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Question

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

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Solution

The correct option is **A** 110

Initially only block A is moving with velocity 2m/s.

Hence initial momentum =3×2=6kgm/s ..................(1)

At time of maximum compression of spring, let the common velocity be vm/s

Hence final momentum =(3+2)×v=5×vkgm/s ................(2)

By conservation of momentum, we equate (1) and (2) to get common velocity v=1.2m/s

Initial kinetic energy =(12)×3×2×2=6J................(3)

Final kinetic energy =(12)×5×1.2×1.2=3.6J ..............(4)

difference in kinetic energy is stored as potential energy in spring,

i.e. (12)kx2=(12)×480×x2=6−3.6=2.4J ..........(5)

from (5), we get the maximum displacement, x=0.1m=110m

Hence,

option (A) is correct answer.

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