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A ring of radius a=1 m and having mass 4 kg lies on a smooth horizontal surface. One end of each of the two massless spring k and 2k (k=250 N/m) are fixed to the diametrically opposite points of ring and the other end of the two springs is attached to a block of mass m=1kg, which lies at the centre of the ring and in this position springs are at their natural length
If the block is displaced by a distance a/2 along the length of springs such that spring of spring constant k is compressed by a/2 and is released then at the instant when each spring achieve their natural length, the velocity of the ring (in m/s) will be

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answer is 3.

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Detailed Solution

Applying energy conservation $\frac{1}{2} k {(\frac{a}{2})}^{2} + \frac{1}{2} 2 k {(\frac{a}{2})}^{2} = \frac{1}{2} m v_{1}^{2} + \frac{1}{2} M v_{1}^{2} ..... (i)$

By law of conservation of momentum $- m v_{1} + M v_{2} = 0$

$v_{1} = 4 v_{2} ...... (i i)$

$\frac{1}{2} k (\frac{a^{2}}{4}) + \frac{k a^{2}}{4} = \frac{1}{2} v_{1}^{2} + \frac{1}{2} \times 4 v_{2}^{2}$

$v_{2}^{2} = \frac{3 k a^{2}}{80}$

$v_{2} = 3 m / s$

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