Physics 9702 · Paper 2

Scalars, Vectors & Resolving

Physical quantities split into two kinds. A has size only. A has size and direction.
A vector can also be split into two parts at right angles. That skill, resolving, is used in every mechanics chapter after this one.

Scalar or vector

The test: does the quantity need a direction to be fully stated? “5 N” is incomplete; “5 N downward” is complete. Force is a vector.

Scalar: size only	Vector: size and direction
distance	displacement
speed	velocity
mass	weight (a force)
energy / work	momentum
time, temperature	acceleration

Your turn— tap to reveal the worked answer (9702/11/M/J/25 Q1)

Which is a vector: pressure, temperature, weight or work? (9702/11/M/J/25 Q1)

Answer: weight. It is the force of gravity, and it has a direction (downward). The others are scalars.

Common mistake

Work and energy are scalars even though force is a vector. Weight is a vector (a force); mass is a scalar. Temperature is a scalar even though it can be negative: a sign is not a direction in space.

Resolving a vector into components

A single vector at an angle has two effects at once: one along each of two directions at right angles. Each effect is called a .
Splitting a vector into its two components is called . The two components are independent: changing one has no effect on the other.

F_x = F\cos\theta \qquad F_y = F\sin\theta

Symbols

$F$ = the vector's magnitude (N (or the vector's own unit))
$\theta$ = angle between the vector and the direction you resolve along (degrees)
$F_x$ = component along that direction (same as F)
$F_y$ = component at right angles to it (same as F)

Where it comes from: the vector is the longest side of a right-angled triangle, and the two components are the other two sides. Cosine and sine of $\theta$ give those sides.
The rule to remember: the cosine goes with the axis the angle is measured from. The sine goes with the other axis.

Drawn to scale: F = 5 N at 37° gives F cos θ = 4 N along the axis the angle is measured from, and F sin θ = 3 N at right angles to it.

Worked example

Smallest case: one velocity, two components

A ball leaves the ground at 20 m s⁻¹, at 60° above the horizontal. Find the horizontal and vertical components of the velocity.

The angle is measured from the horizontal, so the horizontal component takes the cosine: $v_x = 20\cos 60^\circ = 10 \text{ m s}^{-1}$ .
The vertical component takes the sine: $v_y = 20\sin 60^\circ = 17 \text{ m s}^{-1}$ .

Answer

v_x = 10 \text{ m s}^{-1},\ \ v_y = 17 \text{ m s}^{-1}

Drag the magnitude and angle below. $F\cos\theta$ gets smaller as $\theta$ gets closer to 90°:

Vector Resolver

Drag the magnitude and angle. The green leg is F cos θ (horizontal) and the red leg is F sin θ (vertical). Turn on Add a second vector to build the resultant with the parallelogram rule.

Magnitude F8

Angle θ35°

F cos θ

6.6

F sin θ

4.6

magnitude

angle θ

35°

Worked example

One change: the angle is measured from north

A velocity of 75 m s⁻¹ is at 30° to the north direction. Find the northerly and easterly components.

The angle is measured from north, so north takes the cosine: $v_N = 75\cos 30^\circ = 65 \text{ m s}^{-1}$ .
East is the perpendicular direction, so it takes the sine: $v_E = 75\sin 30^\circ = 38 \text{ m s}^{-1}$ .

Answer

v_N = 65 \text{ m s}^{-1},\ \ v_E = 38 \text{ m s}^{-1}

(9702/12/F/M/21 Q3)

Common mistake

The cos/sin swap. Before you write anything, check which axis the angle is measured from: that axis gets the cosine. Here the angle was from north, so north got the cosine, not east.

PreviousCombining Uncertainties NextAdding Vectors

Scalars, Vectors & Resolving

Physical quantities split into two kinds. A has size only. A has size and direction.
A vector can also be split into two parts at right angles. That skill, resolving, is used in every mechanics chapter after this one.

Scalar or vector

The test: does the quantity need a direction to be fully stated? “5 N” is incomplete; “5 N downward” is complete. Force is a vector.

Scalar: size only	Vector: size and direction
distance	displacement
speed	velocity
mass	weight (a force)
energy / work	momentum
time, temperature	acceleration

Your turn— tap to reveal the worked answer (9702/11/M/J/25 Q1)

Which is a vector: pressure, temperature, weight or work? (9702/11/M/J/25 Q1)

Answer: weight. It is the force of gravity, and it has a direction (downward). The others are scalars.

Common mistake

Work and energy are scalars even though force is a vector. Weight is a vector (a force); mass is a scalar. Temperature is a scalar even though it can be negative: a sign is not a direction in space.

Resolving a vector into components

A single vector at an angle has two effects at once: one along each of two directions at right angles. Each effect is called a .
Splitting a vector into its two components is called . The two components are independent: changing one has no effect on the other.

F_x = F\cos\theta \qquad F_y = F\sin\theta

Symbols

$F$ = the vector's magnitude (N (or the vector's own unit))
$\theta$ = angle between the vector and the direction you resolve along (degrees)
$F_x$ = component along that direction (same as F)
$F_y$ = component at right angles to it (same as F)

Where it comes from: the vector is the longest side of a right-angled triangle, and the two components are the other two sides. Cosine and sine of $\theta$ give those sides.
The rule to remember: the cosine goes with the axis the angle is measured from. The sine goes with the other axis.

Drawn to scale: F = 5 N at 37° gives F cos θ = 4 N along the axis the angle is measured from, and F sin θ = 3 N at right angles to it.

Worked example

Smallest case: one velocity, two components

A ball leaves the ground at 20 m s⁻¹, at 60° above the horizontal. Find the horizontal and vertical components of the velocity.

The angle is measured from the horizontal, so the horizontal component takes the cosine: $v_x = 20\cos 60^\circ = 10 \text{ m s}^{-1}$ .
The vertical component takes the sine: $v_y = 20\sin 60^\circ = 17 \text{ m s}^{-1}$ .

Answer

v_x = 10 \text{ m s}^{-1},\ \ v_y = 17 \text{ m s}^{-1}

Drag the magnitude and angle below. $F\cos\theta$ gets smaller as $\theta$ gets closer to 90°:

Vector Resolver

Magnitude F8

Angle θ35°

F cos θ

6.6

F sin θ

4.6

magnitude

angle θ

35°

Worked example

One change: the angle is measured from north

A velocity of 75 m s⁻¹ is at 30° to the north direction. Find the northerly and easterly components.

The angle is measured from north, so north takes the cosine: $v_N = 75\cos 30^\circ = 65 \text{ m s}^{-1}$ .
East is the perpendicular direction, so it takes the sine: $v_E = 75\sin 30^\circ = 38 \text{ m s}^{-1}$ .

Answer

v_N = 65 \text{ m s}^{-1},\ \ v_E = 38 \text{ m s}^{-1}

(9702/12/F/M/21 Q3)

Common mistake

The cos/sin swap. Before you write anything, check which axis the angle is measured from: that axis gets the cosine. Here the angle was from north, so north got the cosine, not east.