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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Work, Power and Energy
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Two identical billiard balls are in contact on a smooth table. A third identical ball strikes them symmetrically and come to rest after impact. The coefficient of restitution is

\[\begin {array} {1 1} (a)\;\frac{2}{3} & \quad (b)\; \frac{1}{2} \\ (c)\;\frac{1}{3} & \quad  (d)\;\sqrt {\frac{2}{3}} \end {array}\]

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By symmetry both the billiard balls will move with the same velocity along their respectively lines of impact .
By momentum conservation along the line of impact.
$mu= 2 mv \cos 30$
$\qquad= 2 mv . \large\frac{\sqrt 3}{2}$
$u= \sqrt 3 v$
Newton's law of restitution along line of impact.
$\large\frac{V-0}{0- u \cos 30}$$=-e$
$\large\frac{V}{\sqrt 3 u/2}$$=e$
$e= \large\frac{2}{3}$
answered Nov 25, 2013 by meena.p
edited Mar 16, 2014 by pady_1
 

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