Physics 9702 / 23 — May/June 2025

(b)

In an experiment, two objects A and B are released from the side of a building, as shown in Fig. 1.1.

Fig. 1.1

Object A is released from rest at a height of $10.0\text{ m}$ above horizontal ground.
Object B is released with an initial upward velocity of $3.0\text{ m s}^{-1}$ at a height $h$ above the ground.
Both objects take the same time to reach the ground and they do not collide with each other.
Air resistance is negligible.

Calculate $h$ .

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

DifficultyMedium-Easy

Worked solution

Working

For A: $u = 0$ , $s = 10.0\ \text{m}$ , $a = g$ .

10.0 = \frac{1}{2}gt^2 \;\;\Rightarrow\;\; t = \sqrt{\frac{20.0}{g}}

For B: take downward as positive. $u = -3.0\ \text{m s}^{-1}$ , $s = h$ , $a = g$ .

h = ut + \frac{1}{2}gt^2 = (-3.0)t + 10.0

With $g = 9.81\ \text{m s}^{-2}$ ,

t = \sqrt{\frac{20.0}{9.81}} = 1.43\ \text{s}

h = 10.0 - 3.0(1.43) = 5.72\ \text{m}

Answer

$h = 5.7\ \text{m}$

Final answer

5.7 m

Detailed explanation

Background Concept

For vertical motion with constant acceleration (free fall, negligible air resistance), we can use the SUVAT equation

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

where:

$s$ is displacement along the chosen axis,
$u$ is initial velocity,
$a$ is acceleration (for free fall, magnitude $g$ ),
$t$ is time.

A key idea here is that if two objects take the same time to reach the ground, we can find that common $t$ from one object and substitute into the other.

Understanding the Question

Object A is dropped from rest from height $10.0\ \text{m}$ .
Object B is released from height $h$ with initial upward speed $3.0\ \text{m s}^{-1}$ .
They take the same time to hit the ground.
Air resistance is negligible, so acceleration is constant $g$ downward.

We need to calculate $h$ .

Approach

Find the time $t$ for object A to fall $10.0\ \text{m}$ from rest.
Use the same $t$ for object B, with its initial velocity, to find the height $h$ .

The cleanest method is to use $s = ut + \frac{1}{2}at^2$ for both objects with a consistent sign convention.

Step-by-Step Reasoning

1) Choose a sign convention
Take downward as positive so $a = +g$ .

2) Object A (dropped from rest)

$u_A = 0$
$s_A = 10.0\ \text{m}$
$a = g$

Apply SUVAT:

10.0 = 0 \cdot t + \frac{1}{2}gt^2

So

\frac{1}{2}gt^2 = 10.0 \quad\Rightarrow\quad t = \sqrt{\frac{20.0}{g}}.

Using $g = 9.81\ \text{m s}^{-2}$ :

t = \sqrt{\frac{20.0}{9.81}} = 1.43\ \text{s}.

3) Object B (thrown upward initially)
Downward positive means an initial upward velocity is negative:

u_B = -3.0\ \text{m s}^{-1}.

It falls from height $h$ to the ground, so the downward displacement is $s_B = h$ .

Apply SUVAT:

h = (-3.0)t + \frac{1}{2}gt^2.

But from object A we already know

\frac{1}{2}gt^2 = 10.0.

So

h = -3.0t + 10.0.

Substitute $t = 1.43\ \text{s}$ :

h = 10.0 - 3.0(1.43) = 5.72\ \text{m} \approx 5.7\ \text{m}.

Key Takeaways

Use one object to determine the common time, then substitute into the other equation.
A consistent sign convention (upward positive or downward positive) avoids sign errors.
For free fall with negligible air resistance, acceleration is constant $g$ .

Common Mistakes

Taking $u_B = +3.0\ \text{m s}^{-1}$ while also taking downward as positive (sign inconsistency).
Forgetting that $\frac{1}{2}gt^2$ for object A equals $10.0$ exactly, which simplifies the algebra.
Rounding $t$ too aggressively before substituting, causing noticeable rounding error in $h$ .

Things to Be Careful About

The displacement $s$ is the vertical drop to the ground: $10.0\ \text{m}$ for A and $h$ for B.
Use $g$ in $\text{m s}^{-2}$ and heights in $\text{m}$ .
Give the final height to a sensible number of significant figures (here $5.7\ \text{m}$ ).

Techniques used

use constant-acceleration kinematics in the vertical directionequate times for two motions to eliminate an unknownsubstitute from one equation into another to solve for a height

(c)

In a second experiment, object B is released from the same height as in (b) but with a speed of $6.0\text{ m s}^{-1}$ at an angle of $60^{\circ}$ to the vertical, as shown in Fig. 1.2.

Fig. 1.2

4M

(i)

State and explain whether the time taken for object B to reach the ground is less than, the same as, or greater than the time taken in the first experiment.

DifficultyMedium-Easy

Worked solution

Answer

The time is the same.

The initial vertical component is

u_y = 6.0\cos 60^\circ = 3.0\ \text{m s}^{-1}$$ which is the same as in the first experiment, and horizontal velocity does not affect the vertical motion/time to reach the ground.

Final answer

The same.

Detailed explanation

Background Concept

In projectile motion (neglecting air resistance):

The horizontal motion has zero acceleration (constant horizontal velocity).
The vertical motion has constant acceleration $g$ downward.

Crucially, the time taken to reach the ground depends only on the vertical motion, because the vertical displacement to the ground is achieved by the vertical component of velocity under vertical acceleration.

To compare situations, we often resolve the initial velocity into components:

u_y = u\cos\theta \quad (\theta \text{ measured from the vertical})

u_x = u\sin\theta.

Understanding the Question

Object B is released from the same height $h$ found in part (b), but now with speed $6.0\ \text{m s}^{-1}$ at $60^\circ$ to the vertical.

You must state and explain whether the time to reach the ground is less than, the same as, or greater than in the first experiment (where B had an initial upward velocity of $3.0\ \text{m s}^{-1}$ ).

Approach

Resolve the $6.0\ \text{m s}^{-1}$ velocity into its vertical component.
Compare this vertical component to the initial vertical velocity in the first experiment.
Use the fact that horizontal motion does not affect time of flight.

Step-by-Step Reasoning

diagram-c-i-components

The angle is given to the vertical, so the upward vertical component is

u_y = 6.0\cos 60^\circ.

Since $\cos 60^\circ = 0.5$ ,

u_y = 6.0 \times 0.5 = 3.0\ \text{m s}^{-1}.

In the first experiment, object B was released with initial upward velocity $3.0\ \text{m s}^{-1}$ . So the initial vertical conditions are identical:

same height $h$ ,
same initial vertical velocity $u_y = 3.0\ \text{m s}^{-1}$ ,
same vertical acceleration $g$ .

Therefore the vertical motion (and hence the time to reach the ground) is the same.

The additional horizontal component only changes how far sideways B travels, not how long it takes to fall.

Key Takeaways

Time of flight is determined by vertical displacement, vertical initial velocity, and $g$ .
Horizontal and vertical motions are independent in projectile motion.
Be careful about whether the angle is to the horizontal or to the vertical when resolving components.

Common Mistakes

Using $u_y = u\sin 60^\circ$ instead of $u\cos 60^\circ$ (because the angle is to the vertical).
Thinking that a larger total speed automatically means a shorter time (not true if the vertical component is unchanged).

Things to Be Careful About

Always write down what the angle is measured from.
State explicitly that horizontal velocity does not affect the vertical fall time; this is usually required for the explanation mark.

Techniques used

resolve an initial velocity into vertical and horizontal componentsapply independence of horizontal and vertical motioncompare vertical components to deduce equal times of flight

(ii)

By considering energy, state and explain whether the speed at which object B reaches the ground is less than, the same as, or greater than in the first experiment.

DifficultyMedium-Easy

Worked solution

Answer

Greater.

From energy conservation,

\frac{1}{2}mv^2 = \frac{1}{2}mu^2 + mgh

The height $h$ is the same in both experiments, so $mgh$ gained is the same, but in experiment 2 $u = 6.0\ \text{m s}^{-1}$ so the initial kinetic energy is larger than in experiment 1 where $u = 3.0\ \text{m s}^{-1}$ . Hence $v$ is larger in experiment 2.

Final answer

Greater.

Detailed explanation

Background Concept

With negligible air resistance, mechanical energy is conserved:

\text{GPE} + \text{KE} = \text{constant}.

Taking the ground as zero gravitational potential energy, an object dropped from height $h$ loses GPE $mgh$ and gains the same amount of KE.

Between release (initial) and impact (final):

\frac{1}{2}mv^2 = \frac{1}{2}mu^2 + mgh

where:

$u$ is the initial speed,
$v$ is the final speed at the ground.

Note: energy depends on speed squared, and does not care about direction of the velocity components — all components contribute to KE.

Understanding the Question

You compare experiment 1 (object B released upward at $3.0\ \text{m s}^{-1}$ from height $h$ ) with experiment 2 (released at speed $6.0\ \text{m s}^{-1}$ from the same height $h$ at an angle).

You must decide whether the speed at the ground is less, same, or greater in experiment 2, using energy reasoning.

Approach

Use conservation of mechanical energy from release to ground.
Recognise that $h$ is the same, so the change in GPE is the same in both experiments.
Compare the initial kinetic energies: experiment 2 has larger initial speed, so larger initial KE.
Conclude the final KE (and speed) must be larger in experiment 2.

Step-by-Step Reasoning

Write energy conservation for each experiment:

\frac{1}{2}mv_1^2 = \frac{1}{2}m(3.0)^2 + mgh

\frac{1}{2}mv_2^2 = \frac{1}{2}m(6.0)^2 + mgh

Since $mgh$ is identical (same mass and same height), the only difference is the initial kinetic energy term.
Compare initial KE:

\frac{1}{2}m(6.0)^2 > \frac{1}{2}m(3.0)^2.

So

\frac{1}{2}mv_2^2 > \frac{1}{2}mv_1^2 \quad\Rightarrow\quad v_2 > v_1.

Therefore, the speed at the ground in the second experiment is greater.

Key Takeaways

With negligible air resistance, use energy conservation to relate speeds and heights.
If the drop height is the same, a larger initial speed means a larger final speed.
KE depends on total speed, not just the vertical component.

Common Mistakes

Using only the vertical component of speed in the energy equation (energy uses total speed).
Thinking that because the vertical component in (c)(i) matches experiment 1, the final speed must also match (that is true for time, but not for KE because there is extra horizontal KE).
Assuming the object “loses” horizontal kinetic energy (it does not; horizontal KE stays as KE if no air resistance).

Things to Be Careful About

The phrase “By considering energy” indicates you should mention conservation of mechanical energy and compare KE and GPE.
State clearly that $h$ is the same, so the GPE change is the same; this is the key comparison point.

Techniques used

apply conservation of mechanical energy between release point and groundcompare initial kinetic energies for two cases at the same heightdeduce the relative final speeds from kinetic energy

Q2MediumForces, Density and PressureDeformation of Solids

(a)

Define the moment of a force about a pivot.

(b)

Three objects A, B and C are placed on a horizontal beam. The beam is in equilibrium, as shown in Fig. 2.1.

Fig. 2.1

The beam is uniform and has length $9.0\text{ m}$ .
A pivot is at the midpoint of the beam.
Object A has mass $90\text{ kg}$ and is at one end of the beam.
Object B has mass $m$ and is a distance of $3.0\text{ m}$ from the pivot.
Object C has mass $150\text{ kg}$ and is at the other end of the beam.

8M

(i)

Calculate $m$ .

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

DifficultyMedium-Easy

Worked solution

Working

Taking moments about the pivot (beam weight acts at pivot so has no moment):

150g(4.5)=90g(4.5)+mg(3.0)

150(4.5)=90(4.5)+3m

675=405+3m \;\Rightarrow\; m=90

Answer

$m = 90\ \text{kg}$

Final answer

90 kg

Detailed explanation

Background Concept

For a rigid body in rotational equilibrium about a pivot:

sum of clockwise moments about the pivot = sum of anticlockwise moments about the pivot.

Using $\tau = Fd$ , each moment is the force (often the weight $mg$ ) multiplied by the perpendicular distance from the pivot.

If a force acts through the pivot, its moment is zero because $d=0$ .

Understanding the Question

A uniform beam of length $9.0\ \text{m}$ is balanced on a pivot at its midpoint, so each end is $4.5\ \text{m}$ from the pivot.

Object A: $90\ \text{kg}$ at one end (left end in the diagram) $\Rightarrow d_A=4.5\ \text{m}$ .
Object B: mass $m$ at $3.0\ \text{m}$ from the pivot on the left side $\Rightarrow d_B=3.0\ \text{m}$ .
Object C: $150\ \text{kg}$ at the other end (right end) $\Rightarrow d_C=4.5\ \text{m}$ .

The beam is uniform, so its own weight acts at its centre, which coincides with the pivot, so it produces no turning effect.

Approach

Choose the pivot as the point about which to take moments (this removes any unknown reaction at the pivot).
Write an equation equating clockwise and anticlockwise moments.
Substitute distances and weights ( $mg$ ) and solve for $m$ .

Step-by-Step Reasoning

Take moments about the midpoint pivot.

Clockwise moment is produced by the weight of C on the right.
Anticlockwise moments are produced by the weights of A and B on the left.

So,

150g(4.5)=90g(4.5)+mg(3.0)

Cancel $g$ (it appears in every term):

150(4.5)=90(4.5)+3m

Compute:

675=405+3m

3m=270 \;\Rightarrow\; m=90\ \text{kg}

Key Takeaways

For equilibrium, total clockwise moment equals total anticlockwise moment.
Choosing the pivot as the moment point often eliminates unknown support forces.
Forces through the pivot have zero moment.

Common Mistakes

Using $9.0\ \text{m}$ instead of $4.5\ \text{m}$ for the end distances.
Including the weight of the beam as a moment (it acts at the pivot so moment is zero).
Forgetting that B is $3.0\ \text{m}$ from the pivot (not from the end).

Things to Be Careful About

Keep a consistent sign convention (clockwise vs anticlockwise) and stick to it.
Use perpendicular distances; here the beam is horizontal and weights are vertical, so the perpendicular distance is just the horizontal distance shown.

Techniques used

take moments about the pivotequate clockwise and anticlockwise moments for equilibriumuse distances from the pivot to compute torques

(ii)

Object A is removed and replaced by a wire fixed to the end of the beam and to the ground, as shown in Fig. 2.2.

Fig. 2.2

After the change, the beam is again horizontal and in equilibrium. The positions of B and C are unchanged.
The wire has a diameter of $1.8 \times 10^{-3}\text{ m}$ and has a strain of $1.2 \times 10^{-3}$ .
The wire is not extended beyond its limit of proportionality.

Calculate the Young modulus of the wire.

Young modulus = ______ $\text{Pa}$

DifficultyMedium

Worked solution

Working

From (i), $m=90\ \text{kg}$ .

Take moments about the pivot:

150g(4.5)=90g(3.0)+T(4.5)

T(4.5)=(675-270)g=405g \;\Rightarrow\; T=90g

Wire cross-sectional area:

A=\pi\left(\frac{1.8\times10^{-3}}{2}\right)^2 =\pi(0.9\times10^{-3})^2=2.54\times10^{-6}\ \text{m}^2

Young modulus:

E=\frac{\text{stress}}{\text{strain}}=\frac{T/A}{1.2\times10^{-3}}

Using $T=90g=8.83\times10^{2}\ \text{N}$ ,

E=\frac{(8.83\times10^{2})/(2.54\times10^{-6})}{1.2\times10^{-3}} =2.9\times10^{11}\ \text{Pa}

Answer

Young modulus $= 2.9\times10^{11}\ \text{Pa}$

Final answer

2.9 × 10^11 Pa

Detailed explanation

Background Concept

1) Moments in equilibrium

A rigid body is in equilibrium when:

net force is zero, and
net moment about any point is zero.

In many beam problems, using moments about the pivot is efficient because the pivot reaction force has zero moment.

2) Young modulus

For a wire within its limit of proportionality,

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

Stress is force per unit area:

\text{stress} = \frac{F}{A}

Strain is extension per original length, given here directly.

For a circular wire of diameter $d$ ,

A = \pi\left(\frac{d}{2}\right)^2

Understanding the Question

Object A is replaced by a vertical wire attached to the left end of the beam and to the ground. The beam is again horizontal and in equilibrium; objects B and C stay where they were.

So, instead of the weight of A producing a moment, the tension in the wire provides a force at the left end. Because the wire is connected to the ground below, the wire pulls downward on the beam (tension acts along the wire, away from the beam).

Given:

diameter $d=1.8\times10^{-3}\ \text{m}$
strain $=1.2\times10^{-3}$
wire is within proportional region $\Rightarrow$ Hooke’s law behaviour $\Rightarrow E$ is constant.

We must find $E$ .

diagram-bii-1

Approach

Use the principle of moments about the pivot to find the tension $T$ in the wire.
Compute the wire’s cross-sectional area $A$ from its diameter.
Compute stress $=T/A$ .
Divide by the given strain to obtain $E$ .

Step-by-Step Reasoning

Step 1: Find the tension using moments

Distances from pivot:

left end: $4.5\ \text{m}$
B: $3.0\ \text{m}$ to the left
C: $4.5\ \text{m}$ to the right

Clockwise moment is due to C (right side). Anticlockwise moments are due to B and the wire tension at the left end.

150g(4.5)=90g(3.0)+T(4.5)

Solve:

T(4.5)=(675-270)g=405g

T=\frac{405g}{4.5}=90g \approx 8.83\times10^{2}\ \text{N}

Step 2: Calculate area

Radius $r=d/2=0.9\times10^{-3}\ \text{m}$ .

A=\pi r^2=\pi(0.9\times10^{-3})^2=2.54\times10^{-6}\ \text{m}^2

Step 3: Stress and Young modulus

\text{stress} = \frac{T}{A} = \frac{8.83\times10^{2}}{2.54\times10^{-6}}\ \text{Pa} \approx 3.47\times10^{8}\ \text{Pa}

Then

E=\frac{\text{stress}}{\text{strain}}= \frac{3.47\times10^{8}}{1.2\times10^{-3}} \approx 2.9\times10^{11}\ \text{Pa}

Key Takeaways

Replace a weight on a beam with another force: you still use moments about the pivot to find that force.
Young modulus links force and extension via stress and strain: $E=(F/A)/(\Delta L/L)$ .
For circular wires, careful area calculation (radius is half the diameter) is essential.

Common Mistakes

Assuming the wire pulls upward on the beam (it pulls down if it is connected to the ground below).
Using $d$ as the radius in $A=\pi r^2$ (forgetting to halve the diameter).
Forgetting to divide by strain (giving stress instead of Young modulus).
Using the full beam length $9.0\ \text{m}$ as a moment arm for end forces instead of $4.5\ \text{m}$ .

Things to Be Careful About

The beam being uniform matters because its own weight acts at the pivot (zero moment).
Keep units consistent: diameter must be in metres for $A$ in $\text{m}^2$ , giving $E$ in $\text{Pa}$ .
Significant figures: the given data (e.g. $1.8\times10^{-3}$ , $1.2\times10^{-3}$ ) supports about 2 s.f. for $E$ .

Techniques used

take moments about the pivot to determine the wire tensioncalculate cross-sectional area from diameteruse stress and strain to compute Young modulus

(iii)

Object B is now moved to a new position closer to the pivot without passing it. The beam is again horizontal and in equilibrium.

State and explain the effect, if any, that this has on the strain in the wire.

DifficultyMedium-Easy

Worked solution

Answer

Strain increases.

Moving B closer to the pivot reduces its anticlockwise moment, so to balance the unchanged clockwise moment from C the wire must provide a larger moment, meaning the tension increases. Since the wire is within the proportional region, strain is proportional to stress (and hence tension), so the strain increases.

Final answer

Strain increases.

Detailed explanation

Background Concept

For rotational equilibrium about the pivot:

\sum \tau_{\text{clockwise}} = \sum \tau_{\text{anticlockwise}}

A moment is $\tau = Fd$ . If a force stays the same but its distance $d$ from the pivot changes, its moment changes.

For a wire within its limit of proportionality:

stress $\propto$ strain
and stress $=F/A$ .

So if the cross-sectional area $A$ is constant, then strain is proportional to the tension $T$ .

Understanding the Question

In the modified arrangement (with the wire), object C stays at the right end and object B stays on the left side but is moved closer to the pivot (without crossing it).

You must state and explain what happens to the strain in the wire.

Approach

Decide what moving B does to the moment produced by B about the pivot.
Use equilibrium of moments to infer how the wire tension must change to keep the beam horizontal.
Use the fact that the wire is within the proportional region to link tension change to strain change.

Step-by-Step Reasoning

The weight of C and its distance from the pivot do not change, so the clockwise moment from C is unchanged.
B remains the same weight, but is moved closer to the pivot, so its distance $d_B$ decreases. Therefore its moment $\tau_B = (mg)d_B$ decreases.
The total anticlockwise moment must still equal the (unchanged) clockwise moment. Since B now contributes less anticlockwise moment, the wire must contribute more anticlockwise moment.
The wire’s moment is $\tau_T = T(4.5)$ because it acts at the end of the beam. The distance $4.5\ \text{m}$ is unchanged, so the only way for $\tau_T$ to increase is for $T$ to increase.
Stress in the wire is $T/A$ , and $A$ is unchanged, so stress increases.
Within the proportional limit, strain $\propto$ stress, so the strain increases.

Key Takeaways

Moving a load closer to the pivot reduces its moment.
If the opposing moment must stay the same, another force (here the wire tension) must increase.
In the linear elastic region, increased tension means increased strain.

Common Mistakes

Saying the strain decreases because B is “closer” (it is the moment balance that matters, not closeness alone).
Forgetting that the clockwise moment from C is unchanged.
Claiming strain is unchanged because the wire’s length is unchanged (strain refers to extension relative to original length; here extension changes because tension changes).

Things to Be Careful About

The phrase “without passing it” means B stays on the same side of the pivot, so its moment keeps the same rotational sense (still anticlockwise), only its magnitude changes.
The proportionality statement is only valid because the question states the wire is not extended beyond its limit of proportionality.

Techniques used

reason qualitatively using changes in moment armsapply rotational equilibrium to infer change in tensionuse proportionality of stress and strain below the limit of proportionality

Q3MediumWork, Energy and PowerPhysical Quantities and UnitsDynamics

A car of mass $1500\text{ kg}$ is travelling along a straight horizontal road at constant velocity $v$ . The car is subject to a total resistive force $F$ , as shown in Fig. 3.1.

Fig. 3.1

(a)

Show that the power $P$ developed by the engine in overcoming the total resistive force is given by the equation

P = Fv

DifficultyMedium-Easy

Worked solution

Working

Power:

P = \frac{W}{t}

Work done against resistive force over distance $s$ :

W = Fs

So

P = \frac{Fs}{t} = F\left(\frac{s}{t}\right)

But $v = \dfrac{s}{t}$ , hence

P = Fv

Answer

$P = Fv$

Final answer

P = Fv

Detailed explanation

Background Concept

Power is the rate at which work is done (or energy is transferred).

P = \frac{W}{t}

If a constant force $F$ acts in the direction of motion and the object moves a distance $s$ , the work done by that force is

W = Fs

Speed is distance per unit time:

v = \frac{s}{t}

Understanding the Question

The engine is providing a driving force that balances the total resistive force $F$ while the car moves at constant velocity $v$ on a horizontal road. The question asks you to show (derive) the expression for the engine power needed to overcome the resistive force.

Approach

Connect power to work done per unit time. The work done in overcoming a resistive force over some distance is $Fs$ . Divide by the time taken, then recognise that $s/t$ is the speed $v$ .

Step-by-Step Reasoning

Start from the definition of power:

P = \frac{W}{t}

The resistive force $F$ opposes motion; to maintain constant speed, the engine must do work against this force. Over a distance $s$ , work done against a constant force is:

W = Fs

Substitute into the power definition:

P = \frac{Fs}{t}

Group as $F(s/t)$ , and use $v = s/t$ :

P = Fv

Key Takeaways

Use $P = W/t$ and $W = Fs$ for constant forces.
Recognise $s/t$ as the speed $v$ , giving $P = Fv$ .

Common Mistakes

Using $P = F/v$ (inverting $v$ incorrectly).
Mixing up distance $s$ with displacement in a different direction than the force (here force and motion are along the same line).

Things to Be Careful About

$P = Fv$ applies when the force is along the direction of motion (here, the resistive force is opposite the velocity, so the engine supplies an equal forward force in magnitude).
Units check: $Fv$ gives $\text{N} \times \text{m s}^{-1} = \text{J s}^{-1} = \text{W}$ .

Techniques used

use the definition of power as work done per unit timeuse work done by a constant force as force multiplied by distance movedsubstitute speed as distance per unit time

(b)

The car now moves up a slope at a constant speed of $30\text{ m s}^{-1}$ .
The slope is at an angle to the horizontal of $6.0^{\circ}$ , as shown in Fig. 3.2.

Fig. 3.2

The total resistive force acting on the car is $1600\text{ N}$ .

4M

(i)

Show that the increase in gravitational potential energy of the car in a time of $1.0\text{ s}$ is $46000\text{ J}$ .

DifficultyMedium-Easy

Worked solution

Working

Distance moved along slope in $1.0\ \text{s}$ :

s = vt = 30 \times 1.0 = 30\ \text{m}

Vertical height gained:

h = s\sin 6.0^\circ = 30\sin 6.0^\circ = 3.14\ \text{m}

Increase in GPE:

\Delta E_p = mgh = 1500 \times 9.81 \times 3.14 = 4.62 \times 10^4\ \text{J} \approx 4.6 \times 10^4\ \text{J}

Answer

$\Delta E_p \approx 46000\ \text{J}$ (in $1.0\ \text{s}$ ).

Final answer

46000 J

Detailed explanation

Background Concept

Gravitational potential energy (GPE) increases when an object gains height $h$ in a gravitational field:

\Delta E_p = mgh

where $m$ is mass and $g$ is gravitational field strength. On a slope, the car’s movement is along the slope, but only the vertical component of that movement contributes to the height gained.

If the car travels distance $s$ up a slope at angle $\theta$ to the horizontal, the vertical height gained is:

h = s\sin\theta

Understanding the Question

The car moves up a $6.0^\circ$ slope at constant speed $30\ \text{m s}^{-1}$ . In a time of $1.0\ \text{s}$ , you must find how much its GPE increases. You are given the mass $1500\ \text{kg}$ and can take $g = 9.81\ \text{m s}^{-2}$ .

Approach

Find the distance travelled along the slope in $1.0\ \text{s}$ using $s = vt$ .
Convert that slope distance into a vertical height using $h = s\sin\theta$ .
Substitute into $\Delta E_p = mgh$ .

Step-by-Step Reasoning

Distance up the slope in $1.0\ \text{s}$ :

s = vt = 30 \times 1.0 = 30\ \text{m}

Height gained is the vertical component of this displacement:

h = 30\sin 6.0^\circ

Numerically, $\sin 6.0^\circ \approx 0.105$ , so $h \approx 3.1\ \text{m}$ .

Increase in GPE:

\Delta E_p = 1500 \times 9.81 \times 3.14 \approx 4.6 \times 10^4\ \text{J}

This rounds to $46000\ \text{J}$ as required.

Key Takeaways

On a slope, height gained is $h = s\sin\theta$ , not $s$ .
Then use $\Delta E_p = mgh$ .

Common Mistakes

Using $h = s\cos\theta$ (wrong component).
Forgetting to multiply by time when finding distance ( $s = vt$ ).
Using $g = 10\ \text{m s}^{-2}$ without being consistent with expected rounding.

Things to Be Careful About

Angle is to the horizontal, so the vertical component of the slope distance uses $\sin\theta$ .
Keep units consistent: $m$ in kg, $h$ in m, $g$ in $\text{m s}^{-2}$ to get joules.

Techniques used

calculate distance travelled from speed and timeresolve displacement into vertical height using trigonometryuse change in gravitational potential energy as mgh

(ii)

Use the information in (b)(i) to determine the power developed by the engine to move the car up the slope.

power = ______ $\text{W}$

DifficultyMedium

Worked solution

Working

Power to overcome resistive force:

P_{\text{res}} = Fv = 1600 \times 30 = 4.8 \times 10^4\ \text{W}

From (b)(i), increase in GPE in $1.0\ \text{s}$ is $46000\ \text{J}$ , so

P_{\text{gpe}} = \frac{\Delta E_p}{t} = \frac{46000}{1.0} = 4.6 \times 10^4\ \text{W}

Total engine power:

P = P_{\text{res}} + P_{\text{gpe}} = 4.8 \times 10^4 + 4.6 \times 10^4 = 9.4 \times 10^4\ \text{W}

Answer

$9.4 \times 10^4\ \text{W}$

Final answer

9.4 × 10^4 W

Detailed explanation

Background Concept

Power is the rate of energy transfer:

P = \frac{\Delta E}{t}

For motion at constant speed with a force in the direction of motion, the rate of doing work against that force is:

P = Fv

When a car climbs a hill at constant speed, the engine must supply energy per second for two main purposes:

to increase the car’s gravitational potential energy (gain height),
to overcome resistive forces (air resistance + friction etc.).

Understanding the Question

The car climbs at constant speed $30\ \text{m s}^{-1}$ up a $6.0^\circ$ slope. The total resistive force is $1600\ \text{N}$ . From part (b)(i), the car gains $46000\ \text{J}$ of GPE every $1.0\ \text{s}$ . We need the total engine power output.

Approach

Calculate two powers and add them:

Resistive power: $P_{\text{res}} = Fv$ .
Power used to gain height: $P_{\text{gpe}} = \Delta E_p / t$ .
Since speed is constant, the engine’s power output must supply both rates of energy transfer.

Step-by-Step Reasoning

diagram-bii-forces

Resistive forces remove energy at a rate

P_{\text{res}} = Fv

Substitute $F = 1600\ \text{N}$ and $v = 30\ \text{m s}^{-1}$ :

P_{\text{res}} = 1600 \times 30 = 48000\ \text{W}

The car’s GPE increases by $46000\ \text{J}$ in $1.0\ \text{s}$ . That is an energy gain per second, so the corresponding power is:

P_{\text{gpe}} = \frac{46000}{1.0} = 46000\ \text{W}

Total power supplied by the engine is the sum:

P = 48000 + 46000 = 94000\ \text{W} = 9.4 \times 10^4\ \text{W}

Key Takeaways

Total engine power on a hill at constant speed = (power against resistance) + (rate of increase of GPE).
Use $P = Fv$ and $P = \Delta E / t$ appropriately.

Common Mistakes

Using only $P = Fv$ with the resistive force and forgetting the power needed to gain GPE.
Treating $46000\ \text{J}$ as the final power (forgetting to divide by time, though here $t=1.0\ \text{s}$ makes it numerically the same).
Mixing up the resistive force with the component of weight down the slope.

Things to Be Careful About

Constant speed means no change in kinetic energy, but it does NOT mean no power is needed.
Ensure the resistive force given is already the total resistive force (so you do not add another friction term).
Keep consistent significant figures; $9.4 \times 10^4\ \text{W}$ is appropriate here.

Techniques used

convert an energy change in a given time to a powercalculate resistive power using P = Fvadd power contributions to find total engine power

(c)

The car picks up a passenger and then continues up the slope at the same speed as in (b).

State and explain the effect, if any, that the passenger has on:

2M

(i)

the air resistance acting on the car

(ii)

the power developed by the engine.

DifficultyMedium-Easy

Worked solution

Answer

Power increases: the passenger increases the mass, so the component of weight down the slope ( $mg\sin\theta$ ) is larger and the rate of gain of GPE is larger; at the same speed a larger driving force is needed, so $P = Fv$ is larger.

Final answer

Power increases.

Detailed explanation

Background Concept

When climbing a slope at constant speed, the engine must supply energy per second to:

overcome resistive forces,
increase gravitational potential energy.

The rate of increase of GPE is

P_{\text{gpe}} = \frac{\Delta E_p}{t} = mg\frac{h}{t}

Also, at constant speed, required power associated with a required forward (driving) force along the slope is

P = Fv

Understanding the Question

A passenger increases the car’s mass, but the car continues up the same slope at the same speed as in (b). You must state and explain what happens to the engine power.

Approach

At fixed speed, power is proportional to the driving force required. Increasing mass increases the component of weight down the slope and increases the GPE gained per second, so the engine must supply more power.

Step-by-Step Reasoning

Speed is unchanged, so if the required driving force increases, engine power must increase because $P = Fv$ .
Adding a passenger increases mass $m$ .
The downslope component of weight is $mg\sin\theta$ , so it increases with $m$ .
The car also gains height at the same rate (same speed up the slope), so the GPE gained each second is proportional to $m$ :

P_{\text{gpe}} \propto m

Therefore the engine must provide a greater rate of energy transfer, so the power developed increases.

Key Takeaways

At constant speed on a hill, engine power is set by the total opposing force and the speed.
Increasing mass increases the opposing gravitational component and the power needed to gain GPE.

Common Mistakes

Saying power is unchanged because speed is unchanged (ignores that the required force changes).
Claiming air resistance increases due to mass (it does not, if speed and shape are unchanged).

Things to Be Careful About

Constant speed means net force is zero, but the driving force must still balance all opposing forces.
If the question had said the engine power stays the same, then speed would have to decrease; here it states speed is the same, so power must increase.

Techniques used

relate increased mass to increased component of weight down a slopeuse constant speed condition to link required driving force to opposing forcesinfer change in power using P = Fv at fixed speed

Q4MediumDynamicsWork, Energy and Power

(a)

State the principle of conservation of momentum.

DifficultyEasy

Worked solution

Answer

The total momentum of a system remains constant (momentum before = momentum after) provided no resultant external force acts on the system (system is isolated).

Final answer

Total momentum remains constant provided no resultant external force acts.

Detailed explanation

Background Concept

Momentum $p$ is defined by

p = mv

where $m$ is mass and $v$ is velocity.

Newton’s second law can be written as

F = \frac{\Delta p}{\Delta t}

If the resultant external force on a system is zero, then $\Delta p = 0$ , so the total momentum of the system cannot change.

Understanding the Question

You are asked to state (not calculate) the principle. That means you should give the core statement and the condition under which it applies (usually worth the second mark): no resultant external force on the system.

Approach

Write a single clear sentence:

“total momentum is constant” (or “momentum before equals momentum after”), and
the condition: “isolated system / no resultant external force”.

Step-by-Step Reasoning

Conservation means a quantity stays the same.
Momentum is conserved for a system only when external influences do not provide a net force (otherwise external impulse changes momentum).
Therefore, state that total momentum before an interaction equals total momentum after, provided the system is isolated (no resultant external force).

Key Takeaways

Momentum conservation is about the total momentum of the system, not individual objects.
The required condition is zero resultant external force (isolated system).

Common Mistakes

Saying “momentum is always conserved” without stating the condition.
Referring to “forces between the objects are equal” (Newton’s third law) but not concluding total momentum is constant.
Writing about kinetic energy conservation (that is only for elastic collisions).

Things to Be Careful About

Use the word total momentum.
Include the condition “no resultant external force” (or “isolated system”).

Techniques used

state the condition for momentum to be conservedexpress conservation of momentum as equality of total momentum before and after

(b)

An object A of mass $4.0\text{ kg}$ travels at a velocity of $6.0\text{ m s}^{-1}$ to the right on a horizontal frictionless surface. It moves towards a second object B of mass $2.0\text{ kg}$ that is moving at a velocity of $3.0\text{ m s}^{-1}$ in the same direction as A, as shown in Fig. 4.1.

Fig. 4.1

Object A collides with object B. The two objects join and move off together with velocity $v$ .

4M

(i)

Calculate:

velocity $v$

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

DifficultyMedium-Easy

Worked solution

Working

Take right as positive.

Initial momentum:

p_i = (4.0)(6.0) + (2.0)(3.0) = 30\ \text{kg m s}^{-1}

Final momentum (joined mass $= 6.0\ \text{kg}$ ):

p_f = (6.0)v

Conservation of momentum $p_i=p_f$ :

30 = 6.0v \Rightarrow v = 5.0\ \text{m s}^{-1}

Answer

$5.0\ \text{m s}^{-1}$

Final answer

5.0 m s⁻1

Detailed explanation

Background Concept

For a collision in one dimension (straight line), if the system is isolated (no resultant external force), total momentum is conserved:

\sum p_\text{before} = \sum p_\text{after}

Momentum is $p=mv$ . When objects join together after impact, the collision is perfectly inelastic: momentum is still conserved, but kinetic energy is not.

Understanding the Question

Object A ( $4.0\ \text{kg}$ ) moves right at $6.0\ \text{m s}^{-1}$ and catches object B ( $2.0\ \text{kg}$ ) moving right at $3.0\ \text{m s}^{-1}$ . After they collide, they stick together and move with a common speed $v$ .

You need to find that common speed using momentum conservation.

Approach

Choose a positive direction (right is easiest).
Calculate total initial momentum: $m_Au_A + m_Bu_B$ .
After collision the combined mass is $(m_A+m_B)$ with speed $v$ , so final momentum is $(m_A+m_B)v$ .
Set initial momentum = final momentum and solve for $v$ .

Step-by-Step Reasoning

Choose right as positive.

Initial momentum:

p_i = m_Au_A + m_Bu_B

Substitute values:

p_i = (4.0)(6.0) + (2.0)(3.0) = 24 + 6 = 30\ \text{kg m s}^{-1}

After collision they stick, so total mass is

m_A+m_B = 4.0+2.0 = 6.0\ \text{kg}

Final momentum:

p_f = (m_A+m_B)v = 6.0v

Conservation of momentum gives:

30 = 6.0v \Rightarrow v = 5.0\ \text{m s}^{-1}

The value is between $3.0$ and $6.0\ \text{m s}^{-1}$ , which makes physical sense: after sticking together, the speed should lie between the two initial speeds.

Key Takeaways

For an isolated system, momentum before = momentum after.
If objects stick together, treat them as one mass after collision.
A quick check: the final speed should lie between the initial speeds (when both move in the same direction).

Common Mistakes

Adding masses incorrectly (using $4.0-2.0$ or using only one mass in $p_f$ ).
Using kinetic energy conservation (not valid when they stick).
Sign errors if choosing left as positive or if one object were moving opposite (not the case here).

Things to Be Careful About

Keep a consistent sign convention (right positive).
Use the correct unit for momentum ( $\text{kg m s}^{-1}$ ) and for velocity ( $\text{m s}^{-1}$ ).
Sticking together implies one final velocity for the combined mass.

Techniques used

apply conservation of momentum to a collisionsum momenta using a consistent sign conventionsolve for the common final velocity after a completely inelastic collision

(ii)

the percentage of the total initial kinetic energy of the two objects that is transferred to other forms of energy during the collision.

percentage = ______ $\%$

DifficultyMedium

Worked solution

Working

Initial total kinetic energy:

E_{k,i} = \frac{1}{2}(4.0)(6.0^2) + \frac{1}{2}(2.0)(3.0^2) = 72 + 9 = 81\ \text{J}

After collision ( $v=5.0\ \text{m s}^{-1}$ , total mass $6.0\ \text{kg}$ ):

E_{k,f} = \frac{1}{2}(6.0)(5.0^2) = 75\ \text{J}

Energy transferred to other forms:

\Delta E = 81 - 75 = 6\ \text{J}

Percentage transferred:

\text{percentage} = \frac{6}{81} \times 100 = 7.4\%

Answer

$7.4\%$

Final answer

7.4 %

Detailed explanation

Background Concept

Kinetic energy is

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

In collisions:

Momentum is conserved if there is no resultant external force.
Kinetic energy is conserved only in elastic collisions.

If objects stick together, the collision is perfectly inelastic, so some kinetic energy is converted into other forms (internal energy/heat, sound, deformation).

Understanding the Question

You are asked for the percentage of the total initial kinetic energy that is transferred to other forms during the collision.

So you must:

find total initial $E_k$ of both objects,
find final $E_k$ of the combined object moving at speed $v$ (from part (i)),
calculate the loss $\Delta E = E_{k,i} - E_{k,f}$ ,
express this loss as a percentage of $E_{k,i}$ .

Approach

Compute $E_{k,i}$ by adding $\frac{1}{2}mv^2$ for A and B.
Compute $E_{k,f}$ using the combined mass $(m_A+m_B)$ and the common velocity $v$ .
Percentage transferred:

\% = \frac{E_{k,i}-E_{k,f}}{E_{k,i}}\times 100

Step-by-Step Reasoning

1) Initial kinetic energy

For A:

E_{kA} = \frac{1}{2}(4.0)(6.0^2) = 2.0 \times 36 = 72\ \text{J}

For B:

E_{kB} = \frac{1}{2}(2.0)(3.0^2) = 1.0 \times 9 = 9\ \text{J}

Total initial kinetic energy:

E_{k,i} = 72 + 9 = 81\ \text{J}

2) Final kinetic energy

After collision they move together. Total mass:

m = 4.0 + 2.0 = 6.0\ \text{kg}

From part (i), $v=5.0\ \text{m s}^{-1}$ , so

E_{k,f} = \frac{1}{2}(6.0)(5.0^2) = 3.0 \times 25 = 75\ \text{J}

3) Energy transferred to other forms

\Delta E = E_{k,i} - E_{k,f} = 81 - 75 = 6\ \text{J}

4) Percentage transferred

\text{percentage} = \frac{6}{81} \times 100 = 7.41\% \approx 7.4\%

Key Takeaways

Perfectly inelastic collision: momentum conserved, kinetic energy decreases.
“Transferred to other forms” means the loss in kinetic energy.
Percentage loss is always relative to the initial kinetic energy.

Common Mistakes

Using the final kinetic energy divided by the initial kinetic energy (that gives percentage remaining, not percentage transferred).
Forgetting to add both objects’ initial kinetic energies.
Using $v=5.0\ \text{m s}^{-1}$ without first obtaining it from momentum conservation.
Treating the loss as negative (you should report a positive percentage transferred away from kinetic energy).

Things to Be Careful About

Use $E_k \propto v^2$ (square the speeds correctly).
Keep enough significant figures during the working so the final percentage rounds correctly (here to $7.4\%$ ).
Ensure you use the combined mass for the final kinetic energy.

Techniques used

calculate kinetic energies before and after a collisionuse the speed from conservation of momentum in an energy calculationcompute a percentage energy transfer using a ratio

Q5MediumWaves

(a)

State why sound waves cannot be polarised.

DifficultyEasy

Worked solution

Answer

Sound waves are longitudinal, so the vibrations are parallel to the direction of travel and there is no transverse plane of oscillation to restrict (so they cannot be polarised).

Final answer

Sound waves are longitudinal, so there is no transverse plane of oscillation to restrict; hence sound cannot be polarised.

Detailed explanation

Background Concept

Polarisation is a property of transverse waves. A transverse wave vibrates in directions perpendicular to the direction it travels, so there are many possible directions (planes) in which the oscillations could occur.

A polariser works by allowing only oscillations in one direction (along its transmission axis). This only makes sense if the wave has oscillations perpendicular to its direction of travel.

Sound in air is (to A-level standard) a longitudinal wave: the particle displacements/pressure variations are parallel to the direction of propagation.

Understanding the Question

The question asks for the reason sound waves cannot be polarised. So we must connect (i) what “polarised” means to (ii) the nature of sound waves.

Approach

State that only transverse waves can be polarised, then state that sound is longitudinal (oscillations along the direction of travel), so there is no perpendicular direction/plane to “select”.

Step-by-Step Reasoning

Polarisation means restricting the oscillations of a wave to one plane/direction perpendicular to propagation.
Sound waves in air are longitudinal, so their oscillations are along the line of travel.
Therefore there is no transverse oscillation direction to restrict, so sound cannot be polarised.

Key Takeaways

Polarisation is only possible for transverse waves.
Sound in air is longitudinal, so it is not polarisable.

Common Mistakes

Saying “sound cannot be polarised because it is not electromagnetic” (not the reason).
Saying “sound waves have no direction” (they do; but the oscillations are parallel to travel).

Things to Be Careful About

In some solids, transverse (shear) waves can exist, but the standard statement for “sound waves” in air is longitudinal, hence not polarisable.

Techniques used

identify the direction of oscillations relative to the direction of propagationlink polarisation to transverse oscillations in a single plane

(b)

A plane-polarised light wave is incident on a polarising filter as shown in Fig. 5.1.

Fig. 5.1

The intensity of the light incident on the filter is $I_0$ .
The light is incident normally on the filter and the transmission axis of the filter is initially perpendicular to the plane of polarisation of the light.
The filter is now rotated through $360^{\circ}$ about the direction of travel of the light wave.

7M

(i)

On Fig. 5.2, sketch the variation of the intensity $I$ of the transmitted light with the angle of rotation $\alpha$ as the filter is rotated through $360^{\circ}$ from its initial position.

Fig. 5.2

DifficultyMedium-Easy

Worked solution

diagram-b-i

Answer

With transmission axis initially perpendicular, $I=0$ at $\alpha=0^\circ$ . Variation follows

I = I_0\sin^2\alpha

So $I=I_0$ at $90^\circ$ and $270^\circ$ , and $I=0$ at $0^\circ,180^\circ,360^\circ$ (smooth $\sin^2$ curve).

Final answer

A (\sin^2) curve: (I=0) at 0°, 180°, 360°; (I=I_0) at 90°, 270°.

Detailed explanation

Background Concept

For plane-polarised light incident on an ideal polarising filter, Malus’s law gives the transmitted intensity:

I = I_0 \cos^2\theta

where:

$I_0$ is the incident intensity,
$I$ is the transmitted intensity,
$\theta$ is the angle between the light’s plane of polarisation (electric field direction) and the filter’s transmission axis.

Because of the square, the intensity repeats every $180^\circ$ (since $\cos^2\theta = \cos^2(\theta+180^\circ)$ ).

Understanding the Question

Initially, the transmission axis is perpendicular to the plane of polarisation.
The filter is rotated by an angle $\alpha$ through $360^\circ$ .
We must sketch $I$ against $\alpha$ .

Key implication: at $\alpha=0^\circ$ , $\theta=90^\circ$ , so no light is transmitted.

Approach

Relate $\theta$ (the Malus angle) to $\alpha$ (the rotation from the initial position).
Use Malus’s law to find how $I$ varies.
Mark the important points ( $0, 90, 180, 270, 360^\circ$ ) and sketch a smooth curve through them.

Step-by-Step Reasoning

Initially (at $\alpha=0^\circ$ ):

transmission axis is perpendicular to polarisation, so $\theta = 90^\circ$ .
hence

I = I_0 \cos^2 90^\circ = 0.

After rotating the filter by $\alpha$ , the angle between the transmission axis and the (fixed) polarisation direction becomes

\theta = 90^\circ - \alpha.

So

I = I_0\cos^2(90^\circ-\alpha) = I_0\sin^2\alpha.

Now read off key values:

$\alpha=0^\circ$ : $I=0$
$\alpha=90^\circ$ : $I=I_0\sin^2 90^\circ = I_0$ (maximum)
$\alpha=180^\circ$ : $I=0$
$\alpha=270^\circ$ : $I=I_0$ (maximum)
$\alpha=360^\circ$ : $I=0$

So the sketch is a smooth $\sin^2\alpha$ curve with maxima at $90^\circ$ and $270^\circ$ .

diagram-b-i

Key Takeaways

Malus’s law: $I \propto \cos^2\theta$ .
If the initial position is perpendicular, the graph is shifted so $I$ starts at zero.
The transmitted intensity pattern repeats every $180^\circ$ .

Common Mistakes

Using $I = I_0\cos\theta$ (missing the square).
Starting the graph at $I=I_0$ at $\alpha=0^\circ$ (would correspond to initially parallel, not perpendicular).
Drawing a $\sin\alpha$ or $\cos\alpha$ curve that goes negative (intensity cannot be negative).

Things to Be Careful About

Distinguish the rotation angle $\alpha$ from the Malus angle $\theta$ ; you must account for the initial $90^\circ$ offset.
Ensure the peak is at $I_0$ (not $I_0/2$ ) because a single ideal polariser transmits $I_0$ when aligned.

Techniques used

apply Malus’s law for transmitted intensityconvert the given initial perpendicular condition into an angular dependencesketch a periodic squared-trigonometric intensity graph over one full rotation

(ii)

The amplitude of the incident light wave is $A_0$ when the intensity of the wave is $I_0$ .

Use Malus’s law to determine, in terms of $A_0$ , the amplitude of the transmitted wave when $\alpha = 20^{\circ}$ .

amplitude = ______ $A_0$

4M Similar

DifficultyMedium

Worked solution

Working

Initially transmission axis is perpendicular, so

I = I_0\sin^2\alpha.

At $\alpha=20^\circ$ :

\frac{I}{I_0}=\sin^2 20^\circ.

Since $I \propto A^2$ ,

\left(\frac{A}{A_0}\right)^2 = \sin^2 20^\circ \;\Rightarrow\; \frac{A}{A_0}=\sin 20^\circ.

Answer

A = (\sin 20^\circ)A_0 = 0.342\,A_0.

Final answer

0.342 A0

Detailed explanation

Background Concept

Two key ideas are used together:

Malus’s law for a polariser:

I = I_0 \cos^2\theta

where $\theta$ is the angle between the polarisation direction and the polariser transmission axis.

Intensity and amplitude:
For electromagnetic waves,

I \propto A^2

So if an intensity changes by a factor, the amplitude changes by the square root of that factor.

Understanding the Question

We are told:

incident intensity is $I_0$ and its amplitude is $A_0$ ,
the transmission axis starts perpendicular to the plane of polarisation,
the filter is rotated by $\alpha=20^\circ$ ,
we must find the transmitted amplitude $A$ as a multiple of $A_0$ .

Approach

Write the transmitted intensity in terms of $\alpha$ given the initial perpendicular condition.
Compute the ratio $I/I_0$ at $\alpha=20^\circ$ .
Convert intensity ratio to amplitude ratio using $I \propto A^2$ .

Step-by-Step Reasoning

1) Express $I$ in terms of $\alpha$ .
At $\alpha=0^\circ$ , the axis is perpendicular, so the Malus angle is $\theta=90^\circ$ . After rotating by $\alpha$ ,

\theta = 90^\circ - \alpha.

Substitute into Malus’s law:

I = I_0\cos^2(90^\circ-\alpha)= I_0\sin^2\alpha.

2) Evaluate the transmitted intensity fraction at $\alpha=20^\circ$ .

\frac{I}{I_0} = \sin^2 20^\circ.

3) Convert intensity fraction to amplitude fraction.
Because $I \propto A^2$ ,

\frac{I}{I_0} = \frac{A^2}{A_0^2} = \left(\frac{A}{A_0}\right)^2.

So

\left(\frac{A}{A_0}\right)^2 = \sin^2 20^\circ \Rightarrow \frac{A}{A_0} = \sin 20^\circ.

Taking the positive root is correct because amplitude is a magnitude.

Numerically,

\sin 20^\circ \approx 0.342

so

A \approx 0.342\,A_0.

Key Takeaways

Malus’s law gives an intensity ratio using a squared trig function.
To get an amplitude, take the square root: $A/A_0 = \sqrt{I/I_0}$ .
The initial perpendicular orientation shifts the dependence to $\sin^2\alpha$ .

Common Mistakes

Forgetting $I \propto A^2$ and writing $A/A_0 = I/I_0$ .
Using $\cos^2 20^\circ$ directly without accounting for the initial perpendicular position.
Giving $A = \sin^2 20^\circ\,A_0$ (missing the square root).

Things to Be Careful About

Ensure the angle used in Malus’s law is the angle between polarisation direction and transmission axis, not just the rotation angle unless the initial alignment is stated.
Keep the result in the requested form “___ $A_0$ ”; either $\sin 20^\circ\,A_0$ or its decimal is acceptable, but match typical exam rounding (3 s.f. is safe here).

Techniques used

apply Malus’s law to find transmitted intensity as a fraction of incident intensityuse the proportionality between intensity and square of amplitudetake a square root to convert an intensity ratio to an amplitude ratio

Q6MediumSuperpositionWaves

(a)

State what is meant by diffraction.

(b)

Light of wavelength $720\text{ nm}$ in a vacuum is incident normally on a diffraction grating as shown in Fig. 6.1.

Fig. 6.1

A screen is parallel to the grating. An interference pattern is seen on the screen and the angle between the second-order maxima is $52^{\circ}$ .

7M

(i)

Calculate the frequency of the light.

frequency = ______ $\text{Hz}$

DifficultyMedium-Easy

Worked solution

Working

f = \frac{c}{\lambda}

\lambda = 720\ \text{nm} = 720 \times 10^{-9}\ \text{m}

f = \frac{3.00 \times 10^{8}}{720 \times 10^{-9}} = 4.17 \times 10^{14}\ \text{Hz}

Answer

$4.17 \times 10^{14}\ \text{Hz}$

Final answer

4.17 × 10^14 Hz

Detailed explanation

Background Concept

For any wave,

v = f\lambda

For electromagnetic waves in a vacuum, the speed is

c = 3.00 \times 10^{8}\ \text{m s}^{-1}

So the frequency is

f = \frac{c}{\lambda}

Understanding the Question

You are given the wavelength of light in vacuum, $\lambda = 720\ \text{nm}$ , and asked for its frequency $f$ . Since it is in vacuum, you should use $c$ .

Approach

Convert $720\ \text{nm}$ into metres.
Apply $f = c/\lambda$ .
Give the answer to an appropriate number of significant figures and include units (Hz).

Step-by-Step Reasoning

Convert the wavelength:

720\ \text{nm} = 720 \times 10^{-9}\ \text{m} = 7.20 \times 10^{-7}\ \text{m}

Now calculate the frequency:

f = \frac{3.00 \times 10^{8}}{7.20 \times 10^{-7}}

Handle the powers of ten and the numerical part:

f = \left(\frac{3.00}{7.20}\right) \times 10^{15} = 0.4167 \times 10^{15} = 4.17 \times 10^{14}\ \text{Hz}

Key Takeaways

Use $c$ for light in vacuum.
Always convert nm to m before substituting.
$f$ is typically very large for visible/near-visible light (around $10^{14}$ to $10^{15}$ Hz).

Common Mistakes

Forgetting to convert nm to m (gives an answer smaller by $10^{9}$ ).
Using $v$ incorrectly (e.g. writing $f = \lambda / c$ ).

Things to Be Careful About

Use $c = 3.00 \times 10^{8}\ \text{m s}^{-1}$ (vacuum specified).
Quote the final answer with unit $\text{Hz}$ and sensible significant figures.

Techniques used

use the electromagnetic wave speed in vacuumconvert wavelength to SI unitsrearrange the wave equation to find frequency

(ii)

Calculate the number of lines per unit length in the diffraction grating.

number per unit length = ______ $\text{m}^{-1}$

DifficultyMedium

Worked solution

Working

Angle to one second-order maximum:

\theta = \frac{52^{\circ}}{2} = 26^{\circ}

Diffraction grating:

d\sin\theta = n\lambda

For $n=2$ and $\lambda = 720\times 10^{-9}\ \text{m}$ ,

d = \frac{2\lambda}{\sin 26^{\circ}} = \frac{2(720\times 10^{-9})}{\sin 26^{\circ}} = 3.28\times 10^{-6}\ \text{m}

Number of lines per unit length:

N = \frac{1}{d} = \frac{1}{3.28\times 10^{-6}} = 3.05\times 10^{5}\ \text{m}^{-1}

Answer

$3.05 \times 10^{5}\ \text{m}^{-1}$

Final answer

3.05 × 10^5 m^-1

Detailed explanation

Background Concept

A diffraction grating has many equally spaced slits. Constructive interference (principal maxima) occurs when the path difference between light from adjacent slits is an integer multiple of the wavelength:

d\sin\theta = n\lambda

where:

$d$ is the grating spacing (distance between adjacent slits),
$\theta$ is the angle from the normal to the grating to the $n$ th order maximum,
$n$ is the order number ( $n=0,1,2,\dots$ ),
$\lambda$ is the wavelength.

The “number of lines per unit length” is the line density:

N = \frac{1}{d}

Understanding the Question

You are told the angle between the two second-order maxima (one on each side of the central maximum) is $52^{\circ}$ . The pattern is symmetric, so each second-order maximum is at the same angle $\theta$ from the central axis (normal). Therefore the given $52^{\circ}$ is actually $2\theta$ .

You must use the grating equation to find $d$ , then invert to get $N$ .

Approach

Convert “angle between the second-order maxima” into the single angle $\theta$ for one side: $\theta = 52^{\circ}/2$ .
Apply $d\sin\theta = n\lambda$ with $n=2$ and $\lambda = 720\ \text{nm}$ .
Compute $d$ , then calculate $N = 1/d$ .

Step-by-Step Reasoning

Symmetry gives:

\theta = \frac{52^{\circ}}{2} = 26^{\circ}

Apply the grating equation for second order ( $n=2$ ):

d\sin 26^{\circ} = 2\lambda

Rearrange:

d = \frac{2\lambda}{\sin 26^{\circ}}

Substitute $\lambda = 720 \times 10^{-9}\ \text{m}$ :

d = \frac{2(720\times 10^{-9})}{\sin 26^{\circ}} = 3.28\times 10^{-6}\ \text{m}

Now invert to find line density:

N = \frac{1}{d} = \frac{1}{3.28\times 10^{-6}} = 3.05\times 10^{5}\ \text{m}^{-1}

Key Takeaways

In grating questions, angles are measured from the normal.
If the question gives the angle between maxima on opposite sides, halve it to get $\theta$ .
Line density is the reciprocal of slit spacing.

Common Mistakes

Using $52^{\circ}$ directly as $\theta$ (gives a wrong $d$ and wrong line density).
Forgetting to convert $720\ \text{nm}$ into metres.
Mixing up $d$ and $N$ (some students report $d$ when asked for lines per unit length).

Things to Be Careful About

Ensure $n=2$ for second order.
Use $\sin\theta$ , not $\tan\theta$ .
Quote $N$ in $\text{m}^{-1}$ and in standard form.

Techniques used

interpret angular geometry of symmetric maximaapply the diffraction grating equationcalculate grating spacing and invert to obtain line density

(iii)

The light in Fig. 6.1 is now replaced with light of a different wavelength $\lambda$ . It is observed that the third-order maxima of this light are at the same positions as the second-order maxima of the light in Fig. 6.1.

Calculate, in nm, the wavelength $\lambda$ .

$\lambda$ = ______ $\text{nm}$

DifficultyMedium-Easy

Worked solution

Working

Same grating and same positions on the screen $\Rightarrow$ same angle $\theta$ , so $d\sin\theta$ is the same.

2(720\ \text{nm}) = 3\lambda

\lambda = \frac{2\times 720}{3} = 480\ \text{nm}

Answer

$480\ \text{nm}$

Final answer

480 nm

Detailed explanation

Background Concept

For a fixed grating (fixed $d$ ), the angle $\theta$ of an $n$ th order maximum is determined by

d\sin\theta = n\lambda

If two different wavelengths produce maxima at the same angle, then the left-hand side $d\sin\theta$ is identical for both cases, so the products $n\lambda$ must be equal.

Understanding the Question

The original light has wavelength $720\ \text{nm}$ and its second-order maxima occur at certain positions on the screen (i.e. certain angles). A new wavelength $\lambda$ is used, and its third-order maxima appear at exactly the same positions. That means the third-order of the new light has the same angle as the second-order of the old light.

So we can equate $n\lambda$ for the two cases.

Approach

Use the grating condition $d\sin\theta = n\lambda$ .
Since $d$ is the same and the positions are the same, $\theta$ is the same, hence $d\sin\theta$ is the same.
Set $2(720\ \text{nm}) = 3\lambda$ and solve.

Step-by-Step Reasoning

Original light (second order):

d\sin\theta = 2(720\ \text{nm})

New light (third order at the same angle):

d\sin\theta = 3\lambda

Equate them:

2(720\ \text{nm}) = 3\lambda

Solve:

\lambda = \frac{2}{3}\times 720\ \text{nm} = 480\ \text{nm}

Key Takeaways

“Same position on the screen” means “same diffraction angle $\theta$ ”.
For the same grating, same $\theta$ implies $n\lambda$ is constant.
Higher order ( $n$ larger) at the same angle requires a smaller wavelength.

Common Mistakes

Using frequency ratios instead of wavelength ratios.
Forgetting that order number matters: equating wavelengths directly instead of $n\lambda$ .

Things to Be Careful About

Keep units consistent (nm is fine here since it cancels in the ratio).
Ensure you match the correct orders: second order for $720\ \text{nm}$ and third order for $\lambda$ .

Techniques used

equate diffraction conditions at the same angleuse proportionality between order number and wavelength for a fixed gratingrearrange to find the unknown wavelength

Q7MediumElectricityD.C. Circuits

A nichrome resistance wire has length $150\text{ cm}$ , cross-sectional area $2.45 \times 10^{-7}\text{ m}^2$ and resistivity $1.12 \times 10^{-6}\ \Omega\text{ m}$ .

(a)

Calculate, to three significant figures, the resistance of the wire.

resistance = ______ $\Omega$

DifficultyMedium-Easy

Worked solution

Working

R = \frac{\rho L}{A}

L = 150\ \text{cm} = 1.50\ \text{m}

R = \frac{(1.12\times 10^{-6})(1.50)}{2.45\times 10^{-7}} = 6.86\ \Omega

Answer

R = 6.86\ \Omega

Final answer

6.86 Ω

Detailed explanation

Background Concept

For a uniform wire, the resistance depends on:

its length $L$ (longer wire (\Rightarrow) larger resistance),
its cross-sectional area $A$ (thicker wire (\Rightarrow) smaller resistance),
and the material property resistivity $\rho$ .

The relationship is

R = \frac{\rho L}{A}

This equation assumes the wire has a constant cross-sectional area and is at (approximately) constant temperature.

Understanding the Question

You are given:

$L = 150\ \text{cm}$ ,
$A = 2.45 \times 10^{-7}\ \text{m}^2$ ,
$\rho = 1.12 \times 10^{-6}\ \Omega\,\text{m}$ .

You must calculate the wire resistance $R$ and quote it to three significant figures.

Approach

Convert the length from cm to m to match the SI units in $\rho$ and $A$ .
Substitute into $R = \rho L/A$ .
Round the final result to 3 s.f. and include the unit $\Omega$ .

Step-by-Step Reasoning

Convert length:

150\ \text{cm} = 1.50\ \text{m}

Substitute into the resistivity equation:

R = \frac{\rho L}{A} = \frac{(1.12\times 10^{-6})(1.50)}{2.45\times 10^{-7}}

Evaluate:

R = \frac{1.68\times 10^{-6}}{2.45\times 10^{-7}} = 6.857\ldots\ \Omega

Round to three significant figures:

R = 6.86\ \Omega

Key Takeaways

Always match units to the formula (here, convert cm to m).
Use $R = \rho L/A$ for uniform wires.
Quote answers to the requested significant figures with units.

Common Mistakes

Using $L = 150\ \text{m}$ instead of $1.50\ \text{m}$ .
Mixing cm with $\Omega\,\text{m}$ and $\text{m}^2$ (unit inconsistency).
Incorrect rounding (e.g. giving 6.9 $\Omega$ or too many s.f.).

Things to Be Careful About

$\rho$ is in $\Omega\,\text{m}$ , so $L$ must be in m.
Significant figures: since data are typically 3 s.f., giving $6.86\ \Omega$ is appropriate.

Techniques used

convert all given quantities into SI unitsapply the resistivity relation to calculate resistanceuse significant figures appropriately in the final answer

(b)

The nichrome wire forms part of a potentiometer circuit together with a cell of electromotive force (e.m.f.) $1.2\text{ V}$ and negligible internal resistance, as shown in Fig. 7.1.

Fig. 7.1

The circuit is used to determine the e.m.f. of cell X.
The galvanometer is used in a null method to find the null point $64\text{ cm}$ from the left-hand end of the nichrome wire.

5M

(i)

Explain what is meant by a null method.

DifficultyEasy

Worked solution

Answer

A null method is one where the result is obtained when the galvanometer shows zero deflection (no current flows in the test circuit at balance).

Final answer

A method where the balance point is found when the galvanometer reads zero (no current in the test circuit).

Detailed explanation

Background Concept

A galvanometer detects current in a circuit branch. In a potentiometer, you slide a contact along a uniform resistance wire to find a point where the potential difference along a length of the wire equals the e.m.f. being measured.

At the balance point, the galvanometer reads zero, meaning there is no current in the branch containing the unknown cell and galvanometer.

Understanding the Question

The circuit is used as a potentiometer to measure the e.m.f. of cell X. You are asked to explain what “null method” means in this context.

Approach

State what “null” refers to (zero galvanometer deflection) and what that implies physically (no current through the measured cell/galvanometer branch).

Step-by-Step Reasoning

You adjust the slider until the galvanometer shows zero deflection.
Zero deflection means the current through the galvanometer is zero.
If no current flows in that branch, the cell X is not supplying current to an external circuit; the potential difference along the selected length of potentiometer wire exactly equals the e.m.f. of cell X.

Key Takeaways

“Null” means a zero reading on the detector.
In potentiometers, null balance implies equality of potential differences and allows e.m.f. measurement without drawing current from the cell being measured.

Common Mistakes

Saying “null method means minimum reading” rather than zero reading.
Not linking the null reading to “no current in the test circuit”.

Things to Be Careful About

The crucial point is no current in the branch with the unknown cell at balance (this is why potentiometers are accurate for e.m.f. measurement).

Techniques used

identify the condition for a null deflection in a galvanometerinterpret zero current as zero potential difference across the galvanometer branch

(ii)

Calculate the e.m.f. of cell X.

e.m.f. = ______ $\text{V}$

DifficultyMedium-Easy

Worked solution

Working

Potential gradient along wire:

\frac{V}{L} = \frac{1.2}{150\ \text{cm}}

At null, e.m.f. of X equals p.d. across $64\ \text{cm}$ :

\mathcal{E}_X = 1.2\times \frac{64}{150} = 0.512\ \text{V}

Answer

\mathcal{E}_X = 0.512\ \text{V}

Final answer

0.512 V

Detailed explanation

Background Concept

In a potentiometer, a known source drives a steady current through a uniform resistance wire. This creates a uniform potential gradient (potential drop per unit length) along the wire, provided the current is constant and the wire is uniform.

If the full wire of length $L$ has a potential difference $V$ across it, then the potential difference across a shorter length $l$ is

V_l = V\frac{l}{L}

At the null point, the unknown e.m.f. equals this potential difference because no current flows in the test branch.

Understanding the Question

The 1.2 V cell is connected across the full $150\ \text{cm}$ nichrome wire.
The null point is at $64\ \text{cm}$ from the left end.
At null, the p.d. along the $64\ \text{cm}$ section equals the e.m.f. of cell X.

Approach

Use the proportionality of potential difference to length on a uniform potentiometer wire:

\mathcal{E}_X = 1.2 \times \frac{64}{150}

Step-by-Step Reasoning

Full wire has p.d. $1.2\ \text{V}$ across $150\ \text{cm}$ .
Potential gradient is constant, so p.d. across $64\ \text{cm}$ is the same fraction of $1.2\ \text{V}$ :

V_{64} = 1.2\times \frac{64}{150}

Calculate:

\frac{64}{150} = 0.426\overline{6}

\mathcal{E}_X = 1.2 \times 0.426\overline{6} = 0.512\ \text{V}

Key Takeaways

Uniform wire + constant current (\Rightarrow) constant potential gradient.
At null, $\mathcal{E}$ of the unknown cell equals the p.d. across the balancing length.
Use ratios of lengths to avoid unnecessary steps.

Common Mistakes

Using $150\ \text{cm} = 1.50\ \text{m}$ but leaving $64$ in cm (mixing units in the ratio). The ratio must use the same unit for both lengths.
Forgetting that it is a fraction of $1.2\ \text{V}$ .
Writing $\mathcal{E}_X = 1.2\times(150/64)$ (inverting the ratio).

Things to Be Careful About

The 1.2 V is across the entire wire, so the p.d. across $64\ \text{cm}$ is less than 1.2 V.
Sensible check: $64/150 < 1$ so $\mathcal{E}_X$ must be less than $1.2\ \text{V}$ .

Techniques used

use potential gradient along a uniform resistance wireapply proportionality of potential difference to balancing lengthcalculate an unknown e.m.f. from a known reference supply

(iii)

The cell of e.m.f. $1.2\text{ V}$ is replaced by a new cell with the same e.m.f. but with an internal resistance that is not negligible.

State and explain the effect, if any, of the internal resistance of the new cell on the position of the null point.

DifficultyMedium

Worked solution

Answer

The internal resistance causes a drop in terminal p.d. across the potentiometer wire, so the current in the wire (and hence the potential gradient) decreases.
Therefore a greater length of wire is needed to balance the same e.m.f. of X, so the null point moves further to the right (beyond $64\ \text{cm}$ ).

Final answer

Null point shifts to a larger length (to the right) because internal resistance reduces the p.d. across the wire and hence the potential gradient.

Detailed explanation

Background Concept

A real cell with e.m.f. $\mathcal{E}$ and internal resistance $r$ delivers a terminal potential difference

V = \mathcal{E} - Ir

when it supplies current $I$ . The term $Ir$ is the “lost volts” inside the cell.

In a potentiometer, the potential gradient along the wire depends on the potential difference across the wire and the wire length:

\text{potential gradient} = \frac{V_{\text{wire}}}{L_{\text{wire}}}

If $V_{\text{wire}}$ decreases, the gradient decreases.

Understanding the Question

Originally, the 1.2 V cell is ideal (negligible internal resistance), so essentially the full 1.2 V appears across the 150 cm wire.

Now it is replaced by another cell with the same e.m.f. (still 1.2 V) but with non-negligible internal resistance. You must state and explain what happens to the balance (null) point position.

Approach

Recognise the driving cell now has internal resistance, so when it drives current through the potentiometer wire, some voltage is lost inside the cell.
That means the terminal p.d. across the wire is less than 1.2 V.
A smaller p.d. across the same 150 cm gives a smaller potential gradient.
To get the same balancing p.d. equal to $\mathcal{E}_X$ , you need a longer length of wire.

Step-by-Step Reasoning

With internal resistance $r$ , current $I$ flows through the wire, so the terminal p.d. is

V_{\text{wire}} = \mathcal{E} - Ir

Because $Ir > 0$ when current flows, then

V_{\text{wire}} < \mathcal{E} = 1.2\ \text{V}

Potential gradient becomes

\frac{V_{\text{wire}}}{150\ \text{cm}}

which is smaller than before.

The unknown e.m.f. $\mathcal{E}_X$ has not changed. At balance,

\mathcal{E}_X = (\text{potential gradient})\times l

If the gradient is smaller, $l$ must be larger for the product to equal the same $\mathcal{E}_X$ .

Therefore the null point moves to the right (a distance greater than 64 cm from the left end).

Key Takeaways

Internal resistance matters only when the cell supplies current.
Lost volts reduce the terminal p.d. across the potentiometer wire.
Smaller potential gradient (\Rightarrow) longer balancing length for the same unknown e.m.f.

Common Mistakes

Saying “no effect because e.m.f. is unchanged” (e.m.f. is unchanged, but the terminal p.d. across the wire is reduced when current flows).
Claiming the null point moves left (this would require the gradient to increase, which does not happen here).
Confusing the internal resistance of the driving cell with that of cell X (at null, cell X has no current so its internal resistance does not affect the balance).

Things to Be Careful About

The potentiometer wire is not directly at the e.m.f. when internal resistance is present; it is at the terminal p.d.
The direction of shift should be justified: reduced gradient means you need more length, so the null point must move further from the left end.

Techniques used

apply the idea of lost volts across internal resistancerelate reduced current to reduced potential gradient along a potentiometer wireinfer the direction of shift of the balance point from a change in potential gradient

Q8MediumParticle Physics

(a)

An antiparticle equivalent of the neutron is called the antineutron. The quarks in the antineutron are the antiparticles of the quarks in a neutron.

The elementary charge is $e$ .

In Table 8.1, state the flavour and charge of the three antiquarks that comprise the antineutron.

Table 8.1

flavour	charge / $e$

DifficultyMedium-Easy

Worked solution

A neutron is $udd$ , so an antineutron is $\bar{u}\bar{d}\bar{d}$ .

flavour	charge / $e$
$\bar{u}$	$-\frac{2}{3}$
$\bar{d}$	$+\frac{1}{3}$
$\bar{d}$	$+\frac{1}{3}$

Final answer

$\bar{u}$ with $-\frac{2}{3}e$ , and two $\bar{d}$ each with $+\frac{1}{3}e$ .

Detailed explanation

Background Concept

Hadrons (such as protons and neutrons) are made of quarks. The relevant quark flavours here are:

up quark $u$ with charge $+\frac{2}{3}e$
down quark $d$ with charge $-\frac{1}{3}e$

An antiquark has:

the same flavour name but with a bar (e.g. $\bar{u}$ )
the opposite electric charge to its corresponding quark

So:

$\bar{u}$ has charge $-\frac{2}{3}e$
$\bar{d}$ has charge $+\frac{1}{3}e$

Understanding the Question

You are told the antineutron is the antiparticle of the neutron, and that its quarks are the antiparticles of the neutron’s quarks. The question asks for the three antiquarks in an antineutron, and their charges (in units of $e$ ).

Approach

Recall the quark composition of a neutron.
Replace each quark by its corresponding antiquark.
Write the charge of each antiquark as the negative of the corresponding quark’s charge.

Step-by-Step Reasoning

Neutron composition is:
- neutron: $udd$ .
Therefore antineutron composition is the antiquarks:
- antineutron: $\bar{u}\bar{d}\bar{d}$ .
Charges:
- since $u$ is $+\frac{2}{3}e$ , $\bar{u}$ is $-\frac{2}{3}e$ .
- since $d$ is $-\frac{1}{3}e$ , $\bar{d}$ is $+\frac{1}{3}e$ .
- there are two $\bar{d}$ antiquarks, so the last two rows are the same.

Key Takeaways

Neutron quark content: $udd$ .
Antiparticles have opposite charge: $q(\bar{q}) = -q(q)$ .
Antineutron quark content: $\bar{u}\bar{d}\bar{d}$ .

Common Mistakes

Writing the antineutron as $\bar{u}\bar{u}\bar{d}$ (that is an antiproton).
Giving $\bar{u}$ the charge $+\frac{2}{3}e$ (forgetting the sign change).
Listing only two quarks instead of three.

Things to Be Careful About

The table wants charge in units of $e$ , so write $\pm\frac{1}{3}$ and $\pm\frac{2}{3}$ (not in coulombs).
Use bars (or the word “anti-”) clearly to distinguish quarks from antiquarks.

Techniques used

recall the quark composition of a neutronreplace quarks by the corresponding antiquarksuse the rule that antiquarks have opposite charge to quarks

(b)

In $\beta^-$ decay, a neutron decays to form a proton.

Theory predicts that an antineutron should decay to form an antiproton. A particle and an antiparticle should also be observed.

Suggest the names of the particle and the antiparticle.

particle: ______
antiparticle: ______