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Homework Help: Cannon ejecting bullets moving on surface with friction

  1. Nov 7, 2012 #1
    1. The problem statement, all variables and given/known data

    A cannon of total mass [itex]m_{0}[/itex] is at rest on a rough horizontal road.It ejects bullets at rate of λ kg/s at an angle θ with the horizontal and at velocity u (constant) relative to the cannon .The coefficient of friction between the cannon and the ground is μ .Find the velocity of the cannon in terms of time t .The cannon moves with sliding .

    Answer : [itex] v = - μgt + ( ucosθ - μusinθ)ln[\frac{m_0}{m_0-λt}][/itex]


    2. Relevant equations



    3. The attempt at a solution

    Taking the positive direction of x axis towards right ,cannon is to move rightwards and eject bullets leftwards

    We define system to comprise of (cannon + bullets)

    Let M be the mass of the system at time t
    then M-dm be the mass of the system at time t+dt
    V be the Velocity of the system at time t
    V + dv be the velocity of the system at time t+dt

    Now [itex]\frac{dm}{dt}=-λ[/itex]

    N=Normal force on the system from ground

    Now in the vertical direction

    Momentum of the system at time t = 0
    Momentum of the system at time t + dt =(M-dm)(0) +(dm)(usinθ)

    [itex]dp=(dm)(usinθ)[/itex]
    [itex]\frac{dp}{dt}=\frac{dm}{dt}(usinθ)[/itex]
    [itex]\frac{dp}{dt}=(-λ)(usinθ)[/itex]

    now [itex]F_{ext}=N-Mg[/itex]

    [itex]\frac{dp}{dt}=F_{ext}[/itex]

    [itex]N-Mg=(-λ)(usinθ)[/itex]

    [itex]N=Mg-λusinθ[/itex] (1)


    In the horizontal direction

    Momentum of the system at time t = MV
    Momentum of the system at time t + dt =(M-dm)(V+dv) +(dm)(-ucosθ+V+dv)

    dp=MV + Mdv - dmV - dmdv - dmucosθ +dmV +dmdv -MV

    dp=Mdv - dmucosθ

    [itex]\frac{dp}{dt}=M\frac{dv}{dt}-\frac{dm}{dt}(ucosθ)[/itex]

    [itex]\frac{dp}{dt}=(-λ)M\frac{dv}{dm}+λucosθ[/itex]

    [itex]F_{ext}=-μN[/itex]
    [itex]F_{ext}=-μ(Mg-λusinθ)[/itex]

    [itex]\frac{dp}{dt}=F_{ext}[/itex]

    [itex]\frac{dp}{dt}=-μ(Mg-λusinθ)[/itex]

    Thus,we have

    [itex]-μ(Mg-λusinθ) = (-λ)M\frac{dv}{dm}+λucosθ[/itex]

    [itex]-μMg + λμusinθ = (-λ)M\frac{dv}{dm}+λucosθ[/itex]

    Diving by -λ throughout,we get

    [itex]\frac{μ}{λ}Mg - μusinθ = M\frac{dv}{dm} - ucosθ[/itex]

    [itex]M\frac{dv}{dm} = \frac{μ}{λ}Mg -μusinθ + ucosθ[/itex]

    putting [itex]λ = -\frac{dm}{dt}[/itex] ,we get

    [itex]M\frac{dv}{dm} = -μMg\frac{dt}{dm} +( ucosθ - μusinθ)[/itex]

    [itex]dv=-μgdt + ( ucosθ - μusinθ)\frac{dm}{M}[/itex]

    [itex]\int_{0}^{v}dv = -μg\int_{0}^{t}dt + ( ucosθ - μusinθ)\int_{m_0}^{m_0-λt}\frac{dm}{m}[/itex]

    [itex] v = - μgt + ( ucosθ - μusinθ)ln[\frac{m_0-λt}{m_0}][/itex]

    which gives

    [itex] v = - μgt - ( ucosθ - μusinθ)ln[\frac{m_0}{m_0-λt}][/itex]

    This is not the correct answer...kindly help me with the problem ...
     
  2. jcsd
  3. Nov 8, 2012 #2

    ehild

    User Avatar
    Homework Helper

    Check the signs. If Δm is the mass of a ball, it is positive. Ejecting a ball, the mass of the cannon decreases. If the cannon ejects balls so the rate of the ejected mass is λ, the mass of the cannon is M(t)=mo-λt.

    By ejecting a ball, the vertical momentum of the cannon-ball system increases, as the ball gets the vertical velocity vy=usin(θ), and the vertical velocity of the cannon does not change. The change of the vertical component of momentum is Δm*usin(θ)=λΔt*usin(θ) and it is equal to the impulse of the external force: (-Mg+N)Δt, so N=Mg+λusin(θ): ejecting balls increases the normal force.

    ehild
     
  4. Nov 8, 2012 #3
    ehild...thanks for the response....Considering what you have suggested , I have amended my work ...but again I arrive at an incorrect answer


    Taking the positive direction of x axis towards right ,cannon is to move rightwards and eject bullets leftwards

    We define system to comprise of (cannon + bullets)

    Let M be the mass of the system at time t
    then M-dm be the mass of the system at time t+dt
    V be the Velocity of the system at time t
    V + dv be the velocity of the system at time t+dt

    Now [itex]\frac{dm}{dt}=λ[/itex]

    N=Normal force on the system from ground

    Now in the vertical direction

    Momentum of the system at time t = 0
    Momentum of the system at time t + dt =(M-dm)(0) +(dm)(usinθ)

    [itex]dp=(dm)(usinθ)[/itex]
    [itex]\frac{dp}{dt}=\frac{dm}{dt}(usinθ)[/itex]
    [itex]\frac{dp}{dt}=(λ)(usinθ)[/itex]

    now [itex]F_{ext}=N-Mg[/itex]

    [itex]\frac{dp}{dt}=F_{ext}[/itex]

    [itex]N-Mg=(λ)(usinθ)[/itex]

    [itex]N=Mg + λusinθ[/itex] (1)


    In the horizontal direction

    Momentum of the system at time t = MV
    Momentum of the system at time t + dt =(M-dm)(V+dv) +(dm)(-ucosθ+V+dv)

    dp=MV + Mdv - dmV - dmdv - dmucosθ +dmV +dmdv -MV

    dp=Mdv - dmucosθ

    [itex]\frac{dp}{dt}=M\frac{dv}{dt}-\frac{dm}{dt}(ucosθ)[/itex]

    [itex]\frac{dp}{dt} = λM\frac{dv}{dm} - λucosθ[/itex]

    [itex]F_{ext}=-μN[/itex]
    [itex]F_{ext}=-μ(Mg + λusinθ)[/itex]

    [itex]\frac{dp}{dt}=F_{ext}[/itex]

    [itex]\frac{dp}{dt}=-μ(Mg + λusinθ)[/itex]

    Thus,we have

    [itex]-μ(Mg + λusinθ) = λM\frac{dv}{dm} - λucosθ[/itex]

    [itex]-μMg - λμusinθ = λM\frac{dv}{dm} - λucosθ[/itex]

    Diving by λ throughout,we get

    [itex]\frac{-μ}{λ}Mg - μusinθ = M\frac{dv}{dm} - ucosθ[/itex]

    [itex]M\frac{dv}{dm} = \frac{-μ}{λ}Mg -μusinθ + ucosθ[/itex]

    putting [itex]λ = \frac{dm}{dt}[/itex] ,we get

    [itex]M\frac{dv}{dm} = -μMg\frac{dt}{dm} +( ucosθ - μusinθ)[/itex]

    [itex]dv=-μgdt + ( ucosθ - μusinθ)\frac{dm}{M}[/itex]

    [itex]\int_{0}^{v}dv = -μg\int_{0}^{t}dt + ( ucosθ - μusinθ)\int_{m_0}^{m_0-λt}\frac{dm}{m}[/itex]

    [itex] v = - μgt + ( ucosθ - μusinθ)ln[\frac{m_0-λt}{m_0}][/itex]

    which gives

    [itex] v = - μgt - ( ucosθ - μusinθ)ln[\frac{m_0}{m_0-λt}][/itex]

    Again I arrive at the same incorrect answer . Where am I getting it wrong??
     
  5. Nov 8, 2012 #4

    ehild

    User Avatar
    Homework Helper

    On system you mean the cannon and the balls still inside.

    So dm is the mass of a ball. And the time derivative of the mass is dM/dt=-λ.


    Beware: dm is the mass of a ball. M is the mass of the system. dM/dt=-λ.

    M is a variable. dm is not. Keep t as independent variable. M=M0-λt.

    [itex]\frac{dp}{dt} = M\frac{dv}{dt} - λucosθ[/itex]


    Keeping t as independent variable,
    [itex]-μMg - λμusinθ = M\frac{dv}{dt} - λucosθ[/itex]

    Substitute M=Mo-λt, collect the terms containing t and integrate.

    ehild
     
  6. Nov 8, 2012 #5
    ehild ....what you have suggested is excatly what i did initially....Kindly look at post #1

    If i write [itex]\frac{dm}{dt}=-λ[/itex] then N =Mg - λusinθ and again we arrive at the same incorrect result ...

    Kindly have a look at the initial post...
     
  7. Nov 8, 2012 #6

    ehild

    User Avatar
    Homework Helper

    I suggested to keep the time as independent variable. dm/dt is not -λ. dm≠dM. Please, read my previous post. You integrate with respect to the mass, which is M. dm is the loss of mass when a ball is ejected. dM=-dm. dM/dt=-λ.

    ehild
     
    Last edited: Nov 8, 2012
  8. Nov 9, 2012 #7
    ehild...Thank you very much :smile:
     
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