# Cannon ejecting bullets moving on surface with friction

1. Nov 7, 2012

### Tanya Sharma

1. The problem statement, all variables and given/known data

A cannon of total mass $m_{0}$ 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 : $v = - μgt + ( ucosθ - μusinθ)ln[\frac{m_0}{m_0-λt}]$

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 $\frac{dm}{dt}=-λ$

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)

$dp=(dm)(usinθ)$
$\frac{dp}{dt}=\frac{dm}{dt}(usinθ)$
$\frac{dp}{dt}=(-λ)(usinθ)$

now $F_{ext}=N-Mg$

$\frac{dp}{dt}=F_{ext}$

$N-Mg=(-λ)(usinθ)$

$N=Mg-λusinθ$ (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

$\frac{dp}{dt}=M\frac{dv}{dt}-\frac{dm}{dt}(ucosθ)$

$\frac{dp}{dt}=(-λ)M\frac{dv}{dm}+λucosθ$

$F_{ext}=-μN$
$F_{ext}=-μ(Mg-λusinθ)$

$\frac{dp}{dt}=F_{ext}$

$\frac{dp}{dt}=-μ(Mg-λusinθ)$

Thus,we have

$-μ(Mg-λusinθ) = (-λ)M\frac{dv}{dm}+λucosθ$

$-μMg + λμusinθ = (-λ)M\frac{dv}{dm}+λucosθ$

Diving by -λ throughout,we get

$\frac{μ}{λ}Mg - μusinθ = M\frac{dv}{dm} - ucosθ$

$M\frac{dv}{dm} = \frac{μ}{λ}Mg -μusinθ + ucosθ$

putting $λ = -\frac{dm}{dt}$ ,we get

$M\frac{dv}{dm} = -μMg\frac{dt}{dm} +( ucosθ - μusinθ)$

$dv=-μgdt + ( ucosθ - μusinθ)\frac{dm}{M}$

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

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

which gives

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

This is not the correct answer...kindly help me with the problem ...

2. Nov 8, 2012

### ehild

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

3. Nov 8, 2012

### Tanya Sharma

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 $\frac{dm}{dt}=λ$

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)

$dp=(dm)(usinθ)$
$\frac{dp}{dt}=\frac{dm}{dt}(usinθ)$
$\frac{dp}{dt}=(λ)(usinθ)$

now $F_{ext}=N-Mg$

$\frac{dp}{dt}=F_{ext}$

$N-Mg=(λ)(usinθ)$

$N=Mg + λusinθ$ (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

$\frac{dp}{dt}=M\frac{dv}{dt}-\frac{dm}{dt}(ucosθ)$

$\frac{dp}{dt} = λM\frac{dv}{dm} - λucosθ$

$F_{ext}=-μN$
$F_{ext}=-μ(Mg + λusinθ)$

$\frac{dp}{dt}=F_{ext}$

$\frac{dp}{dt}=-μ(Mg + λusinθ)$

Thus,we have

$-μ(Mg + λusinθ) = λM\frac{dv}{dm} - λucosθ$

$-μMg - λμusinθ = λM\frac{dv}{dm} - λucosθ$

Diving by λ throughout,we get

$\frac{-μ}{λ}Mg - μusinθ = M\frac{dv}{dm} - ucosθ$

$M\frac{dv}{dm} = \frac{-μ}{λ}Mg -μusinθ + ucosθ$

putting $λ = \frac{dm}{dt}$ ,we get

$M\frac{dv}{dm} = -μMg\frac{dt}{dm} +( ucosθ - μusinθ)$

$dv=-μgdt + ( ucosθ - μusinθ)\frac{dm}{M}$

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

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

which gives

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

Again I arrive at the same incorrect answer . Where am I getting it wrong??

4. Nov 8, 2012

### ehild

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.

$\frac{dp}{dt} = M\frac{dv}{dt} - λucosθ$

Keeping t as independent variable,
$-μMg - λμusinθ = M\frac{dv}{dt} - λucosθ$

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

ehild

5. Nov 8, 2012

### Tanya Sharma

ehild ....what you have suggested is excatly what i did initially....Kindly look at post #1

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

Kindly have a look at the initial post...

6. Nov 8, 2012

### ehild

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
7. Nov 9, 2012

### Tanya Sharma

ehild...Thank you very much