Let $v_1$ and $v_2$ denote the speeds of the first and the second blocks, respectively, after the collision. We are given that $v_1 = v/2$ and we would like to find $v_2$ . In elastic collisions, kinetic energy is conserved. The initial kinetic energy is

$\begin{align*}\frac{1}{2} M v^2\end{align*}$
The final kinetic energy is
$\begin{align*}\frac{1}{2} M v_1^2 + \frac{1}{2} M v_2^2 = \frac{1}{2} M \left(\frac{v}{2}\right)^2 + \frac{1}{2} M v_2^2\end{...$
Setting the initial kinetic energy equal to the final kinetic energy gives
$\begin{align*}\frac{1}{2} M v^2 = \frac{1}{2} M \left(\frac{v}{2}\right)^2 + \frac{1}{2} M v_2^2\end{align*}$
Dividing both sides by $M v^2/2$ simplies this equation to
$\begin{align*}1 = \frac{1}{4} + \left(\frac{v_2}{v}\right)^2\end{align*}$
Solving for $v_2$ gives
$\begin{align*}v_2 = \boxed{\frac{v\sqrt{3}}{2}}\end{align*}$
Therefore, answer (C) is correct.

Solution to 2008 Problem 29