A solid sphere rolling on a rough horizontal surface with a linear speed v collides elestically 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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Solution

when it collides with wall then its linear momentum get reversed but there is no change in angular speed

the speed of center of mass is reversed

just after the collision ,

angular momentum $= m v r - I w = - m u r + 2 m u r / 5 = - 3 m u r / 5$

after a long time, total angular momentum $= m v r + I w$

$= m v r + 2 m v r / 5 = 7 m v r / 5$

since angular mometam is conserved so
$\frac{7 m v r}{5}$ = $\frac{3 m u r}{7}$

$v = \frac{3 u}{7}$

$u = 7$ so

$v = 3 m / s$

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