Scalars and vectors

AS · 10 min

Some quantities need only a size; others need a size and a direction. Knowing which is which, and handling the directional ones properly, underpins kinematics, forces, momentum and fields for the whole course. The skills here are adding and subtracting vectors (by scale drawing or by calculation) and splitting a vector into two perpendicular components. They appear in Paper 1 most sessions and inside longer Paper 2 questions on forces and projectiles.

Scalars and vectors

Definition

A scalar quantity has magnitude only.

A vector quantity has both magnitude and direction.

Scalars	Vectors
distance	displacement
speed	velocity
mass	acceleration (including acceleration of free fall)
time	force (including weight, tension, friction, upthrust)
energy, work	momentum
power	electric field strength
density, volume	moment / torque (about an axis; treated as clockwise or anticlockwise)
pressure
temperature
charge, current*, potential difference, resistance

*Current has a direction around a circuit but does not add like a vector, so it is treated as a scalar.

Some pairs are worth fixing in memory because they are tested together: distance is scalar but displacement is vector; speed is scalar but velocity is vector. Mass is scalar but weight is a force and therefore a vector. Energy and power are scalars even though they often involve forces.

Watch out

Pressure is a scalar, even though it is defined using force. At a point in a fluid, pressure acts equally in all directions; the force it produces on a surface depends on how the surface is oriented, but pressure itself has no direction. Work is also scalar: it is force times displacement, but the product of two vectors in this way gives a scalar.

A vector is drawn as an arrow whose length represents the magnitude (to a chosen scale) and whose direction is the direction of the vector. A negative vector, $-\mathbf{a}$ , has the same magnitude as $\mathbf{a}$ but points the opposite way.

Adding vectors

To add vectors, draw them head to tail: start the second arrow where the first ends. The resultant is the single arrow from the tail of the first to the head of the last.

(0, 0) -> (5, 0) (5, 0) -> (5, 12) (0, 0) -> (5, 12)

Here a $5.0\ \text{N}$ force (to the right) and a $12.0\ \text{N}$ force (upwards) are added head to tail. The resultant is the diagonal arrow from the origin to $(5, 12)$ .

Three ways to get the resultant:

Scale drawing. Choose a scale (for example $1\ \text{cm}$ to $1\ \text{N}$ ), draw the triangle with a ruler and protractor, then measure the resultant's length and angle. Accurate to a few percent if done carefully; Paper 2 sometimes asks for it explicitly.
Pythagoras and trigonometry when the two vectors are perpendicular.
The cosine and sine rules when they are at any other angle.

Perpendicular vectors

Find the resultant of a $5.0\ \text{N}$ force acting due east and a $12.0\ \text{N}$ force acting due north.

Solution

The forces are perpendicular, so the triangle is right-angled.

F = \sqrt{5.0^2 + 12.0^2} = \sqrt{169} = 13.0\ \text{N}

Direction: $\tan\theta = \dfrac{12.0}{5.0}$ , so $\theta = 67.4^\circ$ north of east.

Resultant: $13.0\ \text{N}$ at $67^\circ$ north of east (equivalently, on a bearing of $023^\circ$ ).

Vectors at an angle

Two forces of $8.0\ \text{N}$ and $6.0\ \text{N}$ act at the same point with an angle of $60^\circ$ between them. Calculate the magnitude of the resultant and its angle to the $8.0\ \text{N}$ force.

Solution

When the vectors are drawn head to tail, the angle inside the triangle between them is $180^\circ - 60^\circ = 120^\circ$ . The cosine rule gives

R^2 = 8.0^2 + 6.0^2 - 2(8.0)(6.0)\cos 120^\circ = 64 + 36 + 48 = 148

R = 12.2\ \text{N}

Angle $\alpha$ between $R$ and the $8.0\ \text{N}$ force, by the sine rule:

\frac{\sin\alpha}{6.0} = \frac{\sin 120^\circ}{12.2} \quad\Rightarrow\quad \alpha = 25^\circ

Watch out

The most common slip in the cosine rule is using the angle between the arrows as given ( $60^\circ$ ) instead of the angle inside the head-to-tail triangle ( $120^\circ$ ). A quick check: the resultant of two forces at a small angle must be close to their sum ( $14\ \text{N}$ here), and at a large angle close to their difference ( $2\ \text{N}$ ). $12.2\ \text{N}$ is sensible for $60^\circ$ ; using $60^\circ$ inside the triangle gives $7.2\ \text{N}$ , which is not.

The resultant of two vectors of magnitudes $a$ and $b$ always lies between $a - b$ (vectors opposite) and $a + b$ (vectors in the same direction).

Subtracting vectors

To find $\mathbf{a} - \mathbf{b}$ , add the negative: $\mathbf{a} + (-\mathbf{b})$ . Reverse the arrow for $\mathbf{b}$ , then add head to tail as before.

The most important use is a change in velocity: $\Delta\mathbf{v} = \mathbf{v} - \mathbf{u}$ (final minus initial). The change in velocity determines the acceleration and, through momentum, the force.

Change in velocity

A ball moving at $3.0\ \text{m s}^{-1}$ due east is struck so that it moves at $4.0\ \text{m s}^{-1}$ due north. Find the change in velocity.

Solution

$\Delta\mathbf{v} = \mathbf{v} - \mathbf{u} = (4.0\ \text{north}) + (3.0\ \text{west})$ .

The two parts are perpendicular:

|\Delta\mathbf{v}| = \sqrt{3.0^2 + 4.0^2} = 5.0\ \text{m s}^{-1}

Direction: $\tan\theta = 4.0/3.0$ , so $\theta = 53^\circ$ north of west.

The change in speed is only $1.0\ \text{m s}^{-1}$ , but the change in velocity is $5.0\ \text{m s}^{-1}$ . Direction matters.

Components of a vector

Any vector can be replaced by two perpendicular vectors that add to give it. These are its components. Resolving into components is usually far easier than drawing triangles, because perpendicular directions can be treated completely independently.

(0, 0) -> (4, 3) (0, 0) -> (4, 0) (4, 0) -> (4, 3)

A vector of magnitude $F$ at angle $\theta$ to the horizontal (the diagonal arrow) has a horizontal component along the bottom and a vertical component up the side.

Key result

For a vector of magnitude $F$ at angle $\theta$ to a chosen direction:

F_{\parallel} = F\cos\theta \quad \text{(component along the direction)}

F_{\perp} = F\sin\theta \quad \text{(component perpendicular to it)}

The component adjacent to the angle uses cosine; the component opposite the angle uses sine.

Resolving a velocity

An aircraft climbs at $250\ \text{m s}^{-1}$ at $30^\circ$ above the horizontal. Find the horizontal and vertical components of its velocity.

Solution

v_x = 250\cos 30^\circ = 217\ \text{m s}^{-1}, \qquad v_y = 250\sin 30^\circ = 125\ \text{m s}^{-1}

Check: $\sqrt{217^2 + 125^2} = 250$ .

Resolving on a slope

For an object on a slope inclined at angle $\theta$ to the horizontal, resolve the weight parallel and perpendicular to the slope, not horizontally and vertically. The geometry puts the angle $\theta$ between the weight and the perpendicular to the slope.

Key result

On a slope at angle $\theta$ , the weight $W$ has components

W\sin\theta \ \text{down the slope}, \qquad W\cos\theta \ \text{perpendicular to (into) the slope}

A quick sanity check: when $\theta = 0$ (flat ground) the component down the slope must be zero, so it must be the sine.

Block on a slope

A block of weight $40\ \text{N}$ rests on a ramp inclined at $25^\circ$ to the horizontal. Find the components of its weight parallel and perpendicular to the ramp.

Solution

W_{\parallel} = 40\sin 25^\circ = 16.9\ \text{N} \ \text{down the slope}

W_{\perp} = 40\cos 25^\circ = 36.3\ \text{N} \ \text{into the slope}

If the block is at rest, friction up the slope is $16.9\ \text{N}$ and the normal contact force is $36.3\ \text{N}$ .

Adding several vectors by components

For three or more vectors, or awkward angles, resolve every vector into the same two perpendicular directions, add the components in each direction (taking signs into account), then recombine with Pythagoras.

Method

Resultant by components

Choose two perpendicular directions ( $x$ and $y$ ) and a positive sense for each.
Resolve every vector: $x$ -component $F\cos\theta$ , $y$ -component $F\sin\theta$ , with signs.
Add: $R_x = \sum F_x$ , $R_y = \sum F_y$ .
Magnitude $R = \sqrt{R_x^2 + R_y^2}$ ; direction $\tan\phi = R_y / R_x$ , interpreted using the signs.

Crossing a river

A boat can move at $2.0\ \text{m s}^{-1}$ relative to the water. The river flows at $1.5\ \text{m s}^{-1}$ and is $60\ \text{m}$ wide. The boat must travel straight across, perpendicular to the banks. (a) At what angle to the perpendicular must it be steered? (b) How long does the crossing take?

Solution

(a) The boat's velocity relative to the water must have an upstream component that cancels the current. If it is steered at angle $\theta$ upstream of the perpendicular, its upstream component is $2.0\sin\theta$ :

2.0\sin\theta = 1.5 \quad\Rightarrow\quad \sin\theta = 0.75 \quad\Rightarrow\quad \theta = 48.6^\circ \approx 49^\circ

(b) The remaining component, straight across, is

v = \sqrt{2.0^2 - 1.5^2} = \sqrt{1.75} = 1.32\ \text{m s}^{-1}

t = \frac{60}{1.32} = 45\ \text{s}

Notice the right angle is between the current and the resultant, so the boat's own speed ( $2.0$ ) is the hypotenuse.

Exam tip

"Distinguish between a scalar and a vector" is a definition mark: magnitude only versus magnitude and direction. Giving examples alone does not score.
For any vector answer, state the direction as well as the magnitude: "at $53^\circ$ north of west" or "at $25^\circ$ to the $8.0\ \text{N}$ force". A magnitude alone usually earns only part of the marks.
If a question says "by drawing a vector triangle" or "by scale drawing", you must draw it, state the scale, and show the measured length. A calculated answer alone may not get full credit.
In Paper 1, always estimate first. If the four options include values bigger than the sum or smaller than the difference of the two vectors, eliminate them straight away.

Summary

Scalars have magnitude only; vectors have magnitude and direction.
Know the pairs: distance/displacement, speed/velocity, mass/weight. Energy, power, pressure and work are scalars.
Add vectors head to tail; the resultant joins the start of the first to the end of the last.
Use Pythagoras for perpendicular vectors and the cosine rule (with the angle inside the triangle) otherwise.
Subtract by adding the reversed vector; change in velocity is $\mathbf{v} - \mathbf{u}$ .
Components: $F\cos\theta$ along the direction, $F\sin\theta$ perpendicular to it.
On a slope at $\theta$ : $W\sin\theta$ down the slope, $W\cos\theta$ into it.

Practice

Question

Classify each as scalar or vector: kinetic energy, weight, displacement, pressure, momentum, time, power, acceleration.
Find the resultant of a $3.0\ \text{N}$ force and a $4.0\ \text{N}$ force acting at right angles. Give its direction relative to the $3.0\ \text{N}$ force.
Two forces of $20\ \text{N}$ and $15\ \text{N}$ act at a point. State the largest and smallest possible resultants, and calculate the resultant when they are perpendicular.
A rope pulls a sledge with a force of $50\ \text{N}$ at $40^\circ$ above the horizontal. Find the horizontal and vertical components of the force.
An aircraft heads due north at $200\ \text{m s}^{-1}$ relative to the air. A wind blows towards the east at $50\ \text{m s}^{-1}$ . Find the aircraft's resultant velocity.
A car travelling north at $20\ \text{m s}^{-1}$ turns a corner and travels east at $20\ \text{m s}^{-1}$ . Find the magnitude and direction of the change in velocity.
A box of weight $120\ \text{N}$ rests on a slope at $30^\circ$ to the horizontal. Find the frictional force and the normal contact force acting on it.
Three forces act at a point: $10\ \text{N}$ along $0^\circ$ , $8.0\ \text{N}$ along $90^\circ$ and $6.0\ \text{N}$ along $225^\circ$ (angles measured anticlockwise from the $x$ -axis). Find the magnitude and direction of their resultant.
A ball travelling at $5.0\ \text{m s}^{-1}$ strikes a smooth wall at $30^\circ$ to the wall and rebounds at the same speed and the same angle. Find the magnitude and direction of the change in velocity.
A sign of weight $400\ \text{N}$ hangs from the midpoint of a cable whose two halves each make $35^\circ$ with the horizontal. Find the tension in the cable. Explain why the tension becomes very large if the cable is pulled nearly horizontal.

Answers

Scalar: kinetic energy, pressure, time, power. Vector: weight, displacement, momentum, acceleration.
$\sqrt{3.0^2 + 4.0^2} = 5.0\ \text{N}$ at $\tan^{-1}(4.0/3.0) = 53^\circ$ to the $3.0\ \text{N}$ force.
Largest $35\ \text{N}$ (same direction); smallest $5\ \text{N}$ (opposite). Perpendicular: $\sqrt{20^2 + 15^2} = 25\ \text{N}$ .
Horizontal $50\cos 40^\circ = 38.3\ \text{N}$ ; vertical $50\sin 40^\circ = 32.1\ \text{N}$ .
$\sqrt{200^2 + 50^2} = 206\ \text{m s}^{-1}$ at $\tan^{-1}(50/200) = 14^\circ$ east of north.
$\Delta\mathbf{v} = (20\ \text{east}) + (20\ \text{south})$ , magnitude $\sqrt{20^2 + 20^2} = 28.3\ \text{m s}^{-1}$ , directed towards the south-east ( $45^\circ$ east of south).
Friction $= 120\sin 30^\circ = 60\ \text{N}$ up the slope; normal force $= 120\cos 30^\circ = 104\ \text{N}$ .
$R_x = 10 + 6.0\cos 225^\circ = 10 - 4.24 = 5.76\ \text{N}$ ; $R_y = 8.0 + 6.0\sin 225^\circ = 8.0 - 4.24 = 3.76\ \text{N}$ . $R = \sqrt{5.76^2 + 3.76^2} = 6.9\ \text{N}$ at $\tan^{-1}(3.76/5.76) = 33^\circ$ above the $x$ -axis.
Parallel to the wall the velocity component ( $5.0\cos 30^\circ$ ) is unchanged. Perpendicular to the wall it changes from $5.0\sin 30^\circ = 2.5\ \text{m s}^{-1}$ towards the wall to $2.5\ \text{m s}^{-1}$ away. $|\Delta v| = 2.5 + 2.5 = 5.0\ \text{m s}^{-1}$ , perpendicular to the wall, directed away from it.
The vertical components of the two tensions support the weight: $2T\sin 35^\circ = 400$ , so $T = 349\ \text{N}$ (about $350\ \text{N}$ ). As the angle to the horizontal tends to zero, $\sin\theta \to 0$ , so $T = 400/(2\sin\theta)$ becomes very large: a horizontal cable cannot provide any upward component.