# A large number (n) of identical beads, each of mass m and radius r are strung on a thin smooth rigid horizontal rod of length L (L>>r) and are at rest at random positions. The rod is mounted between two rigid supports (see figure). If one of the beads is now given a speed, the average force experienced by each support after a long time is (assume all collisions are elastic) : Option 1) Option 2) Option 3) Option 4) zero

As we discussed in

Perfectly Elastic Collision -

Law of conservation of momentum and that of Kinetic Energy hold good.

- wherein

$\frac{1}{2}m_{1}u_{1}^{2}+\frac{1}{2}m_{2}u_{2}^{2}= \frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2}$

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$

$m_{1},m_{2}:masses$

$u_{1},v_{1}:initial \: and\: final \: velocity\: of \: the\: mass\ m_{1}$

$u_{2},v_{2}:initial \: and\: final \: velocity\: of \: the\: mass\ m_{2}$

Average time for one collision $=\frac{2L-4nr}{v}$

Since diameter of each bace = 2 r

Change in momentum per collision = 2 mv

$\therefore$    $_{F_{av}} = \frac{\Delta p_{av}}{\Delta t_{av}}=\frac{2mv}{(2L-4nr)/v}=\frac{mv^{2}}{L-2nr}$

$_{F_{av}} =\frac{mv^{2}}{L-2nr}$

Option 1)

This is an incorrect option.

Option 2)

This is the correct option.

Option 3)

This is an incorrect option.

Option 4)

zero

This is an incorrect option.

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