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
On a smooth incline at angle $\theta$ to the horizontal, the acceleration down the plane is $g\sin\theta$. Its vertical component is $g\sin\theta \times \sin\theta = g\sin^2\theta$.
In the figure, A's plane is inclined at $60°$ to the horizontal and B's at $30°$:
$$a_{A,y} = g\sin^2 60° = \frac{3g}{4}, \qquad a_{B,y} = g\sin^2 30° = \frac{g}{4}$$
Relative vertical acceleration of A with respect to B:
$$\frac{3g}{4} - \frac{g}{4} = \frac{g}{2} = 4.9\ \text{m s}^{-2}\ \text{(vertical)}$$
Solution by Sreeraj P, M.Sc Physics