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A solid cylinder of mass $12 k g$ is rolling on a rough 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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$0.4 m$

$0.6 m$

$0.7 m$

$1.2 m$

answer is D.

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Detailed Solution

At maximum compression of spring, total kinetic energy of rolling cylinder will be converted into potential energy of the spring.

so, $\frac{1}{2} m v^{2} (1 + \frac{K^{2}}{R^{2}}) = \frac{1}{2} k x^{2}$

$\Rightarrow \frac{1}{2} \times 12 \times {(4)}^{2} [1 + \frac{1}{2}] = \frac{1}{2} \times 200 \times x^{2}$

on solving, we get $x = 1.2 m$

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