As per the given figure, two blocks each of mass $250 g$ are connected to a spring of spring constant $2 N m^{- 1}$ . If both are given velocity $v$ in opposite directions, then maximum elongation of the spring is

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Let the maximum elongation of the spring be $x$ .
Using energy conservation
Loss in kinetic energy of both blocks $=$ Gain in spring energy
$\frac{1}{2} m v^{2} \times 2 = \frac{1}{2} k x^{2}$
$\Rightarrow (0.25) v^{2} = \frac{1}{2} \times 2 \times x^{2}$
$\Rightarrow \frac{1}{4} v^{2} = \frac{1}{2} \times 2 \times x^{2}$
$∴ x = \frac{v}{2}$

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