M1.1 Modelling and vectors in mechanics
⤓ Presentation⤓ PowerPointPresentation: opens offline in any browser, simulations live. PowerPoint: editable.
Mechanics 1 (AS), specification sections M1 1.1, 2.1 and 2.2. Straight-line kinematics, friction, connected particles and moments have their own M1 pages; closest approach by differentiation belongs to M2.
What this page covers
6 points
- Modelling vocabulary (particle, lamina, rigid body, rod, light, inextensible, smooth, rough, light smooth pulley, bead, wire, peg) and what each assumption ignores
- Magnitude and direction of a vector, the angle with \(\mathbf{i}\), bearings and unit vectors
- Resultants, resolving, vector diagrams and parallel vectors
- Position, displacement and velocity: \(\mathbf{r} = \mathbf{r}_0 + \mathbf{v}t\) for constant velocity
- Constant acceleration in vector form, and \(\mathbf{F} = m\mathbf{a}\) with forces written as vectors
- When one object is due north or due east of another, whether two objects collide, and their least distance apart (without calculus)
In one minute
6 points
- A mechanical model simplifies a real situation with stated assumptions, and vectors in \(\mathbf{i}, \mathbf{j}\) form keep track of size and direction at once, so a ship, a puck or a drone becomes a point moving according to \(\mathbf{r} = \mathbf{r}_0 + \mathbf{v}t\).
- Each modelling word is a promise: particle (ignore size and rotation), light (ignore mass), inextensible (no stretching), smooth (no friction).
- For \(\mathbf{v} = a\mathbf{i} + b\mathbf{j}\): \(|\mathbf{v}| = \sqrt{a^2 + b^2}\); direction from \(\tan\theta = \frac{b}{a}\), with a sketch to fix the quadrant. With \(\mathbf{i}\) east and \(\mathbf{j}\) north, a bearing is measured clockwise from \(\mathbf{j}\).
- Add vectors by adding components. Two vectors are parallel when one is a multiple of the other.
- Constant velocity: \(\mathbf{r} = \mathbf{r}_0 + \mathbf{v}t\). Constant acceleration: \(\mathbf{v} = \mathbf{u} + \mathbf{a}t\), \(\mathbf{r} = \mathbf{r}_0 + \mathbf{u}t + \frac12\mathbf{a}t^2\). Resultant force: \(\mathbf{F} = m\mathbf{a}\).
- Due north of a point: equal \(\mathbf{i}\)-components. Due east: equal \(\mathbf{j}\)-components. Collide: both components equal at the same time.
Contents
12 sections
- What is a mathematical model in mechanics?
- How do you find the magnitude and direction of a vector?
- How do you add and resolve vectors?
- How do you describe motion with constant velocity?
- How do you use F = ma and constant acceleration with vectors?
- When are two moving objects due north, colliding or closest?
- Exam tips for M1.1
- Common misconceptions
- How this topic connects
- Key formulae
- Check your understanding
- Specification checklist
What is a mathematical model in mechanics?
Real objects are complicated: a sledge has a shape, a rope stretches a little, snow has some friction. Mechanics makes progress by deciding which effects are small enough to ignore. The modelling cycle is: simplify (state assumptions) → solve → compare with reality → refine the assumptions if the answer is not good enough.
Every assumption removes an effect. Questions often ask you to state an assumption, explain what it means in the situation, or say how the answer would change if it were dropped. Learn the vocabulary as pairs: what the word means, and what it lets you ignore.
| Term | What it means in the model | What it ignores |
|---|---|---|
| Particle | the whole mass is at a single point | size, shape, rotation; air resistance is usually ignored too |
| Rigid body | an object whose shape never changes | bending, stretching, squashing |
| Rod | a rigid body with length only (one dimension) | its thickness and width |
| Lamina | a flat object with area but negligible thickness | its thickness |
| Uniform | mass spread evenly, so the centre of mass is at the geometric centre | variations in density |
| Light (string, rod, pulley) | its mass is negligible | its weight; so the tension is the same along a light string |
| Inextensible string | it does not stretch | stretching; so particles joined by a taut string have the same speed and the same size of acceleration |
| Smooth surface | no friction acts | friction; the only contact force is perpendicular to the surface |
| Rough surface | friction can act along the surface | — (this is a refinement, not a simplification) |
| Light smooth pulley | a weightless pulley with no friction at its axle | the pulley's mass and friction, so the tension is the same on both sides |
| Bead | a particle with a hole, threaded on a wire or string | its size; it can only move along the wire |
| Wire | a rigid, thin path that a bead is constrained to follow | its thickness |
| Peg | a support of negligible size; if smooth, it pushes at right angles to the object resting on it | its size (and friction, if smooth) |
| Air resistance negligible; \(g\) constant | the only force from the air is ignored; \(g = 9.8\) m s⁻² everywhere | drag, wind, variation of \(g\) with height |
A child of mass 15 kg sits on a sledge of mass 3 kg. A parent pulls the sledge across horizontal snow with a rope inclined at 25° above the horizontal. The tension in the rope is 40 N. The child and sledge are modelled as a single particle, the rope as light and inextensible, and the snow as smooth. (a) Explain what is meant by modelling the child and sledge as a particle. (b) Explain what is meant by the rope being light. (c) Find the horizontal component of the tension and hence the acceleration of the sledge in this model. (d) State how the answer to (c) would change if the snow were modelled as rough.
(a) The mass is concentrated at one point: the size of the sledge and any rotation are ignored.
(b) The mass of the rope is ignored, so the tension is the same all along it.
(c) \(40\cos 25^\circ = 36.25\ldots\) N. The snow is smooth, so this is the only horizontal force: \(a = 36.25 \div 18 = 2.01\) m s⁻². 36.3 N; 2.01 m s⁻²
(d) Friction would oppose the motion, so the resultant force and the acceleration would be smaller.
The vertical component, \(40\sin 25^\circ = 16.9\) N, does not change the horizontal motion; it only reduces the push of the snow on the sledge.
Mixing up light and inextensible. A light string has the same tension throughout; an inextensible string makes the connected particles move with the same speed and the same size of acceleration. Explain the one the question names.
How do you find the magnitude and direction of a vector?
\(\mathbf{i}\) and \(\mathbf{j}\) are unit vectors (magnitude 1) at right angles; in bearing problems \(\mathbf{i}\) points east and \(\mathbf{j}\) points north. The vector \(a\mathbf{i} + b\mathbf{j}\) can also be written as the column vector \(\begin{pmatrix} a \\ b \end{pmatrix}\). Because the components are at right angles, Pythagoras gives the magnitude and right-angled trigonometry gives the direction.
Magnitude, direction and unit vector
\[|a\mathbf{i} + b\mathbf{j}| = \sqrt{a^2 + b^2} \qquad \tan\theta = \frac{b}{a} \qquad \text{unit vector} = \frac{\mathbf{v}}{|\mathbf{v}|}\]\(\theta\) = angle between the vector and \(\mathbf{i}\), measured anticlockwise; a bearing is measured clockwise from north (\(\mathbf{j}\)), as three figures. Not in the formulae list: derive them from Pythagoras and right-angled trigonometry.
A calculator gives \(\tan^{-1}\) between −90° and 90° only. For a vector pointing left (\(a \lt 0\)), add 180°. A sketch shows which angle you need.
A kayak moves with velocity \((-6\mathbf{i} + 2.5\mathbf{j})\) km h⁻¹, where \(\mathbf{i}\) and \(\mathbf{j}\) are unit vectors due east and due north. (a) Find the speed of the kayak. (b) Find the bearing on which it is moving. (c) Write down a unit vector in the direction of its motion. (d) A second kayak moves at 12 km h⁻¹ on a bearing of 150°. Write its velocity in the form \(p\mathbf{i} + q\mathbf{j}\), giving exact values.
(a) Speed = magnitude of the velocity: \(\sqrt{(-6)^2 + 2.5^2} = \sqrt{42.25} =\) 6.5 km h⁻¹
(b) The sketch shows west and slightly north. Angle north of west: \(\tan^{-1}(2.5 \div 6) = 22.62^\circ\), so bearing \(= 270^\circ + 22.62^\circ =\) 293° (292.6°)
(c) \(\frac{1}{6.5}(-6\mathbf{i} + 2.5\mathbf{j}) =\) \(-\frac{12}{13}\mathbf{i} + \frac{5}{13}\mathbf{j}\)
(d) The velocity makes 30° with south, towards east: \(12\sin 150^\circ\,\mathbf{i} + 12\cos 150^\circ\,\mathbf{j} =\) \((6\mathbf{i} - 6\sqrt3\,\mathbf{j})\) km h⁻¹ \(\approx (6\mathbf{i} - 10.39\mathbf{j})\)
Typing \(\tan^{-1}(2.5 \div -6)\) and quoting −22.6° as the direction. That points east-south-east, the opposite way to the kayak. Always draw the vector before you name its angle.
How do you add and resolve vectors?
- Resultant = the single vector with the same effect as the vectors combined. Add the \(\mathbf{i}\)-components, then the \(\mathbf{j}\)-components.
- On a vector diagram, draw the vectors tip to tail: the resultant runs from the first tail to the last tip. Forces in equilibrium close up into a polygon.
- Resolving a vector of magnitude \(F\) at angle \(\theta\) to \(\mathbf{i}\) gives \(F\cos\theta\,\mathbf{i} + F\sin\theta\,\mathbf{j}\).
- Parallel: \(a\mathbf{i} + b\mathbf{j}\) is parallel to \(c\mathbf{i} + d\mathbf{j}\) when \(\frac{a}{c} = \frac{b}{d}\). Parallel to \(\mathbf{i}\): the \(\mathbf{j}\)-component is zero. Parallel to \(\mathbf{j}\): the \(\mathbf{i}\)-component is zero.
- Equilibrium: the resultant force is the zero vector, so the \(\mathbf{i}\)-components and the \(\mathbf{j}\)-components each sum to zero.
Three horizontal forces act on a particle: 8 N in the direction of \(\mathbf{i}\), 5 N at 60° to \(\mathbf{i}\) and 6 N at 150° to \(\mathbf{i}\), both angles measured anticlockwise from \(\mathbf{i}\). Find the magnitude of the resultant force and the angle it makes with \(\mathbf{i}\).
Resolve each force: \(8\mathbf{i}\); \(5\cos 60^\circ\,\mathbf{i} + 5\sin 60^\circ\,\mathbf{j} = 2.5\mathbf{i} + 4.33\mathbf{j}\); \(6\cos 150^\circ\,\mathbf{i} + 6\sin 150^\circ\,\mathbf{j} = -5.20\mathbf{i} + 3\mathbf{j}\).
Add components: \(\mathbf{R} = (8 + 2.5 - 5.196)\mathbf{i} + (4.330 + 3)\mathbf{j} = 5.304\mathbf{i} + 7.330\mathbf{j}\).
Magnitude and direction: \(|\mathbf{R}| = \sqrt{5.304^2 + 7.330^2} =\) 9.05 N; \(\tan^{-1}(7.330 \div 5.304) =\) 54.1° to \(\mathbf{i}\)
Two forces, \(\mathbf{F}_1 = (2\mathbf{i} + 3\mathbf{j})\) N and \(\mathbf{F}_2 = (p\mathbf{i} - 7\mathbf{j})\) N, act on a particle of mass 0.5 kg. Their resultant is parallel to \((3\mathbf{i} - 2\mathbf{j})\). (a) Find \(p\). (b) Find the magnitude of the resultant, in exact form. (c) A third force \(\mathbf{F}_3\) is added so that the particle is in equilibrium. Write down \(\mathbf{F}_3\). (d) Without \(\mathbf{F}_3\), find the magnitude of the acceleration of the particle.
(a) \(\mathbf{R} = (2 + p)\mathbf{i} - 4\mathbf{j}\). Parallel means the components are in the same ratio: \(\frac{2 + p}{3} = \frac{-4}{-2} = 2 \Rightarrow\) \(p = 4\)
(b) \(\mathbf{R} = 6\mathbf{i} - 4\mathbf{j}\), \(|\mathbf{R}| = \sqrt{36 + 16} = \sqrt{52} =\) \(2\sqrt{13}\) N (7.21 N)
(c) Equilibrium needs a zero resultant: \(\mathbf{F}_3 = -\mathbf{R} = (-6\mathbf{i} + 4\mathbf{j})\) N
(d) \(\mathbf{a} = (6\mathbf{i} - 4\mathbf{j}) \div 0.5 = 12\mathbf{i} - 8\mathbf{j}\), \(|\mathbf{a}| = 4\sqrt{13} =\) 14.4 m s⁻²
Write the resultant in components, then set the ratio equal to the ratio of the given direction. Check the signs as well as the sizes: \((3\mathbf{i} - 2\mathbf{j})\) and \((-3\mathbf{i} + 2\mathbf{j})\) are parallel but point opposite ways.
How do you describe motion with constant velocity?
The position vector \(\mathbf{r}\) of a point is its displacement from a fixed origin O. The displacement from A to B is \(\mathbf{r}_B - \mathbf{r}_A\) (“to” minus “from”); the distance AB is its magnitude. With constant velocity \(\mathbf{v}\), the displacement after time \(t\) is \(\mathbf{v}t\), so the position is the starting position plus that displacement.
Constant velocity
\[\mathbf{r} = \mathbf{r}_0 + \mathbf{v}t\]\(\mathbf{r}\) = position vector at time \(t\); \(\mathbf{r}_0\) = position vector when \(t = 0\); \(\mathbf{v}\) = constant velocity; speed = \(|\mathbf{v}|\). Not in the formulae list: must be known. Keep the units consistent: km with km h⁻¹ and hours, or m with m s⁻¹ and seconds.
At noon a ship is at the point with position vector \((-4\mathbf{i} + 2\mathbf{j})\) km relative to a harbour O, where \(\mathbf{i}\) and \(\mathbf{j}\) are due east and due north. It moves with constant velocity \((5\mathbf{i} + 3\mathbf{j})\) km h⁻¹. A buoy is at \((6\mathbf{i} + 3\mathbf{j})\) km. (a) Find the speed of the ship and the bearing on which it moves. (b) Write down the position vector of the ship \(t\) hours after noon. (c) Find the time at which the ship is due north of the buoy, and its distance from the buoy then.
(a) \(|\mathbf{v}| = \sqrt{25 + 9} = \sqrt{34} =\) 5.83 km h⁻¹; bearing \(= \tan^{-1}(5 \div 3) =\) 059.0° (the angle east of north).
(b) \(\mathbf{r} = (-4 + 5t)\mathbf{i} + (2 + 3t)\mathbf{j}\)
(c) Due north of the buoy means the same \(\mathbf{i}\)-component: \(-4 + 5t = 6 \Rightarrow t = 2\). Then \(\mathbf{r} = 6\mathbf{i} + 8\mathbf{j}\), and the distance is \(8 - 3 = 5\) km. At 14:00, 5 km north of the buoy.
Writing \(\mathbf{r} = \mathbf{v}t\) and forgetting the starting position, or adding the speed (a number) instead of the velocity (a vector). The position vector needs \(\mathbf{r}_0\) and both components of \(\mathbf{v}\).
How do you use F = ma and constant acceleration with vectors?
Newton's second law works with whole vectors: the resultant force \(\mathbf{F}\) and the acceleration \(\mathbf{a}\) point in the same direction. If the forces are constant, \(\mathbf{a}\) is constant, and the constant-acceleration formulae hold component by component. Read the wording carefully: “moving parallel to \(\mathbf{i}\)” or “moving due east” refers to the velocity (the \(\mathbf{j}\)-component of \(\mathbf{v}\) is zero); “due east of O” refers to the position (the \(\mathbf{j}\)-component of \(\mathbf{r}\) is zero).
Constant acceleration and Newton's second law in vector form
\[\mathbf{v} = \mathbf{u} + \mathbf{a}t \qquad \mathbf{r} = \mathbf{r}_0 + \mathbf{u}t + \tfrac12\mathbf{a}t^2 \qquad \mathbf{F} = m\mathbf{a}\]\(\mathbf{u}\) = initial velocity; \(\mathbf{v}\) = velocity at time \(t\); \(\mathbf{a}\) = constant acceleration; \(\mathbf{F}\) = resultant force (N); \(m\) = mass (kg). Not in the formulae list: must be known. Only for constant acceleration; if the acceleration varies with time you need calculus (M2).
A puck of mass 2.5 kg moves on smooth horizontal ice. Two forces act on it: \((6\mathbf{i} - 2\mathbf{j})\) N and \((-\mathbf{i} + 7\mathbf{j})\) N. At \(t = 0\) its velocity is \((-3\mathbf{i} + 4\mathbf{j})\) m s⁻¹. (a) Find the acceleration of the puck. (b) Find its speed when \(t = 3\) s. (c) Find the time at which it is moving parallel to \(\mathbf{j}\). (d) Find the time at which it is moving in the direction of \((\mathbf{i} + 2\mathbf{j})\).
(a) \(\mathbf{F} = 5\mathbf{i} + 5\mathbf{j}\), so \(\mathbf{a} = (5\mathbf{i} + 5\mathbf{j}) \div 2.5 =\) \((2\mathbf{i} + 2\mathbf{j})\) m s⁻²
(b) \(\mathbf{v} = (-3 + 2t)\mathbf{i} + (4 + 2t)\mathbf{j}\); at \(t = 3\): \(3\mathbf{i} + 10\mathbf{j}\), speed \(= \sqrt{109} =\) 10.4 m s⁻¹
(c) Parallel to \(\mathbf{j}\): the \(\mathbf{i}\)-component of \(\mathbf{v}\) is zero, \(-3 + 2t = 0 \Rightarrow\) \(t = 1.5\) s (then \(\mathbf{v} = 7\mathbf{j}\)).
(d) Direction \((\mathbf{i} + 2\mathbf{j})\): \(\mathbf{j}\)-component = 2 × \(\mathbf{i}\)-component, both positive. \(4 + 2t = 2(-3 + 2t) \Rightarrow\) \(t = 5\) s (then \(\mathbf{v} = 7\mathbf{i} + 14\mathbf{j}\) ✓).
A particle P of mass 0.6 kg moves on a smooth horizontal plane with constant acceleration \((-2\mathbf{i} + \mathbf{j})\) m s⁻². At \(t = 0\) it is at the point with position vector \((-8\mathbf{i} - 3\mathbf{j})\) m and has velocity \((6\mathbf{i} + \mathbf{j})\) m s⁻¹. (a) Find the magnitude of the resultant force on P. (b) Find the position vector of P at time \(t\). (c) Find the times at which P is due north of O, and its distance from O at each. (d) Find the speed of P at the first of these times.
(a) \(\mathbf{F} = 0.6(-2\mathbf{i} + \mathbf{j}) = -1.2\mathbf{i} + 0.6\mathbf{j}\), \(|\mathbf{F}| =\) 1.34 N (1.342 N)
(b) \(\mathbf{r} = \mathbf{r}_0 + \mathbf{u}t + \frac12\mathbf{a}t^2 =\) \((-8 + 6t - t^2)\mathbf{i} + (-3 + t + \frac12 t^2)\mathbf{j}\)
(c) Due north of O: \(\mathbf{i}\)-component zero (and \(\mathbf{j}\)-component positive). \(t^2 - 6t + 8 = 0 \Rightarrow (t - 2)(t - 4) = 0 \Rightarrow t = 2\) or \(4\). At \(t = 2\): \(\mathbf{j}\)-component \(= -3 + 2 + 2 = 1\); at \(t = 4\): \(-3 + 4 + 8 = 9\) (both positive ✓). \(t = 2\) s, 1 m from O; \(t = 4\) s, 9 m from O.
(d) \(\mathbf{v} = \mathbf{u} + \mathbf{a}t = 2\mathbf{i} + 3\mathbf{j}\) at \(t = 2\), speed \(= \sqrt{13} =\) 3.61 m s⁻¹
After setting the \(\mathbf{i}\)-component to zero, check the sign of the \(\mathbf{j}\)-component. A negative value would mean due south of O, and that time must be rejected.
When are two moving objects due north, colliding or closest?
Write both position vectors in terms of the same \(t\), then form \(\mathbf{r}_B - \mathbf{r}_A\), the position of B relative to A. Everything follows from this one vector:
- B due north of A: \(\mathbf{i}\)-component of \(\mathbf{r}_B - \mathbf{r}_A\) is zero (and the \(\mathbf{j}\)-component positive). Due east: \(\mathbf{j}\)-component zero (and the \(\mathbf{i}\)-component positive).
- Collision: both components of \(\mathbf{r}_B - \mathbf{r}_A\) are zero at the same value of \(t\). Paths that cross at different times do not collide.
- Least distance: \(d^2 = |\mathbf{r}_B - \mathbf{r}_A|^2\) is a quadratic in \(t\). Complete the square: the minimum value of \(d^2\) gives the least distance and the time it happens. No calculus is needed.
[\(\mathbf{i}\) and \(\mathbf{j}\) are unit vectors due east and due north; position vectors are relative to a fixed point O.] At noon ship A is at \((\mathbf{i} + 2\mathbf{j})\) km and ship B is at \((13\mathbf{i} - 2\mathbf{j})\) km. A moves with constant velocity \((3\mathbf{i} + 4\mathbf{j})\) km h⁻¹ and B with constant velocity \((-\mathbf{i} + 6\mathbf{j})\) km h⁻¹. (a) Find the speed of B. (b) Find, in terms of \(t\), the position vector of each ship \(t\) hours after noon. (c) Show that B is due east of A at 14:00 and find the distance between them then. (d) Show that the ships do not collide. (e) Show that \(d^2 = 20t^2 - 112t + 160\), where \(d\) km is the distance AB, and hence find the least distance between the ships and the time when it occurs.
(a) \(|-\mathbf{i} + 6\mathbf{j}| = \sqrt{37} =\) 6.08 km h⁻¹
(b) \(\mathbf{r}_A = (1 + 3t)\mathbf{i} + (2 + 4t)\mathbf{j}\), \(\mathbf{r}_B = (13 - t)\mathbf{i} + (-2 + 6t)\mathbf{j}\).
(c) \(\mathbf{r}_B - \mathbf{r}_A = (12 - 4t)\mathbf{i} + (-4 + 2t)\mathbf{j}\). The \(\mathbf{j}\)-component is zero when \(t = 2\) (14:00); then \(\mathbf{r}_B - \mathbf{r}_A = 4\mathbf{i}\). B is 4 km due east of A.
(d) The \(\mathbf{i}\)-component is zero only at \(t = 3\), the \(\mathbf{j}\)-component only at \(t = 2\): never together, so no collision.
(e) \(d^2 = (12 - 4t)^2 + (2t - 4)^2 = 144 - 96t + 16t^2 + 4t^2 - 16t + 16 = 20t^2 - 112t + 160\) ✓. Completing the square, \(d^2 = 20(t - 2.8)^2 + 3.2\), so the least \(d = \sqrt{3.2} =\) 1.79 km (1.7889) when \(t = 2.8\), at 14:48.
Saying two ships collide because their paths cross. A collision needs \(\mathbf{r}_A = \mathbf{r}_B\) for one value of \(t\).
Exam tips for M1.1
A five-second sketch with \(\mathbf{i}\) east and \(\mathbf{j}\) north fixes the quadrant of every direction, shows whether “due north” is possible, and catches a sign error before it spreads.
“The angle with \(\mathbf{i}\)”, “the angle with \(\mathbf{j}\)” and “the bearing” are different numbers for the same vector. Give the one asked for, and give bearings as three figures (059°, not 59°).
“Due north of O” is about \(\mathbf{r}\); “moving due north” is about \(\mathbf{v}\): say which vector you are using. A speed or distance ends with a magnitude and a unit; a velocity, displacement or force “as a vector” is given in \(\mathbf{i}, \mathbf{j}\) form.
Link an assumption to the object in the question: “the sledge's size is ignored, so it can be treated as a point”, not just a dictionary definition. When asked how dropping it would change an answer, say which way (bigger or smaller) and why.
Common misconceptions
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| A particle must be something small. | “Particle” is a modelling choice, not a size. | A ship or a car can be a particle when its size does not affect the motion being studied. |
| A light string makes connected particles move together. | That comes from the string being inextensible. | Light ⇒ same tension throughout; inextensible ⇒ same speed and size of acceleration. |
| Velocity \(3\mathbf{i} + 4\mathbf{j}\) means a speed of 7. | Components are at right angles, so they do not add as numbers. | Speed \(= \sqrt{3^2 + 4^2} = 5\). |
| \(\tan^{-1}(b/a)\) always gives the direction. | The calculator cannot tell \(a\mathbf{i} + b\mathbf{j}\) from \(-a\mathbf{i} - b\mathbf{j}\). | Sketch the vector; add 180° when \(a \lt 0\). |
| Crossing paths means a collision. | The objects may pass the crossing point at different times. | Collision needs equal position vectors at the same \(t\). |
| The force on a moving particle points the way it moves. | \(\mathbf{F} = m\mathbf{a}\) links force to acceleration, not to velocity. | \(\mathbf{F}\) is parallel to \(\mathbf{a}\); the path bends towards \(\mathbf{a}\). |
How this topic connects
P1 Pure Mathematics
Pythagoras, surds, right-angled trigonometry and completing the square are the algebra behind magnitudes, exact answers and least-distance problems.
M1.2 Kinematics and the rest of M1
Straight-line kinematics is the one-component case of \(\mathbf{r} = \mathbf{r}_0 + \mathbf{u}t + \frac12\mathbf{a}t^2\). Resolving forces, statics, friction and connected particles (M1.3) all use the modelling words and components from this page; rods and laminas return in moments.
M2 Mechanics 2 and IAL Physics Unit 1
Projectiles are constant-acceleration vector motion with \(\mathbf{a} = -g\mathbf{j}\); when \(\mathbf{a}\) varies, \(\mathbf{v}\) and \(\mathbf{r}\) come from calculus. Adding vectors tip to tail and resolving forces are the same skills in IAL Physics.
Key formulae
| Formula | Use | In the formulae list? |
|---|---|---|
| \(|a\mathbf{i} + b\mathbf{j}| = \sqrt{a^2 + b^2}\) | Magnitude: speed, distance, size of a force | No — derive it |
| \(\tan\theta = b/a\) | Angle with \(\mathbf{i}\) (check the quadrant) | No — derive it |
| \(F\cos\theta\,\mathbf{i} + F\sin\theta\,\mathbf{j}\) | Resolving a vector at angle \(\theta\) to \(\mathbf{i}\) | No — derive it |
| \(\mathbf{v}/|\mathbf{v}|\) | Unit vector in the direction of \(\mathbf{v}\) | No — derive it |
| \(\mathbf{r} = \mathbf{r}_0 + \mathbf{v}t\) | Position with constant velocity | No — must be known |
| \(\mathbf{v} = \mathbf{u} + \mathbf{a}t\) | Velocity with constant acceleration | No — must be known |
| \(\mathbf{r} = \mathbf{r}_0 + \mathbf{u}t + \frac12\mathbf{a}t^2\) | Position with constant acceleration | No — must be known |
| \(\mathbf{F} = m\mathbf{a}\) | Resultant force and acceleration | No — must be known |
Check your understanding
1. A load hangs from a light inextensible string. State what each of the two words tells you about the string.
Light: its mass is ignored, so the tension is the same throughout. Inextensible: it does not stretch, so anything attached to it moves with the same speed and the same size of acceleration.
2. Find the magnitude of \((5\mathbf{i} - 3\mathbf{j})\) and its bearing, where \(\mathbf{i}\) is east and \(\mathbf{j}\) is north.
\(\sqrt{34} = 5.83\). It points 31.0° below \(\mathbf{i}\) (south of east), so the bearing is \(090^\circ + 31.0^\circ =\) 121.0°.
3. A force of 20 N acts on a bearing of 240°. Write it in the form \(a\mathbf{i} + b\mathbf{j}\).
\(20\sin 240^\circ\,\mathbf{i} + 20\cos 240^\circ\,\mathbf{j} = -10\sqrt3\,\mathbf{i} - 10\mathbf{j} \approx\) \((-17.3\mathbf{i} - 10\mathbf{j})\) N.
4. A particle is at \((\mathbf{i} + 2\mathbf{j})\) m and moves with constant velocity \((3\mathbf{i} - \mathbf{j})\) m s⁻¹. Find its distance from O after 4 s.
\(\mathbf{r} = (\mathbf{i} + 2\mathbf{j}) + 4(3\mathbf{i} - \mathbf{j}) = 13\mathbf{i} - 2\mathbf{j}\); distance \(= \sqrt{173} =\) 13.2 m.
5. A resultant force \((6\mathbf{i} - 8\mathbf{j})\) N acts on a 4 kg particle. Find the magnitude of its acceleration.
\(\mathbf{a} = (6\mathbf{i} - 8\mathbf{j}) \div 4 = 1.5\mathbf{i} - 2\mathbf{j}\); \(|\mathbf{a}| = \sqrt{2.25 + 4} =\) 2.5 m s⁻².
Specification checklist
- (M1 1.1) Explain what particle, rigid body, rod, lamina, uniform, light, inextensible, smooth, rough, light smooth pulley, bead, wire and peg mean in a model, and what each one ignores.
- (M1 1.1) Say how removing an assumption (adding friction, air resistance or the mass of a string) would change an answer.
- (M1 2.1) Find the magnitude and direction of a vector given in \(\mathbf{i}, \mathbf{j}\) form, as an angle with \(\mathbf{i}\) or as a bearing, and find a unit vector.
- (M1 2.1) Write a vector given by magnitude and direction in \(\mathbf{i}, \mathbf{j}\) form, find resultants by adding components or drawing vector diagrams, and use the condition for parallel vectors.
- (M1 2.2) Use \(\mathbf{r} = \mathbf{r}_0 + \mathbf{v}t\) for constant velocity, and find speeds and distances from vectors.
- (M1 2.2) Use \(\mathbf{v} = \mathbf{u} + \mathbf{a}t\), \(\mathbf{r} = \mathbf{r}_0 + \mathbf{u}t + \frac12\mathbf{a}t^2\) and \(\mathbf{F} = m\mathbf{a}\) with vectors.
- (M1 2.2) Find when one object is due north or due east of another, decide whether two objects collide, and find their least distance apart by completing the square.