Question

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

• A. $0.7\text{m}$
• B. $0.2\text{m}$
• C. $0.5\text{m}$
• D. $0.6\text{m}$

Answer 1

The correct option is D $0.6\text{m}$

Maximum compression in spring will occur when the cylinder comes to rest, $i.e.$, $v=0$ and $\omega =0$.
Work done by frictional force is zero$(W_f=0)$ in case of pure rolling.
Applying the law of conservation of mechanical energy,
Loss in total kinetic energy of cylinder $=$ Gain in potential energy of spring
$\Rightarrow \frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2=\frac{1}{2}k{x}_m^2$
$(x_m=$ maximum compression in spring$)$
For pure rolling, $v=R\omega$
$\Rightarrow \frac{1}{2}m{v}^2+\frac{1}{2}\left(\frac{m{R}^2}{2}\right)\left(\frac{v^2}{R^2}\right)=\frac{1}{2}k{x}_m^2$
$\Rightarrow \frac{m{v}^2}{2}+\frac{m{v}^2}{4}=\frac{1}{2}k{x}_m^2$
$\Rightarrow \frac{3m{v}^2}{4}=\frac{1}{2}k{x}_m^2$
$\Rightarrow \frac{3\times 3\times 4^2}{4}=\frac{1}{2}\times 200\times x_m^2$
$\Rightarrow x_m^2=\frac{9}{25}$
$\Rightarrow x_m=0.6\text{m}$