A ball moving with velocity of 9m / s collides with another similar stationary ball. After the collision both the balls move in directions making an angle of 30° with the initial direction. After the collision their speed will be

Let the mass of the first ball be _m and its initial velocity be $u=9\text{m/s}$ along the x-axis.
The second similar ball has mass _m and is initially stationary ($v=0$).

After the collision, both balls move in directions making an angle of $30°$ with the initial direction (one above the x-axis and one below the x-axis, to conserve momentum perpendicular to the initial line of motion).
Let their speeds after the collision be _v1 and _v2. By symmetry, since they both deflect by the same angle and have equal masses, their speeds after collision will be equal: $v_1=v_2=v\prime$.

Using the law of conservation of linear momentum along the initial direction of motion (x-axis):
$mu=mv\prime \cos (30°)+mv\prime \cos (30°)$

Canceling the mass _m from both sides:
$u=2v\prime \cos (30°)$

Substitute $u=9\text{m/s}$ and $\cos (30°)=\frac{\sqrt{3}}{2}$:
$9=2v\prime \left(\frac{\sqrt{3}}{2}\right)$

This simplifies to:
$9=\sqrt{3}v\prime$

Solving for $v\prime$:
$v\prime =\frac{9}{\sqrt{3}}=3\sqrt{3}\text{m/s}$

Using the value $\sqrt{3}\approx 1.732$:
$v\prime \approx 3\times 1.732=5.196\text{m/s}\approx 5.2\text{m/s}$

Therefore, the speed of each ball after the collision is approximately 5.2 m/s.