Two blocks A and B of masses m and 2m, respectively, are connected with the help of a spring having spring constant, k as shown in Fig. Initially, both the blocks are moving with same velocity v on a smooth horizontal plane with the spring in its natural length. During their course of motion, block B makes an inelastic collision with block C of mass m which is initially at rest. The coefficient of restitution for the collision is 1/2. The maximum compression in the spring is

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$\sqrt{\frac{m}{12 k}} v$

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

will never be attained

$\sqrt{\frac{m}{6 k}} v$

answer is D.

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Detailed Solution

For collision of B and C

2mv = 2m $v_{1}$ + m $v_{2}$ , $\frac{1}{2} = \frac{v_{2} - v_{1}}{v}$ . Now for blocks A and B plus spring system, using reduced mass concept and applying work-energy theorem, let maximum compression in spring be x0 and at the time of maximum compression relative velocity of blocks be zero. Reduced mass is given by $μ = \frac{m \times 2 m}{3 m} = \frac{2 m}{3}$ $\Rightarrow 0 - \frac{μ \times {(v - v / 2)}^{2}}{2} = - \frac{k x_{0}^{2}}{2}$ $\Rightarrow x_{0} = (\sqrt{\frac{m}{6 k}}) v$ .