## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

$\frac{v_A}{v_B} = 4$
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; $\frac{1}{2}(\frac{m}{2})v_A^2 = 8~(\frac{1}{2}mv_B^2)$ $\frac{v_A^2}{v_B^2} = 16$ $\frac{v_A}{v_B} = 4$