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Q.

A ring of radius a=1m and having mass 4kg lies on a smooth horizontal surface. One end of each of the two massless spring k and 2k(k= 240 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.
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If the block is displaced by a distance a2 along  the length of springs such that spring of spring constant k is compressed a2  and is released then at the instant when each spring achieves it’s 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
12k(a2)2+122k(a2)2=12mv12+12Mv22  …(i)
By law of conservation of momentum 
mv1+Mv2= 0 
v1= 4v2    …(ii)
12k(a24)+ka24=12v12+12×4v22

v22=3ka280

v2=3 m/s = 3 m/s.

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