An un-deformed spring of spring constant k is connected to a bead of mass m which can move along a frictionless rod as shown in the figure. If the particle strikes the bead at an angle of $45^{o}$ with the horizontal and sticks to it, then the maximum elongation of the spring after the collision is

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$\frac{v}{2} \sqrt{\frac{m}{k}}$

$\frac{v}{4} \sqrt{\frac{m}{k}}$

$\frac{v}{4} \sqrt{\frac{m}{2 k}}$

$\frac{v}{2} \sqrt{\frac{m}{2 k}}$

answer is B.

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Detailed Solution

Conserving the linear momentum of the system along the rod $mv \cos 45^{0} = 2 {mv}_{1};$

$v_{1} = \frac{v}{2 \sqrt{2}}$

Applying work-energy theorem $- \frac{1}{2} {kx}^{2} = - \frac{1}{2} (2 m) {(\frac{v}{2 \sqrt{2}})}^{2}$

$\Rightarrow x = \frac{v}{2} \sqrt{\frac{m}{k}}$

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