#### Answer

$\frac{v_A}{v_B} = 4$

#### Work Step by Step

Let $m$ be the mass of particle B.
Let $\frac{m}{2}$ be the mass of particle A.
We first write an expression for the kinetic energy of particle B.
$KE_B = \frac{1}{2}mv_B^2$
We then write an expression for the kinetic energy of particle A.
$KE_A = \frac{1}{2}(\frac{m}{2})v_A^2 = 8KE_B$
Next, we equate the two formulas and find the ratio;
$\frac{1}{2}(\frac{m}{2})v_A^2 = 8~(\frac{1}{2}mv_B^2)$
$\frac{v_A^2}{v_B^2} = 16$
$\frac{v_A}{v_B} = 4$