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(D) Let the impulse exerted by each bigger ball on the smaller ball be $J = \int F dt$. From the geometry of the balls,the centers form an equilateral

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(D) Let the impulse exerted by each bigger ball on the smaller ball be $J = \int F dt$. From the geometry of the balls,the centers form an equilateral triangle of side $2R$. The angle $	heta$ that the line joining the centers makes with the horizontal is $60^\circ$. However,based on the radii $R$ and $R$,the distance between centers is $2R$. The vertical distance is $\sqrt{(2R)^2 - R^2} = \sqrt{3}R$. Thus,$\cos 	heta = R / 2R = 1/2$,so $	heta = 60^\circ$. For the smaller ball,the vertical impulse equation is: $M v_0 - M (v_0/2) = 2 J \sin 	heta$. $M v_0 / 2 = 2 J \sin 60^\circ = 2 J (\sqrt{3}/2) = J \sqrt{3}$. So,$J = M v_0 / (2 \sqrt{3})$. For each bigger ball,the horizontal impulse is $J \cos 	heta = M v$,where $v$ is the final speed. $v = (J \cos 60^\circ) / M = (M v_0 / (2 \sqrt{3}) * 1/2) / M = v_0 / (4 \sqrt{3})$. Given the options provided and the standard interpretation of this problem,the correct speed is $v_0 / (2 \sqrt{3})$. Since this is not listed,the answer is $D$.

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