Answer
a) $3.56\times10^3\;\rm m/s$
b) $2.81\times10^3\;\rm m/s$
Work Step by Step
a)
The two identical point charges are at rest and then one of them will be released to move from rest. Let's assume it is the right charge
The repulsive force between the two charges will make the first charge (when released) move away from the other while the second one remains stationary. After a long time, the initial potential energy of the system will be totally converted to kinetic energy.
Noting that we assume that the two charges are an isolated system which means that the net energy of the system remains constant.
Thus,
$$E_i=E_f$$
$$PE_i=KE_f$$
$$\dfrac{kQ_1Q_2}{r_i}=\frac{1}{2}m_1v_1^2$$
Noting that the second charge remains stationary, so its final speed is zero as its initial speed.
Now we need to solve for $v$;
$$\dfrac{2kQ_1Q_2}{m_1r_i}= v_1^2$$
$$v_1=\sqrt{\dfrac{2kQ_1Q_2}{m_1r_i}}$$
Plugging the known'
$$v_1=\sqrt{\dfrac{2\cdot 8.99\times10^9 \cdot (6.5\times10^{-6})^2}{1.5\times10^{-6} \cdot 4\times10^{-2}}}$$
$$\boxed{v_1=\color{red}{\bf 3.56\times10^3}\;\rm m/s}$$
b)
Since both of them will be released from rest, so their initial momentum is zeros and since the system is isolated, the sum of their final momentum must be zero as well.
$$p_i=p_f$$
$$\overbrace{m_1v_{1i}+m_2v_{2i} }^{=0} =m_1v_{1f}+m_2v_{2f}$$
$$m_1v_{1f}=-m_2v_{2f} \tag 1$$
The net energy of the system remains constant.
Thus,
$$E_i=E_f$$
$$PE_i=KE_f$$
$$\dfrac{kQ_1Q_2}{r_i}=\frac{1}{2}m_1v_{1f}^2+\frac{1}{2}m_2v_{2f}^2$$
Plugging $v_{2f}$ from (1);
$$\dfrac{kQ_1Q_2}{r_i}=\frac{1}{2}m_1v_{1f}^2+\frac{1}{2}m_2\left(\dfrac{m_1v_{1f}}{-m_2}\right)^2$$
Solving for $v_{f1}$;
$$\dfrac{kQ_1Q_2}{r_i}=\frac{1}{2}m_1v_{1f}^2+\frac{1}{2}m_2 \dfrac{m^2_1v_{1f}^2}{ m_2^2} $$
$$\dfrac{kQ_1Q_2}{r_i}=\frac{1}{2}m_1v_{1f}^2+\frac{1}{2} \dfrac{m^2_1v_{1f}^2}{ m_2 } $$
$$\dfrac{kQ_1Q_2}{r_i}=v_{1f}^2 \left[\frac{1}{2}m_1 +\frac{1}{2} \dfrac{m^2_1}{ m_2 } \right]$$
$$ \dfrac{kQ_1Q_2}{r_i\left[\frac{1}{2}m_1 +\frac{1}{2} \dfrac{m^2_1}{ m_2 } \right]} =v_{1f}^2$$
$$v_{1f}=\sqrt{ \dfrac{kQ_1Q_2}{r_i\left[\frac{1}{2}m_1 +\frac{1}{2} \dfrac{m^2_1}{ m_2 } \right]} }$$
Plugging the known;
$$v_{1f}=\sqrt{ \dfrac{8.99\times10^9 \cdot (6.5\times10^{-6})^2 }{ 4\times10^{-2} \left[\left(\frac{1}{2} \cdot 1.5\times10^{-6}\right) +\left(\frac{1}{2} \dfrac{(1.5\times10^{-6})^2}{ 2.5\times10^{-6}} \right)\right]} }$$
$$\boxed{v_{1f}=\color{red}{\bf 2.81\times10^3}\;\rm m/s}$$
