A solid cylinder of mass $3 \mathrm{~kg}$ is rolling on a horizontal surface with velocity $4 \mathrm{~m} /…
A solid cylinder of mass $3 \mathrm{~kg}$ is rolling on a horizontal surface with velocity $4 \mathrm{~m} / \mathrm{s}$. It collides with a horizontal spring whose one end is fixed to rigid support. The force constant of material of spring is $200 \mathrm{~N} / \mathrm{m}$. The maximum compression produced in the spring will be (assume collision between cylinder \& spring be elastic)
$\quad 0.7 \mathrm{~m}$
$0.2 \mathrm{~m}$
$0.5 \mathrm{~m}$
$0.6 \mathrm{~m}$
Solution
At maximum compression, the solid cylinder will stop.
So loss in K.E. of cylinder $=$ Gain in P.E. of spring
$\begin{aligned}
& \therefore \quad \frac{1}{2} \mathrm{mv}^2+\frac{1}{2} \mathrm{I \omega}^2=\frac{1}{2} \mathrm{kx}^2 \\
& \therefore \quad \frac{1}{2} \mathrm{mv}^2+\frac{1}{2} \frac{\mathrm{mR}^2}{2}\left(\frac{\mathrm{v}}{\mathrm{R}}\right)^2=\frac{1}{2} \mathrm{kx}^2 \\
& \therefore \quad \frac{3}{4} \mathrm{mv}^2=\frac{1}{2} \mathrm{kx}^2 \\
& \therefore \quad \frac{3}{4} \times 3 \times(4)^2=\frac{1}{2} \times 200 \times \mathrm{x}^2 \\
& \therefore \quad \frac{36}{100}=\mathrm{x}^2 \Rightarrow \mathrm{x}=0.6 \mathrm{~m}
\end{aligned}$