One end of an ideal spring is fixed at point O and other end is attached to a small disc of mass m which is given an initial velocity v0 perpendicular to its length on a smooth horizontal surface. If the maximum elongation of the spring is l04 then ( l0 = natural length, k = stiffness of the spring) (initially the spring is in it’s natural length)
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a
Velocity at maximum elongation is 4v05
b
Velocity at maximum elongation is 3v05
c
v0=5l012km
d
v0=l012km
answer is A.
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Detailed Solution
According to Conservation of Angular Momentum m×v0×l0=m×v×l0+l04⇒v=4v05According to conservation energy 12mv02=12mv2+12kx2⇒mv02=m4v052+kl042⇒v0=5l012km