@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...
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 ?
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.
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? ^^
@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)’
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: $f = u/L$.
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?
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?
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.
Because we have to find the average force on the support and not in between the beads ?
@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.
@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
@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 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
@RaviPrakash hi Ravi. You need to calculate the difference in potential energy between the centre of the sphere and the surface.
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).