Question
A solid sphere rolling on a rough horizontal surface with a linear speed v collides elastically with a fixed, smooth, vertical wall. Find the speed of the sphere after it has started pure rolling in the backward direction.
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Answer
When the solid sphere collides with the wall, it rebounds with velocity ‘v’ towards left but it continues to rotate in the clockwise direction.
So, the angular momentum $=\text{mvR}-\Big(\frac{2}{5}\Big)\text{mR}^2\times\frac{\text{v}}{\text{R}}$
After rebounding, when pure rolling starts let the velocity be v' and the corresponding angular velocity is $\frac{\text{v}'}{\text{R}}$
Therefore angular momentum $=\text{mv}'\text{R}+\Big(\frac{2}{5}\Big)\text{mR}^2\Big(\frac{\text{v}'}{\text{R}}\Big)$
So, $\text{mvR}-\Big(\frac{2}5{}\Big)\text{mR}^2,\ \frac{\text{v}}{\text{R}}=\text{mvR}+\Big(\frac{2}{5}\Big)\text{mR}^2\Big(\frac{\text{v}'}{\text{R}}\Big)$
$\text{mvR}\times\Big(\frac{3}{5}\Big)=\text{mvR}\times\Big(\frac{7}{5}\Big)$
So, the sphere will move with velocity $\frac{\text{3v}}{7}.$
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