Two fixed frictionless inclined plane making an angle `30^(@)` and `60^(@)` with the vertical are shown in the figure. Two block A and B are placed on the two planes. What is the relative vertical acceleration of A with respect to B?

Options

A. `4.9 m//s^(2)` in horizontal direction
B. `9.8 m//s^(2)` in vertical direction
C. zero
D. `4.9 m//s^(2)` in vertical direction

Answer

D

Explanation

`mg sin theta = ma` `:.` `a g sin theta`, where a is long the inclimed plane
Therefore, the vertical component of acceleration is `g sin^(2) theta`. Hence, the relative vertical acceleration of A with respect to B is `g [sin^(2) 60 - sin^(2) 30] = (g)/(2) = 4.9 m//s^(2)` in vertical direction.