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Q. 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 $200\, N\, m^{-1}$. The maximum compression produced in the spring will be

AIPMTAIPMT 2012System of Particles and Rotational Motion

Solution:

$\frac{1}{2} mv ^{2}\left[1+\frac{ K ^{2}}{ R ^{2}}\right]=\frac{1}{2} kx _{\max }^{2}$
Putting $m =3 kg , v =4 m / s , \frac{ K ^{2}}{ R ^{2}}=1 / 2$
(For solid cylinder), $k =200\, Nm ^{-1}$ we get
$x _{\max }=0.6\, m$