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

Options

A. ` 9.8 ms^(2)` in vertical direction
B. Zerp
C. ` 4.9 ms^(-2)` in evrtical direction
D. ` 4.9 ms^(-2)` in evrtical direction

Answer

C

Explanation

Acceleration of the body down the inclined plane of incleinaton ` theta` horizontal direction is,
` a= g= sin theta`
It is along the inclined plane. The vertical component of accleration (a) is
` a_(along vertcal)=a cos (90^@- theta)= a sin thea`
For block A, ` a_A _(along vertical )= g sin^2 60^@`
For block B, ` a_(a (along veratcal)= g sin^2 30^@`
The relative vertical accleration of ` A w.r.t B` is
` a_(AB (along vertical)`
` - a_(B (along vertical)`
`= g sin^2 60^@ -g sin^2 30^@`
` =g [((sqrt 3)/2))^2 - (1/2)^2] = g/2`
` = 4.9 ms^(-2)` in vertical direction.