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  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 speedv, the average force experienced by each support after a long time is (assume all collisions are elastic) :

 

  • Option 1)

    \frac{mv^{2}}{L-nr}

  • Option 2)

    \frac{mv^{2}}{L-2nr}

  • Option 3)

    \frac{mv^{2}}{2\left ( L-nr \right )}

  • Option 4)

    zero

 

Answers (1)

best_answer

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)

\frac{mv^{2}}{L-nr}

This is an incorrect option.

Option 2)

\frac{mv^{2}}{L-2nr}

This is the correct option.

Option 3)

\frac{mv^{2}}{2\left ( L-nr \right )}

This is an incorrect option.

Option 4)

zero

This is an incorrect option.

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