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A narrow but tall cabin is falling freely near the earth's surface. Inside the cabin, two small stones $A$ and $B$ are released from rest (relative to the cabin). Initially $A$ is much above the centre of mass and $B$ much below the centre of mass of the cabin. A close observation of the motion of $A$ and $B$ will reveal that -
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The correct answer is:
A moves slowly upward and $B$ moves slowly downward relative to the cabin

$a_{B}>a_{C}>a_{A}$
$a_{B}=g$
$\vec{a}_{\Lambda / C M}=\vec{a}_{\Lambda}-\vec{a}_{C M}(\uparrow)$
$\vec{a}_{B / C M}=\vec{a}_{B}-\vec{a}_{C M}(\downarrow)$
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