Physics 9702/21 — May/June 2020

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

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Topics Electricity · Work, Energy and Power · Physical Quantities and Units · Measurement Techniques · Dynamics · Kinematics · +7 more

Q1Physical Quantities and UnitsElectricityMeasurement TechniquesFree sample

Answer all the questions in the spaces provided.

(a)

Use an expression for work done, in terms of force, to show that the SI base units of energy are $\text{kg m}^2\ \text{s}^{-2}$ .

DifficultyMedium-Easy

Worked solution

Working

Work done:

W = Fs

[F] = \text{N} = \text{kg m s}^{-2}

[W] = [F][s] = (\text{kg m s}^{-2})(\text{m}) = \text{kg m}^2\ \text{s}^{-2}

Answer

SI base units of energy are $\text{kg m}^2\ \text{s}^{-2}$ .

Final answer

kg m^2 s^-2

Detailed explanation

Background Concept

Energy is measured in joules (J), which is a derived SI unit. A reliable way to find SI base units is to start from a defining equation and replace each quantity by its SI base units.

For work done by a force,

W = Fs

where $W$ is work done (energy transferred), $F$ is force, and $s$ is distance moved in the direction of the force.

Also,

1\ \text{N} = 1\ \text{kg m s}^{-2}

from Newton's second law $F=ma$ .

Understanding the Question

You are asked to show that the SI base units of energy are $\text{kg m}^2\ \text{s}^{-2}$ , using an expression for work done in terms of force. That means you must:

write the work equation,
replace force with its SI base units,
multiply by the metre from distance.

Approach

Use $W=Fs$ . Convert $F$ from newtons to base units using $F=ma$ , then multiply by $s$ (metres) to obtain the base units for $W$ .

Step-by-Step Reasoning

Start with

W = Fs

Force:

F = ma

Units of mass $m$ are $\text{kg}$ . Units of acceleration $a$ are $\text{m s}^{-2}$ . So

[F] = \text{kg} \cdot \text{m s}^{-2} = \text{kg m s}^{-2}

Distance $s$ has units $\text{m}$ . Therefore

[W] = [F][s] = (\text{kg m s}^{-2})(\text{m}) = \text{kg m}^2\ \text{s}^{-2}

This is the SI base-unit form of the joule.

Key Takeaways

Use a defining equation (here $W=Fs$ ) to determine derived units.
Convert derived units like newtons into base units using fundamental definitions (here $F=ma$ ).
Keep careful track of indices when multiplying units.

Common Mistakes

Using the wrong work equation (e.g. confusing with $P=W/t$ ) and getting extra seconds.
Forgetting that $s$ is distance in metres, so missing one factor of $\text{m}$ .
Writing $\text{N} = \text{kg m}^2 \text{s}^{-2}$ (incorrect: that is a joule, not a newton).

Things to Be Careful About

$W=Fs$ assumes the force and displacement are along the same line (or you use the component along the displacement). For units, the same base units result.
Write units in negative-index form (e.g. $\text{s}^{-2}$ ) as expected in Cambridge mark schemes.

Techniques used

use the definition of work done as force times distance moved in the force directionsubstitute SI base units for derived quantitiessimplify unit expressions using index laws

(b)

(i)

The energy $E$ stored in an electrical component is given by

E = \frac{Q^2}{2C}

where $Q$ is charge and $C$ is a constant.

Use this equation and the information in (a) to determine the SI base units of $C$ .

SI base units = ______

DifficultyMedium-Easy

Worked solution

Working

Given

E = \frac{Q^2}{2C}

C = \frac{Q^2}{2E}

Units: $[E]=\text{kg m}^2\ \text{s}^{-2}$ and $[Q]=\text{C}=\text{A s}$ .

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

Answer

$\text{A}^2\ \text{s}^4\ \text{kg}^{-1}\ \text{m}^{-2}$

Final answer

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

Detailed explanation

Background Concept

When a formula links physical quantities, the equation must be dimensionally consistent. You can find the SI base units of an unknown constant by rearranging the formula and substituting SI base units for all known quantities.

Key base-unit facts used here:

Energy $E$ (joule) has base units $\text{kg m}^2\ \text{s}^{-2}$ .
Charge $Q$ is related to current $I$ by

Q = It

So $1\ \text{C} = 1\ \text{A s}$ .

Understanding the Question

You are given

E = \frac{Q^2}{2C}

and told that $C$ is a constant (here it has the same dimensions as capacitance). Using the base units of energy from part (a) and the SI base units of charge, you must determine the SI base units of $C$ .

Approach

Rearrange the equation to make $C$ the subject. Then replace $Q$ with $\text{A s}$ and $E$ with $\text{kg m}^2\ \text{s}^{-2}$ . The factor of 2 has no units, so it does not affect the unit calculation.

Step-by-Step Reasoning

Start with

E = \frac{Q^2}{2C}

Rearrange for $C$ :

C = \frac{Q^2}{2E}

Now substitute units:

$[Q] = \text{A s}$ so $[Q^2] = \text{A}^2\ \text{s}^2$ .
$[E] = \text{kg m}^2\ \text{s}^{-2}$ .

Therefore

[C] = \frac{\text{A}^2\ \text{s}^2}{\text{kg m}^2\ \text{s}^{-2}}

Dividing by $\text{s}^{-2}$ is equivalent to multiplying by $\text{s}^2$ :

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

Key Takeaways

Constants in equations can have units; find them by rearranging and substituting base units.
Use $Q=It$ to express charge in base units ( $\text{A s}$ ).
Pure numbers (like 2) do not affect units.

Common Mistakes

Treating coulomb (C) as a base unit instead of converting to $\text{A s}$ .
Forgetting to square the charge units: $(\text{A s})^2$ .
Handling indices incorrectly when dividing by $\text{s}^{-2}$ (you must add 2 to the power of $s$ ).

Things to Be Careful About

Write the final answer entirely in SI base units (kg, m, s, A).
Keep the order and indices clear: $\text{A}^2\ \text{s}^4\ \text{kg}^{-1}\ \text{m}^{-2}$ is equivalent to $\text{kg}^{-1}\ \text{m}^{-2}\ \text{s}^4\ \text{A}^2$ .

Techniques used

apply dimensional analysis to a given formulause SI base units for charge in terms of current and timerearrange an equation to isolate the unknown constant and its units

(ii)

Measurements of a constant current in a wire are taken using an analogue ammeter.

For these measurements, describe one possible cause of:

a random error
a systematic error.

DifficultyMedium-Easy

Worked solution

Answer

Random error: difficulty judging the pointer position on the analogue scale (limited resolution / pointer thickness), so repeated readings vary slightly.
Systematic error: zero error of the ammeter (pointer not at zero when no current flows), so all readings are offset by the same amount.

Final answer

Random: reading uncertainty of pointer position; Systematic: zero error of ammeter.

Detailed explanation

Background Concept

A measurement error is the difference between a measured value and the true value.

Random errors cause readings to scatter about a mean value. They affect precision (how close repeated readings are to each other). If you repeat measurements and average, the effect of random error is reduced.
Systematic errors shift all readings in the same direction (all too large or all too small). They affect accuracy (how close the mean is to the true value). Repeating and averaging does not remove systematic error; you must correct the cause (e.g. re-zero, recalibrate).

For an analogue ammeter, the main issues come from reading the pointer and from the instrument calibration/zero.

Understanding the Question

The current is stated to be constant, and it is measured using an analogue ammeter. You must give:

one possible cause of a random error in these readings,
one possible cause of a systematic error.

The question is about the measurement process with an analogue meter, not about fluctuations in the circuit design.

Approach

Pick one credible, specific instrument-related source for each type:

Random: something that varies unpredictably from reading to reading (e.g. judgement of pointer position).
Systematic: something that biases every reading the same way (e.g. zero error or miscalibration).

Step-by-Step Reasoning

Random error example (analogue scale reading):

The pointer has a finite thickness and the scale has finite spacing.
When you read the scale, you must estimate between divisions.
Small differences in where your eye is and how you judge the pointer position lead to slightly different values each time.
Result: readings are spread around a mean value.

Systematic error example (zero error):

If the ammeter does not read exactly zero when no current flows, then every current reading is shifted by that zero offset.
For example, if it reads $+0.02\ \text{A}$ at zero, then every measurement is $0.02\ \text{A}$ too high.
Result: repeated readings are consistent (precise) but all wrong by the same amount.

Other acceptable systematic causes could include a miscalibrated ammeter scale or consistently viewing the scale at an angle (consistent parallax).

Key Takeaways

Random error: produces scatter; reduced by repeats and averaging.
Systematic error: produces bias; fixed by correcting the instrument or method.
Analogue meters commonly produce random reading uncertainty and systematic zero/calibration errors.

Common Mistakes

Giving a vague answer like “human error” without specifying what physically causes the error.
Mixing up parallax: if the viewing angle changes each time it can be random; if you always view from the same wrong angle it becomes systematic.
Saying “random error is due to zero error” (zero error is systematic).

Things to Be Careful About

The question specifies a constant current, so do not rely solely on “current fluctuates” unless you clearly link it to measurement noise or pointer fluctuations.
Make sure your systematic error clearly implies all readings shift in the same direction by the same amount.

Techniques used

distinguish between random and systematic errorsidentify instrument-related causes of measurement errorlink an error source to its effect on repeated readings

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