Question 7 of 210

A small block of mass $m$ slides down from the top of a frictionless inclined surface, while the inclined plane is moving towards left with constant acceleration $a_{0}$. The angle between the inclined plane and ground is $\theta$ and its base length is $L$. Assuming that initially the small block is at the top of the inclined plane, the time it takes to reach the lowest point of the inclined plane is $\_\_\_\_$.

A$\sqrt{\frac{2 L}{g \sin \theta-a_{0} \cos \theta}}$B$\sqrt{\frac{4 L}{g \cos ^{2} \theta-a_{0} \sin \theta \cos \theta}}$C$\sqrt{\frac{2 L}{g \sin 2 \theta-a_{0}(1+\cos 2 \theta)}}$D$\sqrt{\frac{4 L}{g \sin 2 \theta-a_{0}(1+\cos 2 \theta)}}$

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Answer: (D) $\sqrt{\frac{4 L}{g \sin 2 \theta-a_{0}(1+\cos 2 \theta)}}$

$mg\sin\theta - ma_0\cos\theta = ma$
$a = g\sin\theta - a_0\cos\theta$
Now using,
$S = ut + \dfrac{1}{2}a_{down}t^2$
$\dfrac{L}{\cos\theta} = \dfrac{1}{2}(g\sin\theta - a_0\cos\theta)t^2$
$t = \sqrt{\dfrac{2L}{g\sin\theta\cos\theta - a_0\cos^2\theta}}$
$t = \sqrt{\dfrac{4L}{g\sin 2\theta - a_0(1 + \cos 2\theta)}}$