A chain of mass $M$ and length $L$ is held on a frictionless table with a portion of length $L/n$ hanging over the edge. If the chain is released, what is the speed of the chain when it just leaves the table?

Solution

1. Initial potential energy $U_i$ with reference at the table top: the hanging part has mass $M/n$ and its CM is at $-L/(2n)$, so $$\begin{array}{rl}{U}_i& =(M/n)g(-L/(2n))\\ & =-MgL/(2n^2)\end{array}$$.
2. Final potential energy $U_f$ when the whole chain is hanging: the CM is at $-L/2$, so $$\begin{array}{rl}{U}_f& =Mg(-L/2)\\ & =-MgL/2\end{array}$$.
3. By energy conservation, $$\begin{array}{rl}\frac{1}{2}M{v}^2& =U_i-U_f\\ & =-MgL/(2n^2)+MgL/2\end{array}$$.
4. Simplifying, $\frac{1}{2}{v}^2=\frac{gL}{2}(1-1/n^2)$, so $v=\sqrt{gL(1-1/n^2)}$.
5. Hence the answer is (A).