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Impact of a car on a wall

  1. Mar 25, 2004 #1
    I am trying to find the kinematics of a car collision with a wall. To calculate the damages, should I take into account the momentum or work done?
    In other words, is this proportional to velocity or square of the velocity?
  2. jcsd
  3. Mar 26, 2004 #2
    i think its something like, velocity squared by mass.

    as in:

    10 kg weight going 10 m/s
    5 kg weight going 40 m/s

    have the same kinetic energy.

    >i think< that's what the squared thingie means.

    also, surface area of inpact makes a big differnce too. the smaller the more damage.

    also note that cars are designed to crumble. the more parts of the cars crumble, the more kinetic energy is "lost" in the inpact, as well as directed around the driver.

    a 1000 kg brick into a car would just smoosh the car. but a 1000 kg car into another car, would do a lot less damage, cos it has crumble zones.
  4. Mar 26, 2004 #3


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    Welcome to the forums, Venkatesh!

    I think you need to define what you mean by "damage"

    Like Gara pointed out, a car's frame is designed to dissipate the energy around the passenger compartment, resulting in a slower deceleration (which is what hurts the soft squishie parts)
  5. Mar 26, 2004 #4


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    How do you know when to use momentum and when to use kinetic energy? I've had problems in the past trying to figure out which one to use.
  6. Mar 26, 2004 #5
    from what i learned usually you are using kinetic energy when the collision is plastic or non-elastic.
  7. Mar 26, 2004 #6


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    What does elastic mean though?

    I can't find a definition on the web other than "Capable of resuming original shape after stretching or compression; springy".
  8. Mar 26, 2004 #7
    There is not much point in using the energy equations in non-elastic collisions because energy is not conserved. It's only useful in fully elastic collisions, in which all energy is conserved.

  9. Mar 26, 2004 #8
    umm momentum = Impulse mathmatically

    momentum = mass*velocity and equal to impulse =force * time
    both unit in Ns

    That The lower the force , the low the impact on car

    and if reduce the velocity would help. mv=Ft F= (mv)/t

    and if increase the time of crash it will experience less force (compare to same amount of momentum) so the use of croumple zone is built to do such thing
    which allow the head of car be squashed so increase the time to reach to the person sits in. P=Ft F=P/t

    P= momentum f=Foce t=time v= velocity

    Mementum = Impluse not work done i apologize and momentum = Mv not Mv^2
    Last edited: Mar 27, 2004
  10. Mar 26, 2004 #9


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    Uh... no it isn't.

    [tex]mv^2 = Ft[/tex]


    [tex](kg)(\frac{m^2}{s^2}) = Ns[/tex]

    [tex](kg)(\frac{m^2}{s^2}) = (kg)(\frac{m}{s^2})(s)[/tex]

    [tex](kg)(\frac{m^2}{s^2}) = (kg)(\frac{m}{s})[/tex]

    That doesn't work out.
  11. Mar 26, 2004 #10
    You got it all wrong, expscv. "Work" is defined as the scalar product of the force and the path along which it works. "Kinetic energy" is defined as the product of the velocity of an object squared and its mass, divided by two. "Momentum" is defined as the product of the velocity of an object and its mass. "Impulse" is defined as the product of the force and the time period.
  12. Mar 27, 2004 #11
    all yeah .... change work to Impulse so sorry I apologise ~~~

    work= Fd impulse = Ft F=force d=distace t=time

    would that work now?

    opps thats a big mistake
  13. Mar 27, 2004 #12
    and momentum = mass*velocity

    relationship between Momentum and impulse is the mass*velocity=force*time
    force = (mass*velocity)/time

    and the formula Kinetic energy = mass * velocity ^2

    is too help understand the damage to the car after the crash which the kinetic energy become 0 that means the KE is converted into other form of energy like sound heat or damage to the car.
  14. Mar 27, 2004 #13


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    It doesn't really matter if you use momentum or kinetic energy

    kinetic energy:

    [tex]Fd = \frac{mv^2}{2}[/tex]

    [tex]F = \frac{mv^2}{2d}[/tex]

    find acceleration

    [tex]Vf^2 = Vi^2 + 2ad[/tex]

    [tex]a = \frac{Vf^2 - Vi^2}{2d}[/tex]


    [tex]F = ma[/tex]

    [tex]F = \frac{m(Vf^2 - Vi^2)}{2d}[/tex]

    How interesting.
    Last edited: Mar 27, 2004
  15. Mar 27, 2004 #14
    Did you really expect to get different results, ShawnD? ;)
  16. Mar 27, 2004 #15
  17. Mar 27, 2004 #16


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    Neither momentum nor energy alone give a good estimate of what's going on here. It matters how quickly the car is decelerated, which depends on the specifics of the car and the rigidity of the wall. You also need to define "damages." Its maybe best to ask what happens to the occupants. In that case, you can make estimates in the following way:

    Assume the car will crush uniformly, and is of length L. There's more than one way to define "uniformly," but I'll take the simplest definition - that the deceleration is constant. Also assume the wall is perfectly solid. The occupant has to completely decelerate within a distance L/4 or so (otherwise he'll be crushed and certainly die). So use work:



    This estimate is very weak, and does not replicate reality well at all. The average acceleration is not necessarily the most useful measure (i.e. the seatbelt's concentrated pressure could crack ribs), acceleration is not necessarily constant, the drivers legs are usually crushed with a far weaker collision than would be required to otherwise cause serious injury, and most importantly the crush distance would in reality be a very complicated function of v, among other things. There's a reason that companies invest large amounts of money physically crashing their cars to improve safety.
  18. Mar 27, 2004 #17
  19. Mar 27, 2004 #18


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    Of course I did

    On that impact question (the first one he asks), momentum and energy give completely different answers. Momentum gives 2600N, energy gives something like 3200N if I remember correctly.
  20. Mar 27, 2004 #19
    Only because the energy method was applied incorrectly.
  21. Mar 27, 2004 #20


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    Wait a second... not it wasnt. That momentum method on the other question did not take gravity into account. Gravity applies force at all time so the force should have been from the change in velocity (from 4 to 0) but also include the force of gravity. The mass of the guy was 65kg. 65 x 9.8 = 637. 2600 + 637 = 3237
    Dang, I think i gotta go post in that other one.
    Last edited: Mar 27, 2004
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