Block ‘A’ is hanging from a vertical spring and is at rest. Block ‘B’ strikes the block ‘A’ with velocity v and sticks to it. Then the value of v for which the spring just attains natural length is

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$\sqrt{\frac{60 {mg}^{2}}{K}}$

$\sqrt{\frac{6 {mg}^{2}}{K}}$

$\sqrt{\frac{10 {mg}^{2}}{K}}$

$\sqrt{\frac{50 {mg}^{2}}{K}}$

answer is B.

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Detailed Solution

Initial extension in spring = $= x_{0} = \frac{mg}{K}$
Just after collision speed of A & B is v/2.
$So \frac{1}{2} (2 m) {(\frac{v}{2})}^{2} + \frac{1}{2} {Kx}_{0}^{2} = 2 {mgx}_{0} Solving we get v = \sqrt{\frac{6 {mg}^{2}}{K}}$

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