In the figure shown,A and B are free to move. | vedclass

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(A) Let $m$ be the mass of block $A$ and $M$ be the mass of wedge $B$. Let $a_0$ be the acceleration of the wedge $B$ to the right.For block $A$,the f

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the acceleration of $A$ will be more than $g \sin 	heta$

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(A) Let $m$ be the mass of block $A$ and $M$ be the mass of wedge $B$. Let $a_0$ be the acceleration of the wedge $B$ to the right.For block $A$,the forces perpendicular to the incline are $mg \cos 	heta$ (downwards) and $N$ (normal force upwards). The component of acceleration of $A$ perpendicular to the incline is $a_0 \sin 	heta$ (downwards).Applying Newton's second law perpendicular to the incline: $mg \cos 	heta - N = m(a_0 \sin 	heta)$.Therefore,$N = mg \cos 	heta - ma_0 \sin 	heta$. Since $a_0 > 0$,it follows that $N For the acceleration of $A$ along the incline,let $a$ be the acceleration of $A$ relative to the ground. The component of $a_0$ along the incline is $a_0 \cos 	heta$. The acceleration of $A$ relative to the wedge is $a' = g \sin 	heta + a_0 \cos 	heta$.The absolute acceleration of $A$ is the vector sum of the acceleration of the wedge and the acceleration of $A$ relative to the wedge. Calculating the magnitude,we find the acceleration of $A$ is greater than $g \sin 	heta$.Thus,option $(A)$ is correct.

In the figure shown,$A$ and $B$ are free to move. All the surfaces are smooth. Then for $(0 < \theta < 90^o)$:

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