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Homework Help: Curious Question on momentum

  1. Nov 5, 2009 #1
    A stream of water travelling horizontally at 30m/s is ejected from a pipe of cross-sectional area 40 cm squared and is directed against a vertical wall. Calculate the force exerted on the wall assuming that the water does not rebound.
    What is the power of the pump needed to give the ejected water the necessary kinetic energy?

    (Density of water = 1000kg/mcubed)

    All i know so far is that due to newtons 3rd law is that if of course there is an exerting force then the forces must be equal and opposite and the value must be negative, but thats all ive got really anyone can anyone help?
  2. jcsd
  3. Nov 5, 2009 #2
    Power as far as I know is work/time=A/t, as you should know: A=Fscos(teta), F - force, s - displacement, teta - angle between force and displacement vectors. in this case i guess teta can be taken as zero. if you divide both sides of A=Fs by time t, you get that power P is: P=Fv, where v is the speed. so you need only force to find power. and force can be found by second Newton's Law:
    [tex]F=\frac{\Delta (mv)}{\Delta t}[/tex]
    which says that force is equal to change rate in the momentum, if v=30m/s, and v2=0m/s (when water hits the wall), then
    [tex]F=\frac{m(v-v_2)}{\Delta t}=\frac{mv}{\Delta t}[/tex]
    to find m, take some value of [tex]\Delta t[/tex], and calculate what volume of water ejects from pipe in that time and multiply this volume by water density. Hope this will help ;]
  4. Nov 5, 2009 #3
    This did help indeed
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