Truck with leaking sand - How can it not have a change in momentum? If we have a truck filled with sand moving at a velocity "v", and the truck is leaking sand vertically downwards at a rate of $\frac{dm}{dt}$, if we ignore friction and air resistance, why is it that the truck will be moving at a constant velocity?
It makes sense if we consider the sand-filled-truck and the sand of mass △m that is falling out as one system. If we do so, the total mass in the system we defined is constant. Since the sand that falls out will have the same horizontal velocity of "V", so change in momentum of the system we defined is zero. Hence $\frac{dp}{dt} = 0$ and so no net force acts on the system.
While this explanation makes sense, I am not sure what is wrong with the following explanation:
If we calculate the momentum of only the sand-filled-truck at two instances in time, the momentum will definitely be smaller over time since the mass of the truck is decreasing. So how is it that $\frac{dp}{dt} = 0?$
Also, if we define our system to be only the sand-filled truck and not include the mass of sand that leaks out, the equation will for this system will be modelled by this differential equation $\frac{dp}{dt} = m\frac{dv}{dt} + v\frac{dm}{dt}$, and since mass is being loss from the truck, $\frac{dm}{dt}$ is negative, and so shouldn't there be a net external force acting on the truck?
The two explanations seem to contradict. Hope someone can help me, I have been stuck with this for days.
Edit: After seeing the comments, the more fundemental question is why can't we use $F = m\frac{dv}{dt} + v\frac{dm}{dt}$ for a system of variable mass? Isn't that what the $\frac{dm}{dt}$ term is for?
Thanks in advance!
 A: I believe the source of your confusion is a full understanding of the requirements for conservation of momentum.
Normally we say that system momentum is conserved if there are no net external forces acting on the system. Implicit in this description is the assumption that the mass of the system is constant. But the more general requirement for conservation of momentum in a system is that the system be isolated. An isolated system exchanges neither mass nor energy with its surroundings. In that respect the truck and its contents is not an isolated system since it loses mass to its surroundings, meaning $\frac{dm}{dt}\lt 0$ and since $\frac{dv}{dt}=0$, $\frac{dp}{dt}\lt0$.

if we ignore friction and air resistance, why is it that the truck
will be moving at a constant velocity?

Because, per Newton's 2nd law, a change in horizontal velocity (acceleration/deceleration) requires a net horizontal force acting on the truck and its contents. Neglecting air resistance and friction there is no net force acting on the truck plus its contents. Since $v$ is constant and mass is decreasing, the momentum of the truck plus its contents is decreasing.

It makes sense if we consider the sand-filled-truck and the sand of
mass △m that is falling out as one system

Yes, if we include the sand falling out of the truck as the system, there is no change in the horizontal momentum of the truck plus total sand. But there is a change in vertical momentum (discussed at the end).

If we calculate the momentum of only the sand-filled-truck at two
instances in time, the momentum will definitely be smaller over time
since the mass of the truck is decreasing. So how is it that
$\frac{dp}{dt} = 0?$

Yes the momentum will be less and for that reason $\frac{dp}{dt}\lt0$ not $\frac{dp}{dt}=0$ . It is decreasing because although the velocity is constant the mass is decreasing.

Also, if we define our system to be only the sand-filled truck and not
include the mass of sand that leaks out, the equation will for this
system will be modelled by this differential equation $\frac{dp}{dt} =
> m\frac{dv}{dt} + v\frac{dm}{dt}$, and since mass is being loss from
the truck, $\frac{dm}{dt}$ is negative, and so shouldn't there be a
net external force acting on the truck?

As stated in the beginning of this answer, while it is necessary that the the net external force on a system be zero for conservation of momentum, it it is not sufficient. The mass of the system must also be constant.
Even if we consider the truck, its contents, and the leaking sand together as the system, total momentum  is still not conserved. Before the sand starts leaking, there is no vertical momentum in the system. Once leaking starts, there is an increase in vertical momentum. Vertical momentum is therefore not conserved.
In order for vertical momentum to be conserved we need to include the Earth in the system. Per Newton's third law the sand and Earth exert an equal and opposite force on one another. The sand  "pulls up" on the earth giving the earth momentum equal and opposite to the sand. The mass of the Earth being so much more than the sand, its upward velocity is too small to observe.
Finally there is the issue of what happens when the sand impacts the ground and stops moving forward. Once again, if we include the Earth in our system momentum will be conserved. In this case angular momentum is conserved. Before colliding with the Earth each particle of sand has angular momentum about the center of the Earth equal to the cross product of its linear momentum and the position vector of the particle measured from the center of the earth (assumes the Earth is a sphere of uniform density and both the particle and Earth are rigid bodies).  When it impacts the ground it angular momentum is transferred to the earth in order that angular momentum is conserved.
Hope this helps.
A: What you have not defined is a system of constant mass which in this case would be the truck, the sand inside the truck and the sand which has fallen out of the truck.
The momentum of that system does not change and hence no external forces act on the system.
If we calculate the momentum of only the sand-filled-truck at two instances in time, the momentum will definitely be smaller over time since the mass of the truck is decreasing. So how is it that $dp/dt=0$?
You are not considering a system of fixed mass so that you cannot use Newton's second law, $\vec F_{\rm external}= \dfrac {d\vec p}{dt}$.
The equation that you have mentioned, $\frac{dp}{dt} = m\frac{dv}{dt} + v\frac{dm}{dt}$, is not valid for such a system as explained in the answers to the post Second law of Newton for variable mass systems.
This chapter on continuous mass transfer might be worth reading?
A: 
If we calculate the momentum of only the sand-filled-truck at two instances in time, the momentum will definitely be smaller over time since the mass of the truck is decreasing. So how is it that dp/dt=0?

The momentum of the truck will decrease, yes, but the momentum of the sand that's not in the truck increases.  The total momentum of the system, $p=p_\mathrm{truck} + p_\mathrm{sand}$ is what is zero (we ignore gravity), which implies that $p_\mathrm{truck} = -p_\mathrm{sand}$.
You can also consider a thought experiment to approach this -- if you have a flying object and cut it in half mid-flight, if no forces accelerate each half, why would their speeds change?  The same can be argued here.

Also, if we define our system to be only the sand-filled truck and not include the mass of sand that leaks out, the equation will for this system will be modelled by this differential equation dp/dt=m dv/dt+v dm/dt, and since mass is being loss from the truck, dmdt is negative, and so shouldn't there be a net external force acting on the truck?

You ought to be careful here -- although it is true that the truck loses momentum (in the form of mass), you cannot apply the $F = v \dfrac{dm}{dt} + m\dfrac{dv}{dt}$ equation here.
So no, a force is not necessarily applied on the truck.  With this approach, all you're really doing is deciding to first consider the truck and sand as your system with some momentum $p_1$, and then a few seconds later are defining your system to be that truck but with less sand in it, so naturally it will have momentum $p_2<p_1$.  Of course, this should come as no surprise, since the "remaining" momentum is in the other part of the system (leaked sand) that you stopped considering as part of the truck momentum.
A: lets look at the equations
momentum at time t
$$p_t=(\Delta m_s+m_T(t))\,v$$
and at $~t+\Delta t$
$$p_{t+\Delta t}=(\Delta m_s+m_T(t))\,(v+\Delta v)$$
where $$m_T(t)=M_T+M_s+\underbrace{\frac{d M_s(t)}{dt}}_{\text{constant}}\,t$$
now the force $~F~$ is
$$F=\lim_{\Delta t\mapsto 0}=\frac{\Delta p}{\Delta t}$$
neglecting $~\Delta m_s\,\Delta v~$ you obtain
$$F=m_T(t)\frac{dv}{dt}\overset{!}{=}0\quad\text{your case}\quad\Rightarrow\\
\frac{dv}{dt}=0\quad ,v=\text{constant}$$

*

*$~T~$ for truck

*$~s~$ for sand

