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Physics collision problem

  1. 40mph-40mph head on

    7 vote(s)
    23.3%
  2. 80mph-0mph head on

    7 vote(s)
    23.3%
  3. no difference

    16 vote(s)
    53.3%
  1. Oct 30, 2005 #1
    Question,

    two cars, equal masses each going 40mph collide head on.... would this do the same damage as a collision with one car going 80mph and the other car at rest?
     
  2. jcsd
  3. Oct 30, 2005 #2
    I'll take the 40 40 collision.
     
  4. Oct 30, 2005 #3
    No. It's not like you are hitting a solid wall. The "other car at rest" has no permanent foundation with the ground. Part of the damage momentum will be transferred to the recoil movement of the second car.
     
  5. Oct 30, 2005 #4

    Mk

    User Avatar

  6. Nov 1, 2005 #5
    A 40mph-40mph head-on collision does the same damage as a 40mph collision with an immovable object.

    An 80mph collision with an object at rest of equal mass and strength object not attached to the ground (eg a car oil an oil slick) would also do the same damage.

    If the car at rest is in gear or has its brakes on, then the 80mph-rest collision does more damage than the 40-40 collision.
     
  7. Nov 1, 2005 #6
    Conservation of momentum / energy.

    In an ideal situation the scenarios will give the same result as long as the stationary car isnt fixed to the spot. (Air track experiment.)

    If you are to include crumple zones and other real model considerations I think the 80mph colision will come off better, perhaps even be inelastic to some degree (assuming that the moving car has to overcome no external forces on the stationary car besides static friction/gravity.)
     
  8. Nov 1, 2005 #7
    i take the 40-40 because all energy is transformed into destruction of the car

    while the 80-0 part of the energy is used on pushing the car that is at 0 speed

    the exact amount of energy that is not used to cause destruction in the cars
    can be calculated by mesuring how much the car at 0 has moved away from its initial position
     
  9. Nov 1, 2005 #8
    if you make the 80-0 collision with 2 jeeps that are ultrataff the cars will behave like billiard balls the 80 car will reach 0 speed and the 0 car will reach 80 speed being the damage minimal(though not for the people inside)

    but if you make the collision 80-0 with two volvos the cars will convert most of the kinetic energy into deformation, better for the passengers, so the 0 car will aquire less speed

    in the case of th 40- 40 collision with the jeeps the damage of the cars will be terribel because they will not be likely to rebound while the 80-0 the damage of the car will be minimal because theyll behave like billiards balls

    thats why volvos are safetier than old jeeps they absorve the impact
     
  10. Nov 1, 2005 #9
    What about kinetic energy, [itex]1/2 m v^{2}[/itex]? The 80-0 crash will have twice as much energy to have to dispose of than the 40-40, so it should be more fun to watch. Think about how race cars disintegrate versus Hondas crumpling.
     
  11. Nov 1, 2005 #10
    Wouldn't it follow the Kinetic Energy equation?

    Kinetic Energy = 1/2 * mass * velocity^2

    As velocity is squared the 80-0 would do more damage. (assuming that both accidents involved the exact same type of cars)
     
  12. Nov 1, 2005 #11

    Astronuc

    User Avatar

    Staff: Mentor

    Does the combined collision move? That is important to compare with the 40-40 crash, where if the mass of the cars is the same, the net momentum is zero.

    In the case of the 80 mph vehicle, the kinetic energy is 4 times greater, but the mass is that of one car, not two. How much kinetic energy is transferred to destructive mechanical energy depends on the whether or not the combined mass recoils or not.
     
  13. Nov 1, 2005 #12
    Reference frames

    40-40 and 80-0 collisions are the exact same event viewed from different reference frames, if you ignore friction etc. They will therefore do precisely the same damage. If you cannot ignore friction (e.g. one car has brakes on, as PeteSF mentioned) then the 80-0 might do slightly more damage.

    The kinetic argument doesn't hold, because you have to look at change in kinetic energy, not just the initial kinetic energy. In the 40-40 case, initial kinetic energy is 2(1/2.40.40.m) = 1,600m, and final is 0 (assuming inelastic collision). In the 80-0 case, initial kinetic energy is 1/2.80.80m = 3200m, and final kinetic energy is 1,600m (because the cars will both be moving at 40mph after an inelastic collision without friction.) Therefore, the kinetic energy lost in both cases is the same, 1,600m.
     
  14. Nov 1, 2005 #13
    in my point of view the kinetic energy argument fails because according to that a collision of 2 cars going in the same direction one at 80 km/h and the other at 160 would tranfer less energy into destruction that if one goes at 1080 and the other at 1000 km/h

    what is important is the relative kinetic energy they have with respect to each other that will always be the difference in speeds
     
  15. Nov 1, 2005 #14
    Fundementally the questions arent comparable, because the combined forces of two cars the momentum will produce deceleration in two directions therefore the damage, reciprocal force will be dubbled. Think about what it would look like slow motion. The cars will both react to the crash in similar ways. Providing the basis of testing is that the situation for both is equal. Alos consider the parts of the car where the most weight is, and where the drive originates. The car into wall test will result in one crash therefore no matter what speed or distance the wreck will be half that of the first test. In answer one would deduce that the greatest impact will be from the cars. There is a difference as previously outlined.
     
  16. Nov 1, 2005 #15
    the impact force equals the change of momentum. Since the latter is invariant under Galilean transformations it does not matter wether they collide 40-40 or 80-0 head on. Also, the nature of the collision (elastic/non-elastic) is entirely irrelevant.
     
  17. Nov 1, 2005 #16
    I would say there is most definitely a difference between these two crashes. It's not a huge difference, but it's there. There is something tantamount to an accordion effect that takes place, wherein the energy is dissipated over time and the length of the cars. An insurance adjuster could give you an answer by looking at the rear of the cars in both scenerios. I'll still stick with the 40 40. Somehow I think I'd like to be seat belted in the trunk of the car not moving. :-)
     
  18. Nov 1, 2005 #17
    There can only arise a difference when both of the following are satisfied:
    (a) the collision occurs in a finite amount of time
    (b) during that time there is a significant interaction with a third object which depends on the relative velocities with respect to the latter.

    It does not seem to matter wether the car is an extended object or not.
     
  19. Nov 1, 2005 #18
    In the 80-0 case, some of the momentum of the 80 car is transfered to the 0 car, which then will accellerate from 0 to its final constant velocity.

    Which pair of cars would cost the most to repair?
    I think the 80-0 collision, when the directions of vectors in the 40-40 case are kept the same, would cost more to fix.

    Do you think the injuries to the drivers and passengers would be exactly the same in the 80-0 and in the 40-40?

    What do auto accident statistics say?

    Some real-life applications of physics look counter-intuitive.
     
    Last edited: Nov 1, 2005
  20. Nov 1, 2005 #19
    I think the difference comes with the way the forces get translated, such as forward, backward, left, right, up , down.
     
  21. Nov 1, 2005 #20
    The way the impact *travels* through the car depends upon the car design and not upon the absolute velocities.
     
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