A conveyor belt is driven at velocity v by a motor. Sand drops vertically on to the belt at a rate of m kg s-1.

Question 35

A conveyor belt is driven at velocity v by a motor. Sand drops vertically on to the belt at a rate of m kg s^-1.

What is the additional power needed to keep the conveyor belt moving at a steady speed when the sand starts to fall on it?

A ½ mv                       B mv               C ½ mv²                            D mv²

Reference: Past Exam Paper – June 2015 Paper 11 Q19

Solution:

Answer: D.

Power = Force × velocity

Force is the rate of change of momentum. So we first need to know force (rate of change of momentum) to find the power.

Sand drops vertically on to the belt at a rate of m kg s^–1.

Taking into account the units, it can be determined that the above represent.

Rate of drop of sand = Δmass / time = m kg s^–1

Consider the momentum.

Momentum p = mass × velocity

Rate of change of momentum = Δp / t = (Δmass × velocity) / t

{We want to manipulate the formula so as to obtain Δmass / time which is known to be equal to m in this question.}

Rate of change of momentum = (Δmass / t) × velocity) = mv

{NOTE that ‘m’ here is not the mass, but it represents the rate of drop of the sand.}

Rate of change of momentum = mv

In other words: in one second, the momentum of mass m of sand increases from zero to mv.

Recall that force is defined as the rate of change of momentum. So the force involved is mv.

Force = rate of change of momentum = mv

Power = Force × velocity = mv × v = mv²