Two blocks A and B, each of mass m, are connected by a massless spring of natural length L and natural spring constant K. The blocks are initially resting on a smooth horizontal floor with the spring as its natural length as shown in the figure. 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 makes a perfectly elastic collision 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 the A-B system, at maximum compression of the spring is $\frac{m v^{2}}{4}$

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

All of the above

answer is B, C.

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Detailed Solution

C collides with A ; velocities interchange because collide is elastic

$\begin{array}{l} L . c L . M \\ m V = (m + m) V^{1} \\ v = 2 v^{1} \Rightarrow V^{1} = \frac{v}{2} \\ L . C . E \\ \frac{1}{2} m V^{2} = \frac{1}{2} K x^{2} + \frac{1}{2} (2 m) {(\frac{V}{2})}^{2} \Rightarrow x = V \sqrt{\frac{m}{k}} \\ a t \max i m u m c o m p r e s s i o n K . E o f A + B = \frac{1}{2} (2 m) {(\frac{V}{2})}^{2} = \frac{m V^{2}}{4} \end{array}$