Two blocks of masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on a friction less horizontal surface. An impulse gives a velocity of 14 m/s to the heavier block in the direction of lighter block. The velocity of center of mass is

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30 m/s

20 m/s

10 m/s

5 m/s

answer is C.

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Detailed Solution

see fig 4 The velocity of center of mass VCM given by the expression vCM=m1v1+m2v2m1+m2 Substituting the given values, we get VCM=10×14+4×010+4=10m/s

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