Question 97 of 210

A block of mass $m$ slides on the wooden wedge, which in turn slides backward on the horizontal surface. The acceleration of the block with respect to the wedge is:
Given $m=8kg,M=16kg$
Assume all the surfaces shown in the figure to be frictionless.

A$\dfrac{3}{5}g$B$\dfrac{4}{3}g$C$\dfrac{6}{5}g$D$\dfrac{2}{3}g$

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Answer: (D) $\dfrac{2}{3}g$

$N\cos 60^{\circ}=M{a}_{1}=16{a}_{1}$
$\Rightarrow N=32{a}_{1}$
$\perp$ to incline
$N=8gcos30^{\circ}-8{a}_{1}\sin 30^{\circ}\Rightarrow 32{a}_{1}=44\sqrt{3}g-4{a}_{1}$
$\Rightarrow {a}_{2}=\dfrac{\sqrt{3}}{9}g$
Along incline
$8gsin30^{\circ}+8{a}_{1}\cos 30^{\circ}=m{a}_{2}=8{a}_{2}$
${a}_{2}=g\times \dfrac{1}{2}+\dfrac{\sqrt{3}}{9}g\cdot \dfrac{\sqrt{3}}{2}=\dfrac{2g}{3}$