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Question

Total $KE$ of sphere

$=\frac{1}{2} \mathrm{Mv}^{2}\left(1+\frac{\mathrm{k}^{2}}{\mathrm{R}^{2}}\right)=\frac{1}{2} \mathrm{Mv}^{2}\left(1+\frac{2}{

Vedclass · Accepted Answer

$v\sqrt {\frac{{7M}}{{5k}}} $

Vedclass · Answer

Total $KE$ of sphere

$=\frac{1}{2} \mathrm{Mv}^{2}\left(1+\frac{\mathrm{k}^{2}}{\mathrm{R}^{2}}\right)=\frac{1}{2} \mathrm{Mv}^{2}\left(1+\frac{2}{5}\right)$

$=\frac{7}{10} \mathrm{Mv}^{2}$

Which is utilised in compression in the spring.

So $\frac{1}{2} \mathrm{kx}_{\mathrm{m}}^{2}=\frac{7}{10} \mathrm{Mv}^{2} \Rightarrow \mathrm{x}_{\mathrm{m}}=\mathrm{v} \sqrt{\frac{7 \mathrm{M}}{5 \mathrm{k}}}$

A solid sphere rolls without slipping and presses a spring of spring constant $k$ as shown in figure. Then, the maximum compression in the spring will be

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