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Question

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.

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Solution

P compressed state of string
Q natural length of spring
From P to Q
aB=KxmB
aCM=mBaBmA+mB=(mB)(Kx)mA+mB
From P to Q, compression x decreases. Therefore, aCM decreases. After Q, A leaves contact with wall, spring comes in its natural length. Net force on system becomes zero.
Therefore, aCM become zero.
From P to Q velocity of B, therefore velocity of COM will increase. After that aCM becomes zero.
Therefore, vCM becomes constant.

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