A solid cylinder of mass 3 , kg is rolling on | vedclass

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(B) At maximum compression,the solid cylinder will stop momentarily.According to the law of conservation of mechanical energy:Loss in kinetic energy o

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$0.6$

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(B) At maximum compression,the solid cylinder will stop momentarily.According to the law of conservation of mechanical energy:Loss in kinetic energy of the cylinder = Gain in potential energy of the spring.The total kinetic energy of a rolling cylinder is $K = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.For a solid cylinder,$I = \frac{1}{2}mR^2$ and for pure rolling,$\omega = \frac{v}{R}$.Substituting these,$K = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{1}{2}mR^2)(\frac{v}{R})^2 = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$.Equating this to the potential energy stored in the spring,$\frac{1}{2}kx^2 = \frac{3}{4}mv^2$.Solving for $x$: $x^2 = \frac{3mv^2}{2k}$.Given $m = 3 \, kg$,$v = 4 \, m s^{-1}$,and $k = 200 \, N m^{-1}$.$x^2 = \frac{3 	imes 3 	imes (4)^2}{2 	imes 200} = \frac{9 	imes 16}{400} = \frac{144}{400} = 0.36$.Therefore,$x = \sqrt{0.36} = 0.6 \, m$.

$A$ 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 ............... $m$.

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