A chain of mass $M$ and length $L$ is held on a frictionless table with $\frac{1}{n}$ of its length hanging over the edge. If the chain is released, what is the velocity of the chain when it just leaves the table?

Solution

1. The initial potential energy of the hanging part of length $L/n$ is $$\begin{array}{rl}{U}_i& =-(\frac{M}{n})g(\frac{L}{2n})\\ & =-\frac{MgL}{2n^2}\end{array}$$.
2. When the chain leaves the table, its center of mass is at a distance $L/2$ below the table, so the final potential energy is $U_f=-Mg(L/2)$.
3. By conservation of energy, the loss in potential energy equals the gain in kinetic energy: $$\begin{array}{rl}\frac{1}{2}M{v}^2& =U_i-U_f\\ & =\frac{MgL}{2}-\frac{MgL}{2n^2}\end{array}$$.
4. Solving for velocity, we get $v^2=gL(1-\frac{1}{n^2})$, which simplifies to $v=\sqrt{gL(1-\frac{1}{n^2})}$.
5. Hence the answer is (A).