Cecilia asked in 科學及數學其他 - 科學 · 1 decade ago


A toy designer has submitted a design for a water pistol with a barrel area 75mm^2 and jet area 1mm^2. The manufacturer required that when the pistol was fired horizontally, the jet should be able to hit a target 3.5m away not more than 1m below the firing line. Given that the average child is able to exert a force of 10N on the plunger, has the designer satisfied the requirements? you may neglect barrel friction and energy loss at the exit jet.


(Atmospheric pressure= 1x 10^5 Pa, density of water= 1x10^3, g=9.8ms^-2)

1 Answer

  • 1 decade ago
    Favorite Answer

    PART 1: to calculate the PRACTICAL length of barrel where the bullet occupies

    as the trace of the water bullet ejected is independent of mass,

    for simplicity, let us consider a bullet of volume,

    V=1*10^ -6 m^3

    mass of the bullet, m=V*density

    =[1*10^ -6 m^3]*[1000 kg/m^3]

    =1*10^ -3 kg

    when the bullet is initially stored in the barrel,

    it occupies the barrel for a length, l,

    =V/ cross sectional area of barrel

    = 1*10^ -6 m^3 / 75*10^ -6 m^2

    = 0.0133m -------------------------------------------@

    PART 2: to calculate the MINIMUM value of length of barrel where the bullet can be accelerate in order to be ejected and reached the target not lower than 1 m.


    consider the trace of the ejected bullet,


    displacement, s=u*t + 0.5*g*t^2

    as the bullet is ejected horizontally, u=0,




    velocity, w, remains unchanged

    and w=horizontal displacement / t

    =3.5m / 0.447sec.

    =7.826 m/s

    speed at the jet equals w=7.826 m/s

    consider cross sectional area and velocity of the jet and barrel,

    by Bernoulli Principle,


    [75*10^ -6 m^2]*[v]=[1*10^ -6 m^2]*[w]

    velocity at the barrel, v=0.104 m/s

    by Newton's second law,


    F-f = m[v - 0]/ T

    as initial velocity of the bullet in the barrel = 0 m/s

    where F is the force exerted by the child and F= 10 N

    f is the force exerted by the atmosphere and

    f = P(atmosphere)*A(jet)

    =[1*10^5 N/m^2]*[1*10^ -6 m^2]

    =0.1 N

    T=m*v / [ F-f ]

    =[1*10^ -3 kg]*[ 0.104 m/s ] / [ 10 N - 0.1N ]

    =1.0505*10^ -5 sec.


    a=v/t=0.104/1.0505*10^ -5


    Length, s, of barrel where the bullet of 1*10^ -3 cm^3 occupies,



    =5.463*10^ -7m ----------------------------------------#


    As @ is smaller than #, the 1*10^ -3 m^3 bullet can reach the target not lower than 1m

    Source(s): myself
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