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(D) Let $v_1$ be the velocity of block $A$ and $v_2$ be the velocity of ball $B$ relative to the ground when the string is vertical.By conservation of

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(D) Let $v_1$ be the velocity of block $A$ and $v_2$ be the velocity of ball $B$ relative to the ground when the string is vertical.By conservation of linear momentum in the horizontal direction: $m v_1 - m v_2 = 0 \implies v_1 = v_2 = v$.By conservation of mechanical energy: $mgL(1 - \cos 	heta) = \frac{1}{2}mv_1^2 + \frac{1}{2}mv_2^2 = \frac{1}{2}mv^2 + \frac{1}{2}mv^2 = mv^2$.Thus,$v^2 = gL(1 - \cos 	heta)$.For the ball $B$,the net force in the vertical direction provides the centripetal acceleration relative to the block $A$. The velocity of $B$ relative to $A$ is $v_{rel} = v_2 - (-v_1) = 2v$.The equation of motion for $B$ in the vertical direction is: $T - mg = \frac{m(2v)^2}{L} = \frac{4mv^2}{L}$.Substituting $v^2$: $T = mg + \frac{4m}{L} \cdot gL(1 - \cos 	heta) = mg + 4mg(1 - \cos 	heta) = mg(1 + 4 - 4\cos 	heta) = mg(5 - 4\cos 	heta)$.

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