Question

A particle slides down a smooth inclined plane of elevation $\theta$ fixed in the elevator going up with an acceleration $a_0$ as shown. The base of the incline has length $L$. Then the time taken by the particle to reach the bottom of the plane is given by:

• A. $\sqrt{\frac{2L}{(g+a_0)\sin \theta }}$
• B. $\sqrt{\frac{2L}{(g+a_0)\sin \theta \cos \theta }}$
• C. $\sqrt{\frac{2L}{g\sin \theta \cos \theta }}$
• D. None of the above

Answer 1

The correct option is B $\sqrt{\frac{2L}{(g+a_0)\sin \theta \cos \theta }}$

Let a be the acceleration of the particle with respect to the incline.

Taking components of the forces parallel to the incline and applying Newton’s law, we get,

$mgsin\theta +m{a}_{0}sin\theta =ma$

or $a$= $(g+a_0)sin\theta$

This is the acceleration with respect to the elevator.

Now, $cos\theta =L/x$, where $x$ is the distance moved by the particle.

$x=L/cos\theta$

From the equation of motion

$S=ut+\frac{1}{2}{at}^2$

u is 0

$S=\frac{1}{2}{at}^2$

$\frac{L}{cos\theta }=\frac{1}{2}{(g+a_0)sin\theta t}^2$

$t=\left(\frac{2L}{(g+a_0)sin\theta cos{\theta }^{}}\right){}^{1/2}$