Question

Sand is being dropped on a conveyor belt at the rate of $M\text{ }kg.{{s}^{-1}}$. The force necessary to keep the belt moving with a constant velocity of $v\text{ }m.{{s}^{-1}}$ will be
A. $Mv\text{ }newton$
B. $2Mv\text{ }newton$
C. $\dfrac{Mv}{2}\text{ }newton$
D. $Zero$

Answer 1

Hint: This problem can be solved by making use of the fact that the external force acting on a system is the rate of change of its momentum. The momentum of a body is the product of its mass and velocity. We can consider the conveyor belt and the sand present on it as the system at each instant. The mass of the system keeps on changing as more and more sand falls on it, and hence the momentum also changes, which will give a value for the external force.

Formula used:
$\overrightarrow{{{F}_{ext}}}=\dfrac{d\overrightarrow{p}}{dt}$
where $\overrightarrow{{{F}_{ext}}}$ is the external force on the system, $\overrightarrow{p}$ is the momentum of the system and $t$ is the time interval for which the force acts.
$\overrightarrow{p}=m\overrightarrow{v}$
where, $\overrightarrow{p}$ is the momentum of the system, $m$ is its mass and $\overrightarrow{v}$ is its velocity.

Complete step by step answer:
We can solve this problem by considering the conveyor belt and the sand on it at a specific time as the system. Then, by finding out the change in momentum of the system, we can find the force required. This is because external force is the rate of change of momentum of a system.
$\overrightarrow{{{F}_{ext}}}=\dfrac{d\overrightarrow{p}}{dt}$ --(1)
where $\overrightarrow{{{F}_{ext}}}$ is the external force on the system, $\overrightarrow{p}$ is the momentum of the system and $t$ is the time interval for which the force acts.
$\overrightarrow{p}=m\overrightarrow{v}$ --(2)
where, $\overrightarrow{p}$ is the momentum of the system, $m$ is its mass and $\overrightarrow{v}$ is its velocity.
Let us analyze the question and only deal with the magnitudes of the vectors.
Let the mass of the ‘conveyor belt and sand on it’ system be $m$ at a specific point of time and it is moving with velocity $v$.
Hence, using (2), we get, the momentum of the system $\left( p \right)$ as
$p=mv$ --(3)
Now, the force required $\left( F \right)$ can be found out using (1). Therefore, using (1), we get,
$F=\dfrac{dp}{dt}$
$\therefore F=\dfrac{d\left( mv \right)}{dt}=m\dfrac{dv}{dt}+v\dfrac{dm}{dt}$ ---(4) $\left( \because d\left( xy \right)=xdy+ydx \right)$
Now, the velocity of the system remains constant at a given $v\text{ }m.{{s}^{-1}}$,
$\therefore \dfrac{dv}{dt}=0$ --(5) $\left( \because d\left( \text{constant} \right)=0 \right)$
It is also given that sand falls on the conveyor belt at $M\text{ }kg.{{s}^{-1}}$. Hence, the mass of the system will increase.
$\therefore \dfrac{dm}{dt}=M$ --(6)
Using (5) and (6) in (4), we get,
$F=m\left( 0 \right)+vM=0+Mv=Mv$
Hence, the force necessary is $Mv\text{ }newton$.
Therefore, the correct option is A) $Mv\text{ }newton$.
The direction of this force is the same as the direction of the velocity $v$, since an opposing force would have decreased the velocity, not helped it remain constant.

Note: Students often forget that the change in momentum can come due to the change in mass of the body also because it is not very intuitive. However, it is a concept that is heavily used in advanced problems. In fact, rocket propulsion is based upon this concept where there is a change in the mass of the rocket due to the fuel being burnt, which provides a momentum change and resultant thrust for the rocket to travel.
Whenever, there is a problem with force and variable mass, it is best to proceed in the way above, and apply the formula relating the external force to the rate of change of momentum of the body.