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Inelastic Collision HW Problem

  1. Dec 13, 2014 #1
    1. The problem statement, all variables and given/known data
    Pete got into into a car accident and collided with another vehicle, which resulted in the death of the second vehicles passenger, David. Whether Pete was speeding or not determines if he goes to jail. The speed limit where he got into the accident was 45mph. How fast was pete going at the point of collision and will he go to jail?

    Given Data :
    Mass of David’s Car(m2)
    1115lbs

    Mass of Pete’s car (m1)
    1755lbs

    Distance the cars skidded
    13m

    Final velocity post skid
    0mph

    Time of skid
    22sec

    Initial velocity of David’s car
    47mph

    Speed limit
    45mph

    2. Relevant equations
    $$ (m1v1) + (m2v2) = (m1 + m2)Vf $$


    3. The attempt at a solution
    First I converted from lbs and mph to kg and m/s for the masses and velocity since we use kg and m/s in my class, I dont know if this was neccesary.
    1151 lbs = 505.8 kg
    1755 lbs = 796.1 kg
    47mph = 21.01 m/s

    Then I used my equation
    $$ (m1v1) + (m2v2) = (m1 + m2)Vf $$
    $$ (796.1*v1) + (505.8*21.01) = (796.1 + 505.8)0 $$
    $$ (796.1*v1) + (10626.858) = 0 $$
    $$ 796.1*v1/796 = -10626.8/796 $$
    $$ v1 = 13.35 m/s $$

    I then converted that from m/s to mph and got 29.86mph, which would mean he would not be speeding and not go to jail.
    Although I got all the way through, i feel im wrong since they gave me the data for time and distance and i never even used them, which makes me feel like im forgetting something. Could someone help me out or check it to see if i did skip a step?
     
  2. jcsd
  3. Dec 13, 2014 #2

    Nathanael

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    Bit of a morbid physics problem ?:)

    But anyway... the [itex]V_f[/itex] in your relevant equation is the speed immediately after the collision. This speed is not zero, otherwise the cars would not have skid.
     
  4. Dec 13, 2014 #3

    Bystander

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    What you want here rather than Vf = 0, is the velocity of the two cars at the beginning of the 13m skid that took 22 seconds. You might want to double check the problem statement for values for time and distance (distance seems a bit short, and time seems like forever).
     
  5. Dec 13, 2014 #4
    Yes it is morbid!! My physics teacher is a former funeral director untill he became a teacher!!
    Back to the problem...based on what u two said I think that might mean I have to do
    $$ V=D/T $$
    $$ V=13/22 $$
    $$ .59 m/s $$

    Then Maybe I can plug that into this which makes sense
    $$ (796.1∗v1)+(505.8∗21.01)=(796.1+505.8).59m/s $$
    $$ (796.1∗v1)+(10626.858)= 768.121 $$
    $$ (796.1∗v1) = -9858.737 $$
    $$ (796.1∗v1)/796.1 = -9858.73/796.1 $$
    $$ -12.38 m/s $$

    I find it weird this is negative which makes me think Im wrong but Ill just ignore the negative and convert... which means $$ 12.38 m/s = 27.69mph $$... meaning he wasnt speeding

    does this seem better? still feels a little strange
     
  6. Dec 13, 2014 #5

    Nathanael

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    You used [itex]V=\frac{total.distance}{total.time}[/itex]

    Total distance divided by total time just gives you the average velocity (over the time that they were skidding) which is not what you want.
     
  7. Dec 13, 2014 #6

    Bystander

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    Have you ever seen anything skid at constant speed and stop instantaneously?
     
  8. Dec 13, 2014 #7
    Ok guess not... what if I try this one
    $$ D=½(VF+VI)T $$
    $$ 13=½(VF+21.01)22 $$
    $$ VF = -19.83 m/s $$

    Then I do this thing again
    $$ (796.1∗v1)+(505.8∗21.01)=(796.1+505.8)-19.83 $$
    $$ (796.1∗v1)+(10626.858)=-25816.677 $$
    $$ (796.1∗v1)=-36443.533 $$
    $$ (796.1∗v1)/796.1=-36443.533/796.1 $$
    $$ V1 = -45.78 m/s $$

    Still getting a negative but Ill ignore it again and that converts into 102.4 mph which is WAY over the speed limit..... I think i might of messed up again?
     
  9. Dec 13, 2014 #8

    Bystander

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    Let's find a value for deceleration of the cars from the total distance and time.
     
  10. Dec 13, 2014 #9
    Ok...... so I did half of that above i think? Because i need to get average velocity to plug into an acceleration equation?
    $$ V=D/T $$
    $$ V=13/22 $$
    $$ .59m/s $$

    Then a step further for acceleration
    $$ A=V/T $$
    $$ A=.59/22 $$
    $$ A = .027 m/s^2 $$

    Im not sure where that would leave me to go next though
     
  11. Dec 13, 2014 #10

    Nathanael

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    Plugging in random formulas is never a good idea; you have to picture the situation.
    In this new equation you wrote, we're looking at the time interval when they're skidding, (that's why "D" is the distance they skid and "T" is the time they skid for) so "VF" is not the same "VF" as in the other equation. So in this case, the initial velocity "VI" corresponds to the velocity right after the collision, and the final speed "VF" is the speed after they finish skidding.

    P.S.
    You are also assuming that the deceleration of the cars after the collision is constant. You need to make this assumption for the problem to be solvable, but it's good to be aware that you're making this assumption.


    Edit:
    Sorry I'm a little late; I didn't notice that bystander already replied
     
  12. Dec 13, 2014 #11

    Nathanael

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    Again, you're using the average velocity. Acceleration is not "average velocity per time" it is "change in velocity per time."
     
  13. Dec 13, 2014 #12
    Ok the the thing you said about their being two different VF's helps me visualize a little better, i was thinking of it as one big problem when really theres the stuff after the collision when its skidding and then the stuff when it collides.

    Ok so let me try to stop messing up on the velocity. If i need change in velocity I think this is how I would do the equation
    $$ A = Vf-Vi/T $$

    Im stuck here though.......... I think the Vf is 0 but where do I get the Vi from...... because all I have is initial velocity
    All I could think of is trying this
    $$ D=½(Vf+Vi)T $$
    $$ 13=½(0+Vi)22 $$
    $$ Vi = 1.18 m/s $$

    Wait maybe Im getting something...... could I then go here
    $$ A = Vf-Vi/T $$
    $$ A = 0-1.18/22 $$
    $$ A = -.054 m/s^2 $$

    which I assume would go positive? Sorry Im trying not to plug in equations... im just using this reply box as like my constant stream of thinking as i contemplate your responses.. I would still need to figure out what to do with acceleration if correct
     
  14. Dec 13, 2014 #13
    Wait I dont think it goes positive actually... since its a negative acceleration as the car slows.. i am just scared of negative numbers i think haha
     
  15. Dec 13, 2014 #14

    Bystander

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    And that deceleration acts for 22 seconds for what change in velocity?
     
  16. Dec 13, 2014 #15
    I think
    $$ A = ΔV/T $$
    $$ -0.54 = ΔV/22 $$
    $$ ΔV = -11.88 m/s $$

    You said this is my change in velocity.... does that mean I cant put this in my VF all the way at the original equation? If so what would I do with this/where would I go now since I thought we needed VF.
     
  17. Dec 13, 2014 #16

    Nathanael

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    I thought it was -0.054 not -0.54, so that gives you 1.188 m/s, but really it should be 1.18 because 0.054 is not the exact value for the acceleration, you rounded it.
    The exact value of the acceleration was 1.18/22, so (1.18/22)*22=1.18
    There's a good lesson on how premature rounding can affect your answer :oldtongue:

    Anyway in post #12 you said Vi=1.18 m/s
    What does this "Vi" represent? It was the velocity when they first started skidding, right? Which means it was the velocity right after the collision, right?
     
  18. Dec 13, 2014 #17

    Bystander

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    This is good.
    This is getting sloppy transcribing numbers.
    Correct your arithmetic error, and this is the "initial" velocity of the wreckage following the collision, from which you do the calculation of Pete's velocity from the conservation of momentum.
     
  19. Dec 13, 2014 #18
    First off, yeah I definitely made an error (Sorry :() the actual answer would be
    $$ Vi = 1.18m/s $$

    Now let me just write were in the post-collision world here so I remember. Next I think I have to get back on track and find the VF after the collision if I have the initial velocity
    So I think Im now allowed to do this?
    $$ D=½(Vf+Vi)T $$
    $$ 13=½(Vf+1.18)22 $$
    $$ Vf = .00182m/s $$

    seems low to me kind of... should i keep moving forward or did i make an error
     
  20. Dec 13, 2014 #19
    I kind of feel like Im getting lost in the problem now.... we are so far down (my fault since I just cant seem to comprehend it)
    I'm gonna restate this.....
    $$ (m1v1)+(m2v2)=(m1+m2)Vf $$

    $$(796.1∗v1)+(505.8∗21.01)=(796.1+505.8)Vf $$

    I really now just need help figuring out if the stuff I did in like my last two posts can help me get the right VF so i can find V1

    I am operating under the assumption this is still the original equation I should be using
     
  21. Dec 13, 2014 #20

    Nathanael

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    If you assume that the acceleration after the collision is constant, then you can say [itex]V_{avg}=\frac{V_f+V_i}{2}[/itex]. Since you know that the final velocity is zero, (they come to a stop) then the initial velocity (assuming constant acc.) is twice the average velocity. Now the average velocity is just the total distance per total time. You wrote it in post #9, it is 0.59 m/s, therefore the initial velocity (a.k.a. the velocity immediately after the collision) is 1.18 m/s

    This would be the velocity you use for the conservation of momentum equation.
     
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