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Mechanics

Please. Solve. It
Please... Plzz,.......................................

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JOY M.M

7 Years agoGrade

1 Answer

Arun

7 Years ago

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

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

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

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