Collision/Crumpling problem possible solution mistake This question is from Physics for scientist and engineers , Ohanian .
Two automobiles of 540 and 1400 kg collide head-on while
moving at 80 kmh in opposite directions. After the collision
the automobiles remain locked together.
(C) The front end of each automobile crumples by 0.60 m
during the collision. Find the acceleration (relative to the
ground) of the passenger compartment of each automobile;
make the assumption that these accelerations are constant
during the collision.
In Walter Lewin 1999 physics class this problem part 6.1 is solved in top of page number 2.
http://www.myoops.org/twocw/mit/NR/rdonlyres/Physics/8-01Physics-IFall1999/8D3EA3C0-149F-4212-B926-AE0DAE87D642/0/sol6.pdf
My doubt is : Since it is a collision , forces are equal hence 
$$\frac{a_1}{a_2}=\frac{m_2}{m_1}$$
However the solution does not satisfy this condition .I have checked the calculations and the reasoning and both are sound .
Where is the mistake ?
 A: 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 $d_2$.
