...

Start by calling the average velocity $u$ - we'll worry about how to calculate $u$ later.

@JohnRennie it is a perfect inellastic collision, so the ball and the rod stick together as one. You agree with the the following statement? ‘There is no torque due to impulse forces cancel with each other (Newton’s third law)’

So the momentum change in a collision is $2mu$. Yes?

@JohnRennie perfect

@JD_PM be with you in a moment ...

@JohnRennie great thanks

@JohnRennie oh ok

If the bead has to travel a distance $L$ between collisions then the time it takes to travel this distance is $t = L/u$. The collision frequency, i.e. the number of collisions per second is then just the reciprocal of this: $f = u/L$.

@JohnRennie yeah

Only, the bead doesn't quite travel a distance $L$ because there are other beads taking up space on the wire. It only has to travel a distance of $L$ minus the total size of all the beads. So the distance actually travelled is $L - 2nr$. Yes?

@JohnRennie Oh , okay got it . Thanks.

So the collision frequency is $$f = \frac{u}{L-2nr}$$

And the momentum change in each collision is $2mu$ so the rate of change of momentum (i.e. the force) is $$ \frac{2mu^2}{L-2nr} $$

@JohnRennie yes

Oops, I've just realised I missed a factor of two. To collide with the left post the bead has to move to the right then back again, so the distance is twice what I've put. OK with that?

So the force is $$F = \frac{mu^2}{L - 2nr}$$

yeah

And we're nearly there. The KE of one bead is $\tfrac{1}{2}mu^2$ so the total KE is $n \tfrac{1}{2}mu^2$

And this has to be equal to the energy that was put into the system $\tfrac{1}{2}mv^2$

@JohnRennie yes

Ah, I've just realised I missed another step ...

I've worked out the force due to one particle, but there are $n$ particles so the force is $n$ times bigger than what I've writtem:

@JohnRennie It's okay , I have understood the problem. Its done. :)

OK :-)

@JD_PM hi, you still around?

@JohnRennie just a small doubt , why didn't we begin this by finding change in momentum per collision ?

Because we have to find the average force on the support and not in between the beads ?

Momentum isn't conserved so I'm not sure it's useful to start with the initial momentum

@JohnRennie why is it not ? Because the supports aren't free to move ?

@JD_PM you really need to stick around. I have to go now and you're nowhere to be see.

@Tanuj External forces are acting on the system

@JohnRennie not because of this

i.e. the two supports

Oh okay ,yea all cool

@JohnRennie another question ? It'll be quick.

@JD_PM the third law tells us that the force the ball exerts on the rod is the same as the force the rod exerts on the ball. The force the ball exerts on the rod produces a torque on the rod. The force the rod exerts on the ball makes the ball slow down. So there is a torque and the rod will start to rotate about the pivot.

@Tanuj quick then

@JohnRennie therefore angular momentum is not conserved? I was told AM was conserved, so I started to think why there was no torque. But now I am baffled xD

Drawing the components for tension , I find out cos component cancelling and sin components adding up

So the net tension is $2T\theta$ , as angle is small

And those forces do not cancel with each other

?

@JD_PM ah, OK, I had misunderstood you slightly. The torque the ball exerts on the rod is equal and opposite to the torque the rod exerts on the ball. So the change in angular momentum of the rod is equal and opposite to the change in angular momentum of the ball. That's why momentum is conserved.

@Tanuj it's not a quick question, and I have to go now.

@JohnRennie perfect, I got it. Thanks

@JohnRennie alright

@JohnRennie I figured out that question :)

hi

why space time bent near black holes?

Sir John Rennei ?

@Akash.B ... Due to gravitational attraction ?

From GR ...

@NehalSamee hii

@RaviPrakash ... Hello .

@JohnRennie Hi . I have to ask you one question

@Tanuj Yes?

@JohnRennie By the way the last option is $\vec{E}=B_{0}c\sin(kx-\omega t) k$

I know $E=Bc$

but how do the phases relate ?

Also , I know for an electromagnetic wave , $\omega$ is same.

The electric and magnetic fields are in phase and at 90 degrees to each other. But this applies to all four options.

@JohnRennie yeah

The electric field is much larger than the magnetic field so that rules out (2)

@JohnRennie @JohnRennie by same phase you mean arguments of $\sin$ are same ?

(4) is ruled out because the two waves are travelling in the same direction.

Also , note that option 4 is $\vec{E}=B_{0}c\sin(kx-\omega t) \hat{k}$

@Tanuj yes, that's a wave travelling in the opposite direction to $B_{0}c\sin(kx+\omega t) \hat{k}$, which is why it is ruled out.

But options (1) and (3) differ only by a sign and I can't remember which is corret

@JohnRennie why is it ruled out ? And $B$ is in $+\hat{j}$ direction

@Tanuj the electric and magnetic components of the wave have to be travelling in the same direction because they're part of the same wave.

@JohnRennie oh okay.

@JohnRennie answer is given as option (1)

hmm , thanks

@Tanuj I don't think there's an easy way to remember the relative signs of E and B

@JohnRennie there’s some controversy between my classmates and I in the following exercise: You have to calcule the solid’s volume delimited by the surface z=x^2+y^2 and the plane z=3-2y

I’ll post what I obtain

imgur.com/gallery/I3EMk

A double integral which result is (9/2)π

But the most important is how you construct the integral. Do you agree with the integral I shared?

@NehalSamee From where are u

Obtained*

@JohnRennie Can you help me in solving this question? I'm stuck at it.

I see that the force to be overcomed is kq²/R²

Is this to be done through momentum conservation, etc

@RaviPrakash is it (B) ?

The kinetic energy of the bullet when it strikes the surface has to be greater than this potential energy difference for the bullet to reach the centre. (Once the bullet reaches the centre the charge on the sphere will push it on and out the other side).

@RaviPrakash ... I'm from Bangladesh ...

@GaurangTandon Does shielding affect Potential ?

Shielding merely means external Electric field cannot influence the Hollow region

We know dq=du+dw

In the question it is given dq=0; So, dw=-du

Since by sign convention dw is work done by the system, so Work done by the force=-dw=du

And Uinitial=0

Ufinal= Summation ((k qi qj)/rij) for all (i,j) present in the system.

Isn't it?

@GaurangTandon