(a) Coherent visible light of a single frequency is incident normally on a double slit. This produces a pattern of bright and dark interference fringes on a screen, as illustrated in Fig. 6.1. There are seven bright fringes. (i) Explain how the pattern of bright and dark interference fringes is formed. (ii) The distance between the centres of bright fringe X and bright fringe Y in the pattern is $10.2\ \text{mm}$. The slit spacing is $1.2\ \text{mm}$. The distance from the…

(a)(i) Bright: path difference = nλ; Dark: path difference = (n + 1/2)λ, due to superposition of coherent waves from the two slits. · (a)(ii) 6.6 × 10^-7 m · (a)(iii) Smaller fringe separation. · (b)(i) Cross at the central node (between the two loops). · (b)(ii) 0° · (b)(iii) 180°

(a) State the name of the class (group) of fundamental particles that contains a neutrino. (b) A hadron P has a charge of $+1e$, where $e$ is the elementary charge. The hadron P is composed of a down antiquark and only one other quark. (i) Identify a possible flavour for this other quark. (ii) State what type of hadron is P. (c) Nucleus Q undergoes radioactive decay to form nucleus R, emitting an antineutrino and another particle X, as shown in the decay equation. $$…

(a) Lepton · (b)(i) Up quark (u) · (b)(ii) Meson · (c)(i) Electron (beta-minus particle, e−) · (c)(ii) Same (AQ = AR) · (c)(iii) R has charge +1e greater than Q

9702/22 Feb/Mar 2024 — Physics Solved Past Paper

Q1Medium-EasyPhysical Quantities and UnitsWaves

(a)

Table 1.1 lists some SI quantities. Complete the table by indicating with a tick (✓) which rows are SI base quantities.

Table 1.1

quantity	base quantity
current
energy
force
mass

1M Similar

(b)

Use the definition of power to determine its SI base units.

SI base units = ______

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Using definition of power:

P = \frac{W}{t}

Work done:

W = Fs

and

F = ma

So

P = \frac{Fs}{t} = \frac{(ma)s}{t}

Units:

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

Answer

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

Final answer

kg m^2 s^-3

Detailed explanation

Background Concept

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

P = \frac{W}{t}

where $W$ is work done (energy transferred) and $t$ is time.

To express a quantity in SI base units, we rewrite it using definitions until only base quantities remain: mass ( $\text{kg}$ ), length ( $\text{m}$ ), time ( $\text{s}$ ), current ( $\text{A}$ ), etc.

Understanding the Question

You are asked to find the SI base units of power. So the final answer must be written only using $\text{kg}$ , $\text{m}$ , $\text{s}$ (and others if needed). For power, it will end up using $\text{kg}$ , $\text{m}$ and $\text{s}$ .

Approach

Start from the definition $P = W/t$ .
Replace $W$ using $W = Fs$ .
Replace $F$ using $F = ma$ .
Substitute the base units for $m$ , $a$ , $s$ , and $t$ , then simplify.

Step-by-Step Reasoning

Start with

P = \frac{W}{t}

Work done is defined by

W = Fs

so

P = \frac{Fs}{t}

Force is defined by Newton's second law:

F = ma

so

P = \frac{mas}{t}

Now put in base units:

mass $m$ has unit $\text{kg}$
acceleration $a$ has unit $\text{m s}^{-2}$
distance $s$ has unit $\text{m}$
time $t$ has unit $\text{s}$

Hence

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

Combine the $\text{m}$ factors and the $\text{s}$ factors:

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

Key Takeaways

Use definition chains to reduce derived quantities to base quantities.
Power in base units is $\text{kg m}^2\text{ s}^{-3}$ .

Common Mistakes

Writing just $\text{W}$ (watt). The question asks for base units, not the named derived unit.
Missing one factor of time and getting $\text{kg m}^2\text{ s}^{-2}$ (that is energy, joule).

Things to Be Careful About

Keep track of the extra division by $t$ in $P=W/t$ .
Acceleration is $\text{m s}^{-2}$ , not $\text{m s}^{-1}$ .

Techniques used

use a definition equation to express a derived quantity in base unitssubstitute definitions of derived quantities into other definitionssimplify units using index laws

(c)

A light meter is used to measure the intensity of light in a classroom. Daylight is incident normally on the sensor of the meter. The sensor has an area of $2.2\ \text{cm}^2$ . The reading on the meter is $950\ \text{W m}^{-2}$ .

Calculate the power of the daylight incident on the sensor.

power = ______ $\text{W}$

3M Similar

DifficultyMedium-Easy

Worked solution

Working

Intensity:

I = \frac{P}{A} \Rightarrow P = IA

Area:

2.2\ \text{cm}^2 = 2.2 \times 10^{-4}\ \text{m}^2

So

P = (950\ \text{W m}^{-2})(2.2 \times 10^{-4}\ \text{m}^2) = 2.09 \times 10^{-1}\ \text{W}

Answer

$0.21\ \text{W}$

Final answer

0.21 W

Detailed explanation

Background Concept

Intensity $I$ is defined as power per unit area:

I = \frac{P}{A}

where:

$I$ is intensity in $\text{W m}^{-2}$
$P$ is power in $\text{W}$
$A$ is area in $\text{m}^2$

Rearranging gives $P = IA$ .

Understanding the Question

A light meter sensor of area $2.2\ \text{cm}^2$ receives daylight normally (so the relevant area is just the sensor area). The meter reads an intensity of $950\ \text{W m}^{-2}$ . You must calculate the total power incident on the sensor.

Given:

$I = 950\ \text{W m}^{-2}$
$A = 2.2\ \text{cm}^2$

Find:

$P$ in watts.

Approach

Convert the area from $\text{cm}^2$ to $\text{m}^2$ .
Use $P = IA$ .
Round suitably (typically 2 s.f. here, matching the area given to 2 s.f.).

Step-by-Step Reasoning

Start from

P = IA

Convert area:

1\ \text{cm} = 10^{-2}\ \text{m} \Rightarrow 1\ \text{cm}^2 = (10^{-2})^2\ \text{m}^2 = 10^{-4}\ \text{m}^2

So

A = 2.2\ \text{cm}^2 = 2.2 \times 10^{-4}\ \text{m}^2

Now substitute:

P = (950)(2.2 \times 10^{-4})\ \text{W} = 2.09 \times 10^{-1}\ \text{W}

Rounded:

P \approx 0.21\ \text{W}

Key Takeaways

Intensity is power per unit area: $P = IA$ .
Converting $\text{cm}^2$ to $\text{m}^2$ requires squaring the length conversion.

Common Mistakes

Using $2.2\ \text{cm}^2 = 2.2 \times 10^{-2}\ \text{m}^2$ (forgetting to square the conversion).
Leaving the area in $\text{cm}^2$ while using $I$ in $\text{W m}^{-2}$ , causing inconsistent units.

Things to Be Careful About

Because the light is incident normally, you do not need to use any cosine factor; the effective area is just the sensor area.
Ensure the final unit is $\text{W}$ (since $\text{W m}^{-2} \times \text{m}^2 = \text{W}$ ).
Significant figures: $2.2$ (2 s.f.) and $950$ (likely 2 or 3 s.f. depending on context), so $0.21\ \text{W}$ is appropriate.

Techniques used

use intensity as power per unit areaconvert area units to SI unitssubstitute into a proportional relationship and calculate

Q2MediumKinematicsDynamicsForces, Density and Pressure

(a)

Define acceleration.

1M Similar

(b)

An Olympic diver stands on a platform above a pool of water, as shown in Fig. 2.1.

Fig. 2.1

When the diver is on the platform his centre of gravity is a vertical height of $9.0\ \text{m}$ above the surface of the water. The diver jumps from the platform with a velocity of $5.9\ \text{m s}^{-1}$ at an angle of $60^{\circ}$ to the horizontal.

Air resistance is negligible.

When the diver hits the surface of the water, his centre of gravity is a vertical height of $1.2\ \text{m}$ above the surface of the water.

Calculate the speed of the diver at the instant he hits the surface of the water.

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

3M Similar

DifficultyMedium-Easy

Worked solution

Working

Vertical drop of centre of gravity:

\Delta h = 9.0 - 1.2 = 7.8\ \text{m}

With negligible air resistance,

v^2 = u^2 + 2g\Delta h

v^2 = 5.9^2 + 2(9.81)(7.8) = 187.8

v = 13.7\ \text{m s}^{-1}

Answer

\boxed{v \approx 14\ \text{m s}^{-1}}

Final answer

14 m s^-1

Detailed explanation

Background Concept

For motion under gravity with air resistance negligible, the acceleration is constant and equal to $g$ vertically downwards. A very useful result is that the final speed depends on the change in vertical height and not on the direction of launch.

You can show this either by energy conservation or by kinematics:

Energy: loss of gravitational potential energy $mg\Delta h$ becomes gain in kinetic energy.
Kinematics: for the vertical component, $v_y^2 = u_y^2 + 2g\Delta h$ , while the horizontal component stays constant; combining gives $v^2 = u^2 + 2g\Delta h$ .

Understanding the Question

The diver’s centre of gravity starts $9.0\ \text{m}$ above the water surface and is $1.2\ \text{m}$ above the water surface at impact. So the centre of gravity has fallen by $7.8\ \text{m}$ . The diver leaves the platform at speed $5.9\ \text{m s}^{-1}$ at $60^\circ$ to the horizontal. You must find the speed (magnitude of velocity) just as he hits the water.

Approach

Use a method that avoids needing time:

Find the vertical drop $\Delta h$ .
Use

v^2 = u^2 + 2g\Delta h

Take the square root to get $v$ .

Step-by-Step Reasoning

Vertical drop:

\Delta h = 9.0 - 1.2 = 7.8\ \text{m}.

Apply the constant-acceleration relation for speed change due to gravity (equivalently energy conservation):

v^2 = u^2 + 2g\Delta h.

Substitute $u = 5.9\ \text{m s}^{-1}$ , $g = 9.81\ \text{m s}^{-2}$ , $\Delta h = 7.8\ \text{m}$ :

v^2 = 5.9^2 + 2(9.81)(7.8) = 34.81 + 153.04 = 187.85.

Hence

v = \sqrt{187.85} = 13.7\ \text{m s}^{-1} \approx 14\ \text{m s}^{-1}.

Key Takeaways

With no air resistance, final speed depends on initial speed and vertical drop only.
The launch angle is not needed if you use energy or the combined-speed kinematics result.

Common Mistakes

Using $\Delta h = 9.0\ \text{m}$ instead of $9.0 - 1.2\ \text{m}$ .
Trying to find time first (unnecessary and often leads to algebra errors).
Using $v = u + gt$ (needs time) or mixing vertical and horizontal components incorrectly.

Things to Be Careful About

Take the drop as positive in $v^2 = u^2 + 2g\Delta h$ (here $\Delta h = 7.8\ \text{m}$ ).
Quote the final speed to sensible significant figures (typically 2 s.f. here).

Techniques used

identify the vertical displacement from given heightsapply a constant-acceleration equation to relate speed and displacementcombine initial speed and gravitational potential change to find final speed

(c)

The diver in (b) enters the water and decelerates.

7M

(i)

Describe and explain the variation of the viscous drag force acting on the diver in the water as he moves downwards.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

The viscous drag force acts upwards (opposite to the downward motion). It is greatest just after entering the water when the speed is largest, and it decreases as the diver slows down because drag depends on speed.

Final answer

Upwards; largest at high speed on entry, then decreases as speed decreases.

Detailed explanation

Background Concept

Drag (resistive) forces oppose motion through a fluid. For viscous drag at relatively low speeds (laminar flow), a common model is

F_d \propto v

At higher speeds (turbulent flow), a common model is

F_d \propto v^2.

In both cases, the key idea is: bigger speed (\Rightarrow) bigger drag, and drag acts opposite to the velocity.

Understanding the Question

After the diver enters the water, he is moving downward but slowing down (decelerating). You must describe how the viscous drag force changes as he moves downward and explain why.

Approach

State the direction of drag (opposite to motion).
Link drag magnitude to speed.
Use the fact that the diver slows down in water to explain the variation.

Step-by-Step Reasoning

Immediately after entry, the diver’s downward speed is large, so the upward drag force is large.
The drag force contributes to an upward resultant force, so the diver decelerates.
As the diver slows, his speed $v$ gets smaller.
Since viscous drag depends on speed (typically $F_d \propto v$ ), the drag force decreases as he continues moving downwards.

Key Takeaways

Drag always acts opposite to the direction of motion.
Drag magnitude reduces as speed reduces (for both $F_d \propto v$ and $F_d \propto v^2$ models).

Common Mistakes

Saying drag acts downwards (wrong: it opposes motion).
Saying drag increases as he slows down (reverses the dependence on $v$ ).

Things to Be Careful About

The question specifies “viscous drag”, so it is safest to mention dependence on speed (not on depth or mass).

Techniques used

relate drag force magnitude to the speed in a resistive mediumuse the fact that drag acts opposite to the direction of motiondescribe how a changing speed changes the drag force

(ii)

The diver has a volume of $7.5 \times 10^{-2}\ \text{m}^3$ . The density of the water is $1.0 \times 10^3\ \text{kg m}^{-3}$ .

Show that the upthrust acting on the diver when he is entirely underwater is $740\ \text{N}$ .

1M Similar

DifficultyEasy

Worked solution

Working

For a fully submerged object,

U = \rho gV

U = (1.0 \times 10^3)(9.81)(7.5 \times 10^{-2}) = 7.36 \times 10^2\ \text{N} \approx 740\ \text{N}

Answer

\boxed{U \approx 740\ \text{N}}

Final answer

740 N

Detailed explanation

Background Concept

Archimedes’ principle: the upthrust (buoyant force) on an object in a fluid equals the weight of the fluid displaced.

For a fully submerged object of volume $V$ in a fluid of density $\rho$ :

U = \rho gV.

Understanding the Question

You are given the diver’s volume $7.5 \times 10^{-2}\ \text{m}^3$ and water density $1.0 \times 10^3\ \text{kg m}^{-3}$ . When the diver is entirely underwater, he displaces his full volume of water. You must show the upthrust is $740\ \text{N}$ .

Approach

Use $U = \rho gV$ and substitute the given values, then round to match $740\ \text{N}$ .

Step-by-Step Reasoning

Substitute:

U = (1.0 \times 10^3\ \text{kg m}^{-3})(9.81\ \text{m s}^{-2})(7.5 \times 10^{-2}\ \text{m}^3).

Calculate:

1.0 \times 10^3 \times 7.5 \times 10^{-2} = 75,

so

U = 75 \times 9.81 = 735.75\ \text{N} \approx 7.4 \times 10^2\ \text{N} = 740\ \text{N}.

Key Takeaways

Upthrust depends on fluid density and displaced volume, not on the object’s mass.
Fully submerged means displaced volume equals object volume.

Common Mistakes

Using the diver’s mass instead of volume.
Using $\rho$ of the diver rather than $\rho$ of the water.

Things to Be Careful About

Keep units consistent: $\rho$ in $\text{kg m}^{-3}$ and $V$ in $\text{m}^3$ gives $U$ in newtons.

Techniques used

apply Archimedes’ principle to a fully submerged objectcalculate upthrust using density, gravitational field strength, and volumeround to an appropriate number of significant figures

(iii)

At a particular instant when the diver is entirely underwater his horizontal velocity is zero. The viscous drag force acting on him at this instant is $950\ \text{N}$ vertically upwards. The diver has mass $78\ \text{kg}$ .

Determine the magnitude and direction of the acceleration of the diver.

acceleration = ______ $\text{m s}^{-2}$
direction ______

4M Similar

DifficultyMedium

Worked solution

Working

Upward forces:

U + F_d = 740 + 950 = 1690\ \text{N}

Weight:

W = mg = 78 \times 9.81 = 765\ \text{N}

Resultant force upward:

F = 1690 - 765 = 925\ \text{N}

Newton’s second law:

a = \frac{F}{m} = \frac{925}{78} = 11.9\ \text{m s}^{-2}

Answer

\boxed{a \approx 12\ \text{m s}^{-2}\ \text{upwards}}

Final answer

12 m s^-2 upwards

Detailed explanation

Background Concept

When an object is in a fluid, the vertical forces can include:

weight $W = mg$ (downwards),
upthrust $U$ (upwards),
drag $F_d$ (opposes motion, so here upwards if the diver is moving down).

Newton’s second law links resultant force to acceleration:

\sum F = ma.

The direction of the acceleration is the direction of the resultant force, not necessarily the direction of motion.

Understanding the Question

At the stated instant the diver is entirely underwater. We are told:

horizontal velocity is zero (so consider vertical forces only),
drag force is $950\ \text{N}$ vertically upwards,
upthrust is $740\ \text{N}$ upwards (from part (ii)),
mass is $78\ \text{kg}$ so weight is $mg$ downwards.

We must find the magnitude and direction of the diver’s acceleration.

Approach

Draw a vertical free-body diagram and choose a positive direction (e.g. upwards).
Add upward forces and subtract downward forces to get the resultant $F$ .
Use $a = F/m$ .
The sign tells you the direction.

Step-by-Step Reasoning

diagram-ciii-fbd

Take upward as positive.
Upward forces:

upthrust $U = 740\ \text{N}$ ,
drag $F_d = 950\ \text{N}$ .

So total upward force:

F_{\uparrow} = 740 + 950 = 1690\ \text{N}.

Downward force (weight):

W = mg = 78 \times 9.81 = 765\ \text{N}.

Resultant force (upwards):

F = F_{\uparrow} - W = 1690 - 765 = 925\ \text{N} \quad (\text{upwards}).

Apply Newton’s second law:

a = \frac{F}{m} = \frac{925}{78} = 11.9\ \text{m s}^{-2} \approx 12\ \text{m s}^{-2}.

Since $F$ is upwards, the acceleration is upwards.

Key Takeaways

Always resolve forces and assign directions before using $\sum F = ma$ .
An object can be moving downwards while accelerating upwards (that means it is slowing down).

Common Mistakes

Forgetting to include upthrust.
Taking weight as $78\ \text{N}$ instead of $78g$ .
Adding weight to the upward forces (sign error).
Giving only a magnitude with no direction.

Things to Be Careful About

State the acceleration direction explicitly (upwards/downwards).
Keep a consistent sign convention throughout; a quick free-body diagram prevents most errors.

Techniques used

draw a free-body diagram for vertical forcescalculate weight from mass and gravitational field strengthapply Newton’s second law using the resultant forcedetermine acceleration direction from the sign of the resultant

Q3MediumDeformation of Solids

A thin metal wire X, of diameter $1.2 \times 10^{-3}\ \text{m}$ , is used to suspend a model planet, as shown in Fig. 3.1.

Fig. 3.1

The variation with strain of the stress for wire X is shown in Fig. 3.2.

Fig. 3.2

(a)

The strain in X is $5.4 \times 10^{-3}$ .

6M

(i)

Use Fig. 3.2 to calculate the force exerted on the wire by the model planet.

force = ______ $\text{N}$

3M Similar

DifficultyMedium-Easy

Worked solution

Working

From Fig. 3.2, straight line through $(8.0\times 10^{-3},\ 1.0\ \text{GPa})$ .

For strain $= 5.4\times 10^{-3}$ ,

\text{stress} = \frac{5.4}{8.0}\times 1.0\ \text{GPa} = 0.675\ \text{GPa} = 6.75\times 10^{8}\ \text{Pa}

Wire diameter $d = 1.2\times 10^{-3}\ \text{m}$ , so

A = \pi\left(\frac{d}{2}\right)^2 = \pi\left(0.60\times 10^{-3}\right)^2 = 1.13\times 10^{-6}\ \text{m}^2

F = \text{stress}\times A = (6.75\times 10^{8})(1.13\times 10^{-6}) = 7.6\times 10^{2}\ \text{N}

Answer

$7.6\times 10^{2}\ \text{N}$

Final answer

7.6 × 10^2 N

Detailed explanation

Background Concept

Stress and strain for a wire are defined by

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

and

\text{strain} = \frac{\Delta L}{L_0}

where $F$ is the tensile force, $A$ is the cross-sectional area, $\Delta L$ is the extension and $L_0$ is the original length.

In the Hooke’s law (linear) region,

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

so a stress–strain graph is a straight line through the origin whose gradient is the Young modulus $E$ .

Understanding the Question

You are given the strain in wire X and a stress–strain graph for X. You must:

use the graph to find the stress corresponding to this strain,
use $\text{stress} = F/A$ to find the force exerted by the model planet.

The diameter of the wire is given, so the cross-sectional area can be calculated.

Approach

Read (or calculate by proportional scaling) the stress at strain $5.4\times 10^{-3}$ from the straight-line graph.
Convert the stress from $\text{GPa}$ to $\text{Pa}$ .
Compute the wire area using $A = \pi (d/2)^2$ .
Rearrange $\text{stress} = F/A$ to $F = \text{stress}\,A$ .

Step-by-Step Reasoning

Use the straight line on the graph

The graph passes through $(\text{strain}=8.0\times 10^{-3},\ \text{stress}=1.0\ \text{GPa})$ and the origin, so stress is directly proportional to strain.

Therefore,

\text{stress at }5.4\times 10^{-3} = \frac{5.4}{8.0}\times 1.0\ \text{GPa} = 0.675\ \text{GPa}.

Convert units:

0.675\ \text{GPa} = 0.675\times 10^{9}\ \text{Pa} = 6.75\times 10^{8}\ \text{Pa}.

Find cross-sectional area

With diameter $d=1.2\times 10^{-3}\ \text{m}$ , radius $r = 0.60\times 10^{-3}\ \text{m}$ .

A = \pi r^2 = \pi(0.60\times 10^{-3})^2 \approx 1.13\times 10^{-6}\ \text{m}^2.

Calculate force

Rearrange $\text{stress} = F/A$ :

F = (\text{stress})A = (6.75\times 10^{8})(1.13\times 10^{-6}) \approx 7.6\times 10^{2}\ \text{N}.

Key Takeaways

In the linear region, you can scale values directly on a stress–strain graph because stress $\propto$ strain.
Stress must be in $\text{Pa}$ (not $\text{GPa}$ ) when using $F = \text{stress}\times A$ .
Cross-sectional area of a circular wire comes from the diameter: $A = \pi(d/2)^2$ .

Common Mistakes

Using the diameter as the radius when calculating area.
Forgetting to convert $\text{GPa}$ to $\text{Pa}$ (a factor of $10^{9}$ ).
Reading the x-axis incorrectly: it is labelled $\text{strain} / 10^{-3}$ , so the value 5.4 on the axis corresponds to $5.4\times 10^{-3}$ .

Things to Be Careful About

Keep consistent powers of ten in the area calculation: $(10^{-3})^2 = 10^{-6}$ .
Quote the final force to sensible significant figures (limited here by the graph reading and given data).

Techniques used

read stress corresponding to a given strain from a stress–strain graphuse proportionality of stress and strain in the Hooke’s law regioncalculate cross-sectional area from diameteruse stress = force / area to determine the force

(ii)

The elastic potential energy of X is $0.31\ \text{J}$ .

Calculate the original length of the wire before the model planet was attached.

original length = ______ $\text{m}$

3M Similar

DifficultyMedium

Worked solution

Working

For Hooke’s law behaviour,

E_{\text{elastic}} = \frac{1}{2}F\Delta L

and

\text{strain} = \frac{\Delta L}{L_0} \Rightarrow \Delta L = (\text{strain})\,L_0

So

E_{\text{elastic}} = \frac{1}{2}F(\text{strain})L_0

L_0 = \frac{2E_{\text{elastic}}}{F\,(\text{strain})} = \frac{2(0.31)}{(7.6\times 10^{2})(5.4\times 10^{-3})}

L_0 \approx 0.15\ \text{m}

Answer

$0.15\ \text{m}$

Final answer

0.15 m

Detailed explanation

Background Concept

For a wire that obeys Hooke’s law, the force is proportional to extension, so a force–extension graph is a straight line through the origin. The elastic potential energy stored is the area under this graph:

E_{\text{elastic}} = \text{area} = \frac{1}{2}F\Delta L

Strain links extension to original length:

\text{strain} = \frac{\Delta L}{L_0} \Rightarrow \Delta L = (\text{strain})\,L_0.

Understanding the Question

You are told the elastic potential energy stored in wire X when the planet is attached, and you already know the strain (and from part (i) you can know the force). The question asks for the original length $L_0$ before the load was attached.

So you need to connect energy $\to$ extension, and extension $\to$ original length via strain.

Approach

Use $E_{\text{elastic}} = \tfrac{1}{2}F\Delta L$ to express extension $\Delta L$ in terms of energy and force.
Replace $\Delta L$ by $(\text{strain})L_0$ .
Rearrange to get $L_0$ .

Step-by-Step Reasoning

Start from the elastic energy formula:

E_{\text{elastic}} = \frac{1}{2}F\Delta L.

Use the definition of strain:

\Delta L = (\text{strain})\,L_0.

Substitute into the energy expression:

E_{\text{elastic}} = \frac{1}{2}F(\text{strain})L_0.

Rearrange for $L_0$ :

L_0 = \frac{2E_{\text{elastic}}}{F(\text{strain})}.

Substitute values ( $E_{\text{elastic}} = 0.31\ \text{J}$ , strain $=5.4\times 10^{-3}$ , and $F$ from part (i)):

L_0 = \frac{2(0.31)}{(7.6\times 10^{2})(5.4\times 10^{-3})} \approx 0.15\ \text{m}.

Key Takeaways

For Hooke’s law, elastic energy is triangular area: $E = \tfrac{1}{2}F\Delta L$ .
Strain is a ratio: $\Delta L = (\text{strain})L_0$ .
Combining these gives $E = \tfrac{1}{2}F(\text{strain})L_0$ .

Common Mistakes

Using $E = F\Delta L$ (rectangle area) instead of the triangular area $\tfrac{1}{2}F\Delta L$ .
Mixing up $L_0$ and $\Delta L$ (original length vs extension).
Forgetting to use the force from part (i), or using stress instead of force.

Things to Be Careful About

The energy formula $E = \tfrac{1}{2}kx^2$ also works, but only if you first find $k$ ; here $\tfrac{1}{2}F\Delta L$ is direct.
Use consistent significant figures: the given energy $0.31\ \text{J}$ suggests an answer around 2 s.f.

Techniques used

use elastic energy for linear force-extension behaviourrelate extension to strain and original lengthrearrange an equation to solve for the original lengthsubstitute a previously determined force into an energy calculation

(b)

Wire X is replaced by a new wire, Y, with the same original length and diameter but double the Young modulus of X. Wire Y also obeys Hooke’s law.

On Fig. 3.2, draw a line representing the variation with strain of the stress for Y.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Since $\text{stress} = E\,\text{strain}$ and $E_Y = 2E_X$ , the stress–strain line for Y is a straight line through the origin with twice the gradient of X.

E.g. it passes through $(\text{strain}=4.0\times 10^{-3},\ \text{stress}=1.0\ \text{GPa})$ (i.e. $(4.0,\ 1.0)$ on the given axes).

Final answer

Straight line through origin with twice the gradient (e.g. through (4.0, 1.0) on the axes).

Detailed explanation

Background Concept

In the Hooke’s law region, stress and strain are related by

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

On a graph of stress (y-axis) against strain (x-axis), the gradient is the Young modulus $E$ .

Larger $E$ means a steeper line (more stress needed for the same strain).
If the material obeys Hooke’s law, the line passes through the origin.

Understanding the Question

Wire Y has:

the same diameter and original length as X (so the graph axes/definitions are unchanged),
double the Young modulus of X,
still obeys Hooke’s law.

You must draw what its stress–strain graph would look like on the same axes.

Approach

Use the fact that the gradient of the stress–strain graph is $E$ .
Doubling $E$ means doubling the gradient.
Keep the line through the origin (Hooke’s law).
Choose a convenient point within the displayed axes to define the new line accurately.

Step-by-Step Reasoning

For wire X, the line goes through $(8.0\times 10^{-3},\ 1.0\ \text{GPa})$ .

If wire Y has $E_Y = 2E_X$ , then for any given strain,

\text{stress}_Y = 2\,\text{stress}_X.

So when $\text{stress}_X = 1.0\ \text{GPa}$ , wire Y would reach $1.0\ \text{GPa}$ at half the strain:

\text{strain}_Y = \frac{1}{2}(8.0\times 10^{-3}) = 4.0\times 10^{-3}.

Therefore, on the provided axes (labelled $\text{strain}/10^{-3}$ ), you can draw a straight line from the origin to the point $(4.0,\ 1.0)$ .

Key Takeaways

Young modulus is the gradient of a stress–strain graph.
Doubling Young modulus doubles the gradient (steeper line).
Hooke’s law implies a straight line through the origin.

Common Mistakes

Drawing a curve (Hooke’s law requires a straight line).
Shifting the line up/down (it must still pass through the origin).
Keeping the same point $(8.0,1.0)$ and just “making it steeper” by eye; a correct reference point such as $(4.0,1.0)$ makes the doubling precise.

Things to Be Careful About

The x-axis scale is $\text{strain}/10^{-3}$ , so the plotted x-coordinate for $4.0\times 10^{-3}$ is 4.0.
If you extend beyond the y-axis maximum, it’s still the same straight line; but using the point $(4.0,1.0)$ keeps the required drawing within the given axes.

Techniques used

use Young modulus as the gradient of a stress–strain graphapply proportional reasoning to compare gradientsconstruct a new straight-line relationship through the origin on given axes

Q4Medium-EasyParticle PhysicsDynamics

A nucleus P undergoes $\alpha$ -decay to form nucleus Q.

(a)

Complete the equation for this decay.

{}^{215}_{84}\text{P} \rightarrow \_\_\_\_\_\_\text{Q} + \_\_\_\_\_\_\alpha

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

In $\alpha$ -decay, $A$ decreases by $4$ and $Z$ decreases by $2$ :

{}^{215}_{84}\text{P} \rightarrow {}^{211}_{82}\text{Q} + {}^{4}_{2}\alpha

Final answer

${}^{211}_{82}\text{Q}$ and ${}^{4}_{2}\alpha$

Detailed explanation

Background Concept

In any nuclear decay equation, two quantities are conserved:

Nucleon number (mass number) $A$ : total number of protons + neutrons.
Proton number (atomic number) $Z$ : number of protons.

An $\alpha$ particle is the nucleus of a helium atom:

{}^{4}_{2}\alpha

So in $\alpha$ -decay, the parent nucleus loses 2 protons and 2 neutrons:

A \to A-4, \qquad Z \to Z-2

Understanding the Question

You are given a parent nucleus $\,{}^{215}_{84}\text{P}\,$ and told it undergoes $\alpha$ -decay to form nucleus $\text{Q}$ . You must fill in the nuclide notation for:

the daughter nucleus $\text{Q}$
the emitted $\alpha$ particle

so that $A$ and $Z$ balance on both sides.

Approach

Write the $\alpha$ particle as $\,{}^{4}_{2}\alpha$ .
Use conservation of $A$ $A$ and $Z$ $Z$ to find the values for nucleus $\text{Q}$ $Q$ :
- $A_Q = 215 - 4$
- $Z_Q = 84 - 2$

Step-by-Step Reasoning

Conservation of nucleon number:

215 = A_Q + 4 \;\Rightarrow\; A_Q = 211

Conservation of proton number:

84 = Z_Q + 2 \;\Rightarrow\; Z_Q = 82

So the completed equation is:

{}^{215}_{84}\text{P} \rightarrow {}^{211}_{82}\text{Q} + {}^{4}_{2}\alpha

Key Takeaways

In nuclear equations, always conserve $A$ and $Z$ .
$\alpha$ -decay reduces $A$ by $4$ and $Z$ by $2$ .

Common Mistakes

Writing the $\alpha$ particle incorrectly (e.g. mixing up $A$ and $Z$ ).
Subtracting $4$ from $Z$ instead of $2$ .
Not checking that both $A$ and $Z$ balance after filling in the blanks.

Things to Be Careful About

Ensure the nuclide notation is in the correct format $\,{}^{A}_{Z}\text{X}$ .
Both conservation conditions must be satisfied; matching only $A$ (or only $Z$ ) is not enough.

Techniques used

apply conservation of nucleon number and proton number in nuclear equationsidentify the emitted alpha particle as a helium nucleus

(b)

4M

(i)

State the principle of conservation of momentum.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

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

Final answer

Total momentum remains constant in an isolated system (no resultant external force).

Detailed explanation

Background Concept

Linear momentum $p$ is defined as:

p = mv

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 does not change.

Understanding the Question

You are asked to state the principle (not to calculate anything). The mark is typically for:

stating momentum is conserved (total momentum before = total momentum after)
stating the condition: no resultant external force / isolated system

Approach

Give a precise definition:

Mention “total momentum” of the system.
Mention it remains constant (before = after).
Include the condition of no external resultant force.

Step-by-Step Reasoning

A complete statement is:

Consider a system of interacting objects.
If no resultant external force acts on the system, the only forces are internal.
Therefore the total momentum of the system is unchanged:

\text{total momentum before} = \text{total momentum after}

Key Takeaways

Momentum conservation applies to a system.
It requires no resultant external force.

Common Mistakes

Forgetting the condition (no external force).
Saying “momentum is conserved” without specifying “total momentum of the system”.
Confusing momentum with energy (they are different conserved quantities under different conditions).

Things to Be Careful About

Use the phrase “resultant external force” (or “isolated system”) to secure the condition mark.
Don’t talk about “forces are equal” unless linking to the system idea; equal-and-opposite internal forces alone is not the full principle.

Techniques used

state the condition for momentum conservationexpress momentum conservation as constant total momentum in an isolated system

(ii)

Before the decay, nucleus P has a speed of $3.2 \times 10^5\ \text{m s}^{-1}$ . After the decay, nucleus Q is stationary.

Calculate the speed of the alpha particle after the decay.

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

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Conservation of momentum:

m_P v_P = m_Q v_Q + m_{\alpha} v_{\alpha}

With $v_Q = 0$ :

m_P v_P = m_{\alpha} v_{\alpha}

Using $m_P : m_{\alpha} = 215 : 4$ :

v_{\alpha} = \frac{215}{4} \left(3.2 \times 10^{5}\right) = 1.72 \times 10^{7}\ \text{m s}^{-1}

Answer

$1.7 \times 10^{7}\ \text{m s}^{-1}$

Final answer

1.7 × 10^7 m s^-1

Detailed explanation

Background Concept

Momentum is conserved in a system if there is no resultant external force. For a decay event, we treat the parent nucleus and the decay products as a system; the forces involved in the decay are internal.

Momentum is a vector quantity:

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

So conservation means:

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

For nuclei, the mass is approximately proportional to nucleon number $A$ (in atomic mass units), so mass ratios can be taken as ratios of $A$ values when high precision is not required.

Understanding the Question

Given:

Before decay: nucleus $\text{P}$ is moving at $v_P = 3.2 \times 10^{5}\ \text{m s}^{-1}$ .
After decay: nucleus $\text{Q}$ is stationary, so $v_Q = 0$ .
Decay is $\alpha$ -decay, so an $\alpha$ particle (mass number $4$ ) is emitted.

Find: speed of the $\alpha$ particle, $v_{\alpha}$ .

The statement “ $\text{Q}$ is stationary” simplifies the momentum equation a lot because $p_Q = 0$ .

Approach

Write momentum conservation for the system: parent momentum = sum of product momenta.
Substitute $v_Q = 0$ .
Use the mass ratio $m_P/m_{\alpha} \approx 215/4$ (from mass numbers).
Solve for $v_{\alpha}$ .

Step-by-Step Reasoning

Momentum conservation:

m_P v_P = m_Q v_Q + m_{\alpha} v_{\alpha}

Since $v_Q = 0$ :

m_P v_P = m_{\alpha} v_{\alpha}

Rearrange:

v_{\alpha} = \frac{m_P}{m_{\alpha}} v_P

Use nucleon-number ratio as a mass ratio:

\frac{m_P}{m_{\alpha}} \approx \frac{215}{4}

So:

v_{\alpha} = \frac{215}{4} \times 3.2 \times 10^{5}

Compute $215/4 = 53.75$ , hence:

v_{\alpha} = 53.75 \times 3.2 \times 10^{5} = 1.72 \times 10^{7}\ \text{m s}^{-1}

To 2 significant figures (matching $3.2$ ):

v_{\alpha} = 1.7 \times 10^{7}\ \text{m s}^{-1}

Key Takeaways

Use $\sum \vec{p}_{\text{before}} = \sum \vec{p}_{\text{after}}$ .
If one product is stationary, its momentum is zero.
For nuclear decay questions, using mass numbers for mass ratios is usually acceptable at this level.

Common Mistakes

Using $211$ (daughter mass number) instead of $215$ for the parent mass.
Forgetting that $v_Q = 0$ makes the $m_Q v_Q$ term zero.
Writing conservation as $m_P v_P = m_Q v_Q$ only (leaving out the alpha particle).
Giving the answer without unit or with incorrect unit.

Things to Be Careful About

Momentum is a vector: in a full treatment, directions matter. Here, because $\text{Q}$ is stated to be stationary, the $\alpha$ must carry the system’s momentum in the same direction as the initial momentum of $\text{P}$ .
Use an appropriate number of significant figures: the limiting value given is $3.2 \times 10^{5}$ (2 s.f.).

Techniques used

apply conservation of momentum to a decay eventuse nucleon numbers as proportional to nuclear massesrearrange to solve for an unknown speed

Q5Medium-EasyWaves

(a)

By reference to the direction of propagation of energy, state what is meant by a transverse wave.

1M Similar

(b)

A space telescope is designed to detect electromagnetic radiation with wavelengths in the range $12\ \mu\text{m}$ to $28\ \mu\text{m}$ .

State the region of the electromagnetic spectrum for this radiation.

1M Similar

DifficultyEasy

Worked solution

Answer

Infrared.

Final answer

Infrared

Detailed explanation

Background Concept

The electromagnetic (EM) spectrum is arranged by wavelength (or frequency). Typical approximate wavelength ranges:

Visible: about $4 \times 10^{-7}$ to $7 \times 10^{-7}\ \text{m}$
Infrared (IR): longer than visible, roughly $7 \times 10^{-7}\ \text{m}$ up to about $10^{-3}\ \text{m}$
Microwaves: about $10^{-3}$ to $1\ \text{m}$ (boundary is approximate)

Also, $1\ \mu\text{m} = 10^{-6}\ \text{m}$ .

Understanding the Question

The telescope detects EM radiation with wavelengths from $12\ \mu\text{m}$ to $28\ \mu\text{m}$ . You must name which part of the EM spectrum covers this range.

Approach

Convert the given wavelengths into metres (or compare directly in micrometres) and match to the standard EM spectrum regions.

Step-by-Step Reasoning

$12\ \mu\text{m} = 12 \times 10^{-6}\ \text{m}$ and $28\ \mu\text{m} = 28 \times 10^{-6}\ \text{m}$ .
These are far longer than visible light ( $\sim 10^{-7}\ \text{m}$ ) but much shorter than microwaves ( $\sim 10^{-3}\ \text{m}$ and above).
Therefore they lie in the infrared region.

Key Takeaways

Micrometre wavelengths ( $\mu\text{m}$ ) are commonly infrared.

Common Mistakes

Choosing “microwave” just because the wavelength is “long”; $10^{-5}\ \text{m}$ is still IR.
Confusing frequency ranges with wavelength ranges without conversion.

Things to Be Careful About

Spectrum boundaries are approximate, but $12$ – $28\ \mu\text{m}$ is unambiguously infrared in A-Level contexts.

Techniques used

convert wavelength into a familiar scale using prefixesidentify electromagnetic spectrum region from wavelength range

(c)

A detector on another space telescope detects an electromagnetic wave. The signal from the detector is transmitted to Earth and displayed on an oscilloscope as shown in Fig. 5.1. The frequency of the signal displayed on the oscilloscope is equal to the frequency of the detected electromagnetic wave.

Fig. 5.1

The time-base setting on the oscilloscope is $5.0 \times 10^{-15}\ \text{s cm}^{-1}$ .

Calculate the wavelength of the detected electromagnetic wave.

wavelength = ______ $\text{m}$

3M Similar

DifficultyMedium-Easy

Worked solution

Working

From the trace, one period spans $4.0\ \text{cm}$ .

Time-base $= 5.0 \times 10^{-15}\ \text{s cm}^{-1}$ , so

T = 4.0 \times 5.0 \times 10^{-15} = 2.0 \times 10^{-14}\ \text{s}

\lambda = cT = (3.0 \times 10^{8})(2.0 \times 10^{-14}) = 6.0 \times 10^{-6}\ \text{m}

Answer

$\lambda = 6.0 \times 10^{-6}\ \text{m}$

Final answer

6.0 × 10^-6 m

Detailed explanation

Background Concept

An oscilloscope displays voltage against time. The horizontal axis is time, set by the time-base (time per cm or per division). If the signal shown has the same frequency as the electromagnetic wave, then the period $T$ read from the oscilloscope is the EM wave period.

Key relations:

f = \frac{1}{T}

For an electromagnetic wave in space, the wave speed is the speed of light $c \approx 3.0 \times 10^{8}\ \text{m s}^{-1}$ , and

c = f\lambda

Combining gives $\lambda = \frac{c}{f} = cT$ .

Understanding the Question

The oscilloscope trace shows a sinusoidal signal on a $1.0\ \text{cm}$ grid. You are told:

time-base setting is $5.0 \times 10^{-15}\ \text{s cm}^{-1}$
the trace has a period of 4 divisions, i.e. $4.0\ \text{cm}$ , from one peak to the next peak.

You must calculate the electromagnetic wavelength $\lambda$ .

Approach

Measure the period in cm from the oscilloscope trace.
Convert this to time using the time-base to get $T$ .
Use $\lambda = cT$ .

Step-by-Step Reasoning

Read period from trace

One complete cycle is the horizontal distance between identical points (e.g. peak to peak).
The diagram indicates this is $4$ divisions, so $\text{period length} = 4.0\ \text{cm}.$

Convert to time period
Time per cm is $5.0 \times 10^{-15}\ \text{s cm}^{-1}$ , so

T = (4.0\ \text{cm})\left(5.0 \times 10^{-15}\ \text{s cm}^{-1}\right) = 2.0 \times 10^{-14}\ \text{s}.

Convert period to wavelength
Using $\lambda = cT$ :

\lambda = (3.0 \times 10^{8})(2.0 \times 10^{-14}) = 6.0 \times 10^{-6}\ \text{m}.

So the wavelength is $6.0\ \mu\text{m}$ (which is $6.0 \times 10^{-6}\ \text{m}$ ).

Key Takeaways

Oscilloscope horizontal scale converts divisions into time using the time-base.
For EM waves, $c = f\lambda$ and $f = 1/T$ , so $\lambda = cT$ is often the quickest route.

Common Mistakes

Using peak-to-trough distance (half a period) instead of peak-to-peak.
Forgetting the time-base is per cm/division and not the total time.
Using $v = 340\ \text{m s}^{-1}$ (speed of sound) instead of $c$ .

Things to Be Careful About

Check that you used one full cycle (same phase point to same phase point).
Keep powers of ten consistent: $10^{8} \times 10^{-14} = 10^{-6}$ .
Give the final answer in metres as requested, typically to 2 or 3 significant figures.

Techniques used

read the period from an oscilloscope trace using divisionsconvert time-base setting into a time perioduse c = f lambda and f = 1/T to determine wavelength

Q6MediumSuperposition

(a)

Coherent visible light of a single frequency is incident normally on a double slit. This produces a pattern of bright and dark interference fringes on a screen, as illustrated in Fig. 6.1.

Fig. 6.1

There are seven bright fringes.

7M

(i)

Explain how the pattern of bright and dark interference fringes is formed.

3M Similar

DifficultyMedium-Easy

Worked solution

Answer

The two slits act as coherent sources so the waves have a constant phase difference.
The waves from the two slits superpose at the screen; the path difference varies with position.
Bright fringes occur where path difference $= n\lambda$ (constructive interference).
Dark fringes occur where path difference $= \left(n+\tfrac{1}{2}\right)\lambda$ (destructive interference).

Final answer

Bright: path difference = nλ; Dark: path difference = (n + 1/2)λ, due to superposition of coherent waves from the two slits.

Detailed explanation

Background Concept

Interference is a superposition effect: when two waves meet at a point, the resultant displacement is the algebraic sum of the individual displacements.

For a double-slit, each slit acts as a source of waves. If the sources are coherent (same frequency and a constant phase difference), a stable pattern forms because the phase relationship at each point on the screen does not drift with time.

Whether a point on the screen is bright or dark depends on the path difference between the two waves arriving there:

Constructive interference (bright): path difference $= n\lambda$ where $n=0,1,2,\dots$
Destructive interference (dark): path difference $= \left(n+\tfrac{1}{2}\right)\lambda$

Understanding the Question

You are asked to explain, in words, why alternating bright and dark fringes appear on the screen when coherent monochromatic light passes through a double slit.

So you must mention:

coherent sources (the slits),
superposition on the screen,
the path difference conditions for maxima and minima.

Approach

Describe the physical chain:

slits behave as coherent sources,
waves overlap on the screen,
different points correspond to different path differences,
specific path differences give maxima (bright) and minima (dark).

Step-by-Step Reasoning

When light reaches the two narrow slits, each slit diffracts the light and acts like a source of circular wavefronts.
Because both slits are illuminated by the same incident wave, the light from the slits has the same frequency and (approximately) a fixed phase relationship: the slits are coherent.
At any point on the screen, two waves arrive (one from each slit). The total amplitude there depends on whether the waves arrive in phase or out of phase.
If the path difference is $n\lambda$ , the phase difference is $2\pi n$ , so the waves arrive in phase and add to give a maximum intensity (bright fringe).
If the path difference is $\left(n+\tfrac{1}{2}\right)\lambda$ , the phase difference is an odd multiple of $\pi$ , so the waves arrive in antiphase and cancel (minimum intensity, dark fringe).
As you move up/down the screen, the geometry changes smoothly, so the path difference alternates through these conditions, producing alternating bright and dark fringes.

Key Takeaways

Stable interference requires coherent sources.
Bright fringes: path difference $= n\lambda$ .
Dark fringes: path difference $= \left(n+\tfrac{1}{2}\right)\lambda$ .

Common Mistakes

Saying “bright where waves meet” without stating the condition on path difference.
Confusing path difference with phase difference (they are related but not identical).
Forgetting to mention coherence/constant phase difference, which is why the pattern is steady.

Things to Be Careful About

Use the correct form for destructive interference: $\left(n+\tfrac{1}{2}\right)\lambda$ , not $n\lambda/2$ without explanation.
The brightness is due to intensity, which depends on the resultant amplitude from superposition (not just on “more light”).

Techniques used

state the requirement for coherent sourcesapply the principle of superposition to overlapping wavesuse path difference conditions for constructive and destructive interference

(ii)

The distance between the centres of bright fringe X and bright fringe Y in the pattern is $10.2\ \text{mm}$ . The slit spacing is $1.2\ \text{mm}$ . The distance from the slits to the screen is $3.1\ \text{m}$ .

Calculate the wavelength of the light incident on the slits.

wavelength = ______ $\text{m}$

3M Similar

DifficultyMedium

Worked solution

Working

Seven bright fringes means there are $6$ fringe separations between the two outer bright fringes.

\text{fringe spacing } x = \frac{10.2\ \text{mm}}{6} = 1.70\ \text{mm} = 1.70 \times 10^{-3}\ \text{m}

For a double slit,

x = \frac{\lambda D}{a}

\lambda = \frac{xa}{D} = \frac{(1.70\times 10^{-3})(1.2\times 10^{-3})}{3.1} = 6.58\times 10^{-7}\ \text{m}

Answer

\lambda \approx 6.6\times 10^{-7}\ \text{m}

Final answer

6.6 × 10^-7 m

Detailed explanation

Background Concept

In double-slit interference, bright fringes are equally spaced (for small angles). The fringe spacing $x$ is related to wavelength by

x = \frac{\lambda D}{a}

where:

$x$ is the separation of adjacent bright fringes on the screen (in m),
$\lambda$ is the wavelength (in m),
$D$ is the distance from the slits to the screen (in m),
$a$ is the slit separation (in m).

This comes from the condition for maxima $a\sin\theta = n\lambda$ and the small-angle approximation $\tan\theta \approx \sin\theta \approx y/D$ .

Understanding the Question

You are given:

distance between the centres of bright fringes $X$ and $Y$ : $XY = 10.2\ \text{mm}$ ,
slit spacing: $a = 1.2\ \text{mm}$ ,
slit-to-screen distance: $D = 3.1\ \text{m}$ ,
and the diagram states there are seven bright fringes.

The key interpretation is that $X$ and $Y$ are the two outer bright fringes in the set of seven, so the distance $XY$ spans multiple equal fringe spacings.

Approach

Use the number of bright fringes to determine how many gaps (fringe spacings) lie between $X$ and $Y$ .
Divide $XY$ by that number to get the fringe spacing $x$ .
Substitute into $x = \lambda D / a$ and rearrange for $\lambda$ .
Convert all mm values to m.

Step-by-Step Reasoning

If there are $7$ bright fringes in a row, the number of separations between adjacent bright fringes from the first to the seventh is $7-1 = 6$ .
So the fringe spacing is

x = \frac{10.2\ \text{mm}}{6} = 1.70\ \text{mm}

Convert to metres:

1.70\ \text{mm} = 1.70 \times 10^{-3}\ \text{m}

Convert slit spacing:

a = 1.2\ \text{mm} = 1.2 \times 10^{-3}\ \text{m}

Rearrange the fringe spacing formula:

\lambda = \frac{xa}{D}

Substitute values:

\lambda = \frac{(1.70\times 10^{-3})(1.2\times 10^{-3})}{3.1}

Multiply top:

(1.70\times 1.2)\times 10^{-6} = 2.04\times 10^{-6}

Then divide:

\lambda = \frac{2.04\times 10^{-6}}{3.1} = 6.58\times 10^{-7}\ \text{m} \approx 6.6\times 10^{-7}\ \text{m}

The magnitude ( $\sim 10^{-7}\ \text{m}$ ) is consistent with visible light.

Key Takeaways

Adjacent bright fringes are equally spaced: use $x = \lambda D/a$ .
If a distance spans several fringes, count the number of gaps (one fewer than the number of bright fringes involved).
Always convert mm to m before substituting.

Common Mistakes

Using $10.2\ \text{mm}$ as one fringe spacing (gives an infrared wavelength).
Dividing by $7$ instead of $6$ .
Forgetting to convert mm to m, causing a $10^3$ error.

Things to Be Careful About

Significant figures: the limiting data ( $1.2\ \text{mm}$ and $3.1\ \text{m}$ ) are 2 s.f., so quote $\lambda$ to 2 s.f.
Ensure you use $x = \lambda D/a$ (not inverted) and keep consistent symbols ( $a$ is slit separation, not amplitude).

Techniques used

count the number of fringe spacings from the number of bright fringesuse the double-slit fringe spacing relationrearrange an equation to solve for wavelengthconvert millimetres to metres consistently

(iii)

The light is replaced by different visible light with a shorter wavelength.

State how the new fringe separation will compare to the original fringe separation.

1M Similar

(b)

A stationary wave is formed on a stretched string AB, as shown in Fig. 6.2.

Fig. 6.2

P, Q and R are points on the string.

3M

(i)

On Fig. 6.2, draw a cross (×) to show the position of a node.

1M Similar

(ii)

State the phase difference between P and Q.

phase difference = ______ $^{\circ}$

1M Similar

(iii)

State the phase difference between P and R.

phase difference = ______ $^{\circ}$

1M Similar

Q7MediumElectricityD.C. Circuits

(a)

Define electric potential difference.

1M Similar

(b)

A cell of electromotive force (e.m.f.) $1.8\ \text{V}$ and internal resistance $r$ is connected in parallel with a resistor of resistance $6.0\ \Omega$ and a filament lamp, as shown in Fig. 7.1.

Fig. 7.1

The switch S is open. The ammeter reading is $0.25\ \text{A}$ .

Determine the internal resistance $r$ of the cell.

$r$ = ______ $\Omega$

3M Similar

DifficultyMedium-Easy

Worked solution

diagram-7b-1

Working

Switch $S$ open $\Rightarrow$ lamp branch open, so external resistance $R = 6.0\ \Omega$ only.

V_{\text{terminal}} = IR = 0.25 \times 6.0 = 1.50\ \text{V}

Lost volts:

\varepsilon - V_{\text{terminal}} = 1.80 - 1.50 = 0.30\ \text{V}

Ir = 0.30 \Rightarrow r = \frac{0.30}{0.25} = 1.2\ \Omega

Answer

$r = 1.2\ \Omega$

Final answer

1.2 Ω

Detailed explanation

Background Concept

A real cell can be modelled as an ideal source of e.m.f. $\varepsilon$ in series with an internal resistance $r$ . When current $I$ flows, some of the e.m.f. is “lost” across $r$ :

\text{lost volts} = Ir

The terminal p.d. across the external circuit is then

V_{\text{terminal}} = \varepsilon - Ir

Also, for an external resistor $R$ ,

V_{\text{terminal}} = IR

Understanding the Question

With switch $S$ open, the lamp branch is disconnected, so the only load connected across the cell is the $6.0\ \Omega$ resistor. The ammeter reads the current supplied by the cell, $I = 0.25\ \text{A}$ . You must find the internal resistance $r$ given $\varepsilon = 1.8\ \text{V}$ .

Approach

With the lamp branch open, treat the circuit as: cell (with $r$ ) in series with the external $6.0\ \Omega$ .
Find the terminal p.d. across the $6.0\ \Omega$ resistor using $V = IR$ .
The difference $\varepsilon - V$ is the p.d. across the internal resistance, equal to $Ir$ .
Rearrange for $r$ .

Step-by-Step Reasoning

diagram-7b-1

Switch open means no current can flow through the lamp branch, so the circuit reduces to a single loop containing $\varepsilon$ , $r$ , and $6.0\ \Omega$ .
The terminal p.d. equals the p.d. across the $6.0\ \Omega$ resistor:

V_{\text{terminal}} = IR = 0.25 \times 6.0 = 1.50\ \text{V}

The e.m.f. is larger than the terminal p.d. because some p.d. is across internal resistance:

\varepsilon - V_{\text{terminal}} = 1.80 - 1.50 = 0.30\ \text{V}

This is the p.d. across $r$ , so $Ir = 0.30$ :

r = \frac{0.30}{0.25} = 1.2\ \Omega

Key Takeaways

With internal resistance, $V_{\text{terminal}} = \varepsilon - Ir$ .
“Lost volts” is $\varepsilon - V_{\text{terminal}}$ and equals $Ir$ .

Common Mistakes

Using $\varepsilon = IR$ with $R = 6.0\ \Omega$ (that ignores internal resistance).
Treating the $6.0\ \Omega$ resistor as in parallel with $r$ (it is not; $r$ is in series inside the cell model).

Things to Be Careful About

The switch being open is crucial: it removes the lamp branch completely.
Keep enough significant figures through the calculation; final value $1.2\ \Omega$ is appropriate.

Techniques used

model the cell as an emf in series with an internal resistanceuse V = IR for the external resistor to find the terminal potential differenceuse lost volts (epsilon − V) = Ir to determine internal resistance

(c)

At time $t_1$ switch S in Fig. 7.1 is closed. Fig. 7.2 shows the variation with time $t$ of the ammeter reading $I$ .

Fig. 7.2

4M

(i)

State whether the e.m.f. of the cell after $t_1$ is greater than, less than or the same as it was before $t_1$ .

1M Similar

DifficultyEasy

Worked solution

Answer

The e.m.f. is the same after $t_1$ .

Final answer

The same.

Detailed explanation

Background Concept

The e.m.f. $\varepsilon$ of a cell is the energy supplied per unit charge by the source. For an idealised cell model, $\varepsilon$ is a characteristic of the cell and does not change just because you change the external circuit.

What does change when the current changes is the terminal p.d.:

V_{\text{terminal}} = \varepsilon - Ir

So changing the load changes $I$ and therefore the lost volts $Ir$ , which changes the terminal p.d., even though $\varepsilon$ stays (approximately) constant.

Understanding the Question

At time $t_1$ the switch is closed, changing the external circuit and therefore the current. You are asked whether the e.m.f. after $t_1$ changes compared with before $t_1$ .

Approach

Decide whether e.m.f. depends on the external resistance (it does not, in this model). Use the distinction between e.m.f. and terminal p.d.

Step-by-Step Reasoning

Closing the switch changes the circuit resistance and hence the current.
The change in current changes the terminal p.d. because $V_{\text{terminal}} = \varepsilon - Ir$ .
The e.m.f. $\varepsilon$ is a property of the cell itself, so it is taken to remain the same.

Key Takeaways

e.m.f. is energy per unit charge supplied by the source; it is not set by the external circuit.
Terminal p.d. changes with current because of internal resistance.

Common Mistakes

Saying e.m.f. increases because current increases (confusing e.m.f. with terminal p.d.).
Saying e.m.f. decreases because more current is drawn (again confusing with terminal p.d. drop).

Things to Be Careful About

In real cells, e.m.f. can vary slightly with temperature/state of charge, but exam questions usually treat it as constant for these time scales and conditions.

Techniques used

interpret emf as a property of the source independent of the external circuitdistinguish between emf and terminal potential difference

(ii)

By considering the effect of the lamp on the total resistance of the circuit, explain the variation of the ammeter reading shown in Fig. 7.2.

3M Similar

DifficultyMedium

Worked solution

Answer

At $t_1$ the switch is closed so the lamp is connected in parallel with the $6.0\ \Omega$ resistor, decreasing the total external resistance, so the current increases suddenly.
Initially the lamp filament is cool so its resistance is small, making the total resistance very small, giving the large peak current.
As current flows the filament heats up and its resistance increases, so the total resistance increases and the current falls to a new steady value (still higher than before $t_1$ because there are now two parallel branches).

Final answer

Current jumps up when the lamp branch is added (lower total resistance); initially lamp has low resistance (cold) giving a large peak; filament heats so its resistance rises, increasing total resistance so current falls to a new steady value still above the original.

Detailed explanation

Background Concept

When components are connected in parallel, the equivalent resistance decreases because the total conductance adds:

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

So adding an extra parallel branch reduces $R_{\text{eq}}$ and tends to increase the current supplied by the source.

A filament lamp is non-ohmic: as current increases, the filament temperature increases, which increases its resistivity. Therefore its resistance increases as it heats up. So the lamp’s resistance is relatively small when it is cold, and larger when it is hot.

With internal resistance $r$ , a larger current also increases the lost volts $Ir$ , reducing the terminal p.d.:

V_{\text{terminal}} = \varepsilon - Ir

Understanding the Question

At time $t_1$ , switch $S$ is closed and the filament lamp becomes connected in parallel with the $6.0\ \Omega$ resistor. The graph shows the total current (ammeter reading) jumping up immediately, then decreasing gradually to a new steady value that is higher than before $t_1$ .

You must explain that shape by discussing how the lamp affects the total resistance of the circuit over time.

Approach

Consider what happens to the circuit immediately when the switch is closed: you add a new parallel path.
Use the fact that the lamp is cold initially, so it has low resistance, giving a large initial current.
As time passes, the filament heats, the lamp resistance rises, so the equivalent resistance increases and current decreases toward a steady value.
Note that final current is still larger than the original (because the lamp branch remains in parallel, so total external resistance is still less than $6.0\ \Omega$ alone).

Step-by-Step Reasoning

diagram-7c2-1

Before $t_1$ : switch open, only $6.0\ \Omega$ is connected externally, so current is constant.
At the instant just after $t_1$ :
- Closing the switch connects the lamp in parallel with the $6.0\ \Omega$ resistor.
- Adding a parallel branch reduces the external equivalent resistance $R_{\text{ext}}$ .
- At that instant the filament is still cool, so the lamp’s resistance is small. This can make $R_{\text{ext}}$ much smaller than $6.0\ \Omega$ .
- Hence the current drawn from the cell increases sharply, producing the sudden jump/peak.
After $t_1$ as time increases:
- The large current heats the filament.
- Heating increases the filament’s resistance.
- As the lamp resistance increases, the parallel combination’s equivalent resistance increases.
- Increased total resistance reduces the circuit current, so the ammeter reading falls.
Long time after $t_1$ (new steady state):
- The filament reaches a steady operating temperature, so its resistance becomes approximately constant.
- The current becomes constant at a new value.
- This final current is still greater than the original current because there are still two parallel paths, so the external equivalent resistance remains lower than $6.0\ \Omega$ alone.

(An additional contributing idea is that when the current is large, the lost volts $Ir$ is larger, which reduces the terminal p.d. and also helps limit the current; as the lamp resistance rises, current falls and the terminal p.d. recovers to a new steady value.)

Key Takeaways

Closing the switch adds a parallel branch, so $R_{\text{eq}}$ decreases and total current increases.
Filament lamp resistance increases with temperature; this causes the current to decrease from its initial surge to a steady value.
Internal resistance means terminal p.d. changes with current, shaping how large the current can become.

Common Mistakes

Treating the filament lamp as having constant resistance (then you cannot explain the gradual decrease).
Claiming the resistance decreases as it heats (true for thermistors, not for metal filaments).
Saying the current falls below the original steady current (it should remain above, because the parallel branch still reduces the external resistance).

Things to Be Careful About

Mention the key time-dependent physics: the filament is cold at first (low $R$ ) and heats up (higher $R$ ).
Make sure your explanation links resistance change to current change using $I = V/R$ (with the idea that adding a parallel branch reduces total resistance).

Techniques used

reason about changes in equivalent resistance for parallel brancheslink filament temperature rise to increase in resistanceconnect resistance changes to current changes using I = V/R and terminal p.d. with internal resistance

Q8Medium-EasyParticle Physics

(a)

State the name of the class (group) of fundamental particles that contains a neutrino.

1M Similar

(b)

A hadron P has a charge of $+1e$ , where $e$ is the elementary charge. The hadron P is composed of a down antiquark and only one other quark.

2M

(i)

Identify a possible flavour for this other quark.

1M Similar

DifficultyMedium-Easy

Worked solution

Working

Down quark has charge $-\frac{1}{3}e$ , so a down antiquark has charge $+\frac{1}{3}e$ .

For total charge $+1e$ , the other quark must have charge

+1e - \left(+\frac{1}{3}e\right) = +\frac{2}{3}e.

So the other quark can be an up quark.

Answer

Up quark ( $u$ ).

Final answer

Up quark (u)

Detailed explanation

Background Concept

Quarks have fractional charges:

$u, c, t$ have charge $+\frac{2}{3}e$
$d, s, b$ have charge $-\frac{1}{3}e$

An antiquark has the opposite charge to the corresponding quark.

The total charge of a hadron is the sum of the charges of its constituent quarks/antiquarks.

Understanding the Question

You are told hadron $P$ has total charge $+1e$ and is made of:

one down antiquark ( $\bar{d}$ ), and
one other quark (only one).

You must name a possible flavour of that quark.

Approach

Write the charge of $\bar{d}$ .
Subtract it from the total charge to find the required charge of the other quark.
Identify a quark flavour that has that charge.

Step-by-Step Reasoning

A down quark $d$ has charge $-\frac{1}{3}e$ .
Therefore the down antiquark $\bar{d}$ has charge $+\frac{1}{3}e$ .
Let the other quark have charge $q$ .
Then

q + \frac{1}{3}e = +1e.

Solve for $q$ :

q = 1e - \frac{1}{3}e = \frac{2}{3}e.

A quark with charge $+\frac{2}{3}e$ could be $u$ (also $c$ or $t$ , but $u$ is the standard AS-level answer).

Key Takeaways

Antiquarks have the opposite charge to the corresponding quark.
Total hadron charge is the sum of constituent charges.
To make $+1e$ with $\bar{d}$ , you need a $+\frac{2}{3}e$ quark (e.g. $u$ ).

Common Mistakes

Using $-\frac{1}{3}e$ for $\bar{d}$ (forgetting the sign change for antiquarks).
Choosing a $d$ or $s$ quark (charge $-\frac{1}{3}e$ ), which cannot produce total $+1e$ with $\bar{d}$ .

Things to Be Careful About

The question says “only one other quark”: do not try to build a baryon with three quarks.
Ensure you express the charge consistently in units of $e$ when adding/subtracting.

Techniques used

use quark charge values to form a required total chargeadd constituent charges to match the particle chargeselect a quark flavour consistent with the required charge

(ii)

State what type of hadron is P.

1M Similar

(c)

Nucleus Q undergoes radioactive decay to form nucleus R, emitting an antineutrino and another particle X, as shown in the decay equation.

\text{Q} \rightarrow \text{R} + \text{X} + \bar{\nu}

3M

(i)

State what particle is represented by X.

1M Similar

(ii)

Compare the nucleon numbers of Q and R.

1M Similar

(iii)

Compare the charges of Q and R.

1M Similar

Physics 9702/22 — February/March 2024

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Working

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Working

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Working

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Working

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Working

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Working

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Working