A solid cylinder of mass 3 kg is rolling on a horizontal surface with velocity 4 ms^-1. It collides with a horizontal spring of force constant 200 Nm^-1. The maximum compression produced in the sprin…

Question

A solid cylinder of mass 3 kg is rolling on a horizontal surface with velocity 4 ms^-1. It collides with a horizontal spring of force constant 200 Nm^-1. The maximum compression produced in the spring will be

Accepted Answer

Correct answer: B. (b): At maximum compression the solid cylinder will stop. According to law of conservation of mechanical energy Loss in kinetic energy =Gain in potential energy of cylinder of spring 12mv^2+ 12I ^2= 12kx^2 12mv^2+ 12 ( mR^22 ) ( vR )^2= 12kx^2 (v= and for solid cylinder, I= 12mR^2) 12mv^2+ 14mv^2= 12kx^2 34mv^2= 12kx^2 or x^2= 32 mv^2k Here, m=3 kg ,v=4ms^-1 ,k=200Nm^-1 Substituting the given values, we get x^2= 3 3 4 42 200 ⇒x^2= 36100 or x=0.6m

$A solid cylinder of mass 3 kg is rolling on a horizontal surface with velocity 4 m s^{- 1} . It collides with a horizontal spring of force constant 200 N m^{- 1} . The maximum compression produced in the spring will be$

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