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Two blocks A and B, each of mass $m,$ are connected by a massless spring of natural length L and spring constant K. The blocks are initially resting on a smooth horizontal floor with the spring at ists natural length, as shown in fig. A third identical block C, also of mass m, moves on the floor with a speed v along the line joining A and B, and collides elastically with A. Then

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The kinetic energy of the A-B system, at maximum compression of the spring, is zero.

The kinetic energy of A-B system, at maximum compression of the spring, is $m v^{2} / 4.$

The maximum compression of the spring is $v \sqrt{(m / k)}$

The maximum compression of the spring is $v \sqrt{(m / 2 K)}$

answer is X.

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Detailed Solution

$m v = m v^{'} = m v^{'}$

$\Rightarrow v^{'} = \frac{v}{2}$

$∴$ K.E. of the system in situation 3 is

$\frac{1}{2} m v^{'}^{2} + \frac{1}{2} m v^{'}^{2} = m v^{'}^{2}$

$= \frac{m v^{2}}{4}$ $(∵ v^{'} = \frac{v}{2})$

This is the kinetic energy possessed by A-B system (since, C is the rest). Let x be the maximum compression of the spring applying energy conservation

$\frac{1}{2} m v^{2} = \frac{1}{2} m v^{'}^{2} + \frac{1}{2} + K x^{2}$

$\Rightarrow \frac{1}{2} m v^{2} = \frac{1}{4} m v^{2} + \frac{1}{2} K x^{2}$

$\Rightarrow \frac{1}{2} K x^{2} = \frac{1}{4} m v^{2}$

$∴ x = v \sqrt{\frac{m}{2 K}}$

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