The solution is actually strange for me. It assumes that the distance traveled by each car (in the CM frame) during the acceleration (crumpling) is also equal to the crumple length of each car, but is not always the case.
Following the assumption of the solution: In the CM frame, the heavier car moves slower than the lighter car. So that at the moment that they will collide, the time needed for the lighter car to decelerate to 0 would be shorter than the time for the heavier car if it assumes that the distance traveled by each car (at CM frame) would be equal to its crumple distance:
$t_1 = \Delta v_1/a_1 = \frac{0-32.1}{-8.6\times10^2} = 0.0373s$
but $t_2 = \Delta v_2/a_2 = \frac{0-(-12.3)}{1.3\times10^2} = 0.0946s$
as a result, the lighter car is already completely crumpled, while the heavier car is still crumpling:
But since the heavier car is still crumpling, wouldnt it cause the lighter car to be accelerated even further and much longer?
In reality, the time for the two cars to decelerate must be equal, which means some of the assumptions of the solution is not applicable in the real world (more precisely, there is a mistake in the solution).
A more prudent solution (in my opinion) would be to use
$$a_1 = \frac{v_1'^2}{2d_1}$$
$$a_2 = \frac{v_2'^2}{2d_2}$$
where $d_1$ and $d_2$ is the unknown distance traveled by car 1 and 2 during deceleration (crumpling), with respect to the CM frame.
$$d_1 + d_2 = 2(0.6) = 1.2 \space\space[1]$$
$$\frac{a_1}{a_2} = \frac{v_1'^2}{2d_1}\frac{2d_2}{v_2'^2} = \frac{v_1'^2 d_2}{v_2'^2 d_1}$$
and finally,
$$\frac{v_1'^2 d_2}{v_2'^2 d_1} = \frac{m_2}{m_1} \space\space[2]$$
as you suggested. Then use equation 1 and 2 to solve for $d_1$ and $d_2$.
