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4-1.Newton's Laws of Motion
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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 $B$?
A$4.9 $ $m/s^2$ in horizontal direction
B$4.9$ $ m/s^{2 }$ in vertical direction
C$9.8 $ $m/s^2$ in vertical direction
DZero
(AIEEE-2010)
Solution
$\begin{array}{l}
mg\,\sin \theta = ma\,\,\,\therefore \,a = g\sin \theta \\
where\,a\,is\,along\,the\,inclined\,plane\\
\therefore \,vertical\,component\,of\,acceleration\,\\
is\,g\,{\sin ^2}\theta \\
\therefore \,relative\,vertical\,acceleration\,of\,A\,with\,\\
\,respect\,to\,B\,is\\
g\,({\sin ^2}60 – {\sin ^2}30]\, = \frac{g}{2} = 4.9\,m/{s^2}\,in\\
vertical\,direction
\end{array}$
mg\,\sin \theta = ma\,\,\,\therefore \,a = g\sin \theta \\
where\,a\,is\,along\,the\,inclined\,plane\\
\therefore \,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]\, = \frac{g}{2} = 4.9\,m/{s^2}\,in\\
vertical\,direction
\end{array}$
Standard 11
Physics
