I'm dealing with the leaking freight car problem:
A freight car is loaded with a water tank. The mass of the freight car and the filled water tank is from the start $m_0$. At time $t=0$ a horisontal force $\vec{F}$ starts pushing the freight car and at the same time a water tap is opened in the bottom of the freight car so that a constant mass of water per unit of time $k= \left|\dfrac{dm}{dt} \right|$ is leaking out.
Assume that $\vec{F}$ is constant, $\vec{F} = F_0\hat{x}$. Determine the freight car's acceleration.
The given solution
The force equation $\vec{F} = \dot{\vec{p}} \implies F_0\hat{x} = \dot{p}_x\hat{x} \implies p_x\hat{x} = (F_0t + C)\hat{x}$
$\vec{v}(t=0) = 0 \implies \vec{p}(t=0) = 0 \implies C = 0 \implies p_x\hat{x} = F_0t\hat{x}$
$\hat{x}\text{-direction}: p_x = mv_x = F_0t \implies v_x(t) = \dfrac{F_0t}{m(t)}$
$$a_x(t) = \dfrac{d}{dt}v_x(t) = \dfrac{d}{dt} \left ( \dfrac{F_0 t}{m(t)} \right) = F_0 \dfrac{m(t) - t\dot{m}(t)}{m^2(t)} = F_0\dfrac{m_0 - kt + tk}{m^2(t)} = \dfrac{F_0}{m_0}\dfrac{1}{\left(1-\frac{k}{m_0}t \right)^2}$$
Clearly this solution differs from my own solution. I cannot really see anything wrong with it, but on the other hand I don't see anything wrong with my own solution either. I do remember reading from Kleppner and Kolenkow that one cannot apply the equation $\vec{p} = m\vec{v}$ blindly to a system of many particles, so maybe there's something going on there. I wonder which solution is correct and what is wrong in either one?
My solution
Usually I've seen this problem framed in terms of sand but the principle should be the same. My reasoning is similar to the one presented for example in these notes (p. 12-5 to 12-7).
Since the motion is one dimensional, I'll write $P(t)$ instead of $\vec{P}(t)$ etc. I imagine that at time $t$, the freight car (including the water tank) has mass $m_c(t) = m_0 - kt$ and the velocity $v_c(t)$ so that the momentum of the freight car at time $t$ is $$ P(t) = m_c(t)v_c(t) $$ Next consider time $t+\Delta t$. During $\Delta t$, the velocity of the freight car has changed to $v_c(t) + \Delta v_c$. Water of mass $\Delta m_w$ has leaked out were we estimate its velocity to also be $v_c(t) + \Delta v_c$. The freight cars mass is $m_c(t) - \Delta m_w$. The momentum of the freight car and the leaked water (the same system as at time $t$) is therefore $$P(t + \Delta t) = (m_c(t) - \Delta m_w + \Delta m_w)[v_c(t) + \Delta v_c] = m_c(t)[v_c(t) + \Delta v_c]$$
Therefore $\Delta P = P(t + \Delta t) - P(t) = m_c(t)\Delta v_c$ so that
$$F_0 = \dot{P}(t) = \lim_{\Delta t \to 0} \dfrac{\Delta P}{\Delta t} = \lim_{\Delta t \to 0} \dfrac{m_c(t)\Delta v_c}{\Delta t} = m_c(t)\dot{v_c}(t)$$
So that the acceleration $\dot{v_c}(t)$ is given by
$$\dot{v_c}(t) = \dfrac{F_0}{m_c(t)} = \dfrac{F_0}{m_0 - kt}$$
Provided $t \leq \dfrac{m_{w_0}}{k}$ where $m_{w_0}$ is the initial mass of the water in the full tank.
