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Q.

A solid sphere is rolling purely on a rough horizontal surface (coefficient of kinetic friction = $μ$ ) with speed of centre =u. It collides inelastically with a smooth vertical wall at a certain moment, the coefficient of restitution being $\frac{1}{2}$ . The sphere will begin pure rolling after a time

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a

(1) $\frac{3 u}{7 μg}$

b

(3) $\frac{3 u}{5 μg}$

c

(4) $\frac{2 u}{5 μg}$

d

(2) $\frac{2 u}{7 μg}$

answer is A.

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Detailed Solution

$a = \frac{μmg}{m} = μg$

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$ω = \frac{u}{R} \Rightarrow α = \frac{(μmg) R}{2 / 5 {mR}^{2}} = \frac{5 μg}{2 R} Pure rolling will start when v = Rω (\frac{u}{2} - at) = R (αt - ω) or (\frac{u}{2} - μgt) = R (\frac{5 μg}{2 R} t - \frac{u}{R}) Solving we get, t = \frac{3 u}{7 μg}$

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