Question

A solid cylinder of mass $3kg$ is rolling on a horizontal surface with velocity $4m{s}^{-1}$. It collides with a horizontal spring of force constant $200N{m}^{-1}$. The maximum compression produced in the spring will be:

• A. $0.7m$
• B. $0.2m$
• C. $0.5m$
• D. $0.6m$

Answer 1

The correct option is D $0.6m$

We know that for the rolling body, the initial kinetic energy is equal to potential energy of the spring

$K.E=P.E$

Potential energy of the spring $=\frac{k\Delta x^2}{2}$

Here, $\Delta x=$ Maximum compression of the spring

Kinetic energy of rolling body $=TranslationalKineticenergy+Rotationalkineticenergy$

$\begin{array}{c}=\frac{1}{2}m{V}^2+\frac{1}{2}I{\omega }^2\\ =\frac{1}{2}m{V}^2+\frac{1}{2}\left(\frac{M{R}^2}{2}\right){\left(\frac{V}{R}\right)}^2\\ =\frac{3}{4}m{V}^2\end{array}$

Now

$\frac{K\Delta x^2}{2}=\frac{3}{4}m{V}^2$

$\begin{array}{c}\Delta x=\sqrt{\frac{3m{V}^2}{2K}}\\ =\sqrt{\frac{3\times 3\times 4^2}{2\times 200}}\\ =0.6m\end{array}$

Hence, Option $D$ is the correct answer.