Question

For the motion of block along inclined plane
( operatorname{mg} sin theta=m a )
( a=g sin theta )
where a is along the inclined plane. The vertical component of acceleration is ( mathrm{g} sin ^{2} theta ) Therefore, the relative vertical acceleration of ( mathrm{A} ) with
respect to ( mathrm{B} ) is
( gleft(sin ^{2} 60-sin ^{2} 30right)=g / 2=4.9 m s^{-2} ) (in vertical
direction)
For the motion of block along inclined plane
( operatorname{mg} sin theta=m a )
( a=g sin theta )
where a is along the inclined plane. The vertical component of acceleration is ( mathrm{g} sin ^{2} theta ) Therefore, the relative vertical acceleration of ( mathrm{A} ) with
respect to ( mathrm{B} ) is
( gleft(sin ^{2} 60-sin ^{2} 30right)=g / 2=4.9 m s^{-2} ) (in vertical
direction)

# 19. 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? 60 30°

Solution