Two smooth particles of masses $m_{1} = m_{2} = m$ are connected by an undeformed massless spring of length $l_{0} .$ The particle $m_{1}$ is given a velocity $v_{0}$ perpendicular to the spring. If the stiffness of the spring is k, and the deformation of the spring is very small compared to its natural length, the maximum deformation of the spring is

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$≃ \frac{5 m v_{0}^{2}}{2 k l_{0}}$

$≃ \frac{2 m v_{0}^{2}}{k l_{0}}$

$≃ \frac{m v_{0}^{2}}{k l_{0}}$

$≃ \frac{m v_{0}^{2}}{2 k l_{0}}$

answer is A.

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Detailed Solution

A.M.C. about C.M of system
$\begin{array}{l} m (\frac{V_{0}}{2}) (\frac{l_{0}}{2}) 2 = m V_{1} (\frac{l_{0} + e}{2}) 2 \\ v_{1} (l_{0} + e) = \frac{l_{0} v_{0}}{2} \\ m.E conservation in C-frame \frac{1}{2} m {(\frac{V_{0}}{2})}^{2} \times 2 = \frac{1}{2} m V_{1}^{2} \times 2 = \frac{1}{2} k e^{2} \\ e ≃ \frac{m V_{0}^{2}}{k l} \end{array}$