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Recoil speed

  1. Mar 14, 2015 #1
    1. The problem statement, all variables and given/known data
    A hunter on a frozen, essentially frictionless pond uses a rifle that shoots 4.20g bullets at 950m/s . The mass of the hunter (including his gun) is 72.5kg , and the hunter holds tight to the gun after firing it.

    Find the recoil speed of the hunter if he fires the rifle at 52.0∘ above the horizontal.

    2. Relevant equations


    3. The attempt at a solution
    Vx = 950cos(52)
    Vy = 950sin(52)

    Conservation of Momentum in the x-Direction:
    m1v1' = -m2v2'
    (72.5)v1' = (-0.0042)(950cos(52))
    v1' = -0.033882611 m/s

    Conservation of Momentum in the y-Direction:
    m1v1' = -m2v2'
    (72.5)(v1') = (-0.0042)(950sin(52))
    v1' = -0.043367764 m/s

    vf = sqrt(0.0338826112+0.0433677642) = 0.055 m/s

    The correct answer is 0.0339 m/s

    What am I doing incorrectly?
     
  2. jcsd
  3. Mar 14, 2015 #2

    gneill

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    Staff: Mentor

    Is the hunter free to recoil in the Y-direction?
     
  4. Mar 14, 2015 #3
    I know that the hunter wouldn't move in the y-direction, but wouldn't there still be an initial velocity in the y-direction? Similar to how if an object is dropped, it has a y-component of velocity right as it hits the ground.
     
  5. Mar 14, 2015 #4

    gneill

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    Staff: Mentor

    The hunter is already on the ice surface. Any velocity in the downward direction would have to include the ice and the rest of the planet.
     
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