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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 its natural length as shown. A third identical block C also of mass m, moves on the floor with a speed v along line joining A and B and collide with A, then

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The maximum compression of the spring is

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

The maximum compression of the spring is

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

The K.E. of the system at maximum compression of the spring is

{mv}^{2} /4

The K.E. of the A-B system, at the maximum compression of the spring is zero

answer is B, D.

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Detailed Solution

According to conservation of energy According to conservation of

energy at maximum compression

$\frac{1}{2} {mu}^{2} = \frac{1}{2} k x^{2} + \frac{1}{2} {mv}^{2} + \frac{1}{2} {mv}^{2}, KE = \frac{1}{2} \times \frac{{mu}^{2}}{2}$

According to LCLM

$mu = (m + m) v \Rightarrow v = \frac{u}{2}$

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

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