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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

A
110
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B
1210
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C
1215
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D
0.15
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Solution

## The correct option is A 110Initially 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/sHence 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/sInitial 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=110mHence,option (A) is correct answer.

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