# Will the momentum be conserved in this scenario?

I faced a problem:

A truck filled with sand has a mass of $6*10^3$ kg. It's moving at a constant speed of 20 $kms^-1$. If at one moment, sand starts to fall through a hole normal to the motion of the truck from the truck with rate 5 kg/s(due to a downward gravitational force); what would be the force needed to keep the truck at same constant initial speed? Neglect air drag, friction etc.

I am only concerned with if the momentum of the truck in the direction of it's motionn would conserve.

My attempt: No, it won't. Because conservation of momentum only works for closed systems. The sand pouring out and the object creating the gravitational field are in a closed system (not in terms of invaraint mass, but in terms of no external force). But, all other sand particle and the truck is out of this closed system. So, according to inertia of motion, all other particles will keep up the same velocity. So, necessary force $0N$

Feel free to point out my mistakes.

• If the sand is falling down while the truck is moving sideways, and there are no frictional (nonconservative) forces acting on the truck, there will be no impact on the velocity of the truck. The momentum of the truck/sand system will decrease due to a loss of mass. In addition, at a constant speed, NO NET FORCE is acting on the truck. So, the questions are: 1) what direction is the sand falling? 2) are there any nonconservative forces (e.g, drag or friction forces) acting on the truck? – David White Jun 20 '17 at 14:09
• @Davif White See my edit. – Mockingbird Jun 20 '17 at 14:17

If you just mean the momentum of the truck, then yes, that's conserved. There's no net forces acting on it so whatever. This seems closest to the question being asked, which is a question about what force is needed on the truck.

If you mean the momentum of the truck plus the sand it happens to be carrying, then no, that's not conserved, because the amount of sand it's carrying happens to be decreasing.

If you mean the momentum of the truck plus the sand it originally carried, then the answer is "probably not, but this technically depends on things that you have not yet told us." So for example the truck could be moving alongside a conveyor belt that moves at exactly the same speed, and the sand could be passed out from a chute onto that conveyor belt. If we assume that these sorts of shenanigans haven't happened yet then yes, when the sand meets the road there will be a net force on it which will change its momentum.

If you mean the momentum of the truck plus the sand it originally carried plus Earth and everything on it, then we again come back to "yes, that's conserved."

The key take-away from all of this is that when you want to ask about "is the momentum conserved?" you have to be really, really clear about what collection of objects you are evaluating the momentum of.

The easy way to see the very first result is just to think about a grain of sand the very instant after it has fallen. It was moving forward with velocity vector $\vec v = [v_0, 0]$ when it was stored in the truck, and the instant after it falls, now it has picked up some small downward speed, $\vec v = [v_0, -\epsilon].$ Because it has not changed in its horizontal speed, no force was needed to speed it up or slow it down, so it cannot put a third-law reaction force on the car. After this instant, it is totally disconnected from the car and cannot put a third-law force on the car. So this force is 0.

But you can also imagine, I'm sure, a little nozzle where we let a tank of fluid spray out of it, and we could aim this little nozzle "backwards" or "forwards" from the truck, having some velocity $[v_0 \pm u, -\epsilon]$. If we aim it straight down we get the same result of course; but we could use it to accelerate the truck (pointing it backwards) or decelerate the truck (pointing it forwards). Sand will do the same thing if it falls out of a chute pointing one way or the other. So that phrasing "a hole normal to the motion of the truck" is very important, too.