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(A) Let the mass of the block $A$ be $m$ and the mass of the ball $B$ be $m$. Since the system (block + ball) is on a smooth horizontal surface,there

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$\frac{L \sin 	heta}{2}$

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(A) Let the mass of the block $A$ be $m$ and the mass of the ball $B$ be $m$. Since the system (block + ball) is on a smooth horizontal surface,there is no external horizontal force acting on the system.Therefore,the position of the centre of mass of the system remains unchanged in the horizontal direction.Let $x$ be the displacement of the block $A$ to the left when the ball $B$ reaches the equilibrium position.In this process,the ball $B$ moves a horizontal distance of $(L \sin 	heta - x)$ to the right relative to the ground.Using the principle of conservation of the centre of mass in the horizontal direction:$m_A \Delta x_A + m_B \Delta x_B = 0$Taking the left direction as negative and the right direction as positive:$m(-x) + m(L \sin 	heta - x) = 0$$-mx + mL \sin 	heta - mx = 0$$2mx = mL \sin 	heta$$x = \frac{L \sin 	heta}{2}$

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