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(C) Let the acceleration of the system be $a$ such that $M_2$ moves down the plane of angle $\beta$ and $M_1$ moves up the plane of angle $\alpha$.For

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$\left( \frac{M_2 \sin \beta - M_1 \sin \alpha}{M_1 + M_2} \right)g$

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(C) Let the acceleration of the system be $a$ such that $M_2$ moves down the plane of angle $\beta$ and $M_1$ moves up the plane of angle $\alpha$.For mass $M_2$,the equation of motion is: $M_2 g \sin \beta - T = M_2 a$  --- $(1)$For mass $M_1$,the equation of motion is: $T - M_1 g \sin \alpha = M_1 a$  --- $(2)$Adding equations $(1)$ and $(2)$,we get:$(M_2 g \sin \beta - T) + (T - M_1 g \sin \alpha) = M_2 a + M_1 a$$M_2 g \sin \beta - M_1 g \sin \alpha = (M_1 + M_2) a$$a = \frac{(M_2 \sin \beta - M_1 \sin \alpha)g}{M_1 + M_2}$

Two masses $M_1$ and $M_2$ are attached to the ends of a string which passes over a pulley attached to the top of a double inclined plane of angles of inclination $\alpha$ and $\beta$. If the system is released,the acceleration $a$ of the system is given by:

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