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Two identical blocks of mass $m$ each connected by a massless spring of force constant $k$ and natural length $l$ are kept on a smooth horizontal surface. When the spring is unstretched, one of the blocks is given horizontal velocity $u$ perpendicular to the length of the spring. When the spring attains maximum elongation $x$ , the speed of each block is $v$ in centre of mass frame. Assuming $x < < l$ , which of the following are correct?

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$2 (l + x) v = l u$

$\frac{1}{2} m u^{2} = \frac{1}{2} k x^{2}$

$m u^{2} = k l x$

$(l + x) v = l u$

answer is B, D.

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Detailed Solution

Conservation of angular momentum about $c m$
$(l + x) = \frac{l u}{2}$
Conservation of energy gives $\frac{1}{2} k x^{2} = m (u^{2} - v^{2})$

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