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Physics

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A 15 kg, 1m wide door which has frictionless hinges is closed but unlocked. A 400 g ball hits the exact middle of the door at a velocity of 35 m/s and bounces off elastically, thereby causing the door to slowly swing open. How long in seconds does it take for the door to fully open (rotate 90 degrees)?

  • Physics - ,

    http://www.jiskha.com/display.cgi?id=1369042520

  • Physics - ,

    1.68?

  • Physics - ,

    i input your formula and false

  • Physics - ,

    The angular momentum of the door ‘p’= the change in amgular momentum of the ball Δp.
    In vector form:
    Δp⃗ =p₂⃗-p₁⃗ =mv₂⃗-mv₁⃗,
    for magnitudes
    Δp =p₂-(-p₁) =mv₂+mv₁= 2mv.
    p=Mu =>
    Mu=2mv
    The speed of the center of the door is
    u=2mv/M=2•0.4•35/15 = 1.37 m/s
    Its angular speed is
    ω=u/r=2u/L =2. •1.87/1 = 3.73 rad/s
    t= φ/ ω =π/3.73 = 0.84 s.

  • Physics - ,

    please write right answer

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