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Two-car accident physics problem

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  1. Apr 8, 2005 #1
    Please help.

    Consider the following two-car accident: Two cars of equal mass m collide at an intersection. Driver E was traveling eastward, and driver N, northward. After the collision, the two cars remain joined together and slide, with locked wheels, before coming to rest. Police on the scene measure the length d of the skid marks to be 9 meters. The coefficient of friction mu between the locked wheels and the road is equal to 0.9.

    Let the speeds of drivers E and N prior to the collision be denoted by v_e and v_n, respectively. Find v^2, the square of the speed of the two-car system the instant after the collision.
    Express your answer terms of v_e and v_n.

    I tried using the vector addition and ended up with v_e^2 + v_n^2, which was incorrect (I also saw the duty tutor, and he said it was correct). I just wanted to find out the way to determine the final velocity of the combined vehicles.

    Thank you
     
  2. jcsd
  3. Apr 8, 2005 #2

    arildno

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    Let the one traveling east with have velocity [tex]\vec{v}_{E}[/tex] the one traveling north having velocity [tex]\vec{v}_{N}[/tex].
    Let the final velocity be [tex]\vec{v}[/tex]
    The speeds are described with [tex]v_{E},v_{N},v[/tex] respectively.
    Since we have an inelastic collision, conservation of momentum says
    [tex]m\vec{v}_{E}+m\vec{v}_{N}=2m\vec{v}[/tex]
    where m is the mass of a car.
    Do you see where you made a mistake now?
     
  4. Apr 9, 2005 #3
    I also have to do this question and got the answer v_e^2 + v_n^2. I don't see where I have made the mistake.
     
  5. Apr 9, 2005 #4

    arildno

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    Well, have you looked over my answer?
    What is it you don't understand there?
     
  6. Apr 9, 2005 #5
    I don't see how i am meant to use that relationship to find v^2
     
  7. Apr 9, 2005 #6

    arildno

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    Well, dividing with 2m, you've got:
    [tex]\vec{v}=\frac{1}{2}(\vec{v}_{E}+\vec{v}_{N})[/tex]
    So, given this vector equation, how can you determine the squared speeds?
     
  8. Apr 9, 2005 #7
    so would it just be
    V^2 = ((v_e+v_n)^2)/4
    ?
     
  9. Apr 9, 2005 #8

    arildno

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    No; you must use the fact that [tex]\vec{v}_{E}\cdot\vec{v}_{N}=0[/tex]
    You are familiar with vectors, right?
     
    Last edited: Apr 9, 2005
  10. Apr 9, 2005 #9
    Oh, ok, so in that cast when i expand the ((v_e+v_n)^2) I actually get v_e^2 + v_n^2. So the answer is actually (v_e^2 + v_n^2)/4 ?
    Sorry, vectors are new to me so I don't really quite know how to work with them yet.
     
  11. Apr 9, 2005 #10

    arildno

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    "So the answer is actually (v_e^2 + v_n^2)/4 ?"
    Correct!
     
  12. Apr 9, 2005 #11
    We want to know the initial speed on the 9m trajectory. So calculating the acceleration: 0.9g. vf^2=(0.9g)(2)(9)=16.2.

    So the total momentum is 16.2m. So now we know:

    v1^2+v2^2=16.2

    v2^2=(16.2^2)-v1^2
    v1^2=(16.2^2)-v2^2

    With the current information you can't determine the velocities of the vehicles before collision. But at the moment of collision, the velocity would be 4 m/s north of east.

    You have calculated algebraically, but vectors always add up geometrically. That is your mistake.
     
    Last edited: Apr 9, 2005
  13. Apr 9, 2005 #12
    Awesome! Thanks for bearing with me
     
  14. Apr 9, 2005 #13

    arildno

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    No problem.
     
  15. Apr 9, 2005 #14
    Oh ok...thanks for your help. :smile:
     
  16. Apr 9, 2005 #15
    If you were to determine the speed of car N (who claims that thye were travellingat less than 14m/s), when car E was travelling at 12m/s, KE = (m*(v_e^2+v_n^2))/4, and a friction force acting in the opposite direction (work done by it = 2*0.9*m*g*d), how would you equate both. Because, after the collision, the car travelled a distance of 9m, so to determine car Ns speed, you have to include, distance, frction, the speed of car E, etc. Just not sure.
    Thanks.
     
  17. Apr 9, 2005 #16
    There is the fact of the locks that is quite disturbing and not counting in my calculations. The locks certainly have a coefficient of friction wich slows down the wheels.
     
  18. Apr 10, 2005 #17

    arildno

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    No; the locked condition is there so that cars can be regarded as strictly SLIDING, rather than sliding&rolling.
    The latter problem is a lot more difficult to solve; hence, the sliding condition.
     
  19. Apr 10, 2005 #18
    Ok thanks! But in the end did I get the right awnser?
     
  20. Apr 10, 2005 #19

    arildno

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    I haven't checked yet..
     
  21. Apr 10, 2005 #20
    I don't think so I did a mistake I think. So the acceleration on the cars after collision is 0.9g=8.829. And vi^2=(0.9g)(2)(9)=159

    vi=12.6 m/s

    Sorry for this.
     
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