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Shortest collision time that will leave engine block intact?

  1. Sep 12, 2013 #1
    1. The problem statement, all variables and given/known data
    In a crash test a solid concrete wall is
    used to test crashworthiness. A particular
    SUV weighs 338.183 kg and is accelerated to
    33.3485 m/s before colliding head on with the
    What impulse must be applied is required
    to stop the SUV?

    This gives Impulse = 11277.896

    The struggle comes with this part:

    Through structural analysis of the SUV you
    find out that 21801 N will destroy the engine
    block. If we ignore the friction between the
    car and the ground and assume the force the
    wall exerts a constant force on the car, What
    is the shortest collision time that will leave
    the engine block intact?

    Answer in units of s2. Relevant equations

    Impulse = net force * change in time

    p(f) = p(i) + Impulse

    3. The attempt at a solution

    So I assumed that a force of 21800.9N force would mean that the engine
    is not broken. I do not know if this assumption holds. However, it seems to make
    sense in my mind. So i figured if the wall exerts a constant force of 11277.896N
    causing the car to have a momentum of 0; the change in time is:

    21800.9N/11277.896N --> 1.933064465s

    But this is apparently wrong.
  2. jcsd
  3. Sep 12, 2013 #2


    User Avatar

    Staff: Mentor

    Yes, you can take that to be the maximum allowable force.
    Look at the units. N divided by N yields a result without units, not seconds.

    Consider your first Relevant Equation, then look at my comment in red above.
  4. Sep 12, 2013 #3
    I would also check 338.183 kg. It is much too little for an SUV. 3381.83 kg is more like it.
  5. Sep 12, 2013 #4
    So I see that my initial move was naive..haha

    I rethought the problem. Right before impact my momentum will be 11277.896N.

    If I don't want to destroy my motor I need to only contact the wall (If i could be so lucky to choose) for a very small amount of time so that my Impulse = 21000.9 N.

    So I get, p(i) = 11277.896 N and I have to stop moving so my final momentum p(f) = 0.

    so I get:

    0 = 11277.896 + (-21800.9N)t

    So if this logic is correct i just rearrange and get t. I made the force negative because it's acting in the opposite direction of motion.
  6. Sep 12, 2013 #5
    The kg is right...problem is clearly not realistic.
  7. Sep 12, 2013 #6
    Does this seems right at all? The system we use docks points off of our total score if we try and fail..So in the spirit of not trying because of fear of failure a little assurance would be great. In other words, if you say No that is wrong then I can go back to the drawing board, else I already did the work so it's ok for me to know it's right.
  8. Sep 12, 2013 #7


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    Staff: Mentor

    Make sure that you carry correct units through your calculations. What are the units of momentum and impulse?

    Your last version is fine (post #4), except you forgot to include the units on the impulse value. What value do you calculate for the time?
  9. Sep 12, 2013 #8
    Units of Impulse are kg*m/s --> N s; and momentum is N*s. Correct? and If the impulse value

    11277.896N*s + (-21800.9N)t = 0 --> t = -11277.896N*s/-21800.9N = t (seconds)

    (kg*m)/s * s*s/(kg*m) --> s

    For some reason I believed that Force and Momentum had the same units...Now I see the error of my ways!

    Thank you for the lending hand.

  10. Sep 12, 2013 #9
    .517 s --> Correct. Thanks again
  11. Sep 12, 2013 #10


    User Avatar

    Staff: Mentor

    You're welcome :smile:
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