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 :

0.6 m

Work - Energy Theorem : \(\frac{1}{2}kx^2 = \frac{1}{2}mv^2[1 + \frac{K^2}{R^2}]\)

k = 200 N/m ; m = 3 kg ; v = 4 m/s

\(\frac{1}{2}(200)x^2 = \frac{1}{2}×3×4^2[1 + \frac{1}{2}]\)

⇒ $ 100x^2 = 36 $

x = 0.6 m