Two fixed frictionless inclined plane 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 ?

Options

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

Answer

D

Explanation

We can write, `mg sin theta = ma`
`therefore " "a=gsintheta`
Where, a is along the inclined plane.
`because` Vertical component of acceleration is `g sin^(2)theta`.
`therefore` Relative vertical acceleration of A with respect to B is
`g (sin^(2) 60^(@) -sin^(2)30^(@))=(g)/(2)=4.9ms^(-2)` [in vertical direction]