# 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 ?

• I couldn't download (and see) the solution you have linked. But, I think you should consider to loss of energy in collision. You cannot treat with whole of cars as rigid bodies during their deformation (according to definition rigid bodies don't deform). The magnitudes of forces are equal just in the contact area, but magnitudes of forces are transmitted to the passenger compartment of each automobile can be different. – lucas Jun 12 '16 at 4:29
• What did the solution use for d? – philip_0008 Jun 12 '16 at 9:11
• also note that the solution uses $\approx$ sign for $a$, which is just a round-off of the exact value. – philip_0008 Jun 12 '16 at 9:35
• Thanks Lucas for this point , I think maybe this is my confusion . Philip_0008 The solution used 0.6 for d which is the crumple zone. – oamer Jun 12 '16 at 9:51

## 1 Answer

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. $$d_1 + d_2 = 2(0.6) = 1.2 \space\space$$ $$\frac{a_1}{a_2} = \frac{v_1'^2}{2d_1}\frac{2d_2}{v_2'^2} = \frac{v_1'^2 d_2}{v_2'^2 d_1}$$ and finally, $$\frac{v_1'^2 d_2}{v_2'^2 d_1} = \frac{m_2}{m_1} \space\space$$ as you suggested. Then use equation 1 and 2 to solve for $d_1$ and $d_2$.

• Thank you Philip for your illustrative solution .Now I think the mistake lies in the assumption that each car deceleration distance is the same as crumple zone . Lucas raised another issue that since these bodies are not rigid bodies , , , , the forces transmitted to passenger compartments need not to be the same hence the acceleration constraint assumption may not be valid . Can I assume they are the approximately the same since both go through same crumpling distance ? – oamer Jun 12 '16 at 21:30
• The acceleration time should be equal on both cars for a collision, because any stopping force (which causes acceleration) applied to one car will directly affect the other (law of action and reaction), and once one car stops accelerating, the other must stop accelerating also. This equal time condition can only be achieved by equating $a_1/a_2 = m_2/m_1$. Note that $v_1'$ and $v_2'$ should remain the same whatever the condition. – philip_0008 Jun 12 '16 at 22:48