Question

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

• A. $0.5\text{m}$
• B. $7\text{m}$
• C. $0.2\text{m}$
• D. $0.6\text{m}$

Answer 1

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

Given: $k=200{\text{Nm}}^{-1},m=3\text{kg},v=4{\text{ms}}^{-1}$
Using energy conservation,
Loss of kinetic enrgy of the cylinder = Gain in potential energy of the spring.
$\Rightarrow \frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2=\frac{1}{2}k{x}^2$

$\Rightarrow \frac{1}{2}m{v}^2+\frac{1}{2}\times \frac{1}{2}m{r}^2\times \frac{v^2}{r^2}=\frac{1}{2}k{x}^2$

$\Rightarrow \frac{3}{4}m{v}^2=\frac{1}{2}k{x}^2$

Substituting all the values we get,

$\Rightarrow \frac{3}{2}\times 3(4{)}^2=200x^2$

$\Rightarrow x=0.6\text{m}$

Hence, option $\left(D\right)$ is correct.