Two small spheres of mass 2 kg each are connected to each other by means of an undeformed spring of force constant 175 N/m and natural length 1m. The system is placed on a smooth horizontal surface and the two spheres are given velocities $v_{0}$ along and perpendicular to the spring, as shown in the figure. The maximum elongation in the spring is found to be 1m during the subsequent motion. Find the value of $(\frac{v_{0}}{5})$ in m/s.

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Detailed Solution

in CM frame initially, $K_{i} = \frac{1}{2} m {v_{0}}^{2}$
At maximum extension velocity along the length of the spring becomes zero. $\Rightarrow$ finally $K_{f} = 2 (\frac{1}{2} m v^{2})$
Conservation of energy $\Rightarrow$ $\frac{1}{2} m {v_{0}}^{2} = m v^{2} + \frac{1}{2} k x^{2}$ ……………..1)
COAM w.r.t cm $\frac{m}{2} . v_{0} . l_{0} = \frac{m}{2} (2 v) (l_{0} + x)$ ………………….2)
$\Rightarrow v_{0} . l_{0} = 2 v (l_{0} + x)$ $\Rightarrow \frac{1}{2} m {v_{0}}^{2} = \frac{m {v_{0}}^{2} {l_{0}}^{2}}{4 {(l_{0} + x)}^{2}} + \frac{1}{2} k x^{2}$
${v_{0}}^{2} [\frac{(2 l_{0} + x) (x)}{{(l_{0} + x)}^{2}}] = k x^{2} {v_{0}}^{2} [\frac{3 \times 1}{4}] = 175 \times 1 \Rightarrow v_{0} = \sqrt{100} = 10 m / s$ .