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Question

# Two blocks of mass m1=2 kg,m2=4 kg connected by weightless spring of stiffness k rest on a horizontal plane as shown. m2 is shifted a small distance x=1 cm to the left and then released. Find the velocity of COM of the system just after m1 break off the wall. Assume k=4 N/m.

A
1 m/s
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B
23 m/s
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C
3 m/s
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D
2 m/s
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Solution

## The correct option is B 23 m/sIf 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 v2, which is given as 12kx2=12m2v22 v2=√km2.x=√44.1=1 m/s And the velocity of the COM at that instant is, vcom=m1v1+m2v2m1+m2 Hence, vcom=2×0+4×√44.12+4=23 m/s

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