Physics 9702/22 — May/June 2023

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

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

Q1Forces, Density and PressurePhysical Quantities and UnitsFree sample

(a)

(i)

Define pressure.

DifficultyEasy

Worked solution

Answer

Pressure is force per unit area (normal to the surface):

p = \frac{F}{A}

Final answer

Pressure is force per unit area (p = F/A).

Detailed explanation

Background Concept

Pressure measures how concentrated a force is over a surface.

It is defined as the normal (perpendicular) force $F$ acting on a surface divided by the area $A$ over which the force acts:

p = \frac{F}{A}

Understanding the Question

You are asked to define pressure. A definition should state what pressure is in terms of measurable quantities (force and area).

Approach

Write the standard definition: force per unit area, with force taken perpendicular to the surface.

Step-by-Step Reasoning

Pressure compares a force with the area it is spread over.
For the same force, a smaller area gives a larger pressure.
Hence the definition is:

p = \frac{F}{A}

Key Takeaways

Pressure is not just force; it is force density over area.
Always use the perpendicular component of force.

Common Mistakes

Defining pressure as just “force” (missing the division by area).
Not stating “per unit area” or not implying the force is normal to the surface.

Things to Be Careful About

If a force is at an angle, only the component perpendicular to the surface contributes to pressure.
The definition must be general (not restricted to liquids).

Techniques used

state a definition in terms of physical quantitiesidentify the relevant force and area in a pressure definition

(ii)

Use the answer to (a)(i) to show that the SI base units of pressure are $\text{kg m}^{-1}\ \text{s}^{-2}$ .

DifficultyMedium-Easy

Worked solution

Working

From (a)(i),

p = \frac{F}{A}

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

[p] = \frac{\text{kg m s}^{-2}}{\text{m}^2} = \text{kg m}^{-1}\ \text{s}^{-2}

Answer

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

Final answer

kg m^-1 s^-2

Detailed explanation

Background Concept

Many quantities in physics are derived from base quantities.

Pressure is derived from force and area:

p = \frac{F}{A}

To get SI base units, we express every quantity using only $\text{kg}$ , $\text{m}$ , $\text{s}$ (and other base units if needed).

Force has SI unit newton:

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

Area has unit $\text{m}^2$ .

Understanding the Question

You must use your definition of pressure to show the SI base units of pressure. That means start from $p = F/A$ and substitute the base units of $F$ and $A$ .

Approach

Write $p = F/A$ .
Replace $F$ with $\text{kg m s}^{-2}$ .
Replace $A$ with $\text{m}^2$ .
Simplify the powers of $\text{m}$ and $\text{s}$ .

Step-by-Step Reasoning

Start with the definition:

p = \frac{F}{A}

Units of force:

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

Units of area:

[A] = \text{m}^2

So units of pressure:

[p] = \frac{\text{kg m s}^{-2}}{\text{m}^2} = \text{kg m}^{1-2}\ \text{s}^{-2} = \text{kg m}^{-1}\ \text{s}^{-2}

Key Takeaways

Derived units come from the defining equation.
Always simplify indices carefully when dividing units.

Common Mistakes

Leaving the unit as $\text{N m}^{-2}$ and not converting newtons to base units.
Incorrect handling of indices, e.g. claiming $\text{m}/\text{m}^2 = \text{m}$ .

Things to Be Careful About

The newton is not a base unit; you must rewrite it as $\text{kg m s}^{-2}$ .
Keep track of negative indices: dividing by $\text{m}^2$ subtracts 2 from the power of $\text{m}$ .

Techniques used

use a definition equation to derive unitsexpress newton in SI base unitsdivide by area units to obtain derived units

(b)

A horizontal pipe has length $L$ and a circular cross-section of radius $R$ . A liquid of density $\rho$ flows through the pipe. The mass $m$ of liquid flowing through the pipe in time $t$ is given by

m = \frac{\pi(p_2 - p_1)R^4 \rho t}{8kL}

where $p_1$ and $p_2$ are the pressures at the ends of the pipe and $k$ is a constant.

Determine the SI base units of $k$ .

SI base units = ______

DifficultyMedium

Worked solution

Working

Given

m = \frac{\pi (p_2-p_1)R^4\rho t}{8kL}

Rearrange:

k = \frac{\pi (p_2-p_1)R^4\rho t}{8Lm}

Use SI base units:

[p]=\text{kg m}^{-1}\ \text{s}^{-2},\ [R]=\text{m},\ [\rho]=\text{kg m}^{-3},\ [t]=\text{s},\ [L]=\text{m},\ [m]=\text{kg}

Numerator units:

(\text{kg m}^{-1}\ \text{s}^{-2})(\text{m}^4)(\text{kg m}^{-3})(\text{s})=\text{kg}^2\ \text{s}^{-1}

Denominator units:

(\text{m})(\text{kg})=\text{kg m}

[k]=\frac{\text{kg}^2\ \text{s}^{-1}}{\text{kg m}}=\text{kg m}^{-1}\ \text{s}^{-1}

Answer

$\text{kg m}^{-1}\ \text{s}^{-1}$

Final answer

kg m^-1 s^-1

Detailed explanation

Background Concept

For any physically correct equation, both sides must have the same dimensions (principle of homogeneity). This allows us to find the units of an unknown constant.

Here, the equation relates mass flow $m$ to a pressure difference, geometry, density and time:

m = \frac{\pi (p_2-p_1)R^4\rho t}{8kL}

Pure numbers such as $\pi$ and $8$ have no units.

Understanding the Question

You are told the formula for $m$ and asked for the SI base units of the constant $k$ .

Known quantities and their SI base units:

$m$ is mass: $\text{kg}$
$p$ is pressure: $\text{kg m}^{-1}\ \text{s}^{-2}$
$R$ and $L$ are lengths: $\text{m}$
$\rho$ is density: $\text{kg m}^{-3}$
$t$ is time: $\text{s}$

Approach

Rearrange the given equation to make $k$ the subject.
Replace each symbol by its SI base units.
Combine the powers of $\text{kg}$ , $\text{m}$ , $\text{s}$ using index laws.

Step-by-Step Reasoning

Start with

m = \frac{\pi (p_2-p_1)R^4\rho t}{8kL}

Rearrange for $k$ (multiply both sides by $kL$ and divide by $mL$ ):

k = \frac{\pi (p_2-p_1)R^4\rho t}{8Lm}

Now substitute units.

Pressure:

[p] = \text{kg m}^{-1}\ \text{s}^{-2}

Geometry and other quantities:

[R] = \text{m},\quad [R^4] = \text{m}^4,\quad [\rho]=\text{kg m}^{-3},\quad [t]=\text{s},\quad [L]=\text{m},\quad [m]=\text{kg}

Numerator:

(\text{kg m}^{-1}\ \text{s}^{-2})(\text{m}^4)(\text{kg m}^{-3})(\text{s})

Combine $\text{kg}$ : $\text{kg}\times \text{kg} = \text{kg}^2$ .

Combine $\text{m}$ powers: $\text{m}^{-1+4-3} = \text{m}^0$ (so the metre cancels out in the numerator).

Combine $\text{s}$ powers: $\text{s}^{-2+1} = \text{s}^{-1}$ .

So numerator is:

\text{kg}^2\ \text{s}^{-1}

Denominator is $Lm$ :

(\text{m})(\text{kg}) = \text{kg m}

Therefore

[k]=\frac{\text{kg}^2\ \text{s}^{-1}}{\text{kg m}}=\text{kg m}^{-1}\ \text{s}^{-1}

Key Takeaways

Rearranging to make the unknown constant the subject makes unit analysis straightforward.
Constants like $\pi$ and numerical factors do not affect units.
Use index laws carefully when combining powers.

Common Mistakes

Forgetting to include the units of $m$ (mass) in the rearranged expression.
Treating $R^4$ as $4R$ instead of a power: units must become $\text{m}^4$ .
Not using base units for pressure (leaving it as $\text{N m}^{-2}$ without converting to $\text{kg m}^{-1}\ \text{s}^{-2}$ ).

Things to Be Careful About

The symbol $m$ here is mass, not metres; avoid confusing it with the unit $\text{m}$ .
Check each quantity has been included with the correct power (especially $R^4$ and the $L$ in the denominator).
Writing the final answer in base units means only $\text{kg}$ , $\text{m}$ , $\text{s}$ should appear.

Techniques used

rearrange an equation to make the target quantity the subjectsubstitute SI base units for each variablesimplify using index laws to obtain final base units

(c)

An experiment is performed to determine the value of $k$ by measuring the values of the other quantities in the equation in (b).

The values of $L$ and $R$ each have a percentage uncertainty of 2%.

State and explain, quantitatively, which of these two quantities contributes more to the percentage uncertainty in the calculated value of $k$ .

DifficultyMedium-Easy

Worked solution

Working

From

m = \frac{\pi (p_2-p_1)R^4\rho t}{8kL}

k \propto \frac{R^4}{L}

So percentage uncertainty from $L$ is $2\%$ .

For $R^4$ , percentage uncertainty is

4 \times 2\% = 8\%

Answer

$R$ contributes more (its contribution is $8\%$ compared with $2\%$ from $L$ ).

Final answer

R contributes more (8% from R compared with 2% from L).

Detailed explanation

Background Concept

When a quantity depends on measured values raised to powers, percentage uncertainties propagate using simple rules.

y \propto x^n,

then the percentage (fractional) uncertainty in $y$ due to $x$ is multiplied by $|n|$ :

\frac{\Delta y}{y} = |n|\frac{\Delta x}{x} \quad \Rightarrow \quad \%\ \Delta y = |n|\, (\%\ \Delta x).

If a variable is in the denominator (e.g. $1/L$ ), it contributes the same percentage uncertainty as $L$ (power $-1$ gives factor $1$ ).

Understanding the Question

You calculate $k$ using measured quantities. You are told both $L$ and $R$ have the same percentage uncertainty of $2\%$ . The question asks which one affects the percentage uncertainty in $k$ more.

The key clue is that in the formula, $R$ appears as $R^4$ while $L$ appears only as $L^1$ in the denominator.

Approach

Rearrange the given equation to see how $k$ depends on $R$ and $L$ .
Use the power rule: multiply the percentage uncertainty in a variable by the magnitude of its power.
Compare the two contributions.

Step-by-Step Reasoning

Start with

m = \frac{\pi (p_2-p_1)R^4\rho t}{8kL}

Rearrange to show dependence of $k$ :

k = \frac{\pi (p_2-p_1)R^4\rho t}{8Lm}

So, focusing only on $R$ and $L$ :

k \propto \frac{R^4}{L}

For $L$ (power $-1$ ): percentage uncertainty contribution is $1 \times 2\% = 2\%$ .
For $R^4$ (power $+4$ ): percentage uncertainty contribution is $4 \times 2\% = 8\%$ .

Since $8\% > 2\%$ , the radius measurement dominates the uncertainty in $k$ .

Key Takeaways

A measurement raised to a power amplifies its percentage uncertainty by that factor.
Even if two measurements have the same percentage uncertainty, the one with the larger power in the formula contributes more.

Common Mistakes

Saying both contribute equally because both are $2\%$ (ignoring the power of 4 on $R$ ).
Thinking that being in the denominator changes the percentage uncertainty magnitude (it does not; it only changes the sign of the power).

Things to Be Careful About

The question asks which contributes more, not the total uncertainty in $k$ .
Make sure you use percentage uncertainty rules (not absolute uncertainty rules) here.
Use the magnitude of the power: $R^4$ gives a factor 4, $L^{-1}$ gives a factor 1.

Techniques used

rearrange to identify powers of variables in a formulause the power rule for percentage uncertainty propagationcompare contributions to decide the dominant uncertainty source

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