Two fixed frictionless inclined planes making angles $30°$ 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?
Answer: (A) $4.9\ \text{m s}^{-2}$ in the vertical direction
Concept. On a smooth incline of angle $\theta$ to the horizontal, a block accelerates at $g\sin\theta$ along the slope; its vertical component is $g\sin\theta\cdot\sin\theta=g\sin^2\theta$.
Step 1. Angles with the horizontal. $30^\circ$ with the vertical means $60^\circ$ with the horizontal (A); $60^\circ$ with the vertical means $30^\circ$ (B).
Step 2. Vertical accelerations.
$$a_A=g\sin^260^\circ=\tfrac34g,\qquad a_B=g\sin^230^\circ=\tfrac14g$$
Step 3. Relative. $a_A-a_B=\tfrac12g=4.9$ m s$^{-2}$, vertical.
Answer. $4.9\ \text{m s}^{-2}$ in the vertical direction (Option A).
Solution by Sreeraj P, M.Sc Physics
