Two blocks of masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives a velocity of 14 m/s to the heavier block in the direction of the lighter block. The velocity of the center of mass is____m/s
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answer is 10.
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Detailed Solution
The velocity vCM of the centre of mass can be obtained by using the principle of conservation of linear momentum, MV=(M+m)vCMvCM=MV(M+m)=10kg×14ms−1(10+4)kg =10ms−1