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Two identical blocks A and B are connected by a spring as shown in figure. Block A is not connected to the wall parallel to y-axis. B is compressed from the natural length of spring and then left. Neglect friction. Match the following two columns.

Column I Column II
(a) Acceleration of centre of mass of two blocks (p) remains constant
(b) Velocity of centre of mass of two blocks (q) first increases then becomes constant
(c) x-coordinate of centre of mass of two blocks (r) first decreases then becomes zero
(d) y-coordinate of centre of mass of two blocks (s) continuously increases

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$(a) \to (r); (b) \to (q); (c) \to (s); (d) \to (p)$

$(a) \to (r); (b) \to (q); (c) \to (p); (d) \to (s)$

$(a) \to (p); (b) \to (q); (c) \to (r); (d) \to (p)$

$(a) \to (q); (b) \to (r); (c) \to (s); (d) \to (p)$

answer is D.

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Detailed Solution

$P \to$ compressed state of spring

$Q \to$ natural length of spring
Question Image
From P to Q
$\begin{array}{r} a_{B} & = \frac{K x}{m_{B}} \\ a_{CM} & = \frac{m_{B} a_{B}}{m_{A} + m_{B}} = \frac{(m_{B}) (K x)}{m_{A} + m_{B}} \end{array}$
From P to Q, compression x decreases. Therefore, $a_{CM}$ decreases. After $Q$ , A leaves contact with wall, spring comes in its natural length. Net force on system becomes zero.
Therefore, $a_{CM}$ becomes zero.
From P to Q, velocity of B, therefore velocity of COM will increase. After that a_CM becomes zero. Therefore, $v_{CM}$ becomes constant.

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