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What formula can I use to calculate momentum transfer?

  1. Nov 1, 2014 #1
    < Moderator Note -- moved from Academic Guidance to Homework Help >

    Hey, I'm new here, I decided to register because one of the threads has helped me in the past and I'm hoping for some answers; to get into it, I'm doing research on the physics of being hit by a car/truck, and I've got most the math done, but to make it better I want to know how I can calculate the exact momentum transfer, because as of right now I'm assuming that the pedestrian being hit by the truck is only taking 1/100th of the momentum, leaving him flying at 12m/s in the vertical direction which seemed reasonable.

    I saw some sort of equation 60*carspeed*Ns, but I don't even begin to grasp that equation. Does anyone have an idea? If I'm saying something completely retarded right now can you help me out so I don't sound this retarded on my presentation?

    (Grade 11 physics)

    Thanks in advance,
    ~Before

    Also, here's my presentation (powerpoint)(google docs) if you are interested in seeing what I'm doing: https://docs.google.com/presentation/d/1eRI3G0aWYOfkoWJlSwbpdsnroQBWIKnADJzvuxl316k/edit?usp=sharing (just started that today, it's not amazing.)



    And also one last little thing, what kind of damage would you take if you were hit by something with a momentum of 60 000 and if you hit the ground with a momentum of 670?
     
    Last edited by a moderator: Nov 1, 2014
  2. jcsd
  3. Nov 1, 2014 #2

    td21

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    One important measure is coefficient of restitution.
     
  4. Nov 1, 2014 #3
    ?????????????
     
  5. Nov 1, 2014 #4

    haruspex

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    It depends on what data you have available to calculate it from. Please be more specific, and show any working so far.
     
  6. Nov 1, 2014 #5
    My sincere apologize, I did not understand the concept but now I do. The equation I was looking for was ab6645eaf09c6da1ba47b0f662615140.png
    where V1 is the new pedestrians velocity, U1 was the pedestrians velocity before being hit, M1 is the pedestrians mass, and the 2s are the trucks stuff.

    Sorry :(

    I can't seem to edit this to SOLVED, could you do that for me please so no one else tries to help me with something solved?

    Thanks,

    ~Before
     
  7. Nov 1, 2014 #6

    haruspex

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    Are you sure? It looks like you are taking the pedestrian and truck as coalescing (reasonable), so v1 = v2. But in that case I would expect to see ##v_1 = \frac{u_1 m_1 + u_2 m_2}{m_1+m_2}##. If you are not taking them as coalescing you must have some other assumption to provide a second equation.
    In the OP you mentioned vertical motion, but that does not seem to be represented in your equation.
     
  8. Nov 2, 2014 #7
    uhm, I'm assuming there is no loss in energy.. Idk, the first equation worked well for me. Let me just test my numbers with your formula; with mine, I got 37.4m/s I think.. Lemme check quick.

    Okay, that equation gave me 19.67 m/s; I got that first equation from Wikipedia and it also explained how to derive it so it seemed reasonable. Your equation makes sense too.. I'm just going to pass them both by my teacher and ask which one is correct. Sorry I think the issue here isn't math, it's my incapability to express myself :D
     
  9. Nov 2, 2014 #8

    haruspex

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    Probably not a good assumption. People and cars are not made of rubber.
    Link?
     
  10. Nov 2, 2014 #9
    Oh, but I don't know if I can calculate that too :S seems hard
    The presentation is supposed to be 3-4 mins, and it's only the grade 11 physics class (first class offered at my school) so we're doing basics, even Momentum isn't covered until next year..
    http://en.wikipedia.org/wiki/Elastic_collision#Equations
     
  11. Nov 2, 2014 #10

    BvU

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    Loved your presentation. But do check your calculations: 38.4 is 4% less than 2 x 20 and
    I would expect not even 2% less ( 3000/3050 = 1 - 0.016 )

    Also check your physics. Talk it over with a few classmates, perhaps.

    And: "kinetic energy = velocity squared" is very wrong. kinetic energy = ½ times mass times velocity squared !
    (You actually use that correctly later on in ½ mv22 - ½ mv12 = mg Δd .)
    [edit] added the squares​


    So when you get hit by the truck your velocity changes from 0 to 40 m/s in a flash according to your elastic collision calculation.
    Well, at least you don't have to worry about being overrun by the truck: at 72 km/h it'll never catch up with your 144 km/h !


    But when you hit the ground your velocity doesn't change from 40 to 0 : you don't stick at that point, but skid forward at the same speed. Only the vertical component of your velocity is changed from -4 m/s to 0 in a flash. That's not so bad: it also happens if you fall over your shoelaces !

    I also liked "I took the negative half of the square root because that suits my purposes". Because you fall down, not up is what you mean to say.
     
    Last edited: Nov 3, 2014
  12. Nov 3, 2014 #11

    haruspex

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    The equation development at that link relies on conservation of momentum, so like it or not, you are using it.

    As I said, you need one other assumption. In general, a collision involves something called the coefficient of restitution. This says how much of the KE is lost (as heat etc.) At one extreme, the coefficient is 1 and no KE is lost. This would give the equation you quoted. At the other extreme, the coefficient is zero. It doesn't mean all KE is lost, but in a straightforward linear collision it does mean the colliding bodies 'coalesce' (move off together). This leads to the equation I gave in post #6. For a vehicle hitting a person in normal clothing, a coefficient of zero would be far nearer reality than a coefficient of 1.

    All of the above is in respect of horizontal motion, but in the OP you mention vertical velocity, and indeed a pedestrian hit by a car may very well be sent skywards to some extent. But that is because of the typical shape of the front of a car, which will act as a wedge. This may be too complex for your purpose.
     
  13. Nov 3, 2014 #12
    Thanks a lot!!!
     
  14. Nov 3, 2014 #13
    Your assumption that the truck's speed is not changed after collision (slide 4) is inconsistent with the equations you use.
    If this were true, there is no momentum transfer to the person. Unless there is some external force acting on the system during the collision. In which case you cannot use conservation of momentum.
    Of course, in reality this the case. The engine keeps working during the collision. But the collision time is so short that we neglect the effect of this force, in first approximation. But then you must allow for some decrease in the momentum of the car. You can't have it both ways.
     
  15. Nov 3, 2014 #14
    I am assuming the there is an effect on the car's momentum, I just don't care about it because it's irrelevant for this exact "simulation" for lack of a better word.
     
  16. Nov 3, 2014 #15
    If you don't use it, why put it on the "assumption" list? It's misleading.
     
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