How to use the concept of variable mass system? The general equation of variable mass motion is as follows:

It's derivation is given below:

Refer wikipedia page for more details and the link of it is given below:
https://en.wikipedia.org/wiki/Variable-mass_system
In the derivation they considered product of dm and dv to be negligible 
Check this line:

I have a question of this concept which is as below:

The answer's first part is given as:

In this answer they considered 
m=m(intital)-dm ,which is equal to
m(initial)-ut

Why are they taking it like that?
The above is possible only when dm.dv is not considered negligible
check the below derivation:

So,why is it not negligible in this problem and why was it considered negligible in the derivation?
 A: You can help your troubles with considering that the equation you don't understand:
$$m=m_0-\mu t$$
comes from the fact that
$\dfrac{dm}{dt}=-\mu\,\,$ (EDITED) 
because of the hypothesis for the problem, the sand is spilling; so the rate of loss of mass $m$ of the sand inside the cart is negative.
Then,
$$m(t)-m_0=\int_{m_0}^{m(t)} dm=\int_{0}^{t}\frac{dm}{dt}dt=-\mu\int_{0}^tdt=-\mu\,t$$
Then, you pick out an inertial frame for the observer at restwhich is studying the problem, who sees that the equation of motion for the cart-sand system has to be:
$$F=(m+M)\dfrac{dv}{dt}=\left(m_0-\mu t+M\right)\dfrac{dv}{dt}$$ 
$\therefore \dfrac{F}{m_0-\mu t+M}=\dfrac{dv}{dt}=a$ and later on, integrate for $v$.
Note that I introduce a mass $M$ for the cart, which does not appears in the problem. This has to be done for not introducing a division by zero for the acceleration! and think about it, the whole system includes this too. You can approximate m>>M if your mount of sand is a great deal bigger, which for the problem it is (can be) done.
The derivation you include above (Wikipedia, I'm afraid) can be vague when saying that you need a product $dm\,dv$. As physicists we can use differentials as tiny increasements, but don't forget their nature as calculus quantities, which in this problem bring us to a nonsense when you worked out your solution in the sheet you included as photo. You cannot say an impulse $p_1$ has to be equal to the sum of $mv+mdv+vdm$ since you are treating functions as differentials and that is not a good way to build up a differential equation.
You have a differential equation for v, and the reason for the other terms not appearing in the equation for $a$, say $dm\,dv$ or even $v\dfrac{dm}{dt}$ is that the acceleration is just contributed by forces acting in that line of movement.
A: Consider that in the expression, (dm)(dt), each term represents an extremely small quantity. Then consider what you get if you multiply, (.0001)(.0001).
