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Physics - Rotation Question with Solution | TestHub

PhysicsRotationRotational KinematicsMedium2 minPYQ_2017

PhysicsMediumsingle choice

A solid sphere of mass $2 kg$ is rolling on a frictionless horizontal surface with velocity $6 m s^{- 1}$ . It collides on the free end of an ideal spring whose other end is fixed. The maximum compression produced in the spring will be (Force constant of the spring = $36 N m^{- 1}$ ).

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Answer:

Solution:

Kinetic energy of rolling solid sphere
$= \frac{1}{2} m V^{2} + \frac{1}{2} I ω^{2}$
$= \frac{1}{2} m V^{2} + \frac{1}{2} \times \frac{2}{5} m r^{2} ω^{2}$
$= \frac{1}{2} m V^{2} + \frac{1}{5} m V^{2}$
$= \frac{7}{10} m V^{2}$
The potential energy of the spring on maximum compression $x$
$= \frac{1}{2} k x^{2}$
$∴ \frac{1}{2} k x^{2} = \frac{7}{10} m V^{2}$
$x^{2} = \frac{14}{10} \frac{m V^{2}}{k}$
$= \frac{14}{10} \times \frac{2 {(6)}^{2}}{36}$
$= 2.8$
$x = \sqrt{2.8} m$

Stream:MHT_CETSubject:PhysicsTopic:RotationSubtopic:Rotational Kinematics

⏱ 2mℹ️ Source: PYQ_2017

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