a head on collision between 2 cars be more damaging to the occupants if the cars stuck together or rebounded?

Question

Thank you for answering! i really need your help with this question!

Answer 1

Unfortunately, your question is not as simple as the other answers suggest.

The damage suffered by the occupants is a function of the energy absorbed by the occupants AND the change in acceleration. Let me explain. Human beings suffer injury because of blunt trauma ( force and thereby energy absorbed ) and what is known as deceleration trauma ( shearing forces on the tissue such as spinal cord ).

In effect, your question ultimately depends on the mass of cars, their respective speeds, materials , safety features, etc.

technically, the vehicles bouncing off would transfer less energy to the occupants ( less blunt trauma ) but increase deceleration injury !.... this is assuming your are belted ! if not , well then you would hit windshield or get ejected !

you see, the damage depends a lot on the details of the scenario and not on the physics only. this is because "damage" is not a physical quantity but a human one -- where more damage may mean a certain biologic injury.

i know it didnt answer your question but i hope it shed some light on the complexity of your question.

thanks :)

Answer 2

It is tough to give a definite answer tothis question. If the cars stick together--i.e. impact and stop, yes they ARE designed to take the energy of the collision and direct it around the occupants. However, this doesn't always work. Remember Dale Earnhardt, Sr? He hit the wall in a vehicle better designed than a regular car and it didn't help him.
See, there are 3 impacts in every vehicular collison.
Impact 1: vehicle vs whatever it hits.
Impact 2: occupant vs part of interior of vehicle
Impact 3: occupants' internal organs with other organs and/or exterior of occupants body.

The best collsion is the one where all 3 impacts are kept to a minimum, or at least energy of collison does not easily make it to impacts 2 & 3.

Answer 3

More damage would occur if the cars stick together. If the cars rebound, then that means it is an elastic collision (like a ball bouncing on the floor). On the other, if the cars stick that suggest an inelastic collision. In an inelastic collision there is permanent deformation. For example, a ball of clay dropped on the floor will deform and not rebound.

Answer 4

Well think of it this way for every action there is an equal but opposite reaction, with this in mind what hurts more to be hit and just stand firm or to be hit and the reaction of your body hitting something else. Put simply I you feel down the stairs it would hurt but if you feel down the stairs and hit a wall you would bounce back and hit the stairs again Reaction tends to do more damage then just action. The worst injury's come from secondary action of the collision not the collision it self.

Answer 5

as long as we predict the passanger compartment continues to be intact, the occupants might go through extra harm if the motor vehicle rebounded because of the fact they had come to an entire supply up in the direction they have been going and then improve up without delay in the different direction inflicting inner harm. whilst the autos stick at the same time they often cave in at the same time and the potential is going into crumpling quite than reversing the acceleration of the autos and occupants.

Answer 6

Hehe. It is more if the cars rebound. When the cars stick together the time taken for the change in momentum is greater, thus the impulse is less. This principle is used for example in boxing where the boxer 'rides' the puches, that is, allows his head to go back with them. The damage is less as he remains in contact with the fist for a greater period of time.

Answer 7

LOL!!! The answers before mine are ridiculous. They get hurt more if the collision were elastic! There's a reason that cars get all smushed when they crash and thats because the damage is intended for the car rather than the person!

Answer 8

let me give you a clue
what is important is momentum change per unit time
think of it
in which case? (Delta p)/ (Delta t) = Impulse (bigger?)
2nd clue you may assume Delta t to be equal

Answer 9

Nevermind I was wrong. D'oh.