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Question

Two blocks A and B of masses 2 kg and 4 kg are connected by a light spring of force constant 4800 N/m . The system is at rest on a smooth horizontal surface with the spring in its natural state without any compression or extension. Now a velocity of 6m/s is given to B in a direction away from A. During the subsequent motion, the maximum extension of the spring is:

A
5 cm
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B
10 cm
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C
15 cm
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D
20 cm
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Solution

The correct option is B 10 cmUsing conservation of momentum principle we have mAvA+mBvBmA+mB=vCM.vCM=2×0+4×62+4=4m/sThus we get vCM as 4m/s.Now let us sit in the centre of mass frame, i.e. moving at 4m/s away in direction of B.B moves with a velocity of 2m/s away from CM and A moves with a velocity of 4m/s away from CM. At maximum extension, in centre of mass frame, both blocks will be stationary ( since total momentum is 0 in centre of mass frame, so if one stops, so must the other.) Now by conserving energy,12mAv2A+12mBv2B=12kx212×2×42+12×4×22=12×4800×x2(16+8)J=2400x2x=0.1m=10cm

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