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

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Correct answer: B. 0.6 ~m

Loss in KE= Gain in spring energy (1)/(2) m v^2[1+(K^2)/(R^2)]=(1)/(2) k x_ ^2 where k is the force constant. Given, v=4 ~m / s, m=3 ~kg, k=200 ~N / m For solid cylinder, (K^2)/(R^2)=(1)/(2) (1)/(2) × 3 ×(4)^2[1+(1)/(2)]=(1)/(2) × 200 × x_ ^2 The maximum compression in the spring x_ max =0.6 ~m

A solid cylinder of mass 3 ~kg is rolling on a horizontal surface...

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