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One end of an ideal spring is fixed at point O and other end is attached to a small disc of mass m which is given an initial velocity $v_{0}$ perpendicular to its length on a smooth horizontal surface. If the maximum elongation of the spring is $\frac{l_{0}}{4}$ then ( $l_{0}$ = natural length, k = stiffness of the spring) (initially the spring is in it’s natural length)

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Velocity at maximum elongation is $\frac{3 v_{0}}{5}$

Velocity at maximum elongation is $\frac{4 v_{0}}{5}$

$v_{0} = \frac{5 l_{0}}{12} \sqrt{\frac{k}{m}}$

$v_{0} = \frac{l_{0}}{12} \sqrt{\frac{k}{m}}$

answer is A, C.

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Detailed Solution

According to Conservation of Angular Momentum
$m \times v_{0} \times l_{0} = m \times v \times (l_{0} + \frac{l_{0}}{4})$

$\Rightarrow v = \frac{4 v_{0}}{5}$
According to conservation energy
$\frac{1}{2} m {v_{0}}^{2} = \frac{1}{2} m v^{2} + \frac{1}{2} k x^{2}$

$\Rightarrow m {v_{0}}^{2} = m {(\frac{4 v_{0}}{5})}^{2} + k {(\frac{l_{0}}{4})}^{2}$

$\Rightarrow v_{0} = \frac{5 l_{0}}{12} \sqrt{\frac{k}{m}}$

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