Let the equal mass of the balls be $m$. We establish a coordinate system where the x-axis is along the initial direction of ball A's velocity. Since the collision occurs on a smooth horizontal surface, the total momentum of the system is conserved in both the x and y directions.

Initial momentum:

$P_x = m v_1$ $P_y = 0$

Final momentum:

$P'_x = m v'_1 \cos\alpha + m v'_2 \cos\beta$ $P'_y = m v'_1 \sin\alpha - m v'_2 \sin\beta$

(Assuming the balls scatter to opposite sides of the x-axis)

Applying the conservation of momentum and canceling the mass $m$: x-direction: $v_1 = v'_1 \cos\alpha + v'_2 \cos\beta$ y-direction: $0 = v'_1 \sin\alpha - v'_2 \sin\beta$

From the y-direction equation, we find the relationship between the final speeds:

$$v'_2 = v'_1 \frac{\sin\alpha}{\sin\beta}$$

Substitute this expression for $v'_2$ into the x-direction equation:

$$v_1 = v'_1 \cos\alpha + \left(v'_1 \frac{\sin\alpha}{\sin\beta}\right) \cos\beta$$ $$v_1 = v'_1 \left( \cos\alpha + \frac{\sin\alpha \cos\beta}{\sin\beta} \right) = v'_1 \left( \frac{\sin\beta \cos\alpha + \cos\beta \sin\alpha}{\sin\beta} \right)$$

Using the trigonometric identity for $\sin(\alpha+\beta)$:

$$v_1 = v'_1 \frac{\sin(\alpha + \beta)}{\sin\beta}$$

Solving for $v'_1$:

$$v'_1 = v_1 \frac{\sin\beta}{\sin(\alpha + \beta)}$$

Substituting the expression for $v'_1$ back into the equation for $v'_2$:

$$v'_2 = \left(v_1 \frac{\sin\beta}{\sin(\alpha + \beta)}\right) \frac{\sin\alpha}{\sin\beta} = v_1 \frac{\sin\alpha}{\sin(\alpha + \beta)}$$

Now, substitute the given values: $v_1 = 30$ m/s, $\alpha = 30^\circ$, and $\beta = 45^\circ$.

$$v'_1 = 30 \frac{\sin(45^\circ)}{\sin(30^\circ + 45^\circ)} = 30 \frac{\sin(45^\circ)}{\sin(75^\circ)} \approx 21.96 \text{ m/s}$$ $$v'_2 = 30 \frac{\sin(30^\circ)}{\sin(30^\circ + 45^\circ)} = 30 \frac{\sin(30^\circ)}{\sin(75^\circ)} \approx 15.53 \text{ m/s}$$