Q107 / 200 · NEET 2012 (Screening) · Physics
A solid cylinder of mass $3 \mathrm{~kg}$ is rolling on a horizontal surface with velocity $4 \mathrm{~ms}^{-1}$. It collides with a horizontal spring of force constant $200 \mathrm{Nm}^{-1}$. The maximum compression produced in the spring will be
- A$0.5 \mathrm{~m}$
- Correct answerB$0.6 \mathrm{~m}$
- C$0.7 \mathrm{~m}$
- D$0.2 \mathrm{~m}$
Solution
Loss in $\mathrm{KE}=$ Gain in spring energy
$\frac{1}{2} m v^2\left[1+\frac{K^2}{R^2}\right]=\frac{1}{2} k x_{\max }^2$
where $k$ is the force constant.
Given, $v=4 \mathrm{~m} / \mathrm{s}, m=3 \mathrm{~kg}, k=200 \mathrm{~N} / \mathrm{m}$
For solid cylinder, $\frac{K^2}{R^2}=\frac{1}{2}$
$\therefore \frac{1}{2} \times 3 \times(4)^2\left[1+\frac{1}{2}\right]=\frac{1}{2} \times 200 \times x_{\max }^2$
The maximum compression in the spring
$x_{\text {max }}=0.6 \mathrm{~m}$
Hence, the correct answer is option: 'B'.