In the given figure,the acceleration of wedge | vedclass

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(A) Let the acceleration of wedge $A$ be $\vec{a}_A = 10 \ m/s^2$ down the incline.Block $B$ is constrained to move vertically downward because it is

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$5$

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(A) Let the acceleration of wedge $A$ be $\vec{a}_A = 10 \ m/s^2$ down the incline.Block $B$ is constrained to move vertically downward because it is placed on the horizontal surface of wedge $A$.Let the acceleration of block $B$ be $\vec{a}_B$ in the downward vertical direction.The acceleration of block $B$ with respect to wedge $A$ is $\vec{a}_{B/A}$.Since block $B$ stays in contact with the horizontal surface of wedge $A$,the component of acceleration of $B$ perpendicular to the horizontal surface of $A$ must be equal to the component of acceleration of $A$ perpendicular to that surface.However,a simpler approach is to use the constraint that the velocity of $B$ relative to $A$ must be along the horizontal surface of $A$.Let $\vec{a}_A = 10 \ m/s^2$ at an angle of $30^\circ$ with the horizontal.$\vec{a}_A = 10 \cos 30^\circ \hat{i} - 10 \sin 30^\circ \hat{j} = 5\sqrt{3} \hat{i} - 5 \hat{j}$.Let $\vec{a}_B = 0 \hat{i} - a_B \hat{j}$.The relative acceleration $\vec{a}_{B/A} = \vec{a}_B - \vec{a}_A = -5\sqrt{3} \hat{i} + (5 - a_B) \hat{j}$.Since $B$ moves on the horizontal surface of $A$,the relative acceleration $\vec{a}_{B/A}$ must be horizontal.Therefore,the vertical component of $\vec{a}_{B/A}$ must be zero.$5 - a_B = 0 \implies a_B = 5 \ m/s^2$.

In the given figure,the acceleration of wedge $A$ is $10 \ m/s^2$ along the inclined plane. There is no friction between $A$ and $B$,and between $A$ and the fixed inclined plane. Then the acceleration of block $B$ will be ............ $m/s^2$.

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