(adsbygoogle = window.adsbygoogle || []).push({}); 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 ...

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# Homework Help: Cannon ejecting bullets moving on surface with friction

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