Physics 9702/22 — October/November 2019

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

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Topics Dynamics · Work, Energy and Power · Physical Quantities and Units · Electric Fields · Kinematics · Forces, Density and Pressure · +5 more

Q1Physical Quantities and UnitsElectric FieldsFree sample

Answer all the questions in the spaces provided.

(a)

Distinguish between vector and scalar quantities.

DifficultyEasy

Worked solution

Answer

A scalar quantity has magnitude only.

A vector quantity has both magnitude and direction.

Final answer

Scalar: magnitude only. Vector: magnitude and direction.

Detailed explanation

Background Concept

A physical quantity is something measurable (e.g. mass, force, velocity). Quantities fall into two main types:

Scalars: completely described by a single number (magnitude) with a unit.
Vectors: need both a magnitude and a direction to be fully described.

Vectors also obey vector addition rules (e.g. head-to-tail addition), whereas scalars add using ordinary arithmetic.

Understanding the Question

The question asks for a clear distinction between scalar and vector quantities. For full credit you must mention the key difference: direction.

Approach

Give the definitions in one or two concise sentences:

define scalar (magnitude only),
define vector (magnitude + direction).

Step-by-Step Reasoning

For a scalar like mass, writing $m = 2.0\ \text{kg}$ fully specifies it; there is no direction involved.
For a vector like force, writing $F = 5.0\ \text{N}$ is incomplete unless you also specify a direction (e.g. “to the right”).
So the distinguishing feature is whether a direction is required.

Key Takeaways

Scalars: magnitude only.
Vectors: magnitude and direction.
In exams, always explicitly mention “direction” when defining a vector.

Common Mistakes

Saying “a vector has magnitude and unit” (scalars also have units).
Giving examples only (e.g. “force is a vector”) without stating the definition.

Things to Be Careful About

Use the word direction (or “acts in a particular direction”) to secure the mark.
Some quantities that sound directional (e.g. speed) are scalars; the vector version is velocity.

Techniques used

state definitions of scalar and vector quantitiescontrast required properties (magnitude-only vs magnitude-and-direction)

(b)

The electric field strength $E$ at a distance $x$ from an isolated point charge $Q$ is given by the equation

E = \frac{Q}{x^2b}

where $b$ is a constant.

(i)

Use the definition of electric field strength to show that $E$ has SI base units of $\text{kg m A}^{-1} \text{ s}^{-3}$ .

DifficultyMedium-Easy

Worked solution

Working

Using the definition

E = \frac{F}{q}

Units of $F$ are $\text{N} = \text{kg m s}^{-2}$ and units of $q$ are $\text{C} = \text{A s}$ .

[E] = \frac{\text{kg m s}^{-2}}{\text{A s}} = \text{kg m A}^{-1}\text{ s}^{-3}

Answer

$E$ has SI base units $\text{kg m A}^{-1}\text{ s}^{-3}$ .

Final answer

kg m A^-1 s^-3

Detailed explanation

Background Concept

Electric field strength $E$ is defined as the force per unit positive charge at a point:

E = \frac{F}{q}

where:

$F$ is the force on the charge (unit: newton, $\text{N}$ ),
$q$ is the charge experiencing the force (unit: coulomb, $\text{C}$ ).

To express units in SI base units, we rewrite derived units:

$\text{N} = \text{kg m s}^{-2}$ ,
$\text{C} = \text{A s}$ .

Understanding the Question

You are asked to use the definition of $E$ (not the given formula involving $Q$ , $x$ and $b$ ) to show that the SI base units of $E$ are $\text{kg m A}^{-1}\text{ s}^{-3}$ .

Approach

Start from $E = F/q$ .
Replace $F$ with $\text{N}$ and then with base units $\text{kg m s}^{-2}$ .
Replace $q$ with $\text{C}$ and then with base units $\text{A s}$ .
Divide the units carefully.

Step-by-Step Reasoning

From the definition:

E = \frac{F}{q}

So the units of $E$ are units of force divided by units of charge:

[E] = \frac{\text{N}}{\text{C}}

Convert each into base units:

$\text{N} = \text{kg m s}^{-2}$ ,
$\text{C} = \text{A s}$ .

Now divide:

[E] = \frac{\text{kg m s}^{-2}}{\text{A s}} = \text{kg m}\, \text{A}^{-1}\, \text{s}^{-3}

because $s^{-2}/s = s^{-3}$ .

Key Takeaways

Always start from the definition $E = F/q$ for unit proofs.
Convert derived units ( $\text{N}$ , $\text{C}$ ) into base units before simplifying.

Common Mistakes

Using $\text{C} = \text{A}/\text{s}$ (incorrect; it is $\text{A s}$ ).
Forgetting that dividing by $s$ makes the power of $s$ more negative (e.g. $s^{-2}/s = s^{-3}$ ).

Things to Be Careful About

Keep track of exponents systematically when dividing units.
Write the final answer explicitly in base units (kg, m, s, A), not in derived units like $\text{N C}^{-1}$ .

Techniques used

use the definition of electric field strengthsubstitute SI units for force and chargereduce derived units to SI base units

(ii)

Use the units for $E$ given in (b)(i) to determine the SI base units of $b$ .

SI base units of $b$ = ______

DifficultyMedium-Easy

Worked solution

Working

From

E = \frac{Q}{x^2 b}

b = \frac{Q}{E x^2}

$[Q] = \text{A s}$ , $[E] = \text{kg m A}^{-1}\text{ s}^{-3}$ , $[x^2] = \text{m}^2$ .

[b] = \frac{\text{A s}}{(\text{kg m A}^{-1}\text{ s}^{-3})(\text{m}^2)} = \frac{\text{A s}}{\text{kg m}^3\text{ A}^{-1}\text{ s}^{-3}} = \text{A}^2\text{ kg}^{-1}\text{ m}^{-3}\text{ s}^4

Answer

SI base units of $b$ = $\text{A}^2\text{ kg}^{-1}\text{ m}^{-3}\text{ s}^4$ .

Final answer

A^2 kg^-1 m^-3 s^4

Detailed explanation

Background Concept

A key idea in units/dimensions is that an equation must be homogeneous: both sides must have the same dimensions. If you know the base units of all but one quantity in an equation, you can find the unknown’s base units by rearranging and substituting units.

Here you are given:

E = \frac{Q}{x^2 b}

and you already found in (b)(i) that:

[E] = \text{kg m A}^{-1}\text{ s}^{-3}

Also:

$Q$ is charge, $[Q] = \text{C} = \text{A s}$ ,
$x$ is distance, $[x] = \text{m}$ so $[x^2] = \text{m}^2$ .

Understanding the Question

You must determine the SI base units of the constant $b$ using the given formula and the unit result for $E$ from part (i).

Approach

Rearrange the equation to make $b$ the subject.
Replace each symbol by its SI base units.
Simplify using index laws (especially handling negative indices correctly).

Step-by-Step Reasoning

Start with:

E = \frac{Q}{x^2 b}

Rearrange for $b$ :

Ex^2 b = Q \quad \Rightarrow \quad b = \frac{Q}{E x^2}

Now substitute units:

$[Q] = \text{A s}$ ,
$[E] = \text{kg m A}^{-1}\text{ s}^{-3}$ ,
$[x^2] = \text{m}^2$ .

[b] = \frac{\text{A s}}{(\text{kg m A}^{-1}\text{ s}^{-3})(\text{m}^2)}

Combine the $m$ terms in the denominator: $m \times m^2 = m^3$ :

[b] = \frac{\text{A s}}{\text{kg m}^3\text{ A}^{-1}\text{ s}^{-3}}

Now deal with the powers when dividing:

dividing by $A^{-1}$ is multiplying by $A^{+1}$ , so $A \times A = A^2$ ,
dividing by $s^{-3}$ is multiplying by $s^{+3}$ , so $s \times s^3 = s^4$ ,
$kg$ and $m^3$ remain in the denominator, giving negative powers.

Hence:

[b] = \text{A}^2\text{ kg}^{-1}\text{ m}^{-3}\text{ s}^4

Key Takeaways

Rearranging to isolate the unknown makes unit analysis straightforward.
Watch negative indices: $1/(A^{-1}) = A$ and $1/(s^{-3}) = s^3$ .

Common Mistakes

Using $[Q] = \text{A s}^{-1}$ (incorrect; charge is current (\times) time).
Forgetting to square $x$ so using $m$ instead of $m^2$ .
Mishandling the negative powers, leading to $A^0$ or $s^{-4}$ incorrectly.

Things to Be Careful About

Combine the length units correctly: $m \times m^2 = m^3$ .
Write the final answer in base units only and with clear indices: $\text{A}^2\text{ kg}^{-1}\text{ m}^{-3}\text{ s}^4$ .

Techniques used

rearrange an equation to make the required quantity the subjectapply dimensional analysis using SI base unitssimplify units using index laws

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