Two fixed frictionless inclined planes making | vedclass

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Question

$\begin{array}{l}
mg\,\sin 	heta  = ma\,\,\,	herefore \,a = g\sin 	heta \
where\,a\,is\,along\,the\,inclined\,plane\
	herefore \,vertica

Vedclass · Accepted Answer

$4.9$ $ m/s^{2 }$ in vertical direction

Vedclass · Answer

$\begin{array}{l}
mg\,\sin 	heta  = ma\,\,\,	herefore \,a = g\sin 	heta \
where\,a\,is\,along\,the\,inclined\,plane\
	herefore \,vertical\,component\,of\,acceleration\,\
is\,g\,{\sin ^2}	heta \
	herefore \,relative\,vertical\,acceleration\,of\,A\,with\,\
\,respect\,to\,B\,is\
g\,({\sin ^2}60 - {\sin ^2}30]\, = \frac{g}{2} = 4.9\,m/{s^2}\,in\
vertical\,direction
\end{array}$

Two fixed frictionless inclined planes making an angle $30^o $ and $60^o $ 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 $B$?

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