On a billiards table, the 13 ball collides elastically with the equal mass 9 ball, which was initially at rest. After the collision they head in different directions at speeds $v$ and $2 v$, as shown. What was the initial speed of the 13 ball? You may ignore rotation. $4 v$ $\sqrt{5} \cdot v$ $5 v / 2$ $3 v$

Step 1 ：Given that the collision is elastic, both momentum and kinetic energy are conserved.

Step 2 ：Let's denote the initial speed of the 13 ball as \( u \).

Step 3 ：The initial momentum of the system is \( 13u \) (since the 9 ball is initially at rest).

Step 4 ：After the collision, the total momentum is \( 13v + 9(2v) = 31v \).

Step 5 ：By conservation of momentum, we have \( 13u = 31v \).

Step 6 ：The initial kinetic energy of the system is \( \frac{1}{2} \cdot 13 \cdot u^2 \) (since the 9 ball is initially at rest).

Step 7 ：After the collision, the total kinetic energy is \( \frac{1}{2} \cdot 13 \cdot v^2 + \frac{1}{2} \cdot 9 \cdot (2v)^2 = \frac{1}{2} \cdot 13 \cdot v^2 + 18v^2 = \frac{31}{2} \cdot v^2 \).

Step 8 ：By conservation of kinetic energy, we have \( \frac{1}{2} \cdot 13 \cdot u^2 = \frac{31}{2} \cdot v^2 \).

Step 9 ：Dividing the kinetic energy equation by the momentum equation, we get \( \frac{u}{v} = \sqrt{\frac{31}{26}} = \sqrt{\frac{2}{13} \cdot 31} = \sqrt{2} \cdot \sqrt{\frac{31}{13}} \).

Step 10 ：Therefore, the initial speed of the 13 ball is \( u = v \cdot \sqrt{2} \cdot \sqrt{\frac{31}{13}} = \sqrt{2} \cdot v \cdot \sqrt{\frac{31}{13}} \).

Step 11 ：Simplifying, we get \( u = \sqrt{\frac{62}{13}} \cdot v \approx 2.19v \).

Step 12 ：So, the initial speed of the 13 ball is approximately \( 2.19v \), which is not one of the given options. Therefore, there might be a mistake in the problem statement or in our calculations.