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Two particles A and B each of mass m are attached by a light inextensible string of length 2l. The whole system lies on a smooth horizontal table with B initially at a distance l from A. The particle at the end B is projected across the table with speed is perpendicular to AB. velocity of ball A just after the **** is


\[\begin {array} {1 1} (a)\;\frac{u \sqrt 3}{4} & \quad (b)\;u \sqrt 3 \\ (c)\;\frac{u \sqrt 3}{2} & \quad  (d)\;\frac{u}{2} \end {array}\]

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Initial velocity of ball along $AB= u \cos \theta$
and perpendicular to to $AB= u \sin \theta$
When the string becomes , but the two bodies would acquire the same velocity along AB
By momentum conservation
$2 mv =mu \cos \theta$
$v= \large\frac{u \cos \theta}{2}$
$\quad= \large\frac{u}{2}$$ \times \large \frac{4 l^2 -l^2}{2l}$
$\quad= \large\frac{u}{2}$$ \times \large\frac{ \sqrt 3 l} {2l}$
$\quad= \large\frac{\sqrt 3}{4}$$u$


answered Nov 25, 2013 by meena.p

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