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A **bullet** of mass $m$, velocity $v$ strikes the lower end of a uniform rod (mass $M$, length $L$, hinged at top) and embeds. The **angular speed** of the rod+bullet immediately after is:
A$3mv/((M + 3m)L)$
B$mv/(ML)$
C$3mv/(ML)$
D$2mv/(ML)$
Answer & Solution
Correct answer: A. $3mv/((M + 3m)L)$
Angular momentum about hinge conserved (impulsive hinge reaction has zero moment arm at hinge). Before: $L = mvL$ (bullet's angular momentum about hinge). After: $I_{\text{tot}} = ML^2/3 + mL^2$. Equate: $mvL = (ML^2/3 + mL^2)\omega$ ⇒ $\omega = mv/((M/3 + m)L) = 3mv/((M + 3m)L)$.
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