Solution to 2008 Problem 29

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.

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