Answer
The original speeds of the two cars were 5.7 m/s and 11.4 m/s.
Work Step by Step
Let the masses of the two cars be m and 2m.
Let the original speeds of the cars be $v_1$ and $v_2$ respectively.
$\frac{1}{2}(2m)v_1^2 = \frac{1}{2}\times\frac{1}{2}mv_2^2$
$4v_1^2 = v_2^2$
$2v_1 = v_2$
When both cars increase their speed by 8 m/s:
$\frac{1}{2}(2m)(v_1+8)^2 = \frac{1}{2}m(v_2+8)^2$
$\frac{1}{2}(2m)(v_1+8)^2 = \frac{1}{2}m(2v_1+8)^2$
$2v_1^2 + 32v_1+128 = 4v_1^2+32v_1+64$
$v_1^2 = 32$
$v_1 = \sqrt{32}$
$v_1 = 4\sqrt{2} = 5.7 ~m/s$
$v_2 = 2v_1 = 11.4 ~m/s$
The original speeds of the two cars were 5.7 m/s and 11.4 m/s.