Physics 9702/22 — October/November 2024

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

questions

marks

minutes

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

Q1Medium-EasyPhysical Quantities and UnitsDynamicsForces, Density and Pressure

(a)

State what is meant by a vector quantity.

1M Similar

(b)

A sphere falls vertically through a liquid that has density $830\ \text{kg m}^{-3}$ . The sphere has radius $r$ and constant velocity $v$ , as shown in Fig. 1.1.

Fig. 1.1

(i)

The drag force $D$ acting on the sphere is given by

D = 6\pi r\eta v

where $\eta$ is a property of the liquid.

Determine the SI base units of $\eta$ .

SI base units = ______

3M Similar

DifficultyMedium-Easy

Worked solution

Working

From

D = 6\pi r\eta v

\eta = \frac{D}{6\pi r v}

Units:

[D] = \text{N} = \text{kg m s}^{-2},\quad [r] = \text{m},\quad [v] = \text{m s}^{-1}

[\eta] = \frac{\text{kg m s}^{-2}}{\text{m}\,(\text{m s}^{-1})} = \text{kg m}^{-1}\text{ s}^{-1}

Answer

SI base units of $\eta$ : $\boxed{\text{kg m}^{-1}\text{ s}^{-1}}$

Final answer

kg m^-1 s^-1

Detailed explanation

Background Concept

If an equation is physically correct, it must be dimensionally consistent (homogeneous): both sides have the same units.

Here,

D = 6\pi r\eta v

The factor $6\pi$ is a pure number (dimensionless), so it does not affect units. The units must therefore satisfy

[D] = [r][\eta][v].

Understanding the Question

You are given a formula for the drag force $D$ on a sphere moving through a liquid. You must find the SI base units of $\eta$ (the liquid property, viscosity).

Known SI units:

force $D$ in newtons (N)
radius $r$ in metres (m)
speed $v$ in metres per second ( $\text{m s}^{-1}$ )

Approach

Rearrange the equation to make $\eta$ the subject.
Replace each quantity with its SI units (in base form).
Simplify to get the base units of $\eta$ .

Step-by-Step Reasoning

Start with

D = 6\pi r\eta v.

Rearrange:

\eta = \frac{D}{6\pi r v}.

Now insert units.

$[D] = \text{N} = \text{kg m s}^{-2}$
$[r] = \text{m}$
$[v] = \text{m s}^{-1}$

[\eta] = \frac{\text{kg m s}^{-2}}{\text{m}\,(\text{m s}^{-1})}.

Combine the denominator: $\text{m}\,(\text{m s}^{-1}) = \text{m}^2\text{ s}^{-1}$ .
Then

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

Key Takeaways

Use dimensional analysis: $[D] = [r][\eta][v]$ .
Remember $1\ \text{N} = 1\ \text{kg m s}^{-2}$ .
Numerical constants like $6\pi$ do not affect units.

Common Mistakes

Leaving the unit as $\text{N s m}^{-2}$ without converting to base units (the question asks for SI base units).
Treating $6\pi$ as having units.
Forgetting that $v$ has $\text{s}^{-1}$ in its unit.

Things to Be Careful About

Always reduce force to $\text{kg m s}^{-2}$ .
Track powers of metres carefully: you divide by $r$ and by $v$ , so you divide by $\text{m}$ and by $\text{m s}^{-1}$ .

Techniques used

rearrange an equation to make the required quantity the subjectsubstitute SI base units for each physical quantitysimplify units by cancelling and collecting powers

(ii)

State an equation showing the relationship between the magnitudes of the weight $W$ , drag force $D$ and upthrust $U$ acting on the sphere.

1M Similar

(iii)

The volume of the sphere is $4.6\ \text{cm}^3$ . The drag force $D$ is $0.32\ \text{N}$ .

Calculate the weight of the sphere.

weight = ______ $\text{N}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Volume:

V = 4.6\ \text{cm}^3 = 4.6\times 10^{-6}\ \text{m}^3

Upthrust:

U = \rho g V = 830 \times 9.81 \times 4.6\times 10^{-6} = 3.74\times 10^{-2}\ \text{N}

Constant velocity $\Rightarrow W = D + U$ :

W = 0.32 + 3.74\times 10^{-2} = 0.357\ \text{N} \approx 0.36\ \text{N}

Answer

$\boxed{0.36\ \text{N}}$

Final answer

0.36 N

Detailed explanation

Background Concept

The upthrust on an object in a fluid is equal to the weight of fluid displaced (Archimedes' principle):

U = \rho g V

where $\rho$ is the fluid density, $g$ is gravitational field strength, and $V$ is the volume displaced.

If the object is moving at constant velocity, then $a=0$ and the forces balance:

W = D + U.

Understanding the Question

You are given:

liquid density $\rho = 830\ \text{kg m}^{-3}$
sphere volume $V = 4.6\ \text{cm}^3$
drag force $D = 0.32\ \text{N}$
the sphere falls at constant velocity

You must calculate the weight $W$ of the sphere.

Approach

Convert the volume to $\text{m}^3$ so it matches the SI density unit.
Use $U = \rho g V$ to find the upthrust.
Use force balance $W = D + U$ to get the weight.

Step-by-Step Reasoning

1) Convert volume
Since $1\ \text{cm} = 10^{-2}\ \text{m}$ ,

1\ \text{cm}^3 = (10^{-2})^3\ \text{m}^3 = 10^{-6}\ \text{m}^3.

V = 4.6\ \text{cm}^3 = 4.6\times 10^{-6}\ \text{m}^3.

2) Calculate upthrust

U = \rho g V = 830 \times 9.81 \times 4.6\times 10^{-6}.

Compute:

U = 3.74\times 10^{-2}\ \text{N}.

This is much smaller than the weight because the volume is only a few cubic centimetres.

3) Use equilibrium to find weight
At constant velocity, resultant force is zero, so

W = D + U = 0.32 + 3.74\times 10^{-2} = 0.357\ \text{N}.

To 2 s.f.,

W \approx 0.36\ \text{N}.

diagram-1

Key Takeaways

Always convert to consistent SI units before using formulas.
Upthrust is $\rho g V$ .
Constant velocity implies force balance: $W = D + U$ .

Common Mistakes

Using $V = 4.6\times 10^{-3}\ \text{m}^3$ (wrong: that would be litres, not $\text{m}^3$ ).
Forgetting to add upthrust, and taking $W = D$ .
Using $U = \rho V$ (missing $g$ ).

Things to Be Careful About

Check the power of ten in the conversion: $\text{cm}^3 \to \text{m}^3$ gives $10^{-6}$ .
Use an appropriate value of $g$ (usually $9.81\ \text{m s}^{-2}$ or $9.8\ \text{m s}^{-2}$ ).
Give the final weight with a sensible number of significant figures consistent with the data.

Techniques used

convert volume from centimetres cubed to metres cubedcalculate upthrust using U = rho g Vapply equilibrium of forces to find weight from drag and upthrust

Q2MediumDynamicsKinematicsWork, Energy and Power

(a)

Define momentum.

1M Similar

(b)

A child stands on a scooter on horizontal ground. The combined mass of the child and the scooter is $16\ \text{kg}$ .

The child starts from rest and pushes once on the ground with her foot which causes her to accelerate. The push lasts for a time of $1.1\ \text{s}$ . The speed of the child and the scooter after the push is $0.60\ \text{m s}^{-1}$ .

Determine the average resultant force acting horizontally on the child and the scooter during the push.

average force = ______ $\text{N}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Change in momentum:

\Delta p = m(v-u) = 16(0.60-0) = 9.6\ \text{kg m s}^{-1}

Average resultant force:

F_{\text{avg}} = \frac{\Delta p}{\Delta t} = \frac{9.6}{1.1} = 8.7\ \text{N}

Answer

$8.7\ \text{N}$

Final answer

8.7 N

Detailed explanation

Background Concept

A resultant force causes a change in momentum. The average resultant force over a time interval is

F_{\text{avg}} = \frac{\Delta p}{\Delta t}

This comes from $F = \frac{\mathrm{d}p}{\mathrm{d}t}$ , and the product $F\Delta t$ is the impulse.

Understanding the Question

Total mass (child + scooter): $m = 16\ \text{kg}$ .
Starts from rest: $u = 0\ \text{m s}^{-1}$ .
Final speed after push: $v = 0.60\ \text{m s}^{-1}$ .
Duration of push: $\Delta t = 1.1\ \text{s}$ .
You must find the average horizontal resultant force during the push.

Approach

Calculate change in momentum $\Delta p = m(v-u)$ .
Divide by the time for which the push acts: $F_{\text{avg}} = \Delta p/\Delta t$ .

Step-by-Step Reasoning

Initial momentum is zero because the system starts from rest.
Final momentum is $mv = 16 \times 0.60$ .
So the change in momentum is

\Delta p = 16(0.60-0) = 9.6\ \text{kg m s}^{-1}.

Average resultant force is

F_{\text{avg}} = \frac{9.6}{1.1} = 8.7\ \text{N} \ \text{(to 2 s.f.)}.

Key Takeaways

Impulse: $F_{\text{avg}}\Delta t = \Delta p$ .
For a given mass, larger $\Delta v$ or smaller $\Delta t$ means a larger average force.

Common Mistakes

Using $F = ma$ without first finding $a$ (which is fine, but more steps and easy to slip).
Forgetting to use the combined mass of child and scooter.
Not dividing by the time interval.

Things to Be Careful About

The force asked for is the average resultant force horizontally, not the force from the foot alone.
Quote the answer with sensible significant figures based on the data (here, 2 s.f.).

Techniques used

use impulse as change in momentumapply average force as rate of change of momentumsubstitute into F = \Delta p / \Delta t

(c)

Later, the child in (b) travels down a slope at a constant angle to the horizontal, as shown in Fig. 2.1.

Fig. 2.1

At point A her speed is $0.60\ \text{m s}^{-1}$ . She has a constant acceleration of $0.85\ \text{m s}^{-2}$ parallel to the slope. After a time of $3.7\ \text{s}$ , she reaches point B.

Calculate the distance $x$ travelled by the child along the slope from A to B.

$x$ = ______ $\text{m}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

With $u = 0.60\ \text{m s}^{-1}$ , $a = 0.85\ \text{m s}^{-2}$ , $t = 3.7\ \text{s}$ ,

x = ut + \frac{1}{2}at^2

x = 0.60(3.7) + \frac{1}{2}(0.85)(3.7)^2 = 8.04\ \text{m}

Answer

$8.0\ \text{m}$

Final answer

8.0 m

Detailed explanation

Background Concept

For motion with constant acceleration $a$ along a straight line, the displacement $s$ after time $t$ is related to initial velocity $u$ by the SUVAT equation

s = ut + \frac{1}{2}at^2.

This applies when the acceleration is constant and the motion is along one direction (here: parallel to the slope).

Understanding the Question

From point A to point B (along the slope):

initial speed at A: $u = 0.60\ \text{m s}^{-1}$
acceleration (constant, along slope): $a = 0.85\ \text{m s}^{-2}$
time taken: $t = 3.7\ \text{s}$
We want the distance $x$ travelled along the slope in this time.

Approach

Use a constant-acceleration equation that links $x$ , $u$ , $a$ , and $t$ . The most direct is $x = ut + \frac{1}{2}at^2$ .

Step-by-Step Reasoning

Write the equation:

x = ut + \frac{1}{2}at^2.

Substitute values:

x = 0.60(3.7) + \frac{1}{2}(0.85)(3.7)^2.

Evaluate:

$0.60 \times 3.7 = 2.22$
$(3.7)^2 = 13.69$
$\frac{1}{2} \times 0.85 \times 13.69 = 5.82$
So

x = 2.22 + 5.82 = 8.04\ \text{m} \approx 8.0\ \text{m}.

Key Takeaways

Choose the SUVAT equation that contains only known quantities and the unknown.
“Parallel to the slope” means treat the motion as 1D along the slope.

Common Mistakes

Using $v^2 = u^2 + 2as$ without knowing $v$ .
Forgetting the $\frac{1}{2}$ in $\frac{1}{2}at^2$ .
Rounding too early (can shift the final value).

Things to Be Careful About

Ensure units are consistent: $\text{m s}^{-1}$ , $\text{m s}^{-2}$ , $\text{s}$ give $\text{m}$ .
Quote the final answer to an appropriate number of significant figures (here, 2 s.f. matches the given data).

Techniques used

select the appropriate SUVAT equationsubstitute values into s = ut + 1/2 at^2maintain consistent units throughout

(d)

At point B, the child in (c) applies the brake with a constant force to maintain a constant velocity. Point C is $18\ \text{m}$ from point B, as shown in Fig. 2.2.

Fig. 2.2

The work done by the braking force between B and C is $250\ \text{J}$ .

(i)

Determine the magnitude of the braking force.

force = ______ $\text{N}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Work done by a constant braking force:

W = Fs

F = \frac{W}{s} = \frac{250}{18} = 13.9\ \text{N}

Answer

$14\ \text{N}$

Final answer

14 N

Detailed explanation

Background Concept

Work done by a force is the energy transferred when the force causes a displacement. For a constant force acting parallel (or antiparallel) to the displacement,

W = Fs.

If the force is opposite the motion (a braking force), the work done by that force on the object is negative. However, questions often give the magnitude of the work done (energy dissipated), which is a positive number.

Understanding the Question

Between B and C:

distance along the slope: $s = 18\ \text{m}$
braking is applied with a constant force to maintain constant velocity
magnitude of work done by braking force: $250\ \text{J}$
We need the magnitude of the braking force.

Approach

Use $W = Fs$ with magnitudes: $F = W/s$ .

Step-by-Step Reasoning

Start with

W = Fs.

Rearrange:

F = \frac{W}{s}.

Substitute:

F = \frac{250}{18} = 13.9\ \text{N} \approx 14\ \text{N}.

Key Takeaways

For constant force parallel to displacement, work is force times distance.
Braking removes energy; the given work is typically the energy dissipated (a magnitude).

Common Mistakes

Using $W = F/ s$ instead of $W = Fs$ .
Forgetting that $18\ \text{m}$ is the distance along the slope between B and C.
Treating $250\ \text{J}$ as negative in the calculation when the question asks for magnitude.

Things to Be Careful About

If direction/sign were required: the braking force is opposite to the motion along the slope.
Significant figures: $250$ (2 s.f.) and $18$ (2 s.f.) justify an answer of $14\ \text{N}$ (2 s.f.).

Techniques used

apply W = Fs for a constant force along the displacementrearrange an equation to find an unknown forceuse magnitudes when work done is given as a positive energy transfer

(ii)

On Fig. 2.3, sketch the variation of the kinetic energy of the child and scooter with distance travelled from point A to point C.

Numerical values for kinetic energy are not required.

Fig. 2.3

3M Similar

DifficultyMedium

Worked solution

Answer

From A to B (distance $0$ to $x$ ): kinetic energy increases linearly with distance (straight line with positive gradient), starting at a non-zero value at A.

From B to C (distance $x$ to $x+18\ \text{m}$ ): kinetic energy is constant (horizontal line).

Final answer

KE increases linearly from A to B, then remains constant from B to C.

Detailed explanation

Background Concept

Kinetic energy is

E_k = \frac{1}{2}mv^2.

The work-energy principle states that the net work done on an object equals the change in its kinetic energy:

W_{\text{net}} = \Delta E_k.

If the resultant force along the direction of motion is constant, then the work done by this resultant force over distance $s$ is $W_{\text{net}} = F_{\text{res}} s$ , so

\Delta E_k \propto s.

That means kinetic energy varies linearly with distance when the resultant force is constant.

Understanding the Question

You must sketch kinetic energy (vertical axis) against distance from A (horizontal axis), from A to C.

From A to B: the child accelerates at constant $a = 0.85\ \text{m s}^{-2}$ , so her speed (and hence $E_k$ ) increases.
From B to C: brakes are applied so that velocity is constant, so speed (and hence $E_k$ ) does not change.
The graph has two regions, separated at distance $x$ .

Approach

Decide the trend of $E_k$ in each region using speed/acceleration information.
Decide the shape (curved or straight) using the work-energy principle:
- constant acceleration implies constant resultant force (since $F_{\text{res}} = ma$ ), so $E_k$ increases linearly with distance.
- constant velocity implies $\Delta E_k = 0$ , so the graph is horizontal.

Step-by-Step Reasoning

From A to B (distance $0$ to $x$ ):

Initial speed at A is $0.60\ \text{m s}^{-1}$ , so $E_k$ at distance $0$ is not zero. The curve must start above the origin.
Acceleration is constant, so resultant force along slope is constant:

F_{\text{res}} = ma = \text{constant}.

Net work done over distance $s$ is $W_{\text{net}} = F_{\text{res}} s$ , so

\Delta E_k = W_{\text{net}} \propto s.

Therefore, $E_k$ increases as a straight line with distance from A up to $x$ .

From B to C (distance $x$ to $x + 18\ \text{m}$ ):

Braking produces constant velocity, so acceleration is zero and kinetic energy is constant.
Hence the graph is a horizontal line from $x$ to $x + 18\ \text{m}$ .

So the sketch should show a straight rising line up to the dashed line at $x$ , then a flat line to the dashed line at $x + 18\ \text{m}$ .

Key Takeaways

Constant acceleration $\Rightarrow$ constant resultant force $\Rightarrow$ $E_k$ increases linearly with distance.
Constant velocity $\Rightarrow$ $E_k$ constant.
Non-zero initial speed means the graph does not start at zero kinetic energy.

Common Mistakes

Starting the kinetic energy at zero at A (even though $v = 0.60\ \text{m s}^{-1}$ ).
Drawing a curved increase between A and B (it is linear with distance if the resultant force is constant).
Drawing a decreasing kinetic energy from B to C: the brakes do negative work, but gravitational potential energy is also decreasing; constant velocity means these effects balance so $E_k$ stays constant.

Things to Be Careful About

The horizontal axis is distance from A, so the “kink” in the graph must occur exactly at $x$ , not at $18\ \text{m}$ .
No numerical values are required, but the shape and piecewise behaviour must be correct (increase then constant).

Techniques used

relate kinetic energy change to work done by the resultant forceinterpret constant acceleration as constant resultant forcesketch a piecewise graph with changing gradient and a constant section

Q3MediumDeformation of SolidsForces, Density and Pressure

(a)

The variation of stress with strain for a metal P is shown in Fig. 3.1.

Fig. 3.1

Point E is the elastic limit of the metal.

(i)

Use Fig. 3.1 to determine the Young modulus for P.

Young modulus = ______ $\text{Pa}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Young modulus $E$ is the gradient of the straight-line part:

Take a point on the straight-line section, e.g. at strain $= 1.0\% = 0.010$ ,
stress $\approx 25 \times 10^{7}\ \text{Pa} = 2.5 \times 10^{8}\ \text{Pa}$ .

E = \frac{\text{stress}}{\text{strain}} = \frac{2.5 \times 10^{8}}{0.010} = 2.5 \times 10^{10}\ \text{Pa}

Answer

$2.5 \times 10^{10}\ \text{Pa}$

Final answer

2.5 × 10^10 Pa

Detailed explanation

Background Concept

Young modulus $E$ measures stiffness:

E = \frac{\text{stress}}{\text{strain}}

Stress $= F/A$ (unit: Pa).
Strain $= \Delta L/L$ (no unit).
This definition applies in the linear (proportional) region of the stress–strain graph (Hooke’s law region), where stress is directly proportional to strain.

On a stress–strain graph, the Young modulus is the gradient of the straight-line part.

Understanding the Question

Fig. 3.1 gives stress (vertical axis) against strain in % (horizontal axis). You must use the straight portion near the origin and determine its gradient.

Key point: the x-axis is in percent, so you must convert e.g. $1.0\%$ to $0.010$ before dividing.

Approach

Choose a convenient point on the straight-line region (not on the curved part).
Read off stress and strain.
Convert strain from % to a decimal.
Calculate $E = \text{stress}/\text{strain}$ .

Step-by-Step Reasoning

From the graph, at about strain $1.0\%$ , stress is about $25$ on the scale of $10^7\ \text{Pa}$ .

Convert values:

Stress: $25 \times 10^7\ \text{Pa} = 2.5 \times 10^8\ \text{Pa}$ .
Strain: $1.0\% = 1.0/100 = 0.010$ .

Now divide:

E = \frac{2.5 \times 10^{8}}{0.010} = 2.5 \times 10^{10}\ \text{Pa}

(Any similar choice of point on the straight line gives a similar value.)

Key Takeaways

Young modulus is the gradient of the straight-line (Hooke’s law) part of a stress–strain graph.
Always convert % strain to a decimal before using $E = \sigma/\varepsilon$ .

Common Mistakes

Using strain as “1.0” instead of “0.010” (gives $100\times$ error).
Using point E (elastic limit) even though the graph is no longer linear there.
Forgetting the $\times 10^7$ factor on the stress axis.

Things to Be Careful About

Pick a point clearly on the straight-line section.
Quote $E$ in Pa and to a sensible number of significant figures (typically 2–3, matching graph reading precision).

Techniques used

use the gradient of the straight-line region of a graphconvert percentage strain to a decimal strainapply Young modulus as stress divided by strain

(ii)

On the line in Fig. 3.1, draw a cross ( $\times$ ) to show the limit of proportionality. Label this point Q.

1M Similar

DifficultyEasy

Worked solution

Answer

Place $Q$ at the end of the initial straight-line section (where the graph first begins to curve away from a straight line).

Final answer

Q at the end of the straight-line (proportional) region.

Detailed explanation

Background Concept

The limit of proportionality is the greatest stress for which stress is directly proportional to strain (Hooke’s law region). On a stress–strain graph, it is the point where the graph first departs from a straight line through the origin.

This is not necessarily the same as the elastic limit (point E), which is the greatest stress for which the material returns to its original length when unloaded.

Understanding the Question

You are asked to mark point $Q$ on Fig. 3.1. The graph is straight from the origin and then begins to curve; $Q$ should be placed at the transition between these behaviours.

Approach

Follow the straight-line region starting from the origin.
Find the first noticeable deviation from that straight-line trend.
Mark that point with a cross and label it $Q$ .

Step-by-Step Reasoning

diagram-a-ii-1

The proportional region is the straight line from the origin.
As strain increases, the plotted line starts to bend (curvature develops).
The point at which curvature begins is the limit of proportionality, so that is where $Q$ is placed.

Key Takeaways

“Proportional” means a straight line through the origin.
The limit of proportionality is where straight-line behaviour ends, not where elastic behaviour ends.

Common Mistakes

Placing $Q$ at point E (elastic limit) instead of earlier.
Placing $Q$ somewhere on the curved region without identifying the start of curvature.

Things to Be Careful About

The limit of proportionality is usually slightly before the elastic limit for metals.
Choose the point where the deviation first becomes clear (not a point far along the curve).

Techniques used

identify the linear (proportional) region on a stress–strain graphlocate the point where the graph first departs from a straight line

(b)

State the conditions necessary for an object to be in equilibrium.

2M Similar

(c)

A wire is used to hold a uniform shelf AB horizontally in equilibrium as shown in Fig. 3.2.

Fig. 3.2

The wire is connected to the midpoint of shelf AB at an angle of $50^\circ$ to the horizontal. The shelf is attached to a wall by a hinge at A. The length of shelf AB is $0.65\ \text{m}$ and its weight is $33\ \text{N}$ .

A cup of weight $1.5\ \text{N}$ rests on the shelf with its centre of gravity at a horizontal distance of $0.12\ \text{m}$ from B.

(i)

By taking moments about A, determine the tension in the wire.

tension = ______ $\text{N}$

3M Similar

DifficultyMedium

Worked solution

Working

Take moments about hinge at $A$ .

Distances from $A$ :

midpoint: $0.65/2 = 0.325\ \text{m}$
cup: $0.65 - 0.12 = 0.53\ \text{m}$

Only the vertical component of tension produces a moment:

T\sin 50^\circ \times 0.325 = 33 \times 0.325 + 1.5 \times 0.53

T\sin 50^\circ = 35.4

T = \frac{35.4}{\sin 50^\circ} = 46\ \text{N}

Answer

$46\ \text{N}$

Final answer

46 N

Detailed explanation

Background Concept

For a rigid body in equilibrium, the total clockwise moment about any point equals the total anticlockwise moment about that point.

Moment magnitude is:

\tau = F \times d_\perp

where $d_\perp$ is the perpendicular distance from the pivot to the line of action of the force.

Choosing the hinge point as the pivot is useful because hinge forces then have zero moment (their line of action passes through the pivot).

Understanding the Question

A horizontal uniform shelf $AB$ of length $0.65\ \text{m}$ is hinged at $A$ . A wire is attached at the shelf’s midpoint, making $50^\circ$ to the horizontal. The shelf weight is $33\ \text{N}$ acting at its midpoint. A cup of weight $1.5\ \text{N}$ acts at a point $0.12\ \text{m}$ from $B$ (so $0.53\ \text{m}$ from $A$ ). You must find the wire tension $T$ using moments about $A$ .

Approach

Draw/visualise forces and their distances from $A$ .
Take moments about $A$ to remove the unknown hinge reaction.
Use only the component of tension perpendicular to the shelf (vertical component) because the horizontal component passes through $A$ along the shelf and gives no turning effect about $A$ .
Solve for $T$ .

Step-by-Step Reasoning

diagram-c-i-1

Distances:

Midpoint from $A$ is $0.65/2 = 0.325\ \text{m}$ .
Cup position from $A$ is $0.65 - 0.12 = 0.53\ \text{m}$ .

Moments about $A$ :

Shelf weight produces clockwise moment: $33 \times 0.325$ .
Cup weight produces clockwise moment: $1.5 \times 0.53$ .
Tension produces anticlockwise moment through its vertical component $T\sin 50^\circ$ at distance $0.325$ :

(T\sin 50^\circ)(0.325) = (33)(0.325) + (1.5)(0.53)

Compute RHS:

$33 \times 0.325 = 10.725\ \text{N m}$
$1.5 \times 0.53 = 0.795\ \text{N m}$
Total $= 11.52\ \text{N m}$

So:

T\sin 50^\circ = \frac{11.52}{0.325} = 35.4

Then:

T = \frac{35.4}{\sin 50^\circ} \approx 46\ \text{N}

Key Takeaways

Take moments about a point that removes unknown forces (here, the hinge).
Use perpendicular components for moments: only $T\sin 50^\circ$ contributes.

Common Mistakes

Using $T\cos 50^\circ$ instead of $T\sin 50^\circ$ .
Using the wrong cup distance (using $0.12\ \text{m}$ from $A$ instead of from $B$ ).
Forgetting that the shelf’s weight acts at its midpoint.

Things to Be Careful About

Be consistent with clockwise vs anticlockwise signs.
Use distances measured from the pivot point $A$ .
Keep units as N and m so moments are in N m.

Techniques used

take moments about a pivot to eliminate unknown hinge forcesresolve a force to find the perpendicular component producing a momentapply the principle of moments for equilibrium

(ii)

The stress in the wire is $1.5 \times 10^7\ \text{Pa}$ .

Determine the radius of the wire.

radius = ______ $\text{m}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

\text{stress} = \frac{F}{A} = \frac{T}{\pi r^2}

A = \frac{T}{\text{stress}} = \frac{46}{1.5 \times 10^{7}} = 3.07 \times 10^{-6}\ \text{m}^2

r = \sqrt{\frac{A}{\pi}} = \sqrt{\frac{3.07 \times 10^{-6}}{\pi}} = 9.9 \times 10^{-4}\ \text{m}

Answer

$9.9 \times 10^{-4}\ \text{m}$

Final answer

9.9 × 10^-4 m

Detailed explanation

Background Concept

Stress is defined as force per unit cross-sectional area:

\sigma = \frac{F}{A}

For a cylindrical wire of radius $r$ :

A = \pi r^2

So:

\sigma = \frac{F}{\pi r^2}

Understanding the Question

You are told the stress in the wire is $1.5 \times 10^7\ \text{Pa}$ and you have found the wire tension $T$ from part (i). You must use $\sigma = F/A$ to find the wire radius.

Approach

Use $\sigma = T/(\pi r^2)$ .
Rearrange to get $r$ .
Substitute values and compute.

Step-by-Step Reasoning

Start from:

\sigma = \frac{T}{\pi r^2}

Rearrange:

\pi r^2 = \frac{T}{\sigma} \quad\Rightarrow\quad r = \sqrt{\frac{T}{\pi \sigma}}

Substitute $T \approx 46\ \text{N}$ and $\sigma = 1.5 \times 10^7\ \text{Pa}$ :

r = \sqrt{\frac{46}{\pi (1.5 \times 10^7)}} \approx 9.9 \times 10^{-4}\ \text{m}

This is about $1.0\ \text{mm}$ , which is a reasonable wire radius.

Key Takeaways

Stress uses the axial force in the wire: here it is the tension $T$ .
For a circular cross-section, use $A = \pi r^2$ and take a square root to get $r$ .

Common Mistakes

Using diameter instead of radius in $A = \pi r^2$ .
Forgetting to use $\pi$ .
Using $T\sin 50^\circ$ instead of $T$ for stress (stress depends on the force along the wire, i.e. the tension itself).

Things to Be Careful About

Units: $\text{Pa} = \text{N m}^{-2}$ , so $A$ comes out in $\text{m}^2$ and $r$ in m.
Significant figures should match the input (typically 2–3 s.f.).

Techniques used

apply stress equals force divided by cross-sectional arearearrange for cross-sectional area of a circular wiresolve for radius using the area of a circle

(iii)

More items are added to the shelf, doubling the stress in the wire. The wire is made of the metal P from (a).

Use Fig. 3.1 to state and explain whether the wire will behave plastically or elastically as the stress doubles.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Doubling the stress gives $3.0 \times 10^{7}\ \text{Pa}$ .

From Fig. 3.1 the elastic limit is at about $26 \times 10^{7}\ \text{Pa}$ , so $3.0 \times 10^{7}\ \text{Pa}$ is well below the elastic limit.

Therefore the wire behaves elastically (no permanent extension).

Final answer

Elastically (stress still below elastic limit).

Detailed explanation

Background Concept

Elastic behaviour: when the load is removed, the material returns to its original length (no permanent deformation).
Plastic behaviour: when the load is removed, there is permanent extension.

The elastic limit is the maximum stress for which deformation is still fully elastic.

A stress–strain graph shows where the elastic limit occurs; beyond it, unloading would not return to the original zero strain.

Understanding the Question

The wire initially has stress $1.5 \times 10^7\ \text{Pa}$ . More items are added so stress doubles to $3.0 \times 10^7\ \text{Pa}$ . The wire is made of metal P, whose elastic limit is given by point E on Fig. 3.1 (about $26 \times 10^7\ \text{Pa}$ ).

You must decide if doubling the stress takes the wire beyond the elastic limit.

Approach

Calculate the doubled stress.
Read the elastic limit stress from the graph.
Compare: if applied stress < elastic limit, behaviour is elastic; if > elastic limit, plastic deformation occurs.

Step-by-Step Reasoning

Doubled stress:

2(1.5 \times 10^7) = 3.0 \times 10^7\ \text{Pa}

From Fig. 3.1, point E is around:

26 \times 10^7\ \text{Pa} = 2.6 \times 10^8\ \text{Pa}

Comparison:

3.0 \times 10^7\ \text{Pa} \ll 2.6 \times 10^8\ \text{Pa}

So the applied stress is far below the elastic limit; the wire will extend under load but return to its original length when the load is removed (elastic behaviour).

Key Takeaways

Plastic deformation starts only when the stress exceeds the elastic limit.
Always compare stresses using consistent powers of ten.

Common Mistakes

Misreading the stress axis scale ( $\times 10^7\ \text{Pa}$ ) and comparing the wrong numbers.
Confusing limit of proportionality with elastic limit.
Concluding “plastic” simply because the stress increased.

Things to Be Careful About

Convert graph values properly: $26$ on the axis means $26 \times 10^7\ \text{Pa}$ .
The question asks “state and explain”: you need both the conclusion (elastic/plastic) and the comparison to the elastic limit.

Techniques used

compare an applied stress to the elastic limit on a stress–strain graphinfer elastic or plastic behaviour from whether the elastic limit is exceeded

Q4Medium-EasyWavesSuperposition

(a)

With reference to the direction of transfer of energy, compare the oscillations of transverse and longitudinal progressive waves.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Energy is transferred in the direction the wave travels.

For a transverse wave, the oscillations are perpendicular to the direction of energy transfer.

For a longitudinal wave, the oscillations are parallel to the direction of energy transfer.

Final answer

Transverse: oscillations ⟂ to energy transfer; Longitudinal: oscillations ∥ to energy transfer.

Detailed explanation

Background Concept

A progressive wave transfers energy from one place to another. The direction of energy transfer is the same as the direction of wave propagation.

Transverse wave: particles of the medium oscillate at right angles to the direction the wave travels.
Longitudinal wave: particles oscillate back-and-forth along the same line as the wave travels.

Understanding the Question

You are asked to compare transverse and longitudinal progressive waves, specifically referring to how the particle oscillations are oriented relative to the direction of energy transfer.

So you must mention:

direction of energy transfer (along propagation), and
whether oscillations are perpendicular or parallel to that.

Approach

State that energy transfer is along the direction of wave travel. Then give the relative direction of oscillation for each wave type.

Step-by-Step Reasoning

In any progressive wave, the disturbance moves through the medium, carrying energy in the direction the wave propagates.
For a transverse wave, the displacement of the medium is at right angles to that propagation direction, so oscillations are perpendicular to energy transfer.
For a longitudinal wave, compressions/rarefactions mean the medium oscillates back and forth along the same line as propagation, so oscillations are parallel to energy transfer.

Key Takeaways

Energy transfer direction = wave propagation direction.
Transverse: oscillation (\perp) propagation.
Longitudinal: oscillation (\parallel) propagation.

Common Mistakes

Saying “energy is transferred perpendicular in a transverse wave” (false: energy transfer is still along propagation).
Only stating “transverse is up-down, longitudinal is left-right” without referencing energy transfer/wave direction.

Things to Be Careful About

Always compare directions relative to the same reference: the wave/energy-transfer direction.
Use the words perpendicular and parallel (or “at right angles” and “along the same direction”) clearly.

Techniques used

identify the direction of energy transfer (wave propagation)compare oscillation direction relative to propagation direction

(b)

A pipe is open at one end and closed at the other with a piston. The piston can slide freely and is at a distance of $4.5 \times 10^{-2}\ \text{m}$ from the open end of the pipe.

A loudspeaker is positioned near the open end of the pipe and emits a sound wave of a single constant frequency. A stationary wave is formed in the pipe, as illustrated in Fig. 4.1.

Fig. 4.1

(i)

On Fig. 4.1, draw a letter A at the position of an antinode.

1M Similar

(ii)

The speed of sound in air is $340\ \text{m s}^{-1}$ .

Determine the frequency of the sound wave.

frequency = ______ $\text{Hz}$

3M Similar

DifficultyMedium-Easy

Worked solution

Working

For an open–closed pipe (node at closed end, antinode at open end),

L = \frac{\lambda}{4}

\lambda = 4L = 4\left(4.5 \times 10^{-2}\right) = 1.8 \times 10^{-1}\ \text{m}

Using $v = f\lambda$ :

f = \frac{v}{\lambda} = \frac{340}{1.8 \times 10^{-1}} = 1.9 \times 10^{3}\ \text{Hz}

Answer

$1.9 \times 10^{3}\ \text{Hz}$

Final answer

1.9 × 10^3 Hz

Detailed explanation

Background Concept

A stationary wave forms when two waves of the same frequency travel in opposite directions and superpose.

For a pipe:

Closed end: displacement must be zero (\rightarrow) displacement node.
Open end: displacement is maximum (\rightarrow) displacement antinode.

The simplest standing-wave pattern in an open–closed pipe is a quarter of a wavelength:

L = \frac{\lambda}{4}

Once (\lambda) is known, use the wave equation:

v = f\lambda

where (v) is wave speed, (f) frequency, (\lambda) wavelength.

Understanding the Question

The piston (closed end) is (4.5 \times 10^{-2}\ \text{m}) from the open end, so the air-column length is

L = 4.5 \times 10^{-2}\ \text{m}.

The diagram shows a node at the piston and an antinode at the open end, i.e. the fundamental open–closed mode (one antinode).
You are given (v = 340\ \text{m s}^{-1}) and must find (f).

Approach

Use the boundary conditions (node at piston, antinode at open end) to relate (L) to (\lambda).
Substitute (\lambda) into (v=f\lambda) to get (f).

Step-by-Step Reasoning

diagram-bii-1

For the mode shown, the distance from a node to the adjacent antinode is (\lambda/4). Here that distance is exactly the pipe length (L).

L = \frac{\lambda}{4} \quad \Rightarrow \quad \lambda = 4L.

Substitute (L = 4.5\times 10^{-2}\ \text{m}):

\lambda = 4(4.5\times 10^{-2}) = 1.8\times 10^{-1}\ \text{m}.

Apply (v=f\lambda):

f = \frac{v}{\lambda} = \frac{340}{1.8\times 10^{-1}} \approx 1.89\times 10^{3}\ \text{Hz}.

Round appropriately (given (L) is 2 s.f.):

f \approx 1.9\times 10^{3}\ \text{Hz}.

Key Takeaways

Open–closed pipe fundamental mode: (L=\lambda/4).
Convert geometry (\rightarrow) wavelength, then use (v=f\lambda).

Common Mistakes

Using (L=\lambda/2) (that applies to an open–open or closed–closed pipe, not open–closed).
Using (f=vL) or (f=v/4L) without first stating the (L)–(\lambda) relationship.
Rounding too aggressively or giving an answer inconsistent with significant figures.

Things to Be Careful About

Ensure (L) is in metres (it already is).
Recognise which harmonic is drawn: node at one end and antinode at the other indicates an odd-quarter-wavelength mode; here it is the simplest quarter-wave.
Keep units consistent so (f) comes out in (\text{Hz}) ((\text{s}^{-1})).

Techniques used

use open-closed pipe boundary conditions to relate length to wavelengthapply the wave equation to calculate frequencysubstitute values with consistent units and round to appropriate significant figures

(iii)

The piston is moved to the left. The frequency of the sound wave emitted by the loudspeaker is then changed so that a stationary wave is formed with same number of antinodes as in Fig. 4.1.

State and explain the change that is made to the frequency of the sound wave.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

When the piston moves left, the length $L$ decreases.

For the same number of antinodes (same mode),

L = \frac{(2n-1)\lambda}{4} \ \Rightarrow\ \lambda \propto L

So $\lambda$ decreases. With $v$ constant and $v=f\lambda$ , the frequency increases.

Final answer

Increase the frequency.

Detailed explanation

Background Concept

In an open–closed pipe, standing-wave patterns must satisfy the boundary conditions:

closed end: displacement node
open end: displacement antinode

Allowed modes are odd quarter-wavelengths:

L = \frac{(2n-1)\lambda}{4}\qquad (n=1,2,3,\dots)

If the wave speed (v) in air is (approximately) constant, then

v = f\lambda \quad \Rightarrow \quad f = \frac{v}{\lambda}.

Understanding the Question

The piston is moved left, so the air-column length (L) becomes smaller. You then adjust the loudspeaker frequency so that a stationary wave forms with the same number of antinodes as before, meaning the same mode number (n).

You must state what happens to (f) and explain it.

Approach

Same number of antinodes (\Rightarrow) same harmonic/mode (n), so the relationship between (L) and (\lambda) stays in the same form.
Decreasing (L) forces (\lambda) to decrease.
With (v) constant, smaller (\lambda) means larger (f).

Step-by-Step Reasoning

For a particular mode (n),

L = \frac{(2n-1)\lambda}{4}.

Rearranging shows

\lambda = \frac{4L}{2n-1}

so for fixed (n), (\lambda \propto L).
3. Moving the piston left decreases (L), therefore (\lambda) must decrease to keep the same pattern (same number of antinodes).
4. Since the speed of sound in air is approximately constant,

f = \frac{v}{\lambda}.

If (\lambda) decreases, (f) increases.

Key Takeaways

Same number of antinodes (\Rightarrow) same mode number (n).
For fixed (n), (\lambda) changes in direct proportion to (L).
With constant (v), decreasing (\lambda) requires increasing (f).

Common Mistakes

Saying frequency decreases because the pipe is “shorter” without using (v=f\lambda) (you must connect length (\rightarrow) wavelength (\rightarrow) frequency).
Assuming (\lambda) stays constant when the pipe length changes (it cannot if the same mode is to fit in the tube).

Things to Be Careful About

The phrase “same number of antinodes” is the key clue that the harmonic number does not change.
Don’t mix up which variable is adjusted: the piston changes (L), and you must change (f) to re-establish resonance.
State the final change clearly: frequency increases (not wavelength).

Techniques used

use the standing-wave condition for a fixed mode number in an open-closed piperelate wavelength change to frequency using constant wave speedstate a qualitative change and support it with proportional reasoning

Q5Medium-EasyElectricityD.C. CircuitsWork, Energy and Power

(a)

Define electric potential difference (p.d.).

1M Similar

(b)

A power supply, three resistors and a component X are connected in the circuit shown in Fig. 5.1.

Fig. 5.1

The power supply has an electromotive force (e.m.f.) of $230\ \text{V}$ and negligible internal resistance. The current in the power supply is $7.0\ \text{A}$ .

11M

(i)

Identify component X.

1M Similar

(ii)

Show that the p.d. across the resistor of resistance $0.86\ \Omega$ is $6.0\ \text{V}$ .

1M Similar

(iii)

Determine the current $I_1$ .

$I_1$ = ______ $\text{A}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

p.d. across parallel network:

V_{\text{parallel}} = 230 - 6.0 = 224\ \text{V}

Current in $170\ \Omega$ branch:

I_2 = \frac{224}{170} = 1.32\ \text{A}

Using $I = I_1 + I_2$ :

I_1 = 7.0 - 1.32 = 5.68\ \text{A} \approx 5.7\ \text{A}

Answer

$5.7\ \text{A}$

Final answer

5.7 A

Detailed explanation

Background Concept

Two key ideas are used in this type of circuit:

Series p.d. sharing (Kirchhoff’s second law): the supply e.m.f. equals the sum of potential drops around a loop.
Junction rule (Kirchhoff’s first law): total current into a junction equals total current out:

I = I_1 + I_2

In parallel branches, the potential difference is the same across each branch.

Understanding the Question

The supply is $230\ \text{V}$ with total current $7.0\ \text{A}$ . A $0.86\ \Omega$ resistor is in series before the circuit splits into two parallel branches:

top branch: $2.4\ \Omega$ in series with X (current $I_1$ )
bottom branch: $170\ \Omega$ (current $I_2$ )

You need the current $I_1$ in the top branch.

Approach

Use the already-found drop across $0.86\ \Omega$ to find the p.d. across the entire parallel section.
Use $I_2 = V/170$ for the lower branch.
Use Kirchhoff’s first law: $I_1 = I - I_2$ .

Step-by-Step Reasoning

Voltage across the series resistor is $6.0\ \text{V}$ , so the remaining voltage across the parallel network is:

V_{\text{parallel}} = 230 - 6.0 = 224\ \text{V}

The $170\ \Omega$ branch has this same p.d. (parallel branches share p.d.):

I_2 = \frac{V_{\text{parallel}}}{170} = \frac{224}{170} = 1.32\ \text{A}

Total current splits at the junction:

7.0 = I_1 + I_2

I_1 = 7.0 - 1.32 = 5.68\ \text{A} \approx 5.7\ \text{A}

Key Takeaways

Find the p.d. across a parallel combination by subtracting series drops from the supply.
Use $I=V/R$ in a single resistor branch.
Use $I = I_1 + I_2$ at a junction.

Common Mistakes

Using $230\ \text{V}$ across the $170\ \Omega$ branch (ignores the series $0.86\ \Omega$ drop).
Adding branch currents incorrectly (e.g. $I_1 = I + I_2$ ).
Forgetting that p.d. is the same across parallel branches.

Things to Be Careful About

Keep track of which voltage is across which part: $224\ \text{V}$ is across the entire parallel section.
Rounding: keep extra figures until the end, then round to sensible s.f.

Techniques used

find the voltage across a parallel network using series voltage dropsapply Kirchhoff’s first law at a junctionuse I = V/R in a parallel branch

(iv)

Calculate the p.d. across component X.

p.d. = ______ $\text{V}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Voltage across parallel network:

V_{\text{parallel}} = 230 - 6.0 = 224\ \text{V}

Voltage across $2.4\ \Omega$ resistor:

V_{2.4} = I_1R = 5.68 \times 2.4 = 13.6\ \text{V}

So p.d. across X:

V_X = 224 - 13.6 = 210.4\ \text{V} \approx 210\ \text{V}

Answer

$210\ \text{V}$

Final answer

210 V

Detailed explanation

Background Concept

In a series branch, the total p.d. across the branch equals the sum of the p.d.s across each component in that branch.

For a resistor, the p.d. is found using

V = IR

Understanding the Question

The top branch has a $2.4\ \Omega$ resistor in series with component X. The p.d. across the whole parallel section is the same as the p.d. across the top branch. You are asked for the p.d. across X only.

Approach

Use Kirchhoff’s second law to get the p.d. across the parallel network: $230\ \text{V}$ minus the series drop on $0.86\ \Omega$ .
In the top branch, calculate the drop across $2.4\ \Omega$ using $V=IR$ with the branch current $I_1$ .
Subtract to get $V_X$ .

Step-by-Step Reasoning

From earlier,

V_{0.86} = 6.0\ \text{V}

so the voltage across the parallel combination is

V_{\text{parallel}} = 230 - 6.0 = 224\ \text{V}

The p.d. across the whole top branch is therefore $224\ \text{V}$ .
The current in the top branch was found as $I_1 = 5.68\ \text{A}$ . The $2.4\ \Omega$ resistor is in series in that branch, so it has the same current. Its p.d. is

V_{2.4} = I_1R = 5.68 \times 2.4 = 13.6\ \text{V}

The rest of the branch p.d. is across X:

V_X = 224 - 13.6 = 210.4\ \text{V}

So $V_X \approx 210\ \text{V}$ .

Key Takeaways

Parallel section sets the branch p.d.
Series components share the same current; their voltage drops add up.

Common Mistakes

Using $I = 7.0\ \text{A}$ through the $2.4\ \Omega$ resistor (it only carries $I_1$ ).
Forgetting to subtract the $0.86\ \Omega$ drop before using the supply voltage.

Things to Be Careful About

Make sure you subtract the correct series drop: $V_X$ is found from the top-branch p.d. ( $224\ \text{V}$ ), not directly from $230\ \text{V}$ .
Keep enough significant figures during intermediate calculations so rounding doesn’t drift.

Techniques used

use the parallel-section voltage as the branch voltagecalculate a series resistor drop using V = IRsubtract series drops to obtain a component voltage

(v)

Calculate the power dissipated in component X.

power = ______ $\text{W}$

2M Similar

(vi)

The purpose of the circuit is to provide power to component X.

Determine the percentage efficiency of the circuit.

efficiency = ______ %

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Total power from supply:

P_{\text{in}} = VI = 230 \times 7.0 = 1.61 \times 10^3\ \text{W}

Useful power to X:

P_{\text{useful}} = P_X = 1.19 \times 10^3\ \text{W}

Efficiency:

\eta = \frac{P_{\text{useful}}}{P_{\text{in}}} \times 100 = \frac{1.19\times 10^3}{1.61\times 10^3}\times 100 = 74\%

Answer

$74\%$

Final answer

74%

Detailed explanation

Background Concept

Efficiency compares how much of the input power is delivered as the intended (useful) output:

\eta = \frac{\text{useful power output}}{\text{total power input}}

To express as a percentage:

\eta(\%) = \eta \times 100

In an electrical circuit, total input power from a supply is typically

P_{\text{in}} = VI

(using the supply p.d. and the supply current).

Understanding the Question

The circuit is designed to provide power to component X, so the useful power is the power in X. All other power dissipated in the resistors is “wasted” for this purpose.

You must find the efficiency as a percentage.

Approach

Calculate total power from the supply using the supply e.m.f./p.d. ( $230\ \text{V}$ ) and total current ( $7.0\ \text{A}$ ).
Use the power in X from part (v) as useful power.
Form the ratio and multiply by $100$ .

Step-by-Step Reasoning

Total supply power:

P_{\text{in}} = VI = 230\ \text{V} \times 7.0\ \text{A} = 1.61\times 10^3\ \text{W}

Useful power is power in X (from part (v)):

P_{\text{useful}} = P_X = 1.19\times 10^3\ \text{W}

Efficiency:

\eta(\%) = \frac{P_X}{P_{\text{in}}} \times 100 = \frac{1.19\times 10^3}{1.61\times 10^3} \times 100 \approx 74\%

Key Takeaways

Efficiency is a power ratio.
For circuits: input power is from the supply; useful power depends on what the circuit is intended to power.

Common Mistakes

Using $P_{\text{in}} = V_X I_1$ (that is power in X, not total input power).
Forgetting to multiply by $100$ to get a percentage.
Taking “useful power” as power in the whole top branch (includes the $2.4\ \Omega$ resistor losses).

Things to Be Careful About

Use the supply values for total input power: $230\ \text{V}$ and $7.0\ \text{A}$ .
Efficiency must be (< 100%); if you get above $100\%$ , you have mixed up powers.

Techniques used

calculate total input power using P = VIcalculate efficiency as useful power output divided by total power inputconvert a ratio to a percentage

(vii)

The resistor of resistance $170\ \Omega$ is removed, leaving an open circuit in the lower branch of the circuit. There is no change to the resistance of component X.

State whether the current in the power supply increases, decreases or remains the same.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

The current in the power supply decreases.

Final answer

Decreases

Detailed explanation

Background Concept

For a fixed supply voltage, the supply current depends on the total circuit resistance:

I = \frac{V}{R_{\text{total}}}

For resistors in parallel, the equivalent resistance is smaller than any individual branch resistance, because adding more parallel paths increases total conductance:

\frac{1}{R_{\text{parallel}}} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots

Removing a parallel branch increases $R_{\text{parallel}}$ .

Understanding the Question

The $170\ \Omega$ resistor forms the lower parallel branch. It is removed so that branch becomes an open circuit (no current in that branch). The resistance of component X is stated not to change. You must state how the supply current changes.

Approach

Decide how the equivalent resistance of the parallel section changes when one branch is removed; then infer how the supply current changes using $I=V/R_{\text{total}}$ .

Step-by-Step Reasoning

Originally, there are two branches in parallel, so the equivalent resistance of the parallel section is relatively low.
Removing the $170\ \Omega$ branch means there is now only one conducting branch in that section.
With fewer parallel paths, the equivalent resistance of that section increases.
The $0.86\ \Omega$ series resistor remains, so the total circuit resistance increases.
With the supply voltage unchanged, an increase in $R_{\text{total}}$ causes the supply current to decrease:

I = \frac{V}{R_{\text{total}}} \downarrow

Key Takeaways

Removing a branch from a parallel network increases the equivalent resistance.
For constant voltage, higher resistance means smaller current.

Common Mistakes

Saying the current increases because “more current goes through the top branch”; while the top-branch current does increase, the supply current decreases because the total resistance increases.
Forgetting that an open circuit means zero current in that branch.

Things to Be Careful About

Distinguish between supply current and branch current: they do not have to change in the same direction.
The statement “no change to the resistance of component X” is there to prevent you assuming X’s resistance changes with current/temperature; treat it as constant here.

Techniques used

predict the effect of removing a parallel branch on equivalent resistanceuse I = V/R to infer change in supply current

Q6MediumParticle PhysicsDynamicsPhysical Quantities and UnitsWork, Energy and Power

(a)

Compare an $\alpha$ -particle with a $\beta^+$ particle in terms of their masses and charges.

3M Similar

DifficultyMedium-Easy

Worked solution

Answer

$\alpha$ -particle: charge $+2e$ , mass (\approx 4u) (very large compared with an electron).
$\beta^+$ particle (positron): charge $+e$ , mass (\approx m_e) (about (\frac{1}{1836},u), much smaller than an $\alpha$ -particle).

Final answer

Alpha: +2e and mass ≈ 4u; beta+ (positron): +e and mass ≈ me (≈ 1/1836 u).

Detailed explanation

Background Concept

In nuclear physics, particles are identified by their charge and their (rest) mass.

An $\alpha$ -particle is a helium nucleus: $2$ protons and $2$ neutrons. So it has charge $+2e$ and a mass close to $4u$ (where $u$ is the atomic mass unit).
A $\beta^+$ particle is a positron (the antiparticle of the electron). It has the same mass as an electron ( $m_e$ ) but opposite charge: $+e$ .

A useful scale comparison is that nuclear particles (made of nucleons) have masses measured in $u$ , while electrons/positrons have masses about $1/1836$ of a proton.

Understanding the Question

You are asked to compare an $\alpha$ -particle and a $\beta^+$ particle in terms of:

their masses (which is much larger/smaller, and typical values), and
their charges (sign and magnitude in units of $e$ ).

Approach

Write down, for each particle:

what it is physically made of, then
deduce its charge from that composition, and
state a standard mass (either in $u$ for $\alpha$ , or $m_e$ / fraction of $u$ for $\beta^+$ ).

Step-by-Step Reasoning

$\alpha$ is $\,^{4}_{2}\text{He}$ nucleus.
- Charge: $2$ protons $\Rightarrow +2e$ .
- Mass: $4$ nucleons $\Rightarrow$ about $4u$ (a little less due to binding energy, but $4u$ is the expected exam value).
$\beta^+$ is a positron.
- Charge: positron has charge $+e$ .
- Mass: positron mass equals electron mass $m_e$ ; compared with $u$ , $m_e \approx \frac{1}{1836}u$ (more precisely $m_e \approx 5.5\times 10^{-4}u$ ).

So the $\alpha$ -particle is far more massive and has twice the positive charge.

Key Takeaways

$\alpha$ : helium nucleus, charge $+2e$ , mass $\approx 4u$ .
$\beta^+$ : positron, charge $+e$ , mass $= m_e$ (tiny compared with nuclear masses).

Common Mistakes

Saying $\beta^+$ is an electron (wrong sign of charge).
Giving $\alpha$ charge as $+e$ instead of $+2e$ .
Confusing nucleon number with charge (neutrons add mass but no charge).

Things to Be Careful About

The question asks for comparison, so you must give values for both particles (mass and charge), not just one.
Use correct sign conventions: $\beta^-$ is negative; $\beta^+$ is positive.

Techniques used

state the charge in terms of the elementary chargecompare particle masses using known rest-mass scalesidentify a beta-plus particle as a positron

(b)

Nucleus P undergoes $\alpha$ -decay to form nucleus Q. Nucleus Q then undergoes a further decay to form nucleus R. The proton and nucleon numbers of P and R are shown in Fig. 6.1.

Fig. 6.1

(i)

On Fig. 6.1, draw a cross ( $\times$ ) to show the proton number and nucleon number of Q. Label your cross Q.

1M Similar

DifficultyEasy

Worked solution

Working

For $\alpha$ -decay: proton number $Z$ decreases by $2$ and nucleon number $A$ decreases by $4$ .

From $P(84,216)$ :

Z_Q = 84 - 2 = 82,\qquad A_Q = 216 - 4 = 212

Answer

$Q$ is at $(82,212)$ (proton number $82$ , nucleon number $212$ ).

Final answer

Q at proton number 82, nucleon number 212.

Detailed explanation

Background Concept

In nuclear decay, the proton number $Z$ and nucleon number $A$ change in specific ways depending on the emitted particle.

For $\alpha$ -decay, the nucleus emits an $\alpha$ -particle ( $\,^{4}_{2}\text{He}$ ). Conservation of nucleon number and charge implies:

$A$ decreases by $4$ (loss of 4 nucleons)
$Z$ decreases by $2$ (loss of 2 protons)

So:

\,^{A}_{Z}P \rightarrow \,^{A-4}_{Z-2}Q + \,^{4}_{2}\alpha

Understanding the Question

You are given point $P$ on a graph of nucleon number (vertical axis) against proton number (horizontal axis):

$P$ is at $Z=84$ , $A=216$ .

Nucleus $P$ undergoes $\alpha$ -decay to form $Q$ . You must locate $Q$ on the same axes.

Approach

Start from $(Z,A)$ for $P$ , then apply the fixed changes for $\alpha$ -decay: $(Z-2, A-4)$ . Plot/identify that coordinate.

Step-by-Step Reasoning

From the graph, $P$ has: $Z_P = 84,\quad A_P = 216$
After $\alpha$ -decay: $Z_Q = Z_P - 2 = 84 - 2 = 82$ $A_Q = A_P - 4 = 216 - 4 = 212$
Therefore $Q$ must be at proton number $82$ and nucleon number $212$ .

Key Takeaways

$\alpha$ -decay always moves you 2 left in $Z$ and 4 down in $A$ .

Common Mistakes

Subtracting $4$ from $Z$ instead of $2$ .
Subtracting $2$ from $A$ instead of $4$ .
Swapping axes (mixing up proton number and nucleon number).

Things to Be Careful About

Ensure you read the axes correctly: proton number is on the horizontal axis, nucleon number on the vertical axis.
The plotted point should be exactly at the correct grid intersection: $(82,212)$ .

Techniques used

apply changes in proton and nucleon numbers for alpha decayread coordinates from a proton-number vs nucleon-number plot

(ii)

State the names of the particles emitted as Q decays to form R.

2M Similar

DifficultyMedium-Easy

Worked solution

Working

From the graph: $Q(82,212)$ and $R(83,212)$ .

Nucleon number unchanged, proton number increases by $1$ $\Rightarrow$ $\beta^-$ decay.

Answer

Particles emitted: $\beta^-$ (electron) and an antineutrino ( $\bar{\nu}_e$ ).

Final answer

Electron (β−) and an electron antineutrino (ν̅e).

Detailed explanation

Background Concept

In $\beta$ decay, the nucleon number $A$ stays the same because no nucleons leave the nucleus; instead, a nucleon changes type.

In $\beta^-$ decay: a neutron becomes a proton.
$n \rightarrow p + e^- + \bar{\nu}_e$
So $Z$ increases by $1$ , $A$ unchanged.
In $\beta^+$ decay: a proton becomes a neutron.
$p \rightarrow n + e^+ + \nu_e$
So $Z$ decreases by $1$ , $A$ unchanged.

Understanding the Question

You have nuclei $Q$ and $R$ with coordinates (proton number, nucleon number). From part (i),

$Q$ is at $(82,212)$ .
The question tells you
$R$ is at $(83,212)$ .

You must state the names of the particles emitted when $Q$ decays to form $R$ .

Approach

Compare $Z$ and $A$ for $Q$ and $R$ :

If $A$ is unchanged, think $\beta$ decay.
Then check whether $Z$ increased or decreased to decide between $\beta^-$ and $\beta^+$ .
Finally, name the emitted beta particle and the (anti)neutrino that accompanies it.

Step-by-Step Reasoning

Compare nucleon numbers:
$A_Q = 212,\quad A_R = 212 \Rightarrow A\ \text{unchanged}$
So this is not $\alpha$ decay (which would change $A$ by $4$ ). It suggests $\beta$ decay.
Compare proton numbers:
$Z_Q = 82,\quad Z_R = 83 \Rightarrow Z\ \text{increased by 1}$
An increase in $Z$ by $1$ corresponds to $\beta^-$ decay (neutron $\to$ proton).
In $\beta^-$ decay, the emitted particles are:
- an electron ( $\beta^-$ )
- an electron antineutrino ( $\bar{\nu}_e$ )

Key Takeaways

$A$ unchanged and $Z$ increases by $1$ $\Rightarrow$ $\beta^-$ decay.
$\beta^-$ decay emits $e^-$ and $\bar{\nu}_e$ .

Common Mistakes

Saying $\beta^+$ (would make $Z$ decrease, not increase).
Forgetting the neutrino/antineutrino.
Writing “neutrino” when it should be “antineutrino” for $\beta^-$ decay.

Things to Be Careful About

Cambridge often expects both particles for 2 marks: the beta particle and the (anti)neutrino.
Use the correct symbol: $\bar{\nu}_e$ for $\beta^-$ decay; $\nu_e$ for $\beta^+$ decay.

Techniques used

compare proton and nucleon numbers to identify the decay typeuse conservation of nucleon number and charge to infer emitted particles

(c)

Before the $\alpha$ -decay, P is travelling at a constant velocity. After the decay, Q has a velocity of $1.3 \times 10^5\ \text{m s}^{-1}$ at an angle of $68^\circ$ to the original path of P.

The $\alpha$ -particle has a velocity of $150 \times 10^5\ \text{m s}^{-1}$ at an angle of $\theta$ to the original path of P, as shown in Fig. 6.2.

Fig. 6.2

(i)

Use the principle of conservation of momentum to determine $\theta$ .

$\theta$ = ______ $^\circ$

3M Similar

DifficultyMedium

Worked solution

Working

Initial momentum perpendicular to the original path is zero, so after decay:

m_{\alpha} v_{\alpha}\sin\theta = m_Q v_Q \sin 68^{\circ}

Using masses $\propto$ nucleon number: $m_{\alpha}:m_Q = 4:212$ .

\sin\theta = \frac{212\,(1.3\times 10^5)\sin 68^{\circ}}{4\,(150\times 10^5)}

\sin\theta \approx 0.426 \Rightarrow \theta \approx 25^{\circ}

Answer

$\theta = 25^{\circ}$

Final answer

25°

Detailed explanation

Background Concept

Momentum is a vector quantity:

\vec{p} = m\vec{v}

If no external resultant force acts on a system, the total momentum is conserved:

\sum \vec{p}_{\text{before}} = \sum \vec{p}_{\text{after}}

Because momentum is a vector, we often conserve it separately in perpendicular directions (e.g. horizontal and vertical components).

In nuclear decay occurring in free space (or where external forces during the decay are negligible), the nucleus and emitted particle(s) form an isolated system during the decay, so momentum conservation applies.

Understanding the Question

Before decay, nucleus $P$ moves along a straight line (take this as the $x$ -direction). After decay:

nucleus $Q$ has speed $1.3 \times 10^5\ \text{m s}^{-1}$ at $68^{\circ}$ to the original path (downwards in the figure),
the $\alpha$ -particle has speed $150 \times 10^5\ \text{m s}^{-1}$ at angle $\theta$ (upwards in the figure).

You must find the unknown angle $\theta$ using conservation of momentum.

c-i-momentum-components

Approach

Use conservation of momentum in the direction perpendicular to the initial motion (the $y$ -direction):

Initially, the $y$ -component of momentum is zero (since $P$ moves purely along the original path).
Therefore, after decay, the upward $y$ -momentum of the $\alpha$ must equal the downward $y$ -momentum of $Q$ .

To form momenta you need masses. In these questions it is standard to take nuclear mass $\propto A$ (nucleon number), so

$m_{\alpha} \propto 4$ ,
$m_Q \propto 212$ (since $P$ had $A=216$ and $\alpha$ decay reduces $A$ by $4$ ).

Step-by-Step Reasoning

Choose axes:

$x$ along original motion of $P$ .
$y$ perpendicular to it.

Write $y$ -momentum conservation.
Initial $y$ -momentum:

p_{y,\text{before}} = 0

After decay, taking magnitudes and equating upward and downward components:

m_{\alpha} v_{\alpha} \sin\theta = m_Q v_Q \sin 68^{\circ}

Use mass ratio from nucleon numbers.

\frac{m_Q}{m_{\alpha}} \approx \frac{212}{4}

\sin\theta = \frac{m_Q v_Q \sin 68^{\circ}}{m_{\alpha} v_{\alpha}} = \frac{212\,(1.3\times 10^5)\sin 68^{\circ}}{4\,(150\times 10^5)}

Evaluate.

$\sin 68^{\circ} \approx 0.927$ .
Numerator: $212\times 1.3\times 10^5 \approx 2.756\times 10^7$ ; then $\times 0.927 \approx 2.56\times 10^7$ .
Denominator: $4\times 150\times 10^5 = 600\times 10^5 = 6.0\times 10^7$ .
So

\sin\theta \approx \frac{2.56\times 10^7}{6.0\times 10^7} \approx 0.426

Then

\theta \approx \sin^{-1}(0.426) \approx 25^{\circ}

Key Takeaways

Momentum conservation works in each perpendicular direction.
Using the perpendicular component is often simplest when initial motion is along a straight line.
In nuclear questions, mass ratios can often be taken from nucleon numbers ( $m \propto A$ ).

Common Mistakes

Conserving momentum using speeds without resolving into components.
Using $\cos$ instead of $\sin$ for the component perpendicular to the original direction.
Forgetting to use the mass ratio (treating $m_{\alpha}=m_Q$ ).
Mistyping $150 \times 10^5$ as $150 \times 10^{-5}$ or $150 \times 10^5 = 1.5\times 10^6$ (it is $1.5\times 10^7$ ).

Things to Be Careful About

Angles are given to the original path, so identify carefully which component is perpendicular.
Ensure your calculator is in degrees.
The $y$ -components must cancel because the initial $y$ -momentum is exactly zero.

Techniques used

apply conservation of momentum in two perpendicular directionsresolve velocities into components using trigonometryuse mass ratio from nucleon numbers to relate momenta

(ii)

Calculate the kinetic energy of the $\alpha$ -particle.

kinetic energy = ______ $\text{J}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

E_k = \frac{1}{2}mv^2

For an $\alpha$ -particle, $m = 4u = 4(1.66\times 10^{-27}) = 6.64\times 10^{-27}\ \text{kg}$ .

$v = 150\times 10^5 = 1.50\times 10^7\ \text{m s}^{-1}$ .

E_k = \frac{1}{2}(6.64\times 10^{-27})(1.50\times 10^7)^2 = 7.5\times 10^{-13}\ \text{J}

Answer

$7.5\times 10^{-13}\ \text{J}$

Final answer

7.5 × 10^-13 J

Detailed explanation

Background Concept

Kinetic energy of a particle moving with speed $v$ is

E_k = \frac{1}{2}mv^2

This comes from work done to accelerate the mass from rest. It depends on the square of the speed, so getting powers of ten correct is crucial.

For nuclear particles, the mass is often given via the atomic mass unit $u$ :

1u \approx 1.66\times 10^{-27}\ \text{kg}

An $\alpha$ -particle has mass about $4u$ .

Understanding the Question

You are given the $\alpha$ -particle speed after decay:

v = 150\times 10^5\ \text{m s}^{-1}

You must calculate its kinetic energy in joules using an appropriate mass for the $\alpha$ -particle.

Approach

Convert the $\alpha$ mass into kg: $m = 4u$ .
Convert the speed into standard form.
Substitute into $E_k = \frac{1}{2}mv^2$ and keep track of indices.

Step-by-Step Reasoning

Mass of $\alpha$ :

m = 4u = 4(1.66\times 10^{-27}) = 6.64\times 10^{-27}\ \text{kg}

Speed:

v = 150\times 10^5 = 1.50\times 10^7\ \text{m s}^{-1}

Square the speed:

v^2 = (1.50\times 10^7)^2 = 2.25\times 10^{14}

Substitute:

E_k = \frac{1}{2}(6.64\times 10^{-27})(2.25\times 10^{14})

Multiply the numbers and add indices:

$6.64\times 2.25 = 14.94$ ,
$10^{-27}\times 10^{14} = 10^{-13}$ .
So

E_k = \frac{1}{2}(14.94\times 10^{-13}) = 7.47\times 10^{-13}\ \text{J}

To appropriate s.f.:

E_k \approx 7.5\times 10^{-13}\ \text{J}

Key Takeaways

Use $E_k = \frac{1}{2}mv^2$ .
Convert $u$ to kg correctly.
Squaring a speed in standard form doubles the power of ten.

Common Mistakes

Using $m = 4\times 10^{-27}\ \text{kg}$ (forgetting the factor $1.66$ ).
Not converting $150\times 10^5$ correctly (it equals $1.50\times 10^7$ , not $1.50\times 10^6$ ).
Forgetting to square the power of ten when squaring $v$ .

Things to Be Careful About

Keep units consistent: kg and m s $^{-1}$ give joules.
Give the final answer in standard form with a sensible number of significant figures.

Techniques used

use the kinetic energy formulaconvert particle mass to kilograms using the atomic mass unitapply significant figures and standard form