Two particles, each with a mass of 50 g , possess opposite charges...

By the principle of conservation of mechanical energy:

$U_i + K_i = U_f + K_f$

Since the particles are released from rest and have equal masses, they will have equal speeds $v$ at any instant to conserve linear momentum.

$\dfrac{1}{4\pi\epsilon_0} \dfrac{q_1 q_2}{r_i} + 0 = \dfrac{1}{4\pi\epsilon_0} \dfrac{q_1 q_2}{r_f} + \dfrac{1}{2}mv^2 + \dfrac{1}{2}mv^2$

Substituting the given values $q_1 = 4.0 \times 10^{-5} \text{ C}$, $q_2 = -4.0 \times 10^{-5} \text{ C}$, $r_i = 1.0 \text{ m}$, $r_f = 0.5 \text{ m}$, and $m = 0.05 \text{ kg}$:

$mv^2 = \dfrac{1}{4\pi\epsilon_0} q_1 q_2 \left( \dfrac{1}{r_i} - \dfrac{1}{r_f} \right)<br><br>$0.05 v² = (9 $\times$10^9) (4.0 $\times$10^{-5}) (-4.0 $\times$10^{-5}) \left( \dfrac{1}{1.0} - \dfrac{1}{0.5} \right)$<br><br>$0.05 v^2 = (9 \times 10^9) (-16 \times 10^{-10}) (1 - 2)$<br><br>$0.05 v^2 = -14.4 \times (-1)$<br><br>$0.05 v^2 = 14.4$<br><br>$v^2 = \dfrac{14.4}{0.05} = 288$<br><br>$v = \sqrt{288} = 12\sqrt{2} \approx 16.97 \text{ m/s} \approx 17 \text{ m/s}