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Complete Inelastic collision & Force

  1. Apr 21, 2009 #1
    1. The problem statement, all variables and given/known data

    Raindrops of mass 0.99 mg fall vertically at a constant speed of 9.0 m/s, striking a horizontal skylight at the rate of 1000 drops/s and draining off. The window size is 17.0 cm x 25.0 cm. Assume the collisions of the drops with the window are completely inelastic.

    (a) Calculate the magnitude of the average force of the raindrops on the window.
    (b) Calculate the resulting pressure developed by the raindrop collisions.

    2. Relevant equations
    P=F/A
    F=dp/dt=Impulse/(delta-time)
    Impulse=delta-p=pf-pi


    3. The attempt at a solution

    so we know that every 1/1000 second, 1 raindrop falls and this is the delta-time
    I thought since momentum is conserved the impulse would just be twice mv=2(9.9E-7 kg)(9 m/s)=1.782E-5 Ns
    then the Faverage per raindrop= 1.782E-5/(1/1000)=.01782 N...but this is not right

    and I know part b is dependent on (a) since then the pressure is simply Faverage/Area= Faverage/(.0425 m^2), where .0425 m^2=area of window=.17 m x .25 m

    I'm not sure what to make of the fact that it's completely inelastic and it drains off?
    thank you in advance
     
  2. jcsd
  3. Apr 21, 2009 #2

    dx

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    Gold Member

    The momentum transferred would be 2mv if the drops hit the glass and bounced back with the same speed they came in with, i.e. if the collision was elastic. But in this case they just hit it and stop, so how much momentum is transferred to the glass?
     
  4. Apr 21, 2009 #3
    just mv? then
    Fave=mv/delta-t?=8.91E-6/(1/1000s)=0.00891 N
     
  5. Apr 21, 2009 #4
    ah thank you! what a stupid question.
     
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