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A billiard ball of radius $R$, mass $m$ at rest is struck horizontally by an impulse $J$ at height $h$ above its centre. The **height $h$** at which the ball begins **pure rolling immediately** (no slipping) is:
A$h = R$
B$h = R/2$
C$h = 2R/5$
D$h = 7R/5$ above the table (i.e., $2R/5$ above centre)
Answer & Solution
Correct answer: C. $h = 2R/5$
Impulse $J$ at height $h$ above centre: $v = J/m$, $\omega = J\cdot h / I = Jh/(2mR^2/5) = 5Jh/(2mR^2)$. Pure rolling immediately: $v = R\omega$ ⇒ $J/m = R \cdot 5Jh/(2mR^2)$ ⇒ $h = 2R/5$ (above centre). Above table-level: $R + 2R/5 = 7R/5$.
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