Question

Two blocks of masses m1 and m2 connected by a massless spring of spring constant 'k' rest on a smooth horizontal plane as shown in figure. Block 2 is shifted a small distance 'x' to the left and the released. Find the velocity of centre of mass of the system after block 1 breaks off the wall.

A
m2Km1+m2x
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B
m1Km1+m2x
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C
(m1+m2)Km1+m2x
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D
zero
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Solution

The correct option is A √m2Km1+m2xIf m2 is shifted by a distance x and released, the mass m1 will break off from the wall when the spring restores its natural length and m2 will start going towards right. At the time of breaking m1, m2 will be going towards right with a velocity v, which is given as 12kx2=12m2v2⇒v=(√km2)x and the velocity of the centre of mass at this instant is vCM=m1×0+m2×vm1+m2=(√m2k)xm1+m2

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