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Question

# Two blocks A(3kg) and B(2kg) resting on a smooth horizontal surface is connected by a spring of stiffness 480N/m. Initially the spring is undefined and a velocity of 2m/s is imparted to A along the line of the spring away from B .The max extension in meters of the spring subsequent motion is.

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Solution

## Initially, only block-A is moving with velocity 2ms. Hence initial momentum =3×2=6kg−ms..................(1) At time of maximum compression of spring, let the common velocity be v m/s Hence final momentum =(3+2)×v=5×vkg−ms................(2) By conservation of momentum, we equate (1) and (2) to get common velocity v=1.2ms Initial kinetic energy =(12)×3×2×2=6Joules................(3) Final kinetic energy =(12)×5×1.2×1.2=3.6Joules..............(4) difference in kinetic energy is stored as potential energy in spring, i.e., (12)kx2=(12)×480×x2=6−3.6=2.4Joules..........(5)from (5), we get the maximum displacement, x=0.1m

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