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(D) Let the impulse exerted by the falling ball on each of the two stationary balls be $J = F dt$.By the impulse-momentum theorem for the falling ball

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(D) Let the impulse exerted by the falling ball on each of the two stationary balls be $J = F dt$.By the impulse-momentum theorem for the falling ball (mass $M$): $2J \sin 	heta = M v_0$ ... $(i)$where $	heta$ is the angle the line of impact makes with the horizontal.For one of the stationary balls (mass $M$),the horizontal component of the impulse causes it to move with velocity $v$:$J \cos 	heta = M v$ ... (ii)From $(i)$ and (ii),we get $	an 	heta = v_0 / (2v)$.Considering the geometry of the balls at the moment of impact,the centers form an equilateral triangle of side $2R$. The vertical distance between the center of the top ball and the line joining the centers of the bottom balls is $\sqrt{(2R)^2 - R^2} = \sqrt{3}R$.Thus,$	an 	heta = \frac{\sqrt{3}R}{R} = \sqrt{3}$.Equating the two expressions for $	an 	heta$:$\sqrt{3} = \frac{v_0}{2v} \implies v = \frac{v_0}{2\sqrt{3}}$.Since this option is not listed,the correct choice is $(D)$.

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