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प्रश्न
A solid sphere rolling on a rough horizontal surface with a linear speed ν collides elastically with a fixed, smooth, vertical wall. Find the speed of the sphere after it has started pure rolling in the backward direction.
उत्तर
Consider two cases:-
(a) Just after the collision with the wall, the sphere rebounds with velocity \[\nu\] towards left but it continues to rotate in the clockwise direction (as shown in figure)
Angular momentum of the sphere about centre,
\[L = mvR - I\omega\]
\[ = mvR - \frac{2}{5}m R^2 \times \frac{\nu}{R}\]
\[ = \frac{3}{5}mvR\]
(b) When pure rolling starts
Let the velocity be \[\nu'\] and the corresponding angular velocity be \[\frac{v'}{R}.\]
\[L' = mv'R + I\omega'\]
\[ = mv'R + \frac{2}{5}m R^2 \times \frac{v'}{R}\]
\[ = \frac{7}{5}mv'R\]
Angular momentum is constant.
Therefore, we have
\[L = L'\]
\[ \Rightarrow v' = \frac{3v}{7}\]
So, the sphere will move with velocity \[\frac{3v}{7}.\]
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