Understanding variable mass equation for mass ablation/ejection The "Variable-mass system" entry in Wikipedia, under the "Mass ablation/ejection" title, shows:
$$p_2 = (m+dm)(v+dv) + u(-dm)$$
The parameter $p_2$ is for the momentum at $t+dt$, meaning $p_2 = p(t+dt)$.  
I don't understand the following:


*

*Why in the term $(m+dm)$, the parameter $dm$ is positive, while in the term $u(-dm)$ it is treated as negative?

*Why in the first term, $(m + dm)$, it is thought that the mass $m$ is  gaining mass?
Shouldn't it be $(m - dm)$ ? Because that it loses mass.

 A: Everything in the equation is treated as a function of time.
$p(t)$ is defined to be the momentum of the rocket at time $t$.
$m(t)$ is defined to be the mass of the rocket at time $t$.
$v(t)$ is defined to be the speed of the rocket at time $t$.
("Rocket" here is the system that is actually losing mass. This is where the whole discussion originates from; and it helps intuition, so I am talking "rocket".)
Now, if $m(t)$ is the mass of the rocket as a function of time, then $\frac{dm(t)}{dt}$ will be the rate of change of the mass of the rocket as a function of time.
As such, $dm$ (or $dm(t)$, the same thing, really) is the change in the rocket's mass. Note that if the rocket is actually losing mass (which it is), the value of $dm$ is negative, that is $dm < 0$ and $\frac{dm(t)}{dt} < 0$. So, $dm$ is not the amount of mass lost, it is the change of mass of the rocket. Now, if the rocket mass changes by $dm$, how much mass is added to "not rocket", i.e., ejected? Since mass must be conserved, "mass change in rocket" + "mass change in not-rocket" must be zero. So, that ends up as $-dm$. In the end result $dm$ is negative, and $-dm$ is positive. 
The exact reason why there is no explicit minus sign is because $d(anything)$ is the change in that $(anything)$. Even if we know it is negative (that is, we know that $(anything)$ is decreasing, there is no explicit negative sign, because it will then be the change in $(-anything)$... And that will mess up the solution.
A: 
I don't understand the following:
  
  
*
  
*Why in the term (m+dm), the parameter dm is positive, while in the term u(−dm) it is treated as negative?
  
*Why in the first term, (m+dm), it is thought that the mass m is gaining mass?
  Shouldn't it be (m−dm) ? Because that it loses mass.
  

The convention used in that article is that $m(t)$ denotes the mass of the main body as some function of time and that the differential $dm$ is given by $dm = \frac{dm(t)}{dt} dt$. If the main body is losing mass then $dm$ will be negative. You'll still have $m(t+dt) = m(t) + dm$.
The symbols $\vec p_1$ and $\vec p_2$ used in the article represent the momentum of the constant mass system that comprises the main body and the ejected mass. The ejected mass is still a part of the main body prior to the ejection. The momentum of the system prior to ejection is simply $\vec p_1 = m\vec v$.
After ejection, the momentum of the main body is $(m+dm)(\vec v + d\vec v)$. We need to add the momentum of the ejected mass to obtain the momentum of the main body plus ejected mass system. So what is the momentum of the ejected mass? The ejected mass is moving at some velocity $\vec u$. The mass of the ejected mass is $-dm$ ($dm$ is negative, so $-dm$ is positive). The momentum of the ejected mass is thus $-dm\,\vec u$.
The momentum of the main body and ejected mass system is thus $\vec p_2 = (m+dm)(\vec v + d\vec v) - \vec u\,dm$. Ignoring the $dm\,d\vec v$ term, this becomes $\vec p_2 = m\vec v + m\,d\vec v +(\vec v -\vec u)\,dm$. Subtracting the momentum prior to the ejection yields the change in momentum of the system,
$$d\vec p = \vec p_2 - \vec p_1
  = m\,d\vec v +(\vec v -\vec u)\,dm
  = m\,d\vec v - (\vec u -\vec v)\,dm
  = m\,d\vec v - \vec v_\text{rel}\,dm$$
where $\vec v_\text{rel} \equiv \vec u - \vec v$. Rearranging the above, dividing by $dt$, taking the limit as $dt\to0$, and applying Newton's second law yields
$$\vec F_\text{ext} + \vec v_\text{rel}\frac{dm}{dt} = m\frac {d\vec v}{dt}$$
Note that this is the same result as obtained for an object that accretes mass.
