Answer
a) $v_f=7.51\frac{m}{s}$
b) $r=1.83m$
Work Step by Step
(a) We know that
$v_f=\sqrt{\frac{2Kq^2}{mr_i^2}}$
We plug in the known values to obtain:
$v_f=\sqrt{\frac{2(8.99\times 10^9)(3.05\times 10^{-6})^2}{(0.00216)(\sqrt{(1.25)^2+(0.570)^2}}}$
$v_f=7.51\frac{m}{s}$
(b) We know that
$As \frac{Kq}{r}=\frac{3}{4}\frac{Kq}{r_i}$
$\implies r=\frac{4}{3}r_i$
We plug in the known values to obtain:
$r=\frac{4}{3}\sqrt{(1.25)^2+(0.570)^2}$
$r=1.83m$
