A massless elastic cord (that obeys Hooke's Law) will break if the tension in the cord exceeds $T_{max}$. One end of the cord is attached to a fixed point, the other is attached to an object of mass $3m$. If a second, smaller object of mass m moving at an initial speed $v_0$ strikes the larger mass and the two stick together, the cord will stretch and break, but the final kinetic energy of the two masses will be zero. If instead the two collide with a perfectly elastic one-dimensional collision, the cord will still break, and the larger mass will move off with a final speed of $v_f$. All motion occurs on a horizontal, frictionless surface.

Find $v_f/v_0$.

$ \textbf{(A)}\ 1/\sqrt{12}\qquad\textbf{(B)}\ 1/\sqrt{2}\qquad\textbf{(C)}\ 1/\sqrt{6} \qquad\textbf{(D)}\ 1/\sqrt{3}\qquad\textbf{(E)}\ \text{none of the above} $
1. **Understanding the Problem:**
   - We have a massless elastic cord that obeys Hooke's Law and will break if the tension exceeds \( T_{max} \).
   - One end of the cord is attached to a fixed point, and the other end is attached to an object of mass \( 3m \).
   - A second object of mass \( m \) moving at an initial speed \( v_0 \) strikes the larger mass, and the two scenarios are considered: inelastic collision and elastic collision.

2. **Inelastic Collision:**
   - **Conservation of Momentum:**
     \[
     mv_0 = (m + 3m)v_1 \implies v_1 = \frac{v_0}{4}
     \]
   - **Conservation of Mechanical Energy:**
     \[
     K_i + U_i = \frac{1}{2}(4m)v_1^2 + 0 = K_f + U_f = 0 + U_{max}
     \]
     \[
     \frac{1}{2}(4m)\left(\frac{v_0}{4}\right)^2 = U_{max}
     \]
     \[
     \frac{1}{2}(4m) \cdot \frac{v_0^2}{16} = U_{max}
     \]
     \[
     U_{max} = \frac{mv_0^2}{8}
     \]

3. **Elastic Collision:**
   - Let the speed of the smaller mass after the collision be \( v_a \) and for the larger mass \( v_b \).
   - Define \( a = \frac{v_a}{v_0} \), \( b = \frac{v_b}{v_0} \), and \( c = \frac{v_f}{v_0} \).
   - **Conservation of Momentum:**
     \[
     mv_0 = mv_a + 3mv_b \implies 1 = a + 3b
     \]
   - **Conservation of Mechanical Energy:**
     \[
     \frac{1}{2}mv_0^2 = \frac{1}{2}mv_a^2 + \frac{1}{2}(3m)v_b^2
     \]
     \[
     1 = a^2 + 3b^2
     \]
   - Solving the system of equations:
     \[
     1 = a + 3b
     \]
     \[
     1 = a^2 + 3b^2
     \]
     Substitute \( a = 1 - 3b \) into the second equation:
     \[
     1 = (1 - 3b)^2 + 3b^2
     \]
     \[
     1 = 1 - 6b + 9b^2 + 3b^2
     \]
     \[
     0 = 12b^2 - 6b
     \]
     \[
     0 = 6b(2b - 1)
     \]
     \[
     b = 0 \quad \text{or} \quad b = \frac{1}{2}
     \]
     - If \( b = 0 \), then \( a = 1 \) (extraneous solution).
     - If \( b = \frac{1}{2} \), then \( a = -\frac{1}{2} \).

4. **Final Speed Calculation:**
   - After the collision, apply conservation of mechanical energy:
     \[
     K_i + U_i = \frac{1}{2}(3m)v_b^2 + 0 = K_f + U_f = \frac{1}{2}(3m)v_f^2 + U_{max}
     \]
     \[
     \frac{1}{2}(3m)\left(\frac{v_0}{2}\right)^2 = \frac{1}{2}(3m)v_f^2 + \frac{mv_0^2}{8}
     \]
     \[
     \frac{3m}{8}v_0^2 = \frac{3m}{2}v_f^2 + \frac{mv_0^2}{8}
     \]
     Divide by \( \frac{mv_0^2}{8} \):
     \[
     3 = 12c^2 + 1
     \]
     \[
     2 = 12c^2
     \]
     \[
     c^2 = \frac{1}{6}
     \]
     \[
     c = \frac{1}{\sqrt{6}}
     \]

The final answer is \( \boxed{ \frac{1}{\sqrt{6}} } \)