Block A is constrained to move perpendicular to the inextensible cord. The cord makes an angle $\theta$ with the vertical. Therefore, the path of A is tangent to a circle, directed downward at an angle $\theta$ with the horizontal.

Let $a_{tan}$ be the tangential acceleration of A (down-right). Its horizontal component must match the acceleration of block B ($a_B$) to maintain contact: $$ a_{Ax} = a_{tan} \cos \theta = a_B $$ $$ \implies a_{tan} = \frac{a_B}{\cos \theta} $$

We analyze the forces on A along the direction of its motion (tangential to the arc). The forces are:

• Gravity ($mg$): Component along tangent = $mg \sin \theta$.
• Normal Force from B ($N$): This force acts horizontally to the left. Its component along the tangent direction (down-right) is $-N \cos \theta$.

Applying Newton’s Second Law along the tangent: $$ mg \sin \theta – N \cos \theta = m a_{tan} $$ Substituting $a_{tan} = a_B / \cos \theta$: $$ mg \sin \theta – N \cos \theta = \frac{m a_B}{\cos \theta} \quad \dots(1) $$

Block B accelerates horizontally to the right due to the normal force $N$ exerted by A. $$ N = M a_B \quad \dots(2) $$

Substitute equation (2) into equation (1): $$ mg \sin \theta – (M a_B) \cos \theta = \frac{m a_B}{\cos \theta} $$ Multiply the entire equation by $\cos \theta$ to clear the denominator: $$ mg \sin \theta \cos \theta – M a_B \cos^2 \theta = m a_B $$ $$ mg \sin \theta \cos \theta = a_B (m + M \cos^2 \theta) $$