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Question

# Two blocks of masses m1 and m2 are connected by a weightless spring of stiffness k, rests on a smooth horizontal plane. Mass m2 is shifted by a small distance x0 to the left and released. The velocity of the center of mass of the system when m1 breaks off the wall is,

A
(km2m1+m2)x0
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B
(kx0m2m1+m2)
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C
(m1+m2m2)x0k
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D
(km1m1+m2)x0
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Solution

## The correct option is A (√km2m1+m2)x0Since, the system is free from external forces, hence momentum is conserved. Using the relation for velocity of COM we get, vcom=m1v1+m2v2m1+m2 mass m1 is at rest. ∴v1=0 vcom=m2v2m1+m2 ......(1) Using conservation of energy, Ei=Ef 12kx20=12m2v22 ∴v2=x0√km2 Putting v2 in eq. (1) we get, vcom=(√km2m1+m2)x0 <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, option (A) is the correct answer.

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