Two fixed frictionless inclined planes making an angle $30 °$ and $60 °$ with the vertical are shown in the figure. Two blocks $A$ and $B$ are placed on the two planes. What is the relative vertical acceleration of $A$ with respect to $B$ ?

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$4.9 m s^{- 2}$ in horizontal direction

$9.8 m s^{- 2}$ in vertical direction

$4.9 m s^{- 2}$ in vertical direction

Zero

answer is A.

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Detailed Solution

$m g \sin θ = m a ∴ a = g \sin θ$

where $a$ is along the inclined plane

$∴$ vertical component of acceleration is $g \sin^{2} θ$
$∴$ relative vertical acceleration of $A$ with respect to $B$ is $g (\sin^{2} 60 - \sin^{2} 30) = \frac{g}{2} = 4.9 m / s^{2}$ in vertical direction

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