* It may be reasonable to assume the brakes produce the same force on both cars. With both cars weighing the same, this produces the same acceleration. Then the speeds decrease at the same rate, which means the faster car is going 100 - 70 = 30 kph when the slower car stops.
* The faster car must dissipate more energy to do this. We may assume that braking is limited by heat dissipation and that the brakes shed energy at the same rate. Then the answer is sqrt(100^2 - 70^2) ~ 71 kph.
Adding the obstacle:
* As noted in the text, brakes don't shed energy at the "same rate per unit distance". So the faster car hits the obstacle first, at greater speeds than above in each case. We aren't given the information required to calculate the stopping time or distance of the slower car, so the problem is underspecified.