MCQ
A solid sphere of mass $2 kg$ is rolling on a friction horizontal surface with velocity $6 m / s$. It collides on the free end of an ideal spring whose other end is fixed. The maximum compression produced in the spring will be (Force constant of the spring - $36 N / m$ ).
  • A
    $\sqrt{14} m$
  • $\sqrt{2.8} m$
  • C
    $\sqrt{1.4} m$
  • D
    $\sqrt{0.7} m$

Answer

Correct option: B.
$\sqrt{2.8} m$
(b) : Kinetic energy of rolling solid sphere
$
\begin{aligned}

& =\frac{1}{2} m v^2+\frac{1}{2} I \omega^2=\frac{1}{2} m v^2+\frac{1}{2} \times \frac{2}{5} m r^2 \omega^2 \\

& =\frac{1}{2} m v^2+\frac{1}{5} m v^2=\frac{7}{10} m v^2 \\
& =\frac{7}{10} \times 2 \times(6)^2=\frac{7}{10} \times 2 \times 36=\frac{252}{5}
\end{aligned}
$
The potential energy of the spring of maximum compression $x=\frac{1}{2} k x^2$

or $\frac{1}{2} k x^2=\frac{252}{5}$ or $k x^2=\frac{252 \times 2}{5}$

or $\quad x^2=\frac{252 \times 2}{5 \times 36}=2.8 \quad$ or $\quad x=\sqrt{2.8} m$

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