System State ($t=0$): The bar $M$ is on a frictionless incline of angle $\theta$. The ball $m$ hangs by a vertical cord.
Constraint: The cord is inextensible. The vertical acceleration of the ball must equal the vertical component of the bar’s acceleration because the cord remains taut and vertical initially.

Let $A$ be the acceleration of the bar $M$ down the incline. The vertical component of this acceleration is: $$ A_y = A \sin\theta \quad (\text{downwards}) $$ Since the cord connects the ball to the bar and remains vertical (and taut), the ball must accelerate downwards with this same magnitude: $$ a_{\text{ball}} = A \sin\theta $$

For the Ball ($m$):
Forces: Gravity ($mg$ down) and Tension ($T$ up). $$ mg – T = m a_{\text{ball}} $$ Substituting kinematic constraint: $$ mg – T = m (A \sin\theta) \implies T = m(g – A\sin\theta) \quad \text{…(1)} $$

For the Bar ($M$):
Forces along the incline:

• Component of Gravity: $Mg \sin\theta$
• Component of Tension: Since the string is vertical, the angle between string and incline normal is $\theta$. The angle between string and incline plane is $90-\theta$. The component of Tension pulling the bar down the slope is $T \cos(90-\theta) = T \sin\theta$.

Substitute $T$ from (1) into (2): $$ Mg \sin\theta + [m(g – A\sin\theta)] \sin\theta = M A $$ $$ Mg \sin\theta + mg \sin\theta – mA \sin^2\theta = M A $$ $$ (M + m) g \sin\theta = A (M + m \sin^2\theta) $$ Solving for $A$ (Bar’s acceleration): $$ A = \frac{(M+m)g \sin\theta}{M + m \sin^2\theta} \quad (\text{down the slope}) $$

Solving for $a_{\text{ball}}$ (Ball’s acceleration): $$ a_{\text{ball}} = A \sin\theta $$ $$ a_{\text{ball}} = \frac{(M+m)g \sin^2\theta}{M + m \sin^2\theta} \quad (\text{vertically downwards}) $$