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/sAs m2 is shifted by a distance x and released, the mass m1 will break off from the wall when the spring restores to its natural length and m2 starts moving 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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