Answer
$v=7.0 \times 10^5 \mathrm{~m} / \mathrm{s}
$
Work Step by Step
The electric potential energy in the presence of the dipole is
$$
U=q V_{\text {dipole }}=\frac{q p \cos \theta}{4 \pi \varepsilon_0 r^2}=\frac{(-e)(e d) \cos \theta}{4 \pi \varepsilon_0 r^2} .
$$
Noting that $\theta_i=\theta_f=0^{\circ}$, conservation of energy leads to
$$
K_f+U_f=K_i+U_i \Rightarrow v=\sqrt{\frac{2 e^2}{4 \pi \varepsilon_0 m d}\left(\frac{1}{25}-\frac{1}{49}\right)}\\=7.0 \times 10^5 \mathrm{~m} / \mathrm{s} .
$$
