Solution

Our system is closed and isolated, so the laws of conservation of momentum and energy are applicable to it. Since the charges are distributed uniformly over the surfaces of the balls, the fields created by the balls are the same as those created by point charges $q_1$ and $q_2$ placed at the centers of the balls. As the second ball approaches, under the action of the Coulomb repulsive force ($q_1q_2 > 0$), the first ball will start moving. At the moment of collision, the speeds of the balls will be the same ($V$), and the distance between the centers of the balls will be $R_1 + R_2$.

Let us write the law of conservation of momentum in projection onto the axis of motion of the balls:

$$m_2v = (m_1 + m_2)V$$
$$V = \frac{m_2v}{m_1 + m_2} \quad (1)$$

Let us write the law of conservation of energy:

$$\frac{1}{2}m_2v^2 = \frac{1}{2}(m_1 + m_2)V^2 + \frac{1}{4\pi\varepsilon_0} \cdot \frac{q_1q_2}{R_1 + R_2} \quad (2)$$

Substitute $V$ into $(2)$:

$$\frac{1}{2}m_2v^2 = \frac{1}{2}(m_1 + m_2)\left(\frac{m_2v}{m_1 + m_2}\right)^2 + \frac{1}{4\pi\varepsilon_0} \cdot \frac{q_1q_2}{R_1 + R_2}$$

From here we express $v$:

$$v = \sqrt{\frac{q_1q_2(m_1 + m_2)}{2\pi\varepsilon_0m_1m_2(R_1 + R_2)}}$$