Conserving Momentum in Emptying a Freight Car

[bob012345]

The entire argument is over notation.

You can draw a diagram of a system of particles in two different states and define p1 as its momentum in one state and p2 as its momentum in the other.
You can then write $F.\Delta t=p_2-p_1$, and if the expressions for p1 and p2 involve such as $v$ and $v+\Delta v$ then you may well obtain a valid differential equation.

The confusion arises because you denoted the momenta of these states generically as a function of time, p(t). That implies a definition that can be interpreted at any given time t, but I cannot see a way to do that with your p1 and p2.
In any algebraic development, if you write $f(t)=x(t)y(t)$ then it follows that $f(t+\Delta t)=x(t+\Delta t)y(t+\Delta t)$, but your equations do not satisfy that.

Fine. Pick some time between 0 and $\frac{m_s}{b}$, call it $t'$. Then call the first equation $p_1$ and the second equation at a $Δt$ later $p_2$.

 

[hutchphd]

The problem at hand consists only of everything at time t and everything at time t+Δt. I kept track of that interval only not all Δt′s.

I disagree. Tell me again the definition of your system at any time t

 

[bob012345]

I disagree. Tell me again the definition of your system at any time t

At time $t$ the system is the left hand side of this figure. It consists of the mass of cart plus however much sand is in the cart at that moment traveling with velocity $v_c$. Here it is in all it's glory;

 

[haruspex]

Fine. Pick some time between 0 and $\frac{m_s}{b}$, call it $t'$. Then call the first equation $p_1$ and the second equation at a $Δt$ later $p_2$.

So your working becomes

Let p(t) be the momentum of the system consisting of the cart and its remaining sand at time t:
$\vec P(t ) = (m_c + m_s -bt)\vec v_c(t)$
Thus $\vec P(t + Δt) = (m_c + m_s - b(t +Δt))(\vec v_c(t) + Δ\vec v_c))$.
But the momentum that had been in $\vec P(t )$ is now shared between $\vec P(t + Δt)$ and the sand that was lost in interval $(t, t + Δt)$. That lost momentum is $-Δm_s(v_c(t) + Δ\vec v_c)$. ($Δm_s$ being negative).
[We can simplify to $Δm_sv_c(t)$ since the other part is a second order small quantity.]
Were there no external force, conservation of momentum gives
$\vec P(t ) = \vec P(t + Δt)- Δm_s(v_c(t) + Δ\vec v_c)$, or $\Delta \vec P(t ) = Δm_sv_c(t)$.
But in the same interval, a further $\vec F.\Delta t$ has been injected into the system, so
$\Delta \vec P(t ) = Δm_sv_c(t) +\vec F.\Delta t$.

Looks ok to me.

 

[bob012345]

So your working becomes

Let p(t) be the momentum of the system consisting of the cart and its remaining sand at time t:
$\vec P(t ) = (m_c + m_s -bt)\vec v_c(t)$
Thus $\vec P(t + Δt) = (m_c + m_s - b(t +Δt))(\vec v_c(t) + Δ\vec v_c))$.
But the momentum that had been in $\vec P(t )$ is now shared between $\vec P(t + Δt)$ and the sand that was lost in interval $(t, t + Δt)$. That lost momentum is $-Δm_s(v_c(t) + Δ\vec v_c)$. ($Δm_s$ being negative).
[We can simplify to $Δm_sv_c(t)$ since the other part is a second order small quantity.]
Were there no external force, conservation of momentum gives
$\vec P(t ) = \vec P(t + Δt)- Δm_s(v_c(t) + Δ\vec v_c)$, or $\Delta \vec P(t ) = Δm_sv_c(t)$.
But in the same interval, a further $\vec F.\Delta t$ has been injected into the system, so
$\Delta \vec P(t ) = Δm_sv_c(t) +\vec F.\Delta t$.

Looks ok to me.

I like the conserve momentum approach except for or up to $\vec FΔt$.