Year 1 Mechanics · Chapter 1

Modelling in Mechanics

Specification: Edexcel Paper 3 §6.1 — Quantities and units in mechanics · AQA section P

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In this chapter

1. Mathematical models
2. Modelling assumptions
3. Quantities and units
4. The forces you’ll meet
5. Scalars, vectors and velocity
Practice

Key facts — this chapter

A mathematical model is a simplified version of a real situation. You make assumptions, solve, compare with reality, and refine if you need to.
The SI base units are the metre (m), the kilogram (kg) and the second (s). Every other unit is derived from them: velocity $\text{m s}^{-1}$ , acceleration $\text{m s}^{-2}$ , force $1\text{ N} = 1\text{ kg m s}^{-2}$ .
Weight is a force: $W = mg$ with $g = 9.8\text{ m s}^{-2}$ . Mass (kg) is not weight (N).
Particle: all the mass sits at one point, so size, shape and rotation are ignored. Light: mass ignored. Inextensible: doesn’t stretch. Smooth: no friction. Rigid: doesn’t bend. Uniform: mass spread evenly.
A scalar has size only (distance, speed, mass, time). A vector has size and direction (displacement, velocity, acceleration, force).
For a velocity $a\mathbf{i} + b\mathbf{j}$ , the speed is $\sqrt{a^2 + b^2}$ , and its direction comes from $\tan\theta = \dfrac{b}{a}$ .

Mechanics is the mathematics of things that move and the forces that move them. Real objects are messy: a football spins, deforms and drags air around it. Before any equation can say something useful, you have to decide what to ignore. That decision is called modelling, and this chapter is about making it well.

1. Mathematical models

A model keeps the features of a situation that matter most and deliberately throws away the rest. The throwing away is done by stating assumptions. A good model is simple enough to solve and accurate enough to be useful. No model is perfect.

Key result

The modelling cycle

Start from the real-world problem.
Make simplifying assumptions.
Write the mathematical model (a diagram and equations).
Solve it and make a prediction.
Compare the prediction with reality. If it’s good enough, use it. If not, refine the assumptions and go round again.

Exam questions test the last step as much as the others. Expect to be asked to criticise a model: say which assumption is least realistic and how it affects the answer. Then refine it: say what you would change.

2. Modelling assumptions

Mechanics has its own vocabulary of assumptions. Each word means something precise, and each one lets you leave something out of the maths.

Word	What it means	What it lets you ignore
Particle	the object’s mass is concentrated at a single point	size, shape and rotation; air resistance too, usually
Light (string, rod, pulley)	its mass is negligible	the weight of the string; tension is the same all the way along it
Inextensible (string)	it does not stretch	changes in length; objects joined by it move with the same speed and acceleration
Smooth (surface, pulley)	there is no friction	friction forces
Rough (surface)	there is friction	nothing: friction must be included
Rigid (rod, body)	it does not bend or deform	changes of shape
Uniform (rod, body)	its mass is evenly spread	uneven mass: the weight acts at the centre
Bead	a particle threaded on a wire or string	its size, and the wire’s thickness
Wire	a rigid, thin, light path an object moves along	its thickness
Peg	a fixed support a string passes over	its size; if smooth, friction too

Left: a crate on the floor, 1.2 metres wide, pulled by a rope that sags slightly. Right: the model, a single particle on the floor with four force arrows: reaction R up, weight W down, tension T forward and friction F backward — Modelling a crate as a particle on a light rope. Four forces are left, all acting at one point.

Modelling assumption

Treating the crate as a particle means ignoring its size, so it can’t tip or rotate, and every force acts at one point. Treating the rope as light means the rope has no weight, doesn’t sag, and pulls equally hard all along its length.

A crate is dragged across a floor by a rope. Switch each modelling assumption on and watch what it removes from the diagram, and from the maths.

Common mistake

An assumption is not a fact, so don’t write “the crate is a particle”. Write “model the crate as a particle”. When you’re asked to criticise a model, name one specific way the assumption fails for this object. “It isn’t realistic” earns nothing.

Worked example 1.1 · Stating and criticising assumptions

A child on a sledge is pulled across snow by a rope. The sledge and child are modelled as a particle, the rope as light and inextensible, and the snow as smooth. Explain what each assumption means here, and say which is the least realistic.

Particle: the child and sledge are treated as a single point, so their size, the way the weight is spread, and any rotation or tipping are ignored.

Light, inextensible rope: the rope’s weight is ignored, so the tension is the same all along it. It doesn’t stretch, so the sledge moves exactly as the end of the rope does.

Smooth snow: there is no friction between the sledge and the snow.

The least realistic is “smooth”. Snow does resist a sledge, and ignoring friction means the model will predict a larger acceleration than the real sledge has.

In context

Refine the model by treating the snow as rough and including a friction force.

3. Quantities and units

Every quantity in mechanics is a number with a unit. The SI system builds every unit from three base units.

Key result

Quantity	SI unit	Built from
length, displacement	metre, $\text{m}$	base unit
mass	kilogram, $\text{kg}$	base unit
time	second, $\text{s}$	base unit
speed, velocity	$\text{m s}^{-1}$	length ÷ time
acceleration	$\text{m s}^{-2}$	velocity ÷ time
force, weight	newton, $\text{N}$	$1\text{ N} = 1\text{ kg m s}^{-2}$ (mass × acceleration)

Exam papers write units as $\text{m s}^{-1}$ rather than m/s. The negative index means “per”, exactly as in Chapter 1 of Pure: $\text{s}^{-1} = \dfrac{1}{\text{s}}$ .

To convert, multiply by a fraction equal to 1, so the unwanted unit cancels.

Worked example 1.2 · Converting units

(a) Convert $72\text{ km h}^{-1}$ to $\text{m s}^{-1}$ . (b) The density of a material is $1.5\text{ g cm}^{-3}$ . Write this in $\text{kg m}^{-3}$ .

(a) There are $1000$ m in a km and $3600$ s in an hour:

72\text{ km h}^{-1} = \frac{72 \times 1000\text{ m}}{3600\text{ s}} = 20\text{ m s}^{-1}

(b) $1\text{ g} = 10^{-3}\text{ kg}$ and $1\text{ cm}^3 = (10^{-2}\text{ m})^3 = 10^{-6}\text{ m}^3$ :

1.5\text{ g cm}^{-3} = \frac{1.5 \times 10^{-3}\text{ kg}}{10^{-6}\text{ m}^3} = 1500\text{ kg m}^{-3}

Quick check: units as algebra

Units obey the index laws, so you can check a formula by checking its units. In $s = ut + \tfrac12 at^2$ , the term $ut$ has units $\text{m s}^{-1} \times \text{s} = \text{m}$ , and $at^2$ has units $\text{m s}^{-2} \times \text{s}^2 = \text{m}$ . Both match $s$ , so the formula is at least consistent. A formula whose terms have different units must be wrong.

Mass and weight

Mass measures how much matter there is (kg). Weight is the force of gravity on that mass (N), and it acts vertically downwards.

Key result

W = mg, \qquad g = 9.8\text{ m s}^{-2}

Use $g = 9.8\text{ m s}^{-2}$ unless a question says otherwise. An answer that uses $g$ should be given to 2 or 3 significant figures, because $g$ is only known to 2.

A 5 kg mass has weight $5 \times 9.8 = 49\text{ N}$ . A 750 g mass has weight $0.75 \times 9.8 = 7.35 \approx 7.4\text{ N}$ . Convert grams to kilograms first.

Common mistake

“Weighs 5 kg” is fine in a shop but wrong in mechanics: 5 kg is a mass, and its weight is 49 N. And $g$ is an acceleration, not a force. The force is $mg$ .

4. The forces you’ll meet

A force diagram shows every force acting on one object, drawn as arrows from that object. Each type of force has its own letter by convention, and knowing the types is most of the work of drawing one.

The forces of Year 1: weight and reaction, tension, thrust, and the resistances.

Weight $W$ (or $mg$ ): acts vertically downwards, from the centre of mass.
Normal reaction $R$ : a surface pushes back on an object resting on it. It acts at right angles to the surface.
Tension $T$ : a taut string or rope pulls on each object it’s attached to.
Thrust (compression): a rod pushes on the objects at its ends.
Friction $F$ : acts along a rough surface, opposing motion or any tendency to move.
Air resistance $D$ (or $R$ ): opposes motion through the air.

Chapter 3 of this course turns force diagrams into equations with Newton’s laws. For now, the skill is to spot every force and to draw each one in the right direction.

5. Scalars, vectors and velocity

A scalar has size only. A vector has size and direction. Several mechanics words come in pairs, one of each:

Scalar (size only)	Vector (size and direction)
distance	displacement
speed	velocity
mass	weight (a force)
time	acceleration

A runner who completes one 400 m lap has run a distance of 400 m, but their displacement is zero: they finish where they started.

Vectors in mechanics use the $\mathbf{i}$ , $\mathbf{j}$ notation from Chapter 15 of Pure. $\mathbf{i}$ usually points east (or right) and $\mathbf{j}$ north (or up).

Key result

For a velocity $\mathbf{v} = a\mathbf{i} + b\mathbf{j}$ :

\text{speed} = |\mathbf{v}| = \sqrt{a^2 + b^2}, \qquad \tan\theta = \frac{b}{a}

where $\theta$ is the angle $\mathbf{v}$ makes with $\mathbf{i}$ . A body moving with constant velocity $\mathbf{v}$ from position $\mathbf{r}_0$ is at $\mathbf{r} = \mathbf{r}_0 + \mathbf{v}t$ after time $t$ .

The velocity vector 6i + 8j drawn from the origin, with dashed components 6 across and 8 up, the angle theta to i marked, and speed 10 metres per second — Velocity 6i + 8j m s⁻¹: speed √(6² + 8²) = 10 m s⁻¹, at 53.1° to i.

Worked example 1.3 · Speed and direction

A particle moves with velocity $(5\mathbf{i} - 12\mathbf{j})\text{ m s}^{-1}$ . Find its speed, and the angle its direction of motion makes with $\mathbf{i}$ .

The speed is the magnitude of the velocity:

|\mathbf{v}| = \sqrt{5^2 + (-12)^2} = \sqrt{169} = 13\text{ m s}^{-1}

Sketch it: 5 right, 12 down. The angle below $\mathbf{i}$ satisfies

\tan\theta = \frac{12}{5} \quad\Rightarrow\quad \theta = 67.4^\circ

In context

The particle moves at $13\text{ m s}^{-1}$ , in a direction $67.4^\circ$ below the $\mathbf{i}$ direction.

Worked example 1.4 · Bearings and position

A boat leaves a harbour $O$ with constant velocity $(3\mathbf{i} + 4\mathbf{j})\text{ km h}^{-1}$ , where $\mathbf{i}$ points east and $\mathbf{j}$ north. Find (a) its speed, (b) its bearing, and (c) its position vector after 4 hours if it started at $(1\mathbf{i} - 2\mathbf{j})$ km from $O$ .

(a) $\sqrt{3^2 + 4^2} = 5\text{ km h}^{-1}$ .

(b) A bearing is measured clockwise from north, so use the angle between the velocity and $\mathbf{j}$ . It is $3$ east for every $4$ north:

\tan\alpha = \frac{3}{4} \quad\Rightarrow\quad \alpha = 36.9^\circ, \text{ a bearing of } 036.9^\circ

(c) $\mathbf{r} = \mathbf{r}_0 + \mathbf{v}t = (\mathbf{i} - 2\mathbf{j}) + 4(3\mathbf{i} + 4\mathbf{j}) = (13\mathbf{i} + 14\mathbf{j})\text{ km}$ .

Average velocity

If a body moves from $A$ to $B$ in time $t$ , its average velocity is $\dfrac{\overrightarrow{AB}}{t}$ and its average speed is $\dfrac{\text{distance travelled}}{t}$ . For a straight journey from $(2, 1)$ to $(14, 6)$ in 4 s, the displacement is $12\mathbf{i} + 5\mathbf{j}$ . That gives an average velocity of $(3\mathbf{i} + 1.25\mathbf{j})\text{ m s}^{-1}$ and an average speed of $\dfrac{13}{4} = 3.25\text{ m s}^{-1}$ . On a curved or out-and-back journey, the two can be very different, as the 400 m lap shows.

Practice

Twelve questions in three tiers. Try each one properly before opening the solution. The solutions show the full method, not just answers. Take $g = 9.8\text{ m s}^{-2}$ throughout.

Warm-up

1 [3 marks]

Convert:

$54\text{ km h}^{-1}$ to $\text{m s}^{-1}$
$450\text{ g}$ to $\text{kg}$
$6\text{ m s}^{-1}$ to $\text{km h}^{-1}$

Show solution

(a) $\dfrac{54 \times 1000}{3600} = 15\text{ m s}^{-1}$ .

(b) $450 \div 1000 = 0.45\text{ kg}$ .

(c) In one hour the body goes $6 \times 3600 = 21\,600$ m $= 21.6$ km, so the speed is $21.6\text{ km h}^{-1}$ .

2 [3 marks]

Find the weight of a $3\text{ kg}$ mass.
Find the weight of a $250\text{ g}$ mass.
A box has weight $196\text{ N}$ . Find its mass.

Show solution

(a) $W = mg = 3 \times 9.8 = 29.4\text{ N}$ .

(b) Convert to kilograms first: $W = 0.25 \times 9.8 = 2.45\text{ N}$ .

(c) $m = \dfrac{W}{g} = \dfrac{196}{9.8} = 20\text{ kg}$ .

3 [3 marks]

A particle hangs at rest on the end of a string. The string is modelled as light and inextensible. Explain what each of these words means, and what each one allows you to ignore.

Show solution

Light: the string’s mass is negligible. You can ignore its weight, so the tension is the same all the way along it.

Inextensible: the string does not stretch. You can ignore any change in its length, so anything attached to it moves exactly as its end does.

4 [3 marks]

A particle moves with velocity $(-5\mathbf{i} + 12\mathbf{j})\text{ m s}^{-1}$ . Find its speed, and the angle between its direction of motion and the vector $\mathbf{i}$ .

Show solution

Speed $= \sqrt{(-5)^2 + 12^2} = \sqrt{169} = 13\text{ m s}^{-1}$ .

Sketch it: 5 left, 12 up. The acute angle with $-\mathbf{i}$ is $\tan^{-1}\dfrac{12}{5} = 67.4^\circ$ , so the angle with $\mathbf{i}$ is $180^\circ - 67.4^\circ = 112.6^\circ$ .

Standard

5 [4 marks]

An athlete runs exactly one lap of a $400\text{ m}$ track in $80\text{ s}$ , finishing where she started.

State the distance she ran and her displacement.
Find her average speed and her average velocity.
Use your answers to explain the difference between a scalar and a vector.

Show solution

(a) Distance $= 400$ m. Displacement $= 0$ m, because she ends where she began.

(b) Average speed $= \dfrac{400}{80} = 5\text{ m s}^{-1}$ . Average velocity $= \dfrac{\text{displacement}}{\text{time}} = \dfrac{0}{80} = 0\text{ m s}^{-1}$ .

(c) Distance and speed are scalars. They measure how much, whatever the direction, so every metre counts. Displacement and velocity are vectors. They account for direction, so running back to the start undoes the outward journey.

6 [4 marks]

A particle $P$ starts at the point with position vector $(-2\mathbf{i} + 5\mathbf{j})$ m and moves with constant velocity $(3\mathbf{i} - 2\mathbf{j})\text{ m s}^{-1}$ .

Find the position vector of $P$ after 6 seconds.
Find the distance of $P$ from the origin at that time.

Show solution

(a) $\mathbf{r} = \mathbf{r}_0 + \mathbf{v}t = (-2\mathbf{i} + 5\mathbf{j}) + 6(3\mathbf{i} - 2\mathbf{j}) = (16\mathbf{i} - 7\mathbf{j})$ m.

(b) $|\mathbf{r}| = \sqrt{16^2 + (-7)^2} = \sqrt{305} = 17.5$ m (3 s.f.).

7 [5 marks]

A crate is dragged across a warehouse floor by a rope. In a first model, the crate is a particle, the rope is light, and the floor is smooth.

Draw a force diagram for the crate in this model, naming each force.
State which assumption is least realistic, and the effect it has on the acceleration the model predicts.
Suggest how the model could be refined.

Show solution

(a) Three forces act at a single point: the weight $W$ vertically down, the normal reaction $R$ vertically up, and the tension $T$ along the rope. There is no friction arrow, because the floor is smooth.

(b) “Smooth” is least realistic. A crate on a warehouse floor meets significant friction. With the resistance to motion left out, the model predicts a larger acceleration than the crate really has.

(c) Model the floor as rough, and include a friction force $F$ opposing the motion.

In context

With friction included, the model’s predicted acceleration drops towards what is actually observed.

8 [4 marks]

Using $1\text{ N} = 1\text{ kg m s}^{-2}$ :

Momentum is mass × velocity. Find its SI units.
Work done is force × distance. Write its units in terms of kg, m and s.
Show that every term in $v^2 = u^2 + 2as$ has the same units, where $u$ and $v$ are velocities, $a$ an acceleration and $s$ a displacement.

Show solution

(a) $\text{kg} \times \text{m s}^{-1} = \text{kg m s}^{-1}$ .

(b) $\text{N} \times \text{m} = \text{kg m s}^{-2} \times \text{m} = \text{kg m}^2\text{ s}^{-2}$ .

(c) $v^2$ and $u^2$ have units $(\text{m s}^{-1})^2 = \text{m}^2\text{ s}^{-2}$ . The term $2as$ has units $\text{m s}^{-2} \times \text{m} = \text{m}^2\text{ s}^{-2}$ (the 2 is a pure number). All three terms match. ∎

Exam style

9 [6 marks]

In this question $\mathbf{i}$ points due east and $\mathbf{j}$ due north. A boat leaves a harbour $O$ and moves with constant velocity $(4\mathbf{i} + 3\mathbf{j})\text{ km h}^{-1}$ .

Find the speed of the boat.
Find the bearing on which the boat moves.
Find how far the boat is from $O$ after 2 hours 30 minutes.

Show solution

(a) $\sqrt{4^2 + 3^2} = 5\text{ km h}^{-1}$ .

(b) The bearing is measured clockwise from north. The boat goes 4 east for every 3 north, so the angle $\alpha$ from north satisfies $\tan\alpha = \dfrac{4}{3}$ , giving $\alpha = 53.1^\circ$ . The bearing is $053^\circ$ (to the nearest degree).

(c) After $2.5$ hours, $\mathbf{r} = 2.5(4\mathbf{i} + 3\mathbf{j}) = (10\mathbf{i} + 7.5\mathbf{j})$ km, so the distance is $\sqrt{10^2 + 7.5^2} = 12.5$ km.

(Or: the speed is constant at $5\text{ km h}^{-1}$ along a straight line, so the distance is $5 \times 2.5 = 12.5$ km.)

10 [6 marks]

At time $t = 0$ , particle $A$ is at $(2\mathbf{i} + \mathbf{j})$ m and moves with constant velocity $(3\mathbf{i} + 2\mathbf{j})\text{ m s}^{-1}$ . At the same moment, particle $B$ is at $(10\mathbf{i} - 7\mathbf{j})$ m and moves with constant velocity $(\mathbf{i} + 4\mathbf{j})\text{ m s}^{-1}$ .

Write the position vectors of $A$ and $B$ at time $t$ seconds.
Show that $A$ and $B$ collide, and find the time and the position of the collision.

Show solution

(a) $\mathbf{r}_A = (2 + 3t)\mathbf{i} + (1 + 2t)\mathbf{j}$ and $\mathbf{r}_B = (10 + t)\mathbf{i} + (-7 + 4t)\mathbf{j}$ .

(b) They collide if both components are equal at the same time.

$\mathbf{i}$ : $2 + 3t = 10 + t \Rightarrow t = 4$ .

$\mathbf{j}$ : $1 + 2t = -7 + 4t \Rightarrow t = 4$ .

Both components give $t = 4$ , so the particles are at the same point at the same time.

In context

They collide after 4 seconds, at the point with position vector $(14\mathbf{i} + 9\mathbf{j})$ m.

11 [6 marks]

A stone is dropped from rest from the top of a cliff $45$ m high. A student models the stone as a particle falling freely under gravity, which gives a time to reach the sea of $t = \sqrt{\dfrac{2h}{g}}$ , where $h$ is the height fallen.

Find the time the model predicts.
The stone actually takes $3.4$ s. Comment on the model, naming the assumption most likely to be responsible.
Another student uses $g = 10\text{ m s}^{-2}$ . Find their prediction, and explain why changing $g$ is not a sensible way to improve the model.

Show solution

(a) $t = \sqrt{\dfrac{2 \times 45}{9.8}} = \sqrt{9.18\ldots} = 3.03$ s (3 s.f.).

(b) The model predicts a shorter time than was observed, so it underestimates the time by about 0.4 s. The most likely cause is ignoring air resistance: treating the stone as a particle falling freely leaves out the drag that slows it down. A refined model would include a resistance force.

(c) $t = \sqrt{\dfrac{90}{10}} = 3$ s. This moves the prediction the wrong way, further from 3.4 s. More importantly, $g$ is a measured physical constant, not a number to tune. The mismatch comes from a missing force, so that is what the refinement should add.

12 [7 marks]

A particle $P$ moves with constant velocity. At $t = 0$ it is at the point $A$ with position vector $(-3\mathbf{i} + 4\mathbf{j})$ m. At $t = 5$ s it is at $B$ , with position vector $(12\mathbf{i} - 4\mathbf{j})$ m.

Find the velocity of $P$ .
Find the speed of $P$ .
Find the angle between the direction of motion and the vector $\mathbf{i}$ .
Find the time at which $P$ is due east of the origin (i.e. its $\mathbf{j}$ -component is zero), and its position then.

Show solution

(a) $\mathbf{v} = \dfrac{\overrightarrow{AB}}{5} = \dfrac{(12\mathbf{i} - 4\mathbf{j}) - (-3\mathbf{i} + 4\mathbf{j})}{5} = \dfrac{15\mathbf{i} - 8\mathbf{j}}{5} = (3\mathbf{i} - 1.6\mathbf{j})\text{ m s}^{-1}$ .

(b) $\sqrt{3^2 + 1.6^2} = \sqrt{11.56} = 3.4\text{ m s}^{-1}$ .

(c) The velocity points 3 right and 1.6 down, so the angle below $\mathbf{i}$ is $\tan^{-1}\dfrac{1.6}{3} = 28.1^\circ$ .

(d) $\mathbf{r} = (-3 + 3t)\mathbf{i} + (4 - 1.6t)\mathbf{j}$ . Set the $\mathbf{j}$ -component to zero: $4 - 1.6t = 0 \Rightarrow t = 2.5$ s. Then the $\mathbf{i}$ -component is $-3 + 3(2.5) = 4.5$ .

In context

$P$ is due east of the origin after 2.5 s, at $4.5\mathbf{i}$ m, i.e. 4.5 m east of $O$ .