1. Set up the situation

Two identical balls collide.

• Before collision:
  • Ball 1 has velocity $9\,\text{m/s}$ along the original direction (take this as the $x$-axis).
  • Ball 2 is at rest.
• After collision:
  • Let their speeds be $v_1$ and $v_2$.
  • Each moves at an angle of $30^\circ$ with the original direction, on opposite sides of the original line so that transverse momentum cancels.

We need the ratio $v_1 : v_2 = x : y$ and must find $x$.

2. Apply conservation of momentum in the perpendicular direction

Initially, perpendicular momentum is zero.

After collision, since one ball goes at $+30^\circ$ and the other at $-30^\circ$,

$m v_1 \sin 30^\circ - m v_2 \sin 30^\circ = 0$

$m\sin 30^\circ (v_1 - v_2)=0$

So,

$v_1 = v_2$

Thus the ratio is

$v_1 : v_2 = 1:1$

Hence, $x=1$.

3. Check using momentum along original direction

Initial momentum along $x$-axis:

$m\cdot 9$

Final momentum along $x$-axis:

$m v_1 \cos 30^\circ + m v_2 \cos 30^\circ$

Since $v_1=v_2=v$,

$9m = 2mv\cos 30^\circ$

$9 = 2v\cdot \frac{\sqrt{3}}{2}$

$v = \frac{9}{\sqrt{3}} = 3\sqrt{3}\,\text{m/s}$

This is consistent, so the ratio remains $1:1$.

4. Compare with stored answer

Derived answer: $x=1$.

Stored correct answer: $1$.

They match.