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Water pistol physics

  1. Dec 29, 2013 #1
    1. The problem statement, all variables and given/known data

    B7 State the relationships between (i) force and momentum, (ii) force and energy, (iii) force and power, (iv) power and energy. [2]

    Find expressions for the momentum and kinetic energy transported per unit time in a jet of fluid having cross-sectional area A, density ρ, and speed u.[2]

    In a water pistol, a trigger drives a piston at constant speed along a cylinder with cross-sectional area S forcing water out through a hole as a narrow jet with cross-sectional area A. By equating the power supplied to the piston to that transported by the jet show that the net force F needed on the piston to generate a jet with speed u is
    F = 1/2 x pSu^2
    where ρ is the density of water. [4]

    A pistol has S = 1cm2 and A = 4mm2. A target is placed at a height h = 2.5m above the pistol and at a distance w = 5m in front of it. Find the speed of the jet needed, the direction in which the pistol must be aimed, and the force needed on the piston, to produce a jet that hits the target horizontally. [5]

    Is the force of the jet sufficient to dislodge a target with a mass of 30g on a surface with a coefficient of friction of 0.5? [2] [ You may assume that the density of water is ρ = 1,000kgm−3 ]


    2. Relevant equations



    3. The attempt at a solution
    Ok for the first four questions I put : i) Force x time = momentum
    ii) Force x distance = Energy
    iii) (Force x distance)/time = Power
    iv) Energy/time = Power

    For the next part I used the logic that momentum per unit time would have units kgms^(-2) and by using area= m^2 , density= kgm^(-3) and speed = ms^(-1) I arrived at the correct units using density x speed^2 x area = pAu^2
    I used a similar method for kinetic energy per unit time time and arrived at ke per time = 1/2 x pAu^3

    I'm not sure if these are correct.

    For the next part I started with the power transported by the jet = KE/time = 1/2pAu^3

    But then I don't know how to find the power supplied to the piston, I know that using the equation from iii) I can get something in terms of F but I don't know the time or distance nor how I get S involved.

    Help would be much appreciated!
     
    Last edited: Dec 29, 2013
  2. jcsd
  3. Dec 29, 2013 #2
    Rearrange equation iii). What is distance/time?
     
  4. Dec 29, 2013 #3
    Distance/time = speed , If I use u as the speed of the piston I get power(piston) = F x u and use S instead of A for the power transported by the jet I get the correct answer after equating the two. However how would I know that the speed of the piston is u? And I don't see how they involved S?
     
  5. Dec 29, 2013 #4
    The jet speed is u. What must be the speed of the piston? Remember, the volume of water displaced by the piston in unit time is the volume of water ejected from the pistol in unit time.
     
  6. Dec 29, 2013 #5
    Ok, so the volume of water ejected per unit time = area x speed = Au

    Volume of water displaced by piston per unit time = Sv (where v is the speed of the piston)

    Therefore v = (A x u)/S

    Then when I sub that in to the power equation I get the correct answer.

    Thanks!
     
  7. Dec 29, 2013 #6
    Can you proceed with the rest of the problem?
     
  8. Dec 29, 2013 #7
    Ok, I have attempted the next part, I went about it using a standard projectile motion method and ended up with ucos(theta) = root(5g) and then ended up with u = root(10g) and theta = 45 degrees

    Using the value of u I get the force on the piston to be 1.96 x 10^(-5) N which seems pretty small, I'm also not sure how to translate the force on the piston to the force the jet would exert on the target.
     
  9. Dec 29, 2013 #8
    What is the relationship between force and momentum?
     
  10. Dec 29, 2013 #9
    Force = momentum/time, so the force carried by the jet is equal to pAu^2 which was the answer to one of the first questions, converting A into metres squared this would give force of 0.392N, I calculated max friction to be 0.147N and so the force is sufficient to dislodge the target, hopefully this is correct.

    Thanks
     
  11. Dec 29, 2013 #10
    I have not checked the numbers, but you seem to understand the method.
     
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