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(A) Let $v_A$ be the velocity of block $A$ and $v_B$ be the velocity of ball $B$ relative to the ground at the equilibrium position. Since there is no

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$[gl (1-\cos	heta )]^{1/2}$

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(A) Let $v_A$ be the velocity of block $A$ and $v_B$ be the velocity of ball $B$ relative to the ground at the equilibrium position. Since there is no external horizontal force,the horizontal momentum of the system is conserved: $m v_A + m v_B = 0 \implies v_A = -v_B$. Let $v$ be the magnitude of the velocity of the block $(v_A = v)$ and the ball relative to the block $(v_{B/A} = u)$. Then the velocity of the ball relative to the ground is $v_B = v - u$ (in the opposite direction). From momentum conservation: $m v + m(v - u) = 0 \implies 2v = u \implies v = u/2$. By conservation of energy,the potential energy lost by the ball equals the kinetic energy gained by the system: $mgL(1-\cos	heta) = \frac{1}{2} m v^2 + \frac{1}{2} m (v-u)^2$. Substituting $u = 2v$: $mgL(1-\cos	heta) = \frac{1}{2} m v^2 + \frac{1}{2} m (v-2v)^2 = \frac{1}{2} m v^2 + \frac{1}{2} m v^2 = m v^2$. Thus,$v^2 = gL(1-\cos	heta) \implies v = [gL(1-\cos	heta)]^{1/2}$.

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