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Please help with momentum questions

  1. May 2, 2004 #1
    These are two questions that I am having problems with. Here is the first.

    1. The air in a 200km/h wind stikes a 30m by 20m face of building at the rate of 5.4 * 10^4 kg/s. Find the net force on the building, assuming the air comes to rest on impact.

    Now for this I found the force by mutiplying the velocity by the mass per second. But I don't know where to go from there. Wouldn't the net force just be 0?

    2. Calculate the impulse suffered when a 70 kg person lands on firm on the ground after jumping from a height of 5.0m.

    I don't think I have enough info to answer this question.
  2. jcsd
  3. May 2, 2004 #2


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    1. Looks to me like the force you calculated is the answer. At first I thought the problem was asking for "pressure" (force divided by area) but that's not the case.
    Surely they don't mean "net force" in the sense that "since the wall is not moving, there is no force"- that would make the problem trivial.

    2. "Impulse" is change in momentum. You know mass and can calculate the speed just before the the person hits the ground so you can calculate momentum. Since the person comes to a stop, that momentum is the impulse.
  4. May 3, 2004 #3
    For number 2 I used the formula Vf^2=Vi^2 + 2ad. I made Vf = 0 since the person comes to rest. and used a = 9.8. From this I got 9.9 m/s as the initial velocity. Does this sound right?
  5. May 3, 2004 #4

    For number 2 I used the formula Vf^2=Vi^2 + 2ad. I made Vf = 0 since the person comes to rest. and used a = 9.8. From this I got 9.9 m/s as the initial velocity. Does this sound right?
  6. May 3, 2004 #5

    Doc Al

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    You can use that kinematic formula to find how fast the jumper will hit the ground. (Vi = 0, since we assume he jumps starting from rest.) Once you have the the speed as the jumper hits the ground (that's the speed you found), then calculate the impulse that the ground exerts on the person as that person is brought to rest: Δmv
  7. May 3, 2004 #6
    what is the kinematic equation? I don't think I have learned that yet, or my course refers to it by a different name.
  8. May 3, 2004 #7

    Doc Al

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    I was referring to the equation that you used (Vf^2=Vi^2 + 2ad). "Kinematic equations" are just equations that describe motion without regard to the causes of that motion: they relate speed, distance, acceleration, and time.
  9. May 4, 2004 #8


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    No, it doesn't sound right! For one thing, the problem didn't ask for the intial velocity! Although the problem says "jump" I think we are expected to assume that the person's initial velocity was 0. The FINAL velocity is NOT 0 since the formula you use is falling under gravity- it isn't gravity that causes him to stop, it is that impulse that you are asked to find.

    If a person falls from rest at acceleration 9.8 m/s2 then his speed at time t is 9.8t (I am taking + downward) and the distance fallen is
    4.9t2. He will have fallen 5 m (just before he hits the ground) when 4.9t2= 5 or t= 1.01 seconds. At that time his speed will be
    9.8(1.01)= 9.90 m/s and his momentum will be 70(9.9)= 693 kg m/s . That's the "impulse suffered".
  10. May 5, 2004 #9


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    Simple conversion of momentum with a catch, you equate each side as a rate of time.
    Take your standard formula:

    [tex]Ft = m \Delta v[/tex]

    Now suppose you don't know what "t" is, just divide it out.

    [tex]F = \frac{m}{t} \Delta v[/tex]

    You want F, you know what m/t is (5.4 x 10^4 kg/s), and you can calculate what delta v is (convert 200km/h into m/s).

    [tex]V_f = V_i + 2ad[/tex]

    Vi is 0 so ignore it

    [tex]V_f = 2ad[/tex]

    Now impulse is just change in momentum (mv).

    [tex]P = mv[/tex]

    [tex]P = m(2ad)[/tex]
    Last edited: May 5, 2004
  11. May 5, 2004 #10
    another ingredient for the mix

    We were just giving an extension to the question and it is boggling my mind.
    This is an extension from question 2 from my first post.

    Determine if the person will suffer a fractured tibia (lower log bone; cross-sectional area of 3.0 cm^2) if the ultimate compressive strength is about 170 * 10^6 N/m^2. The impulse for a stiff-legged landing takes 0.002 s while a bent-legged landing takes 0.05 s.

    I don't even know where to begin and where to take it.
  12. May 5, 2004 #11

    Doc Al

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    You've already found the impulse, now you get to use it. Remember that impulse = FΔt, which means that to provide a given impulse the less time the force is applied, the greater the force must be. So figure out the force in each case and see if it's enought to fracture the tibia. As the numbers will show, a stiff-legged landing brings to you a stop much quicker, therefore it creates greater force.
  13. May 6, 2004 #12
    Alright I got it thanks
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