Laws of MotionHard

Question

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 $\_\_\_\_$ .

Options

A.$\sqrt{\frac{2L}{gsin2\theta - a_{0}(1 + cos2\theta)}}$
B.$\sqrt{\frac{4\text{ }L}{\text{ }gsin2\theta - a_{0}(1 + cos2\theta)}}$
C.$\sqrt{\frac{4\text{ }L}{\text{ }g\cos^{2}\theta - a_{0}sin\theta cos\theta}}$
D.$\sqrt{\frac{2L}{gsin\theta - a_{0}cos\theta}}$

Solution

$${mgsin\theta - {ma}_{0}cos\theta = ma }{a = gsin\theta - a_{0}cos\theta }$$Now using,

$${S = ut + \frac{1}{2}a_{\text{down~}}t^{2} }{\frac{L}{cos\theta} = \frac{1}{2}\left( \text{ }gsin\theta - a_{0}cos\theta \right)t^{2} }{t = \sqrt{\frac{2\text{ }L}{\text{ }gsin\theta cos\theta - a_{0}\cos^{2}\theta}} }{t = \sqrt{\frac{4\text{ }L}{\text{ }gsin2\theta - a_{0}(1 + cos2\theta)}}}$$

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