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Hailstones and Momentum

  1. Jun 27, 2007 #1
    1. The problem statement, all variables and given/known data
    I'm working on a similar problem and someone else solved the problem (along with the steps). I looked at what they did but I'm really confused on 2 of the steps.

    Here is the problem:

    In a 30-s interval, 500 hailstones strike a glass window of area 0.60 m2 at an angle of 45o to the window surface. Each hailstone has a mass of 5.0 g (0.005 kg) and a speed of 8.0 m/s. If the collisions are elastic, find the average force.

    3. The attempt at a solution

    vx = v cos 45o (Why is it not sin 45?)

    vx = (8.0 m/s)(0.707) = 5.66 m/s

    px = m vx = (0.005 kg)(5.66 m/s) = 2.83 x 10 - 2 kg-m/s

    I understand everything up to this next step. What I don't understand is why is the momentum in the x direction multipied by two?

    px = 2 px = 5.66 x 10 - 2 kg-m/s

    F = pxtot/t

    F = (500)(5.66 x 10 - 2 kg-m/s)/30 s

    F = 0.943 kg-m/s2

    F = 0.943 N
  2. jcsd
  3. Jun 28, 2007 #2
    Sin(45) = Cos(45).

    A substantial understanding of trig is going to be crucial throughout your physics course(s). Review the unit circle, if you're having trouble with this concept.
  4. Jun 28, 2007 #3
    OH, I see! But what is the reason for multiplying Px two times?
  5. Jun 28, 2007 #4

    Doc Al

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    Staff: Mentor

    Because what you really need in order to calculate the force is the change in momentum of each hailstone. Realize that the initial and final values for Px have opposite signs, since they are in opposite directions. If Pxi = Px and Pxf = -Px, then [itex]\Delta Px = (-Px) - (Px) = -2Px[/itex]. Make sense?
  6. Jun 28, 2007 #5
    perfect sense. thank you!
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