Q.
Two fixed frictionless in inclined plane making an angle $30{^\circ}$ and $60{^\circ}$ 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?
A) $4.9\ {ms}^{- 2}\ $in horizontal direction
B) $9.8\ \ {ms}^{- 2}$ in vertical direction
C) Zero
D) $4.9\ {ms}^{- 2}\ $in vertical direction
Explanation
(d) $mg\sin{\theta = ma}$
$\therefore\ \ \ a = g\sin\theta$
Where $a$ is long the inclined plane.
$\therefore\ \ \ $Vertical component of acceleration is $g\ \sin^{2}\theta.$
$\therefore\ $ Relative vertical acceleration of A with respect to B is
$g\left( \sin^{2}\ 60{^\circ} - \sin^{2}30{^\circ} \right) = \frac{g}{2} = 4.9\ {ms}^{- 2}$
(in vertical direction)
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.