Motion when mass varies

Question

I am reading text on motion when mass varies. The equation of motion comes out to be:

$$\frac{d}{dt}(mv) = F + V\frac{dm}{dt}$$

where $V$ is the velocity with which the incremented mass was moving.

Now I have two exercise questions:

1. Trailer full of sand is pulled by constant force F and sand is leaking at constant rate.

2. Snow slides off a (inclined) roof clearing away a part of uniform breadth. All snow slides at once.

What will be the value of $V$ in both cases.

In texts it gives:

1. V = v (velocity of trailer)
2. V = 0

I understand the first part that sand is initially moving with velocity of trailer. But I dont understand second part. The snow that is sliding off must have the velocity of snow at that instant(??). Please explain.

Edit: Here is the snippet:

And I will reframe the second question if I said it wrong. Quoting from the tests, "Snow slides off a roof clearing away a part of uniform breadth. If it all slide at once, find the time in which the roof will be cleared."

Here is snippet for that too:

Answer 1

The below answer was written before significant changes were made to the OP and may no longer be appropriate to its contents.

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What a mess. If this is independent study, you might want to consider obtaining a new textbook by a different author.

Temporarily forget all physics. Our variables are just variables. Go back to the product rule for derivatives:

$$(mv)' = mv'+ vm'$$

for any variables $m, v$.

In physics the function $(mv)'$ is called Force and represented with the variable $F$. We obtain the elementary physics form $F = mv'$ only if $m' = 0$. The equation in your textbook appears to define $F := mv'$ in the context of changing mass, which is wrong for physics (although strictly speaking you can define any variable however you like). Then it does the math wrong by introducing a different variable $V$.

$(mv)' = mv' + Vm'$ only if $V=v$, which makes introducing the new variable pointless and any answers other than $V=v$ false.

Your first scenario (sand leaving a truck accelerating under constant net force) should read:

$F = (mv') = mv' + vm'$

We then examine the scenario to determine what our terms actually represent, noting that the changing mass that is relevant to this equation must be the mass that the force $F$ is pushing on and is traveling at $v$. We find that it is the mass of the sand in the truck (which is decreasing at a constant rate) which satisfies these conditions.

Your second scenario is unphysical. Snow doesn't leave roofs all at once. If it did (perhaps a wizard zaps it out of existence) we'd have $m$ go from $m=m_0$ to $m=0$ in $\Delta t = 0$ hence $vm' \to -\infty$ with $F$ and $mv'$ undefined, which is unphysical. If we replace the scenario with a fixed quantity of snow sliding down a long section of otherwise bare roof in a solid block, then we just have a block sliding down an inclined plane.