Answer all the questions in the spaces provided. (a) State what is meant by a scalar quantity and by a vector quantity. scalar: \_\_\_\_\_\_ vector: \_\_\_\_\_\_ (b) Complete Fig. 1.1 to indicate whether each of the quantities is a vector or a scalar. quantity vector or scalar --- --- power temperature momentum (c) An aircraft is travelling in wind. Fig. 1.2 shows the velocities for the aircraft in still air and for the wind. The velocity of the aircraft in still air is $95\…

(a) Scalar: magnitude only. Vector: magnitude and direction. · (b) power: scalar; temperature: scalar; momentum: vector · (c)(i) Resultant velocity direction is between the two given vectors (north of west). · (c)(ii) 1.10 × 10^2 m s^-1

Physics 9702/21 — May/June 2018

Cambridge AS Level · AS Level Structured Questions · worked solutions for every part, with the mark scheme

questions

marks

minutes

Topics Physical Quantities and Units · Kinematics · Dynamics · Work, Energy and Power · Superposition · Deformation of Solids · +5 more

Q1Physical Quantities and UnitsKinematicsFree sample

Answer all the questions in the spaces provided.

(a)

State what is meant by a scalar quantity and by a vector quantity.

scalar: ______

vector: ______

DifficultyEasy

Worked solution

Answer

Scalar: a quantity with magnitude only.

Vector: a quantity with magnitude and direction.

Final answer

Scalar: magnitude only. Vector: magnitude and direction.

Detailed explanation

Background Concept

A physical quantity is something that can be measured and expressed with a number and a unit.

A scalar quantity is fully described by its magnitude (size) only.
A vector quantity requires both magnitude and direction to be fully described.

This distinction matters because vectors must be added using vector rules (e.g. triangle/parallelogram methods or components), while scalars add by ordinary arithmetic.

Understanding the Question

You are asked to state what is meant by:

a scalar quantity
a vector quantity

So you need short, precise definitions (no examples are required for full credit here).

Approach

Give the defining feature of each:

scalar → magnitude only
vector → magnitude + direction

Step-by-Step Reasoning

Identify what information is needed to specify the quantity.
If only a size is needed, it is a scalar.
If you must also specify which way it acts/points, it is a vector.

Key Takeaways

Scalars: magnitude only.
Vectors: magnitude and direction.
Vector operations (addition/resolution) depend on direction.

Common Mistakes

Saying a vector is “something that moves” (not a definition).
Giving only examples (e.g. “speed is scalar”) without defining the terms.
Forgetting to mention direction for a vector.

Things to Be Careful About

Use the word magnitude explicitly.
For vectors, include direction (or “has direction”).

Techniques used

recall definitions of scalar and vector quantitiesdistinguish quantities by presence or absence of direction

(b)

Complete Fig. 1.1 to indicate whether each of the quantities is a vector or a scalar.

quantity	vector or scalar
power
temperature
momentum

Fig. 1.1

DifficultyEasy

Worked solution

Answer

power: scalar
temperature: scalar
momentum: vector

Final answer

power: scalar; temperature: scalar; momentum: vector

Detailed explanation

Background Concept

A scalar has magnitude only; a vector has magnitude and direction.

Some common classifications:

Temperature has no direction → scalar.
Power is the rate of energy transfer; energy transfer rate has no direction → scalar.
Momentum is defined by $\vec{p} = m\vec{v}$ , and velocity is a vector → momentum is a vector.

Understanding the Question

You must complete the table by stating whether each listed quantity is a vector or a scalar:

power
temperature
momentum

Approach

For each quantity, ask: “Do I need a direction to describe it fully?”

If no → scalar
If yes → vector

Step-by-Step Reasoning

Power: measured in $\text{W}$ , describes how fast work is done/energy is transferred; no direction is needed → scalar.
Temperature: measured in $^\circ\text{C}$ or $\text{K}$ ; no direction → scalar.
Momentum: $\vec{p} = m\vec{v}$ , so it points in the same direction as velocity → vector.

Key Takeaways

Link momentum to velocity: vectors multiplied by scalars remain vectors.
Rates like power are generally scalars unless explicitly defined with direction (not the case here).

Common Mistakes

Thinking “power involves motion so it must be a vector” (it does not have a direction).
Confusing momentum with speed (speed is scalar, velocity is vector).

Things to Be Careful About

Momentum must be treated as a vector in later questions (e.g. collisions), so always attach direction when required.

Techniques used

classify quantities as scalar or vector based on whether they have directionuse definitions of physical quantities (e.g. momentum as a vector)

(c)

An aircraft is travelling in wind. Fig. 1.2 shows the velocities for the aircraft in still air and for the wind.

Fig. 1.2

The velocity of the aircraft in still air is $95\ \text{m s}^{-1}$ to the west.
The velocity of the wind is $28\ \text{m s}^{-1}$ from $65^\circ$ south of east.

(i)

On Fig. 1.2, draw an arrow, labelled R, in the direction of the resultant velocity of the aircraft.

DifficultyMedium-Easy

Worked solution

diagram-c-i

Answer

$\vec{R}$ is drawn from the common origin in a direction between the two given velocity vectors (towards the north of west).

Final answer

Resultant velocity direction is between the two given vectors (north of west).

Detailed explanation

Background Concept

Velocities are vectors, so the resultant (ground) velocity of the aircraft is found by vector addition:

\vec{v}_{\text{resultant}} = \vec{v}_{\text{aircraft relative air}} + \vec{v}_{\text{wind}}

Graphically, vector addition is done by the head-to-tail method (or parallelogram method). The resultant points from the start of the first vector to the end of the second (when placed head-to-tail).

Understanding the Question

The diagram shows:

aircraft velocity in still air: $95\ \text{m s}^{-1}$ to the west
wind velocity: $28\ \text{m s}^{-1}$ in the direction shown

You must draw an arrow labelled $R$ showing the direction of the resultant aircraft velocity (relative to the ground).

Approach

On the diagram, the resultant must be:

the vector sum of the two given vectors
drawn from the same origin (if using the parallelogram method), or from the origin to the final point after a head-to-tail construction

Its direction will lie between the directions of the two vectors being added.

Step-by-Step Reasoning

Treat both velocities as vectors starting at the same point.
Construct the parallelogram (or imagine placing one vector at the head of the other).
The diagonal from the common origin gives the resultant direction.
Since the aircraft velocity is due west and the wind has a component towards the north of west (as stated/drawn), the resultant points slightly north of west.

diagram-c-i

Key Takeaways

Velocities add as vectors, not scalars.
The resultant direction is found by a vector diagram (triangle/parallelogram).

Common Mistakes

Adding magnitudes $95 + 28$ and using that as the resultant direction/magnitude.
Drawing the resultant along one of the given vectors rather than between them.
Drawing the resultant from the wrong point (not from the common origin).

Things to Be Careful About

The arrow must be clearly labelled $R$ .
The direction must correspond to the correct vector sum (diagonal of the parallelogram / start-to-end of head-to-tail addition).

Techniques used

add velocity vectors head-to-tailidentify the direction of a resultant vector from a vector diagram

(ii)

Determine the magnitude of the resultant velocity of the aircraft.

magnitude of velocity = ______ $\text{m s}^{-1}$

DifficultyMedium

Worked solution

Working

Resolve wind velocity ( $28\ \text{m s}^{-1}$ ) into components:

v_{\text{W, west}} = 28\cos 65^\circ = 11.8\ \text{m s}^{-1}

v_{\text{W, north}} = 28\sin 65^\circ = 25.4\ \text{m s}^{-1}

Resultant components:

v_{\text{west}} = 95 + 11.8 = 106.8\ \text{m s}^{-1}

v_{\text{north}} = 25.4\ \text{m s}^{-1}

Magnitude:

|\vec{v}| = \sqrt{(106.8)^2 + (25.4)^2} = 1.10 \times 10^2\ \text{m s}^{-1}

Answer

$1.10 \times 10^2\ \text{m s}^{-1}$

Final answer

1.10 × 10^2 m s^-1

Detailed explanation

Background Concept

To find the magnitude of the sum of two vectors, a reliable method is to resolve each vector into perpendicular components (often East–West and North–South).

If a vector $V$ makes an angle $\theta$ to the horizontal, then:

horizontal component: $V\cos\theta$
vertical component: $V\sin\theta$

After adding components in each perpendicular direction, the magnitude of the resultant is found using Pythagoras:

R = \sqrt{R_x^2 + R_y^2}

Understanding the Question

Given (from the stem):

aircraft velocity in still air: $95\ \text{m s}^{-1}$ to the west
wind velocity: $28\ \text{m s}^{-1}$ from $65^\circ$ south of east (so the wind velocity has components towards west and north, as represented in the vector diagram)

You must determine the magnitude of the aircraft’s resultant velocity (relative to the ground).

Approach

Choose axes: West–East and North–South.
Resolve the wind velocity into west/east and north/south components using $\cos$ and $\sin$ .
Add the wind components to the aircraft’s westward component.
Use Pythagoras to find the magnitude of the resultant velocity.

Step-by-Step Reasoning

Take west as positive in the horizontal direction and north as positive vertically.

Aircraft velocity components:

West component: $95\ \text{m s}^{-1}$
North component: $0$

Wind velocity components (magnitude $28\ \text{m s}^{-1}$ , angle $65^\circ$ ):

v_{\text{W, west}} = 28\cos 65^\circ

v_{\text{W, north}} = 28\sin 65^\circ

Numerically:

v_{\text{W, west}} = 28\cos 65^\circ = 11.8\ \text{m s}^{-1}

v_{\text{W, north}} = 28\sin 65^\circ = 25.4\ \text{m s}^{-1}

Resultant components:

v_{\text{west}} = 95 + 11.8 = 106.8\ \text{m s}^{-1}

v_{\text{north}} = 25.4\ \text{m s}^{-1}

Resultant magnitude:

|\vec{v}| = \sqrt{(106.8)^2 + (25.4)^2} = 109.8\ \text{m s}^{-1}

Rounded to 3 s.f.:

|\vec{v}| = 1.10 \times 10^2\ \text{m s}^{-1}

Key Takeaways

Use components to add vectors cleanly.
Add components in each perpendicular direction separately.
Use Pythagoras to find the magnitude of the resultant.

Common Mistakes

Using $\sin$ and $\cos$ the wrong way round for the chosen angle.
Adding $95$ and $28$ directly (treating vectors as scalars).
Forgetting that the resultant magnitude needs Pythagoras after finding components.

Things to Be Careful About

Check that your angle is measured from the axis you are using (horizontal vs vertical).
Keep units throughout and give the final answer in $\text{m s}^{-1}$ .
Round appropriately (typically 2–3 significant figures).

Techniques used

resolve a vector into perpendicular componentsadd vector components to obtain a resultantuse Pythagoras to find the magnitude of a resultant vector

The rest of this paper

6 more questions

Q2Dynamics · Kinematics · Work, Energy and Power · Deformation of Solids15M
Q3Dynamics9M
Q4Waves · Superposition8M
Q5Superposition5M
Q6Electricity · D.C. Circuits10M
Q7Electric Fields · Work, Energy and Power · Physical Quantities and Units · Particle Physics6M

Loading the full paper…