Physics 9702/22 — October/November 2022

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

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

Q1Work, Energy and PowerPhysical Quantities and UnitsFree sample

(a)

State what is meant by work done.

DifficultyEasy

Worked solution

Answer

Work done is the product of the force and the displacement in the direction of the force:

W = Fs\cos\theta.

Final answer

Work done is force multiplied by displacement in the direction of the force, i.e. W = Fs cosθ.

Detailed explanation

Background Concept

Work done $W$ is the energy transferred when a force causes a displacement. If a force $F$ acts and the object moves a distance $s$ , only the component of the force along the displacement contributes.

Mathematically:

W = Fs\cos\theta

where $\theta$ is the angle between the force and the displacement.

Understanding the Question

You are asked to state what “work done” means, i.e. give a definition/formula that connects work done to force and displacement.

Approach

Use the standard definition: work done equals the force component in the direction of motion multiplied by the displacement.

Step-by-Step Reasoning

Resolve the force along the direction of displacement: component is $F\cos\theta$ .
Multiply by displacement $s$ to get work done:

W = (F\cos\theta)s = Fs\cos\theta.

Key Takeaways

Work done is energy transferred by a force.
Only the force component parallel to the displacement does work.

Common Mistakes

Stating $W = Fs$ without mentioning “in the direction of the force” (only true when $\theta = 0$ ).
Confusing work done with power.

Things to Be Careful About

The angle $\theta$ is between force and displacement, not between force and some axis.
If the force is perpendicular to motion ( $\theta = 90^\circ$ ), then $W = 0$ .

Techniques used

state the definition of work donerelate work done to force and displacement in the direction of the force

(b)

Use the answer to (a) to determine the SI base units of power.

SI base units = ______

DifficultyMedium-Easy

Worked solution

Working

Power:

P = \frac{W}{t}

Work:

W = Fs

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

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

Hence

[P] = \frac{\text{kg m}^2\text{ s}^{-2}}{\text{s}} = \text{kg m}^2\text{ s}^{-3}.

Answer

$\text{kg m}^2\text{ s}^{-3}$

Final answer

kg m^2 s^-3

Detailed explanation

Background Concept

Power $P$ is the rate of doing work (rate of energy transfer):

P = \frac{W}{t}.

Work done $W$ by a constant force is:

W = Fs\cos\theta.

For units, the cosine factor is dimensionless, so it does not affect the unit calculation.

Understanding the Question

You must use the definition of work done from part (a) and then use $P = W/t$ to find the SI base units for power.

Approach

Start with $P = W/t$ .
Replace $W$ with $Fs$ .
Convert newtons into base units ( $\text{kg m s}^{-2}$ ).
Simplify to get base units for $P$ .

Step-by-Step Reasoning

From definitions:

P = \frac{W}{t}, \quad W = Fs.

P = \frac{Fs}{t}.

Unit of force:

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

Therefore unit of work:

\text{J} = \text{N m} = (\text{kg m s}^{-2})\,\text{m} = \text{kg m}^2\text{ s}^{-2}.

Now divide by time ( $\text{s}$ ):

\text{W} = \frac{\text{J}}{\text{s}} = \frac{\text{kg m}^2\text{ s}^{-2}}{\text{s}} = \text{kg m}^2\text{ s}^{-3}.

Key Takeaways

Use definition equations to obtain units.
Convert derived units (N, J) into SI base units.
Power in base units is $\text{kg m}^2\text{ s}^{-3}$ .

Common Mistakes

Leaving the answer as $\text{J s}^{-1}$ instead of SI base units.
Using $W = F/s$ or other incorrect rearrangements.

Things to Be Careful About

Ensure you divide by time once more at the end (to get $\text{s}^{-3}$ , not $\text{s}^{-2}$ ).
Keep track that newton already contains $\text{s}^{-2}$ .

Techniques used

use a definition equation to derive unitsconvert derived units into SI base unitsuse dimensional analysis to simplify units

(c)

The maximum useful output power $P$ of a car travelling on a horizontal road is given by

P = v^3b

where $v$ is the maximum speed of the car and $b$ is a constant.

For the car,

$P = 84\text{ kW} \pm 5\%$
and $b = 0.56 \pm 7\%$ in SI units.

(i)

Calculate the value of $v$ .

$v$ = ______ $\text{m s}^{-1}$

DifficultyMedium-Easy

Worked solution

Working

P = bv^3 \;\Rightarrow\; v = \left(\frac{P}{b}\right)^{1/3}

P = 84\,\text{kW} = 8.4 \times 10^4\,\text{W}

v = \left(\frac{8.4 \times 10^4}{0.56}\right)^{1/3} = \left(1.5 \times 10^5\right)^{1/3} \approx 53\,\text{m s}^{-1}

Answer

$53\,\text{m s}^{-1}$

Final answer

53 m s^-1

Detailed explanation

Background Concept

Many models in mechanics relate power to speed. Here you are given a specific relationship:

P = bv^3

where $P$ is power in watts, $v$ is speed in $\text{m s}^{-1}$ , and $b$ is a constant (with SI units chosen so the equation is dimensionally consistent).

To find $v$ , you rearrange the equation and substitute numerical values.

Understanding the Question

You are told:

$P = 84\,\text{kW}$ (ignore the uncertainty for part (i))
$b = 0.56$ in SI units

You must calculate the maximum speed $v$ using the given formula.

Approach

Rearrange $P = bv^3$ to make $v$ the subject.
Convert $84\,\text{kW}$ to watts.
Substitute into the formula and take the cube root.

Step-by-Step Reasoning

Rearrange:

P = bv^3 \Rightarrow v^3 = \frac{P}{b} \Rightarrow v = \left(\frac{P}{b}\right)^{1/3}.

Convert power to SI:

84\,\text{kW} = 84 \times 10^3\,\text{W} = 8.4 \times 10^4\,\text{W}.

Substitute:

v = \left(\frac{8.4 \times 10^4}{0.56}\right)^{1/3}.

Compute inside the brackets:

\frac{8.4 \times 10^4}{0.56} = 1.5 \times 10^5.

Take cube root:

v = (1.5 \times 10^5)^{1/3} \approx 53\,\text{m s}^{-1}.

Key Takeaways

Always convert to SI units before substituting.
For $v^3$ relationships, solve by taking the cube root.

Common Mistakes

Using $84\,\text{kW} = 84 \times 10^4\,\text{W}$ (wrong by a factor of 10).
Forgetting the cube root and instead dividing by 3.

Things to Be Careful About

Check the final unit is $\text{m s}^{-1}$ (speed).
Do not include uncertainties in part (i) unless asked; they are used in part (ii).

Techniques used

rearrange an equation to make the required variable the subjectconvert between prefixed and SI unitsevaluate a cube root numerically

(ii)

Determine the absolute uncertainty in the value of $v$ .

absolute uncertainty = ______ $\text{m s}^{-1}$

DifficultyMedium

Worked solution

Working

v = \left(\frac{P}{b}\right)^{1/3}

For $\dfrac{P}{b}$ , fractional uncertainties add:

\frac{\Delta (P/b)}{(P/b)} = \frac{\Delta P}{P} + \frac{\Delta b}{b} = 5\% + 7\% = 12\%

For a power $1/3$ :

\frac{\Delta v}{v} = \frac{1}{3} \times 12\% = 4\%

Using $v \approx 53\,\text{m s}^{-1}$ :

\Delta v = 0.04 \times 53 \approx 2.1\,\text{m s}^{-1}

Answer

$2.1\,\text{m s}^{-1}$

Final answer

2.1 m s^-1

Detailed explanation

Background Concept

When a quantity is calculated from measured values, its uncertainty depends on how those values are combined.

Key rules (Cambridge A Level standard):

For multiplication or division, percentage (fractional) uncertainties add.
If $y = x^n$ , then the percentage uncertainty in $y$ is $|n|$ times the percentage uncertainty in $x$ .

Here,

v = \left(\frac{P}{b}\right)^{1/3}.

So we first find the percentage uncertainty in $P/b$ , then multiply by $1/3$ .

Understanding the Question

You are given percentage uncertainties:

$P = 84\,\text{kW} \pm 5\%$
$b = 0.56 \pm 7\%$

You must find the absolute uncertainty in $v$ in $\text{m s}^{-1}$ .

Approach

Write $v$ in terms of $P$ and $b$ .
Add percentage uncertainties for the division $P/b$ .
Apply the power rule for the cube root (power $1/3$ ).
Convert the percentage uncertainty in $v$ into an absolute uncertainty using the calculated value of $v$ .

Step-by-Step Reasoning

Start with:

v = \left(\frac{P}{b}\right)^{1/3}.

Uncertainty in the bracketed term $P/b$ (division):

\%\,\text{uncertainty in }\frac{P}{b} = \%\,\text{uncertainty in }P + \%\,\text{uncertainty in }b = 5\% + 7\% = 12\%.

Now apply the power rule for $v = (P/b)^{1/3}$ :

\%\,\text{uncertainty in }v = \frac{1}{3} \times 12\% = 4\%.

Convert to absolute uncertainty using $v \approx 53\,\text{m s}^{-1}$ from part (i):

\Delta v = 0.04 \times 53 \approx 2.1\,\text{m s}^{-1}.

Key Takeaways

Add percentage uncertainties for a quotient $P/b$ .
Multiply the resulting percentage uncertainty by $1/3$ because of the cube root.
Absolute uncertainty is found by multiplying the percentage uncertainty (as a fraction) by the value.

Common Mistakes

Subtracting uncertainties because it is a division (they still add).
Forgetting to apply the $1/3$ factor for the cube root.
Giving the answer as a percentage instead of an absolute uncertainty in $\text{m s}^{-1}$ .

Things to Be Careful About

Use the value of $v$ you calculated in (i) (or an ECF value if your (i) differs).
Convert $4\%$ to $0.04$ before multiplying.
Quote the uncertainty to a sensible number of significant figures (usually 2 s.f. is fine here).

Techniques used

propagate percentage uncertainties through a power lawcombine percentage uncertainties for multiplication or divisionconvert percentage uncertainty to absolute uncertainty

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