Quantities, Units and Modelling – A Level Maths Year 1

Mechanics begins with measurement (DfE content area P · Quantities and units in mechanics). Every quantity you meet is built from just three SI base units.

Quantity	SI unit	Symbol
Length (distance, displacement)	metre	m
Time	second	s
Mass	kilogram	kg

If every value in a calculation is in SI units, the answer comes out in SI units automatically. So the safest first step in any mechanics problem is to convert everything to metres, seconds and kilograms.

The kilogram is the base unit, not the gram. A mass of $500$ g must be written as $0.5$ kg before you use it in $F = ma$.

Other units are derived by combining the base units in the same way as the quantities combine. A Level notation uses negative indices rather than a slash: m s$^{-1}$ rather than m/s.

Quantity	Definition	Unit
Velocity / speed	displacement ÷ time	m s$^{-1}$
Acceleration	change in velocity ÷ time	m s$^{-2}$
Force	mass × acceleration	newton, N $=$ kg m s$^{-2}$
Weight	mass × $g$	N

The newton

$1\ \text{N}$ is the force that gives a mass of $1$ kg an acceleration of $1$ m s$^{-2}$: $1\ \text{N} = 1\ \text{kg m s}^{-2}$

Worked Example 1 — Converting speeds

Convert (a) $108$ km h$^{-1}$ and (b) $45$ km h$^{-1}$ to m s$^{-1}$.

①$1$ km h$^{-1}$ $= \dfrac{1000\ \text{m}}{3600\ \text{s}} = \dfrac{1}{3.6}$ m s$^{-1}$.

②(a) $108 \div 3.6 = 30$ m s$^{-1}$

③(b) $45 \div 3.6 = 12.5$ m s$^{-1}$

Sense check: m s$^{-1}$ values are always smaller than the km h$^{-1}$ values for the same speed — a car at motorway speed ($\approx 110$ km h$^{-1}$) travels about $30$ m every second.

Worked Example 2 — Checking units of a formula

Show that $\tfrac12 a t^2$ has units of length, where $a$ is an acceleration and $t$ is a time.

①Units of $a t^2$: $\text{m s}^{-2} \times \text{s}^2 = \text{m}$.

②The $\tfrac12$ is a pure number with no units, so $\tfrac12 at^2$ is measured in metres — a length ✓

Checking units like this is a quick way to spot an error in a rearranged formula: every term in an equation must have the same units.

A scalar has size (magnitude) only. A vector has both magnitude and direction. Many mechanics words come in scalar–vector pairs.

Scalar	Vector partner	Difference
Distance	Displacement	Displacement is distance in a stated direction from a fixed origin; it can be negative.
Speed	Velocity	Speed is the magnitude of velocity; it is never negative.
Mass	Weight	Mass (kg) is the amount of matter; weight (N) is the gravitational force on it, acting downwards.
Time	—	Always a scalar.
—	Acceleration, force	Both are vectors.

In one dimension, direction is shown by a sign: choose a positive direction and stick to it. In two dimensions, use $\mathbf{i}$ and $\mathbf{j}$ components (see Vectors in Two Dimensions).

Worked Example 3 — Distance and displacement

A runner goes $300$ m east along a straight road, then $500$ m west. Taking east as positive, find her final displacement and the total distance travelled.

①Displacement $= +300 - 500 = -200$ m, i.e. $200$ m west of the start.

②Distance $= 300 + 500 = 800$ m.

Worked Example 4 — Speed from a velocity vector

A boat has velocity $(6\mathbf{i} - 8\mathbf{j})$ m s$^{-1}$. Find its speed.

①Speed $= |6\mathbf{i} - 8\mathbf{j}| = \sqrt{6^2 + (-8)^2} = \sqrt{100} = 10$ m s$^{-1}$.

②The direction is $\tan^{-1}\!\left(\tfrac{8}{6}\right) = 53.1^\circ$ below the $\mathbf{i}$ direction.

Weight

$W = mg$, with $g = 9.8$ m s$^{-2}$ unless a question says otherwise

Near the Earth's surface every object in free fall (ignoring air resistance) accelerates downwards at the same rate, $g$. The value varies slightly with location; A Level questions use $g = 9.8$ m s$^{-2}$. Because $g$ is given to 2 significant figures, final answers that use it should normally be given to 2 or 3 significant figures.

Worked Example 5 — Weight and mass

(a) Find the weight of a $65$ kg student. (b) A crate has weight $11\,760$ N. Find its mass.

①(a) $W = 65 \times 9.8 = 637$ N.

②(b) $m = \dfrac{W}{g} = \dfrac{11\,760}{9.8} = 1200$ kg.

Do not say a student "weighs $65$ kg". In mechanics, $65$ kg is the mass; the weight is a force measured in newtons.

Real objects are complicated. A mathematical model simplifies them so the maths is manageable. Each simplifying word has a precise meaning and a precise consequence.

Word	Meaning	Consequence in the model
Particle	The object's size is negligible; all its mass is at one point.	No rotation; air resistance often ignored; forces all act at one point.
Light (string, rod, pulley)	Its mass is negligible.	Tension is the same throughout a light string; a light object has no weight.
Inextensible (string)	It does not stretch.	Objects joined by it have the same speed and the same magnitude of acceleration.
Smooth (surface, pulley)	There is no friction.	No frictional force; tension is the same on both sides of a smooth pulley.
Rough	Friction acts.	A frictional force opposes motion (or the tendency to move).
Rigid (rod, body)	It does not bend or deform.	Its shape and length stay fixed.
Uniform	Its mass is evenly spread.	Its weight acts at its geometric centre.
Lamina, wire, bead, peg	Flat object; one-dimensional object; particle threaded on a wire; fixed support.	Thickness / width ignored; bead moves along the wire only.

Worked Example 6 — Stating and criticising assumptions

A cricket ball is hit and its path is modelled using $g = 9.8$ m s$^{-2}$ and no other forces. (a) State two modelling assumptions. (b) Suggest one way the model could be improved.

①(a) The ball is modelled as a particle (so it does not spin); air resistance is negligible.

②(b) Include air resistance, which would reduce the range and maximum height. (Or: allow for the spin of the ball, or a more accurate value of $g$.)

Make improvements specific to the context. "Make it more realistic" earns nothing; "include air resistance, which will slow the ball" earns the mark.

Set up a model: draw a diagram, list assumptions, choose axes and a positive direction.
Solve the mathematical problem.
Interpret the answer in context, with units and sensible accuracy.
Evaluate: is the answer reasonable? Which assumption is weakest? Refine the model if needed.

Base units

m, s, kg.

Derived units

m s$^{-1}$, m s$^{-2}$, N $=$ kg m s$^{-2}$.

km h$^{-1}$ to m s$^{-1}$

Divide by $3.6$.

Weight

$W = mg$, $g = 9.8$ m s$^{-2}$.

Scalars

Distance, speed, mass, time.

Vectors

Displacement, velocity, acceleration, force, weight.

Particle

No size: no rotation, all forces at a point.

Light, inextensible string

Same tension throughout; connected objects share the same acceleration.

Smooth

No friction.

Take $g = 9.8$ m s$^{-2}$ throughout.

Question 1

Write down the SI units of (a) velocity, (b) acceleration, (c) force, in base units.

▶ Show solution

(a) m s$^{-1}$ (b) m s$^{-2}$ (c) kg m s$^{-2}$ (the newton)

Question 2

Convert $72$ km h$^{-1}$ to m s$^{-1}$, and $15$ m s$^{-1}$ to km h$^{-1}$.

▶ Show solution

$72 \div 3.6 = 20$ m s$^{-1}$.

$15 \times 3.6 = 54$ km h$^{-1}$.

Question 3

Classify each as a scalar or a vector: speed, weight, mass, displacement, time, acceleration.

▶ Show solution

Scalars: speed, mass, time.

Vectors: weight, displacement, acceleration.

Question 4

Find the weight of a bag of flour of mass $1.5$ kg, and the mass of a box of weight $49$ N.

▶ Show solution

$W = 1.5 \times 9.8 = 14.7$ N.

$m = 49 \div 9.8 = 5$ kg.

Question 5

A particle has velocity $(-2\mathbf{i} + 5\mathbf{j})$ m s$^{-1}$. Find its speed, and the angle its direction of motion makes with the $\mathbf{j}$ direction.

▶ Show solution

Speed $= \sqrt{(-2)^2 + 5^2} = \sqrt{29} = 5.39$ m s$^{-1}$ (3 s.f.).

The velocity has $5$ units along $\mathbf{j}$ and $2$ units against $\mathbf{i}$, so the angle with $\mathbf{j}$ is $\tan^{-1}\!\left(\tfrac{2}{5}\right) = 21.8^\circ$.

Question 6

A bus travels $4$ km north, then $1.5$ km south. Taking north as positive, find its displacement and the distance it has travelled.

▶ Show solution

Displacement $= 4 - 1.5 = 2.5$ km, i.e. $2500$ m north.

Distance $= 4 + 1.5 = 5.5$ km $= 5500$ m.

Question 7

Explain what is meant by modelling a string as (a) light, (b) inextensible, and give the consequence of each.

▶ Show solution

(a) Light: the mass of the string is negligible, so the tension is the same all the way along it.

(b) Inextensible: the string does not stretch, so particles joined by it move with the same speed and acceleration (in magnitude).

Question 8

A student writes the formula $s = ut + \tfrac12 a t$ for displacement. Use units to show the formula must be wrong.

▶ Show solution

$ut$ has units m s$^{-1}$ × s $=$ m ✓.

$at$ has units m s$^{-2}$ × s $=$ m s$^{-1}$, a velocity, not a length.

You cannot add a length to a velocity, so the formula is wrong. (The correct term is $\tfrac12 at^2$.)

Question 9

A density of $2.7$ g cm$^{-3}$ is given for aluminium. Convert this to kg m$^{-3}$.

▶ Show solution

$1$ g $= 10^{-3}$ kg and $1$ cm$^3$ $= (10^{-2})^3$ m$^3$ $= 10^{-6}$ m$^3$.

$2.7$ g cm$^{-3}$ $= \dfrac{2.7 \times 10^{-3}}{10^{-6}}$ kg m$^{-3}$ $= 2700$ kg m$^{-3}$.

Question 10

A child of mass $25$ kg slides down a playground slide. The motion is modelled with the child as a particle on a smooth, straight, rigid slide.

(a) Find the weight of the child. (b) Explain what each of the words particle, smooth and rigid means here. (c) The model predicts a speed at the bottom of $7.7$ m s$^{-1}$, but the child is measured at $5.2$ m s$^{-1}$. Suggest which assumption is most responsible and how to refine the model.

▶ Show solution

(a) $W = 25 \times 9.8 = 245$ N.

(b) Particle: the child's size is ignored, so all forces act at one point and there is no rotation. Smooth: there is no friction between the child and the slide. Rigid: the slide does not bend under the child's weight.

(c) The measured speed is much lower, so a resistive force has been ignored. The "smooth" assumption is most responsible: in reality friction (and some air resistance) acts. Refine the model by treating the slide as rough and including a frictional force.