(a) State Hooke’s law. (b) The variation of the applied force with the extension for a sample of a material is shown in Fig. 3.1. The sample behaves elastically up to an extension of $80\ \text{mm}$ and breaks at point X. (i) On the line in Fig. 3.1, draw a cross ($\times$) to show the limit of proportionality. Label this cross with the letter P. (ii) On the line in Fig. 3.1, draw a cross ($\times$) to show the elastic limit. Label this cross with the letter E. (c) The…

(a) Extension is proportional to force, provided the limit of proportionality is not exceeded. · (b)(i) At 80 mm extension (≈ 6 N), i.e. the end of the straight-line region. · (b)(ii) At 80 mm extension (≈ 6 N), since the sample is elastic up to 80 mm. · (c)(i) 75 N m⁻¹ · (c)(ii) 6.0 × 10^8 Pa · (d) ≈ 1.0 J · (e) Larger; because for same material and length, k ∝ A (k = YA/L).

A progressive transverse wave travelling from left to right is shown at an instant in time in Fig. 4.1. R and T are points on the wave. (a) State the phase difference between the points R and T. phase difference = \_\_\_\_\_\_ $^{\circ}$ (b) On Fig. 4.1, draw an arrow at point T to show the direction of movement of point T at the instant shown. (c) The horizontal distance between R and T is $0.62\ \text{cm}$, as shown in Fig. 4.2. The speed of the wave is $0.27\ \text{m…

(a) 90° · (b) Downwards · (c) 11 Hz · (d) Resultant amplitude at P is 0 (destructive interference).

(i) State Kirchhoff’s second law. (ii) State the conservation law that gives rise to Kirchhoff’s second law. (b) A circuit contains a cell of internal resistance $r$ and two resistors of resistances $R_1$ and $R_2$, as shown in Fig. 5.1. The potential difference (p.d.) across the two resistors is $V$. The current in the cell is $I$. (i) Use Kirchhoff’s laws to show that the total resistance $R_T$ of the external circuit is given by $$ \frac{1}{R_T} = \frac{1}{R_1} +…

(a)(i) In any closed loop, the algebraic sum of the e.m.f.s equals the algebraic sum of the potential drops (sum of p.d.s around a loop is zero). · (a)(ii) Conservation of energy. · (b)(i) 1/RT = 1/R1 + 1/R2 · (b)(ii) 0.52 Ω · (c)(i) The current in the cell increases. · (c)(ii) The voltmeter reading (terminal p.d.) decreases.

9702/23 May/June 2024 — Physics Solved Past Paper

Q1Medium-EasyPhysical Quantities and UnitsDynamicsForces, Density and Pressure

The drag force $F_D$ acting on a sphere falling through a liquid is given by

F_D = 6\pi\eta rv

where $r$ is the radius of the sphere,
$v$ is the speed of the sphere in the liquid and
$\eta$ is a property of the liquid called the viscosity.

(a)

Show that the SI base units of viscosity are $\text{kg m}^{-1}\ \text{s}^{-1}$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Working

From

F_D = 6\pi\eta rv

the constant $6\pi$ is dimensionless, so

\eta = \frac{F_D}{rv}

Units:

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

[r] = \text{m}, \qquad [v] = \text{m s}^{-1}

So

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

Answer

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

Final answer

kg m^-1 s^-1

Detailed explanation

Background Concept

Dimensional analysis uses the fact that equations must be homogeneous: both sides must have the same units. If a formula is correct, you can deduce the SI base units of an unknown quantity by rearranging the formula and substituting base units for the other quantities.

In Stokes' law,

F_D = 6\pi\eta rv

$6\pi$ is a pure number (dimensionless), so it does not affect the units.

Understanding the Question

You are given an expression for the drag force $F_D$ on a sphere moving through a liquid. The question asks you to show, using SI base units, that viscosity $\eta$ has units $\text{kg m}^{-1}\ \text{s}^{-1}$ .

Approach

Rearrange the equation to make $\eta$ the subject.
Replace each quantity with its SI units written in base units.
Simplify carefully, cancelling powers of metres and seconds.

Step-by-Step Reasoning

Start with

F_D = 6\pi\eta rv.

Rearrange for viscosity:

\eta = \frac{F_D}{6\pi rv}.

Since $6\pi$ is dimensionless,

[\eta] = \frac{[F_D]}{[r][v]}.

Now write each in SI base units:

Force: $\text{N} = \text{kg m s}^{-2}$ .
Radius: $\text{m}$ .
Speed: $\text{m s}^{-1}$ .

So

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

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

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

Key Takeaways

Physical equations must be dimensionally consistent.
Dimensionless constants (like $6\pi$ ) do not affect units.
Converting to SI base units ( $\text{kg}$ , $\text{m}$ , $\text{s}$ ) makes cancellations clear.

Common Mistakes

Treating $6\pi$ as if it has units.
Using $\text{N}$ without converting to base units.
Incorrectly combining indices when dividing (especially with seconds).

Things to Be Careful About

Write $\text{N}$ as $\text{kg m s}^{-2}$ before simplifying.
When dividing by $\text{s}^{-1}$ , remember it is equivalent to multiplying by $\text{s}$ .

Techniques used

rearrange an equation to make the required quantity the subjectsubstitute SI base units for each physical quantityuse dimensional analysis to simplify units

(b)

The sphere has a radius of $3.0\ \text{cm}$ and is falling vertically downwards at a terminal velocity of $2.0\ \text{m s}^{-1}$ through the liquid. The drag force acting on the sphere is $0.096\ \text{N}$ .

Calculate the viscosity of the liquid.

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

2M Similar

DifficultyMedium-Easy

Worked solution

Working

F_D = 6\pi\eta rv \;\Rightarrow\; \eta = \frac{F_D}{6\pi rv}

$r = 3.0\ \text{cm} = 3.0 \times 10^{-2}\ \text{m}$ .

\eta = \frac{0.096}{6\pi (3.0 \times 10^{-2})(2.0)}

\eta = 8.5 \times 10^{-2}\ \text{kg m}^{-1}\ \text{s}^{-1}

Answer

$8.5 \times 10^{-2}\ \text{kg m}^{-1}\ \text{s}^{-1}$

Final answer

8.5 × 10^-2 kg m^-1 s^-1

Detailed explanation

Background Concept

Stokes' law for the viscous drag force on a sphere moving through a liquid at low Reynolds number is

F_D = 6\pi\eta rv,

where $F_D$ is the drag force, $\eta$ is the viscosity, $r$ is the sphere radius, and $v$ is the speed. If three of these are known, the fourth can be found by rearranging the equation.

Understanding the Question

You are told the sphere has radius $3.0\ \text{cm}$ and terminal speed $2.0\ \text{m s}^{-1}$ , and the drag force is $0.096\ \text{N}$ . You must calculate $\eta$ and give it in $\text{kg m}^{-1}\ \text{s}^{-1}$ .

A key detail is that the radius is in centimetres, so it must be converted to metres for SI.

Approach

Convert $r$ from cm to m.
Rearrange $F_D = 6\pi\eta rv$ to $\eta = F_D/(6\pi rv)$ .
Substitute the numerical values and calculate.
Round to a sensible number of significant figures (typically matching the data, here 2 s.f.).

Step-by-Step Reasoning

Rearrange:

\eta = \frac{F_D}{6\pi rv}.

Convert radius:

3.0\ \text{cm} = 3.0 \times 10^{-2}\ \text{m}.

Substitute:

\eta = \frac{0.096}{6\pi (3.0 \times 10^{-2})(2.0)}.

Compute the denominator first:

6\pi (3.0 \times 10^{-2})(2.0) = 6\pi (6.0 \times 10^{-2}) = 0.36\pi \approx 1.13.

So

\eta \approx \frac{0.096}{1.13} \approx 0.085\ \text{kg m}^{-1}\ \text{s}^{-1} = 8.5 \times 10^{-2}\ \text{kg m}^{-1}\ \text{s}^{-1}.

Key Takeaways

Always convert to SI units before substitution.
Rearranging a formula cleanly reduces algebra errors.
Significant figures should reflect the least precise given data.

Common Mistakes

Forgetting to convert $3.0\ \text{cm}$ to $0.030\ \text{m}$ (this causes a factor of $100$ error).
Using diameter instead of radius.
Dropping the factor $6\pi$ .

Things to Be Careful About

Keep track of powers of ten when converting cm to m.
Include the unit $\text{kg m}^{-1}\ \text{s}^{-1}$ in the final answer.

Techniques used

convert given values to SI unitsrearrange a formula to make the required variable the subjectsubstitute values into an equation and evaluatequote an answer to an appropriate number of significant figures

(c)

The sphere is shown in Fig. 1.1.

Fig. 1.1

On Fig. 1.1, draw and label arrows to represent the directions of the three forces acting on the sphere as it falls at terminal velocity through the liquid.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

diagram-c-1

Weight $mg$ acts vertically downwards.

Upthrust acts vertically upwards.

Drag force $F_D$ acts vertically upwards (opposes the downward motion).

Final answer

Arrows: weight down; upthrust up; drag up.

Detailed explanation

Background Concept

When an object moves through a fluid, several forces can act:

Weight $mg$ acts downward due to gravity.
Upthrust (buoyancy) acts upward due to pressure being greater at lower depth; by Archimedes' principle it equals the weight of fluid displaced.
Drag force $F_D$ acts opposite to the direction of motion through the fluid.

At terminal velocity, the acceleration is zero, so the resultant force is zero (Newton’s first/second law): upward forces balance downward forces.

Understanding the Question

The sphere is falling vertically downward at terminal velocity in a liquid. The question asks you to draw and label arrows showing the directions of the three forces acting on the sphere.

So you must identify the three forces and draw them on the provided sphere diagram.

Approach

Recognise “terminal velocity” means constant velocity, so forces balance.
List the forces: weight, upthrust, drag.
Decide directions: weight down; the other two oppose the motion or act upward.
Draw three arrows on the sphere itself and label each clearly.

Step-by-Step Reasoning

The sphere is moving downward. Drag always opposes motion, so drag must be upward.
Upthrust is the buoyant force from the liquid; it always acts upward.
Weight acts toward the centre of the Earth; in this vertical set-up that is downward.

At terminal velocity:

mg = U + F_D,

where $U$ is upthrust.

diagram-c-1

Key Takeaways

Terminal velocity implies zero acceleration and zero resultant force.
Drag acts opposite the motion; upthrust acts upward in a fluid.
A correct force diagram needs correct directions and clear labels.

Common Mistakes

Drawing drag downward (it must oppose the downward motion).
Omitting upthrust and only drawing weight and drag.
Drawing arrows not starting on the sphere or failing to label them.

Things to Be Careful About

The question asks for directions, not magnitudes; however, arrows must be clearly up/down.
Do not confuse upthrust with “reaction force”; there is no contact surface here.

Techniques used

identify all forces acting on an object moving through a fluiduse the fact that terminal velocity implies zero resultant forcedraw a labelled free-body force diagram with correct directions

(d)

4M

(i)

The density of the liquid is $920\ \text{kg m}^{-3}$ .

Show that the upthrust acting on the sphere is $1.0\ \text{N}$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Upthrust $U$ equals weight of liquid displaced:

U = \rho gV

For a sphere,

V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (3.0 \times 10^{-2})^3 = 1.13 \times 10^{-4}\ \text{m}^3

U = (920)(9.81)(1.13 \times 10^{-4}) = 1.02\ \text{N} \approx 1.0\ \text{N}

Answer

$U \approx 1.0\ \text{N}$

Final answer

1.0 N

Detailed explanation

Background Concept

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

U = \rho g V,

where $\rho$ is the fluid density, $g$ is gravitational field strength, and $V$ is the volume of fluid displaced. For a fully submerged object, the displaced volume equals the object’s volume.

Understanding the Question

The sphere (radius $3.0\ \text{cm}$ from earlier) is submerged in a liquid of density $920\ \text{kg m}^{-3}$ . You must show the upthrust is $1.0\ \text{N}$ . So you calculate the sphere’s volume (in $\text{m}^3$ ) and then use $U = \rho gV$ .

Approach

Convert radius to metres.
Find volume of the sphere: $V = \frac{4}{3}\pi r^3$ .
Substitute into $U = \rho gV$ .
Round to $1.0\ \text{N}$ as requested.

Step-by-Step Reasoning

Convert radius:

r = 3.0\ \text{cm} = 3.0 \times 10^{-2}\ \text{m}.

Sphere volume:

V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (3.0 \times 10^{-2})^3.

Calculate $r^3$ :

(3.0 \times 10^{-2})^3 = 27 \times 10^{-6} = 2.7 \times 10^{-5}.

So

V = \frac{4}{3}\pi (2.7 \times 10^{-5}) \approx 1.13 \times 10^{-4}\ \text{m}^3.

Now apply Archimedes:

U = \rho gV = (920)(9.81)(1.13 \times 10^{-4}) \approx 1.02\ \text{N}.

To 2 significant figures this is $1.0\ \text{N}$ , matching the value you were asked to show.

Key Takeaways

Upthrust depends on fluid density and displaced volume, not the object’s mass.
For a fully submerged object, displaced volume = object volume.
Always use SI units: $\rho$ in $\text{kg m}^{-3}$ and $V$ in $\text{m}^3$ .

Common Mistakes

Using diameter instead of radius in the volume formula.
Forgetting to convert $\text{cm}$ to $\text{m}$ , leading to a $10^6$ error in volume.
Using $U = mg$ of the sphere (wrong: it is the weight of displaced liquid).

Things to Be Careful About

The cube on the radius makes unit conversion critical.
Rounding: you can keep extra digits (like $1.02\ \text{N}$ ) but then state it is $\approx 1.0\ \text{N}$ as required.

Techniques used

apply Archimedes' principle for upthrustcalculate the volume of a spheresubstitute values into U = rho g V and evaluate

(ii)

Calculate the mass of the sphere.

mass = ______ $\text{kg}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

At terminal velocity, resultant force is zero:

mg = U + F_D

mg = 1.0 + 0.096 = 1.096\ \text{N}

m = \frac{1.096}{9.81} = 0.112\ \text{kg}

Answer

$0.11\ \text{kg}$

Final answer

0.11 kg

Detailed explanation

Background Concept

Terminal velocity means constant velocity, so acceleration $a = 0$ . By Newton’s second law,

\sum F = ma = 0,

so the forces balance.

For a falling sphere in a liquid:

Weight $mg$ acts downward.
Upthrust $U$ acts upward.
Drag force $F_D$ acts upward (opposes downward motion).

Therefore at terminal velocity:

mg = U + F_D.

Understanding the Question

You have already found the upthrust in (i) as $1.0\ \text{N}$ (to 2 s.f.). You are also told earlier that the drag force at terminal velocity is $0.096\ \text{N}$ . The question asks for the mass of the sphere.

Approach

Write the equilibrium equation at terminal velocity: $mg = U + F_D$ .
Substitute $U$ and $F_D$ to find $mg$ .
Divide by $g$ to get $m$ .
Round appropriately.

Step-by-Step Reasoning

At terminal velocity,

mg = U + F_D.

Substitute the known upward forces:

mg = 1.0 + 0.096 = 1.096\ \text{N}.

Now divide by $g = 9.81\ \text{m s}^{-2}$ :

m = \frac{1.096}{9.81} = 0.112\ \text{kg}.

Given that $U$ was shown as $1.0\ \text{N}$ (2 s.f.), quoting $m$ as $0.11\ \text{kg}$ is appropriate.

Key Takeaways

Terminal velocity implies forces balance (resultant force zero).
For a falling object in a fluid: $mg$ is balanced by upthrust plus drag.
Mass comes from $m = (U + F_D)/g$ .

Common Mistakes

Using $mg = F_D - U$ (wrong sign; both $U$ and $F_D$ act upward).
Forgetting that drag is upward when the motion is downward.
Dividing by $9.8$ or $9.81$ is fine, but keep consistent rounding with earlier parts.

Things to Be Careful About

Use the same rounded value of upthrust the question has guided you to ( $1.0\ \text{N}$ ) unless instructed otherwise.
Ensure the final unit is $\text{kg}$ , not newtons.

Techniques used

apply force balance at terminal velocityrearrange mg = U + F_D to find masssubstitute numerical values and compute

Q2MediumPhysical Quantities and UnitsKinematicsWork, Energy and Power

(a)

Define displacement from a point.

1M Similar

(b)

An object is projected horizontally at a speed of $6.0\ \text{m s}^{-1}$ from a slope, as shown in Fig. 2.1.

Fig. 2.1

The slope is at an angle $\theta$ to the horizontal. Air resistance is negligible.

The object lands on the slope a time of $0.71\ \text{s}$ later and stops without rolling or bouncing.

10M

(i)

Determine the horizontal distance travelled by the object.

distance = ______ $\text{m}$

1M Similar

(ii)

Determine the vertical distance travelled by the object.

distance = ______ $\text{m}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Vertical motion: $u_y = 0$ , $a_y = g = 9.81\ \text{m s}^{-2}$ , $t = 0.71\ \text{s}$ .

y = u_y t + \frac{1}{2}gt^2 = 0 + \frac{1}{2}(9.81)(0.71)^2 = 2.47\ \text{m}

Answer

$2.5\ \text{m}$

Final answer

2.5 m

Detailed explanation

Background Concept

For vertical motion in a projectile (no air resistance), acceleration is constant and equal to the gravitational field strength $g$ downward.

Using upward as positive, $a_y = -g$ . Many candidates instead work with magnitudes and state “vertical distance fallen”. Either is fine if consistent.

A key kinematics equation is:

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

Understanding the Question

The object is projected horizontally, so its initial vertical velocity is zero ( $u_y = 0$ ). After $0.71\ \text{s}$ it lands on the slope. Part (ii) asks for the vertical distance travelled in that time (i.e. how far it falls).

Approach

Treat the vertical motion as free fall:

Set $u_y = 0$ .
Use $y = \frac{1}{2}gt^2$ for the distance fallen.

Step-by-Step Reasoning

Initial vertical component is zero because the launch is horizontal:

u_y = 0

Distance fallen in time $t$ under constant acceleration $g$ is:

y = \frac{1}{2}gt^2

Substitute $g = 9.81\ \text{m s}^{-2}$ and $t = 0.71\ \text{s}$ :

y = \frac{1}{2}(9.81)(0.71)^2 = 2.47\ \text{m}

So the vertical distance is about $2.5\ \text{m}$ (2 s.f.).

Key Takeaways

Horizontal launch implies $u_y = 0$ .
Vertical displacement depends only on time: $y = \tfrac{1}{2}gt^2$ .

Common Mistakes

Using $u = 6.0\ \text{m s}^{-1}$ as the vertical initial velocity.
Forgetting the factor $\tfrac{1}{2}$ .
Mixing sign conventions (getting a negative “distance”).

Things to Be Careful About

Decide whether you are finding a signed displacement (could be negative) or the magnitude “distance fallen”. The question asks for distance travelled, so a positive magnitude is expected.
Use $g$ to 2–3 s.f. consistently; rounding only at the end avoids errors.

Techniques used

treat vertical motion as uniformly accelerated with $a = g$apply $s = ut + \frac{1}{2}at^2$ in the vertical direction

(iii)

Use your answers in (b)(i) and (b)(ii) to calculate $\theta$ .

$\theta$ = ______ $^{\circ}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Using (i) $x = 4.26\ \text{m}$ and (ii) $y = 2.47\ \text{m}$ .

\tan\theta = \frac{y}{x} = \frac{2.47}{4.26} = 0.580

\theta = \tan^{-1}(0.580) = 30.1^{\circ}

Answer

$30^{\circ}$

Final answer

30°

Detailed explanation

Background Concept

If a line makes an angle $\theta$ to the horizontal, then for a right triangle formed by a horizontal run $x$ and vertical drop $y$ :

\tan\theta = \frac{y}{x}

This is purely geometry; it does not depend on the object’s speed, only on the relative vertical and horizontal displacements between the two points.

Understanding the Question

The object starts at the top of the slope and lands somewhere down the slope. The straight line between start and landing lies along the slope, which is at angle $\theta$ to the horizontal. You already found:

horizontal distance $x$ in (b)(i)
vertical distance $y$ in (b)(ii)

You must use these to calculate $\theta$ .

Approach

Treat $x$ and $y$ as the adjacent and opposite sides of a right triangle.
Use $\tan\theta = y/x$ .
Take inverse tangent to get $\theta$ .

Step-by-Step Reasoning

From earlier parts:

x = 4.26\ \text{m}, \qquad y = 2.47\ \text{m}

Form the tangent ratio:

\tan\theta = \frac{y}{x} = \frac{2.47}{4.26} = 0.580

Now find the angle:

\theta = \tan^{-1}(0.580) = 30.1^{\circ}

Rounded appropriately, $\theta \approx 30^{\circ}$ .

Key Takeaways

The slope angle comes from the ratio of vertical to horizontal displacement.
Use $\tan\theta$ when you know opposite and adjacent sides.

Common Mistakes

Using $\sin\theta$ or $\cos\theta$ with the wrong sides.
Swapping the ratio (doing $x/y$ gives the complement angle).
Rounding $x$ and $y$ too early, leading to a noticeably different angle.

Things to Be Careful About

Ensure your calculator is in degree mode, not radians.
Use consistent values from parts (i) and (ii) (ideally unrounded) before taking the inverse tangent.

Techniques used

relate components of displacement using trigonometryuse $\tan\theta = \frac{\text{vertical}}{\text{horizontal}}$ from a right triangle

(iv)

Determine the magnitude of the displacement of the object from its original position.

displacement = ______ $\text{m}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Using $x = 4.26\ \text{m}$ and $y = 2.47\ \text{m}$ .

|\vec{s}| = \sqrt{x^2 + y^2} = \sqrt{(4.26)^2 + (2.47)^2} = 4.93\ \text{m}

Answer

$4.9\ \text{m}$

Final answer

4.9 m

Detailed explanation

Background Concept

Displacement is a vector from the starting point to the ending point. If its horizontal and vertical components are perpendicular (at right angles), the magnitude of the displacement is the resultant:

|\vec{s}| = \sqrt{x^2 + y^2}

This is Pythagoras’ theorem applied to vector components.

Understanding the Question

You have already found the components of the object’s change in position after $0.71\ \text{s}$ :

horizontal distance $x$
vertical distance $y$

Part (iv) asks for the magnitude of the displacement from its original position (the straight-line distance from launch point to landing point).

Approach

Treat $x$ and $y$ as perpendicular components of the displacement vector.
Use Pythagoras to find the resultant magnitude.

Step-by-Step Reasoning

From earlier parts:

x = 4.26\ \text{m}, \qquad y = 2.47\ \text{m}

Magnitude of displacement:

|\vec{s}| = \sqrt{x^2 + y^2}

Substitute:

|\vec{s}| = \sqrt{(4.26)^2 + (2.47)^2} = \sqrt{18.1 + 6.10} = \sqrt{24.2} = 4.93\ \text{m}

So $|\vec{s}| \approx 4.9\ \text{m}$ .

Key Takeaways

Displacement magnitude is the straight-line separation between initial and final positions.
Perpendicular components combine via Pythagoras.

Common Mistakes

Adding $x + y$ instead of using Pythagoras.
Confusing displacement magnitude with distance travelled along the curved projectile path.

Things to Be Careful About

Use the components from (i) and (ii) consistently.
Report the answer with appropriate significant figures and unit (metres).

Techniques used

treat displacement as the resultant of perpendicular componentsuse Pythagoras’ theorem to combine horizontal and vertical components

(v)

By considering energy, calculate the speed of the object just before it lands.

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

3M Similar

DifficultyMedium

Worked solution

Working

Vertical drop $h = 2.47\ \text{m}$ .

Conservation of mechanical energy:

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

v^2 = u^2 + 2gh = (6.0)^2 + 2(9.81)(2.47) = 84.5

v = \sqrt{84.5} = 9.19\ \text{m s}^{-1}

Answer

$9.2\ \text{m s}^{-1}$

Final answer

9.2 m s⁻1

Detailed explanation

Background Concept

When air resistance is negligible, the only significant force doing work is gravity, which is conservative. Therefore mechanical energy is conserved:

E_k + E_p = \text{constant}

Kinetic energy:

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

Gravitational potential energy change for a vertical drop of height $h$ :

\Delta E_p = -mgh

So the loss of GPE becomes a gain in KE.

Understanding the Question

The object is projected horizontally at $u = 6.0\ \text{m s}^{-1}$ . After $0.71\ \text{s}$ it lands on the slope. You found in (ii) that it drops vertically by $h = 2.47\ \text{m}$ . The question asks for the speed just before it lands, using energy.

Approach

Take initial kinetic energy from the launch speed $u$ .
Add the gain in kinetic energy due to falling height $h$ (equal to $mgh$ ).
Solve the energy equation for the final speed $v$ .

Step-by-Step Reasoning

Initial energy at launch:

E_{k,i} = \frac{1}{2}mu^2

During flight it falls a vertical height $h$ , so GPE decreases by $mgh$ and KE increases by the same amount:

E_{k,f} = E_{k,i} + mgh

So:

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

Cancel $m$ and multiply by 2:

v^2 = u^2 + 2gh

Substitute $u = 6.0\ \text{m s}^{-1}$ , $g = 9.81\ \text{m s}^{-2}$ , $h = 2.47\ \text{m}$ :

v^2 = 36 + 2(9.81)(2.47) = 36 + 48.5 = 84.5

v = \sqrt{84.5} = 9.19\ \text{m s}^{-1} \approx 9.2\ \text{m s}^{-1}

This is larger than $6.0\ \text{m s}^{-1}$ because gravity has increased the vertical component of velocity while the horizontal component remains $6.0\ \text{m s}^{-1}$ .

Key Takeaways

With negligible air resistance, use conservation of mechanical energy.
Falling through height $h$ increases speed according to $v^2 = u^2 + 2gh$ .

Common Mistakes

Using displacement along the slope instead of the vertical height in $mgh$ .
Forgetting to square-root at the end (leaving $v^2$ ).
Adding $mgh$ to speed directly instead of to energy.

Things to Be Careful About

In $mgh$ , $h$ must be the vertical drop (here from part (ii)).
Keep units consistent so $2gh$ has units of $\text{m}^2\text{s}^{-2}$ .
Quote the final speed to an appropriate number of significant figures (typically matching the given $6.0$ and $0.71$ ).

Techniques used

apply conservation of mechanical energy (no air resistance)equate loss of gravitational potential energy to gain in kinetic energyrearrange to solve for speed

Q3MediumDeformation of Solids

(a)

State Hooke’s law.

1M Similar

(b)

The variation of the applied force with the extension for a sample of a material is shown in Fig. 3.1.

Fig. 3.1

The sample behaves elastically up to an extension of $80\ \text{mm}$ and breaks at point X.

2M

(i)

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

1M Similar

(ii)

On the line in Fig. 3.1, draw a cross ( $\times$ ) to show the elastic limit. Label this cross with the letter E.

1M Similar

(c)

The sample in (b) has a cross-sectional area of $0.40\ \text{mm}^2$ and an initial length of $3.2\ \text{m}$ .

For deformations within the limit of proportionality of the sample, determine:

5M

(i)

the spring constant of the sample

spring constant = ______ $\text{N m}^{-1}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Within the proportional region,

F = kx

Using $F = 6.0\ \text{N}$ at $x = 80\ \text{mm} = 0.080\ \text{m}$ :

k = \frac{F}{x} = \frac{6.0}{0.080} = 75\ \text{N m}^{-1}

Answer

$k = 75\ \text{N m}^{-1}$

Final answer

75 N m⁻¹

Detailed explanation

Background Concept

In the Hooke’s law (proportional) region:

F = kx

$F$ is the applied force (N)
$x$ is the extension (m)
$k$ is the spring constant (N m $^{-1}$ )

On a force–extension graph, the gradient of the straight-line section is $k$ .

Understanding the Question

You are told to consider deformations within the limit of proportionality, so you must use the straight-line part of the graph. From the graph, a convenient point on the straight section is $x = 80\ \text{mm}$ and $F = 6\ \text{N}$ .

Approach

Pick a point in the linear region, convert extension to metres, then use $k = F/x$ (or gradient $\Delta F/\Delta x$ ).

Step-by-Step Reasoning

Choose the point at the end of the straight-line region: $F = 6.0\ \text{N}$ when $x = 80\ \text{mm}$ .
Convert $80\ \text{mm}$ to metres:

80\ \text{mm} = 80 \times 10^{-3}\ \text{m} = 0.080\ \text{m}

Rearrange Hooke’s law:

k = \frac{F}{x}

Substitute:

k = \frac{6.0}{0.080} = 75\ \text{N m}^{-1}

Key Takeaways

Spring constant is the gradient of the straight-line section.
Always use $x$ in metres to get $k$ in N m $^{-1}$ .

Common Mistakes

Using $80\ \text{mm}$ as $80\ \text{m}$ or $0.80\ \text{m}$ .
Using a point beyond the proportional region (where the gradient is not constant).

Things to Be Careful About

Units: if $x$ is in mm, your $k$ would come out in N mm $^{-1}$ , not the required N m $^{-1}$ .
Use a point clearly on the straight line to avoid reading errors.

Techniques used

use the gradient of a force–extension graph to find spring constantconvert millimetres to metres before substitution

(ii)

the Young modulus of the material from which the sample is made.

Young modulus = ______ $\text{Pa}$

3M Similar

DifficultyMedium

Worked solution

Working

E = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{x/L} = \frac{FL}{Ax}

$A = 0.40\ \text{mm}^2 = 0.40 \times 10^{-6}\ \text{m}^2 = 4.0 \times 10^{-7}\ \text{m}^2$

Using $F = 6.0\ \text{N}$ at $x = 80\ \text{mm} = 0.080\ \text{m}$ and $L = 3.2\ \text{m}$ :

E = \frac{6.0 \times 3.2}{\left(4.0 \times 10^{-7}\right)\left(0.080\right)} = 6.0 \times 10^{8}\ \text{Pa}

Answer

$E = 6.0 \times 10^{8}\ \text{Pa}$

Final answer

6.0 × 10^8 Pa

Detailed explanation

Background Concept

Young modulus $E$ measures stiffness of a material and is defined (in the linear elastic region) by

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

where

\text{stress} = \frac{F}{A}, \qquad \text{strain} = \frac{x}{L}.

So, combining:

E = \frac{F/A}{x/L} = \frac{FL}{Ax}.

This is valid only when the stress–strain graph is linear, i.e. within the limit of proportionality.

Understanding the Question

You are given:

cross-sectional area $A = 0.40\ \text{mm}^2$
original length $L = 3.2\ \text{m}$
from the proportional region of the force–extension graph: at $x = 80\ \text{mm}$ , $F \approx 6\ \text{N}$

You must calculate $E$ in pascals.

Approach

Use

E = \frac{FL}{Ax}

and be very careful converting $\text{mm}^2$ to $\text{m}^2$ and mm to m.

Step-by-Step Reasoning

Convert the area:

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

So

A = 0.40\ \text{mm}^2 = 0.40 \times 10^{-6}\ \text{m}^2 = 4.0\times 10^{-7}\ \text{m}^2.

Convert the extension:

x = 80\ \text{mm} = 0.080\ \text{m}.

Substitute into

E = \frac{FL}{Ax}

giving

E = \frac{6.0\times 3.2}{(4.0\times 10^{-7})(0.080)}.

Evaluate the denominator:

(4.0\times 10^{-7})(0.080) = 3.2\times 10^{-8}

and numerator:

6.0\times 3.2 = 19.2

So

E = \frac{19.2}{3.2\times 10^{-8}} = 6.0\times 10^{8}\ \text{Pa}.

Key Takeaways

$E = \dfrac{FL}{Ax}$ in the proportional (linear) region.
Converting $\text{mm}^2 \to \text{m}^2$ is essential and commonly tested.

Common Mistakes

Converting $0.40\ \text{mm}^2$ to $0.40\times 10^{-3}\ \text{m}^2$ (wrong: that is for mm, not mm $^2$ ).
Using extension in mm, giving $E$ wrong by a factor of $10^3$ .
Using a point outside the proportional region where $E$ would not be constant.

Things to Be Careful About

Units: Pa is N m $^{-2}$ , so $A$ must be in m $^2$ .
Significant figures: your final value should be consistent with the graph readings (usually 2–3 s.f.).

Techniques used

apply Young modulus definition (stress divided by strain)convert cross-sectional area from mm^2 to m^2substitute data from a force–extension graph

(d)

Determine an estimate of the work done on the sample as it is extended from zero extension to its breaking point. Explain your reasoning.

work done = ______ $\text{J}$

2M Similar

DifficultyMedium

Worked solution

Working

Work done $W$ is the area under the $F$ –extension graph.

From $0$ to $80\ \text{mm}$ : triangle,

W_1 = \frac{1}{2} \times 6.0 \times 0.080 = 0.24\ \text{J}

From $80\ \text{mm}$ to breaking at $x \approx 195\ \text{mm}$ , take trapezium with forces $6.0\ \text{N}$ and $7.2\ \text{N}$ :

W_2 = \frac{(6.0+7.2)}{2} \times (0.195-0.080) = 6.6 \times 0.115 = 0.759\ \text{J}

W \approx W_1+W_2 \approx 1.00\ \text{J}

Answer

$W \approx 1.0\ \text{J}$ (area under the graph).

Final answer

≈ 1.0 J

Detailed explanation

Background Concept

The work done in stretching a material by a variable force is

W = \int_0^x F\,\mathrm{d}x

On a force–extension graph, this integral is the area under the graph between the two extensions.

If $F$ increases linearly from $0$ to $F$ , the area is a triangle: $W = \tfrac{1}{2}Fx$ .
If $F$ changes roughly linearly between two values, the area can be estimated by a trapezium: $W \approx \tfrac{(F_1+F_2)}{2}\,\Delta x$ .

Understanding the Question

You must estimate the work done stretching the sample from zero extension to the breaking point $X$ . The graph is straight up to $80\ \text{mm}$ and then curves before breaking at about $x \approx 195\ \text{mm}$ with $F \approx 7.2\ \text{N}$ .

Approach

Split the area into simple shapes:

A triangle for the straight-line (Hooke’s law) part from $0$ to $80\ \text{mm}$ .
An approximate trapezium for the curved part from $80\ \text{mm}$ to $195\ \text{mm}$ using the average of the end forces.

diagram-d1

Step-by-Step Reasoning

Linear region (0 to 80 mm):
- At $80\ \text{mm}$ , $F \approx 6\ \text{N}$ .
- Convert $80\ \text{mm} = 0.080\ \text{m}$ .
- Triangle area:

W_1 = \frac{1}{2}Fx = \frac{1}{2}(6.0)(0.080) = 0.24\ \text{J}.

Non-linear region (80 mm to 195 mm):
- Read breaking point approximately: $x \approx 195\ \text{mm} = 0.195\ \text{m}$ , $F \approx 7.2\ \text{N}$ .
- Width of this section:

\Delta x = 0.195 - 0.080 = 0.115\ \text{m}.

Approximate area under curve using a trapezium (average force):

W_2 \approx \frac{(6.0 + 7.2)}{2} \times 0.115 = 6.6 \times 0.115 = 0.759\ \text{J}.

Total work:

W \approx W_1 + W_2 \approx 0.24 + 0.759 = 0.999\ \text{J} \approx 1.0\ \text{J}.

This is an estimate because the second part is curved; using a trapezium is a reasonable approximation based on the graph shape.

Key Takeaways

Work done = area under the $F$ –extension curve.
For curved sections, estimate using trapezia/rectangles based on the graph.

Common Mistakes

Using $W = Fx$ with $F$ taken as the final force (overestimates when force starts from zero).
Forgetting to convert mm to m, giving an answer too large by a factor of $1000$ .
Calculating only the triangle up to $80\ \text{mm}$ and ignoring the extra area to breaking.

Things to Be Careful About

Always use SI units for the area: $\text{N} \times \text{m} = \text{J}$ .
Because the curve is not straight after $80\ \text{mm}$ , any single numerical answer must be presented as an estimate and justified by the “area under graph” reasoning.

Techniques used

estimate work done as area under a force–extension graphapproximate a curved region using a trapezium (average force method)convert millimetres to metres before calculating area

(e)

A second sample of the same material has a larger cross-sectional area than the original sample but the same initial length. The two samples are each deformed with the limit of proportionality.

State and explain qualitatively how the spring constant of the second sample compares with that of the original sample.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

The second sample has a larger spring constant (is stiffer).

For the same material and length,

k = \frac{YA}{L}

so increasing cross-sectional area $A$ increases $k$ (and gives a smaller extension for a given force).

Final answer

Larger; because for same material and length, k ∝ A (k = YA/L).

Detailed explanation

Background Concept

For a uniform wire/rod in the linear elastic region:

E = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{x/L} \Rightarrow x = \frac{FL}{EA}

Comparing with Hooke’s law $F = kx$ , we can write:

F = kx = k\left(\frac{FL}{EA}\right) \Rightarrow k = \frac{EA}{L}.

So for the same material (same $E$ ) and same length $L$ , the spring constant is proportional to cross-sectional area $A$ .

Understanding the Question

Two samples are made of the same material (so same Young modulus $E$ ) and have the same initial length $L$ , but the second has a larger cross-sectional area. Both are stretched only within the limit of proportionality, so the linear relationships apply.

You must state how the spring constant changes and explain qualitatively why.

Approach

Use the relationship $k = EA/L$ (or argue from $x = FL/(EA)$ ) to deduce how changing $A$ affects stiffness.

Step-by-Step Reasoning

Since $k = EA/L$ , keeping $E$ and $L$ constant means:

k \propto A.

A larger area gives a larger $k$ .
Interpreting physically: for a given force, extension

x = \frac{F}{k}

so a larger $k$ means a smaller extension; the thicker sample is harder to stretch.

Key Takeaways

For a given material and length, increasing cross-sectional area makes the sample stiffer.
Useful link: $k = EA/L$ .

Common Mistakes

Claiming $k$ is unchanged because it is “the same material” (geometry also matters).
Saying extension increases with area (it decreases).

Things to Be Careful About

The argument only applies within the limit of proportionality (linear elastic behaviour).
State both the comparison (larger/smaller) and the reason (dependence on $A$ ).

Techniques used

relate spring constant to Young modulus and sample dimensionspredict qualitative effect of changing cross-sectional area

Q4MediumWavesSuperposition

A progressive transverse wave travelling from left to right is shown at an instant in time in Fig. 4.1.

Fig. 4.1

R and T are points on the wave.

(a)

State the phase difference between the points R and T.

phase difference = ______ $^{\circ}$

1M Similar

DifficultyEasy

Worked solution

Answer

Peak to equilibrium position corresponds to a quarter cycle.

Phase difference $= 90^{\circ}$ .

Final answer

90°

Detailed explanation

Background Concept

For a progressive sinusoidal wave, points along the wave can be described by a phase angle. One complete cycle corresponds to $360^{\circ}$ (or $2\pi$ rad).

Key phase positions on a sinusoid:

crest (maximum positive displacement): $90^{\circ}$ after a chosen zero crossing,
equilibrium (zero displacement): can be $0^{\circ}$ , $180^{\circ}$ , etc.

The phase difference between two points is the fraction of a cycle separating them, multiplied by $360^{\circ}$ .

Understanding the Question

The diagram shows a transverse wave at one instant. Point $R$ is at a crest. Point $T$ is at the equilibrium line (zero displacement). The question asks for the phase difference between $R$ and $T$ .

Approach

Relate each marked point to a standard position on a sine wave and count how much of a cycle separates them (in terms of fractions like $\tfrac{1}{4}$ , $\tfrac{1}{2}$ , etc.). Convert that fraction into degrees.

Step-by-Step Reasoning

A crest is one quarter of a cycle away from the nearest equilibrium crossing.
Therefore the separation in phase from equilibrium to crest is

\Delta \phi = \frac{1}{4} \times 360^{\circ} = 90^{\circ}.

So the phase difference between $R$ (crest) and $T$ (equilibrium) is $90^{\circ}$ .

Key Takeaways

$360^{\circ}$ corresponds to one wavelength (one full cycle).
Crest and equilibrium are separated by $90^{\circ}$ .

Common Mistakes

Writing $180^{\circ}$ (confusing crest-to-trough with crest-to-zero).
Giving the answer in radians when the question asks for degrees.

Things to Be Careful About

There are two types of zero crossing (upward slope and downward slope). Both are still $90^{\circ}$ away from a crest; the sign/direction only affects whether you call it “leading” or “lagging”, not the magnitude here.

Techniques used

identify relative phase from positions on a sinusoidal waveuse the fraction of a cycle between two points to state phase difference

(b)

On Fig. 4.1, draw an arrow at point T to show the direction of movement of point T at the instant shown.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

Draw the arrow at $T$ vertically downwards.

Final answer

Downwards

Detailed explanation

Background Concept

In a transverse progressive wave, the wave travels horizontally but each point of the medium oscillates perpendicular to that direction (here, vertically).

For a right-travelling wave $y(x,t)$ , a useful relationship is that the particle velocity is related to the local slope:

\frac{\partial y}{\partial t} = -v\,\frac{\partial y}{\partial x}

So, for a wave travelling to the right ( $v>0$ ):

if the graph slopes upward as $x$ increases ( $\partial y/\partial x > 0$ ), then $\partial y/\partial t < 0$ so the particle is moving downward,
if the graph slopes downward as $x$ increases ( $\partial y/\partial x < 0$ ), then the particle is moving upward.

Understanding the Question

You must indicate the instantaneous direction of motion of the water at point $T$ on the snapshot of the wave, given the wave travels left to right.

Approach

Look at the local slope of the wave at $T$ on the snapshot. Use the fact that for a right-moving wave, the particle moves in the opposite direction to the slope.

Step-by-Step Reasoning

At $T$ , the displacement is zero (on the equilibrium line).
From the diagram, the curve at $T$ is crossing the equilibrium with a positive slope (rising as you move to the right).
Because the wave travels to the right, particle motion is opposite to the slope:

\frac{\partial y}{\partial x} > 0 \;\Rightarrow\; \frac{\partial y}{\partial t} < 0

So the point $T$ is moving downward at that instant.

Key Takeaways

In a transverse wave, the medium moves up/down, not along the direction of travel.
For a wave travelling right, particle velocity direction is opposite to the graph slope at that point.

Common Mistakes

Drawing the arrow to the right (confusing wave speed with particle motion).
Using the wrong sign rule (thinking particle motion is the same direction as the slope for a right-travelling wave).

Things to Be Careful About

The direction is instantaneous: it depends on the slope at that exact point and the given travel direction (left-to-right here).

Techniques used

use the direction of wave travel to relate local slope to particle motioninfer instantaneous particle velocity direction at a point on a progressive wave

(c)

The horizontal distance between R and T is $0.62\ \text{cm}$ , as shown in Fig. 4.2.

Fig. 4.2

The speed of the wave is $0.27\ \text{m s}^{-1}$ .

Calculate the frequency of the wave.

frequency = ______ $\text{Hz}$

3M Similar

DifficultyMedium-Easy

Worked solution

Working

From (a), $R$ to $T$ is a phase difference of $90^{\circ}$ , so separation $= \lambda/4$ .

\lambda = 4 \times 0.62\,\text{cm} = 2.48\,\text{cm} = 2.48 \times 10^{-2}\,\text{m}

Using $v = f\lambda$ :

f = \frac{v}{\lambda} = \frac{0.27}{2.48 \times 10^{-2}} = 10.9\,\text{Hz}

Answer

$11\,\text{Hz}$

Final answer

11 Hz

Detailed explanation

Background Concept

For a progressive wave:

v = f\lambda

where

$v$ is wave speed ( $\text{m s}^{-1}$ ),
$f$ is frequency ( $\text{Hz}$ ),
$\lambda$ is wavelength ( $\text{m}$ ).

Phase difference is linked to separation along the wave:

$360^{\circ}$ corresponds to a separation of $1\lambda$ ,
$90^{\circ}$ corresponds to $\tfrac{1}{4}\lambda$ .

Understanding the Question

You are given:

the horizontal distance between $R$ (a crest) and $T$ (equilibrium point) is $0.62\ \text{cm}$ ,
wave speed $v = 0.27\ \text{m s}^{-1}$ .

You must find the frequency $f$ . To do that, you need the wavelength $\lambda$ .

Approach

Use the phase relationship between $R$ and $T$ to decide what fraction of a wavelength $RT$ represents.
Use the measured distance $RT$ to calculate $\lambda$ .
Convert $\lambda$ to metres.
Use $v = f\lambda$ to find $f$ .

Step-by-Step Reasoning

From the wave shape, $R$ (crest) and $T$ (equilibrium) are separated by $90^{\circ}$ (a quarter cycle).

So:

RT = \frac{\lambda}{4}

Hence:

\lambda = 4RT = 4 \times 0.62\,\text{cm} = 2.48\,\text{cm}

Convert to SI units:

2.48\,\text{cm} = 2.48 \times 10^{-2}\,\text{m}

Apply the wave equation:

f = \frac{v}{\lambda} = \frac{0.27}{2.48 \times 10^{-2}} = 10.9\,\text{Hz}

Rounded appropriately, $f \approx 11\,\text{Hz}$ .

Key Takeaways

Use phase difference to relate a measured separation to a fraction of $\lambda$ .
Always convert cm to m before using $v = f\lambda$ .

Common Mistakes

Taking $0.62\,\text{cm}$ as the full wavelength (missing the phase/fraction step).
Forgetting to convert cm to m, giving an answer $100$ times too big.
Using $f = v\lambda$ instead of $f = v/\lambda$ .

Things to Be Careful About

Significant figures: the inputs are typically 2 s.f., so quoting $11\,\text{Hz}$ is appropriate.
Ensure you use the horizontal separation along the direction of travel (which is what is given).

Techniques used

convert phase difference into a fraction of a wavelengthcalculate wavelength from a measured separationuse v = f lambda to determine frequencyconvert units consistently before substitution

(d)

The wave is a water wave produced by a dipper $S_1$ attached to a vibrator in a ripple tank.

An identical dipper $S_2$ is attached to the same vibrator. The two dippers produce an interference pattern on the water in the tank, as shown in Fig. 4.3.

Fig. 4.3

The wave crests from each source are represented by solid lines on Fig. 4.3 and the wave troughs are represented by dashed lines.

At point P in Fig. 4.3, the wave from $S_1$ has the same amplitude $A$ as the wave from $S_2$ .

Describe and explain the amplitude of the resultant wave at point P.

3M Similar

DifficultyMedium

Worked solution

Answer

At $P$ , a crest from $S_1$ meets a trough from $S_2$ , so the waves arrive $180^{\circ}$ out of phase (path difference $= \left(n+\tfrac{1}{2}\right)\lambda$ ).

With equal amplitudes $A$ from each source, the displacements cancel by superposition, so the resultant amplitude at $P$ is $0$ (destructive interference).

Final answer

Resultant amplitude at P is 0 (destructive interference).

Detailed explanation

Background Concept

When two waves overlap, the resultant displacement is the algebraic sum of their individual displacements (principle of superposition).

For two coherent sources:

If the waves arrive in phase (phase difference $0^{\circ}$ , path difference $n\lambda$ ), you get constructive interference and the amplitude is maximum.
If the waves arrive in antiphase (phase difference $180^{\circ}$ , path difference $\left(n+\tfrac{1}{2}\right)\lambda$ ), you get destructive interference and the amplitude is minimum.

If the two waves at a point have equal amplitudes $A$ :

in phase: resultant amplitude $= 2A$ ,
$180^{\circ}$ out of phase: resultant amplitude $= 0$ .

Understanding the Question

At point $P$ , the diagram shows that a crest (maximum positive displacement) from $S_1$ coincides with a trough (maximum negative displacement) from $S_2$ . Also, each wave has the same amplitude $A$ at $P$ . You are asked to describe and explain the amplitude of the resultant wave at $P$ .

Approach

Use the crest/trough information to identify the phase relationship at $P$ .
Use superposition to add the displacements (or add phasors) given equal amplitudes.
State the resultant amplitude and name the type of interference.

Step-by-Step Reasoning

A crest corresponds to displacement $+A$ and a trough corresponds to displacement $-A$ at the same instant.
Crest meeting trough means the two waves at $P$ are in antiphase, i.e. phase difference $180^{\circ}$ (equivalently, path difference $\left(n+\tfrac{1}{2}\right)\lambda$ ).
Since each has amplitude $A$ , at the instant shown the displacements are equal and opposite.

By superposition:

y_{\text{resultant}} = (+A) + (-A) = 0

Thus the resultant amplitude at $P$ is zero: this point lies on a nodal line (destructive interference).

A phasor view shows the same result: two equal vectors $180^{\circ}$ apart add to zero.

diagram-4d-1

Key Takeaways

Interference depends on phase difference (or path difference).
Crest + trough with equal amplitudes gives complete cancellation and zero resultant amplitude.

Common Mistakes

Saying the resultant amplitude is $A$ (forgetting superposition is an algebraic sum).
Confusing amplitude with displacement: amplitude is the maximum possible displacement; at destructive interference with equal amplitudes the amplitude becomes zero.
Stating constructive interference even though the diagram shows crest meeting trough.

Things to Be Careful About

The question states the amplitudes from each source at $P$ are the same; without that, cancellation might be partial.
Use correct terminology: “destructive interference”, “antiphase”, “phase difference $180^{\circ}$ ”, “path difference $\left(n+\tfrac{1}{2}\right)\lambda$ ”.

Techniques used

apply the principle of superposition to add displacementsidentify constructive versus destructive interference from crest/trough meetinguse phase difference of 180 degrees to infer cancellation of equal amplitudes

Q5MediumD.C. CircuitsWork, Energy and Power

(a)

2M

(i)

State Kirchhoff’s second law.

1M Similar

DifficultyEasy

Worked solution

Answer

In any closed loop, the algebraic sum of the e.m.f.s equals the algebraic sum of the potential drops (equivalently the sum of p.d.s around the loop is zero).

Final answer

In any closed loop, the algebraic sum of the e.m.f.s equals the algebraic sum of the potential drops (sum of p.d.s around a loop is zero).

Detailed explanation

Background Concept

Kirchhoff’s second law is an expression of energy conservation for electric circuits. When a charge $q$ goes once around a complete closed loop, it can gain energy from sources (cells) and lose energy in components (resistors, internal resistance). The total energy gained per unit charge equals the total energy lost per unit charge.

Potential difference is energy transferred per unit charge, so p.d. and e.m.f. can be combined using an algebraic (signed) sum around the loop.

Understanding the Question

You are asked to state Kirchhoff’s second law: a general statement that applies to any closed loop in a circuit.

Approach

Write the standard wording: either “sum of e.m.f.s = sum of potential drops” or “algebraic sum of p.d.s around a closed loop is zero”.

Step-by-Step Reasoning

Choose any closed loop in a circuit.
Consider a test charge going around the loop: it gains energy across sources (e.m.f.) and loses energy across resistive components (p.d. drops).
Since it returns to the starting point with no net change in energy, the net energy change per unit charge is zero.
Therefore, the algebraic sum of voltages around the loop is zero (or gains = losses).

Key Takeaways

Kirchhoff’s second law is a loop (voltage) law.
It is fundamentally an energy conservation statement for a complete loop.

Common Mistakes

Stating Kirchhoff’s first law (currents at a junction) instead of the second.
Forgetting that it must be a closed loop.

Things to Be Careful About

Use “algebraic sum” (signs matter) if you write it as a sum equals zero.
e.m.f. is a rise in potential; resistor p.d. is a drop, depending on traversal direction.

Techniques used

state Kirchhoff’s second law in terms of potential differences and e.m.f.suse an algebraic sum around a closed loop

(ii)

State the conservation law that gives rise to Kirchhoff’s second law.

1M Similar

(b)

A circuit contains a cell of internal resistance $r$ and two resistors of resistances $R_1$ and $R_2$ , as shown in Fig. 5.1.

Fig. 5.1

The potential difference (p.d.) across the two resistors is $V$ .

The current in the cell is $I$ .

5M

(i)

Use Kirchhoff’s laws to show that the total resistance $R_T$ of the external circuit is given by

\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2}

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Let currents in $R_1$ and $R_2$ be $I_1$ and $I_2$ .

Kirchhoff 1st law at the junction:

I = I_1 + I_2

In parallel, p.d. across each resistor is $V$ :

I_1 = \frac{V}{R_1},\qquad I_2 = \frac{V}{R_2}

So

I = V\left(\frac{1}{R_1} + \frac{1}{R_2}\right)

But $R_T = \frac{V}{I}$ , hence

\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2}

Answer

\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2}

Final answer

1/RT = 1/R1 + 1/R2

Detailed explanation

Background Concept

For resistors in parallel:

The potential difference across each branch is the same (because both ends of each resistor connect to the same two junctions).
The total current entering the junction equals the total current leaving (Kirchhoff’s first law: conservation of charge).

Ohm’s law relates current and p.d. for each resistor:

I = \frac{V}{R}

The equivalent (total) resistance $R_T$ of the parallel combination is defined by the same $V$ across the combination and the total current $I$ through it:

R_T = \frac{V}{I}

Understanding the Question

The external circuit is a parallel network of $R_1$ and $R_2$ , with a voltmeter measuring the p.d. $V$ across the parallel combination. You must use Kirchhoff’s laws (current at junctions and/or loop rule) to show the standard parallel-resistance formula.

Approach

Label branch currents $I_1$ and $I_2$ .
Use Kirchhoff’s first law: $I = I_1 + I_2$ .
Use that each branch has the same p.d. $V$ and apply Ohm’s law to write $I_1$ and $I_2$ .
Substitute into the junction equation and use $R_T = V/I$ .

Step-by-Step Reasoning

diagram-b-i-1

At the junction where the current splits, Kirchhoff’s first law gives:

I = I_1 + I_2

Because $R_1$ and $R_2$ are connected in parallel, the p.d. across each is the same as the voltmeter reading $V$ .
Apply Ohm’s law to each branch:

I_1 = \frac{V}{R_1}, \qquad I_2 = \frac{V}{R_2}

Substitute into the junction equation:

I = \frac{V}{R_1} + \frac{V}{R_2} = V\left(\frac{1}{R_1} + \frac{1}{R_2}\right)

The equivalent resistance $R_T$ is defined by $I = V/R_T$ , so:

\frac{V}{R_T} = V\left(\frac{1}{R_1} + \frac{1}{R_2}\right)

Cancel $V$ (non-zero), giving:

\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2}

Key Takeaways

Parallel branches share the same p.d.
Junction currents add: $I = I_1 + I_2$ .
Combining those with $I=V/R$ leads directly to the reciprocal-resistance formula.

Common Mistakes

Adding resistances directly for a parallel network (that is for series).
Writing $I_1 = I_2 = I$ (only true for certain symmetric cases, not generally).

Things to Be Careful About

Use the same $V$ for both branches (that is the key property of parallel).
Remember $R_T$ is defined from the total current and the common p.d.: $R_T=V/I$ .

Techniques used

apply Kirchhoff’s first law at a junction to relate currentsuse equality of potential difference across parallel componentssubstitute Ohm’s law for each branch currentdefine equivalent resistance from total current and common potential difference

(ii)

The electromotive force (e.m.f.) of the cell is $1.50\ \text{V}$ .

When the values of $R_1$ and $R_2$ are $10\ \Omega$ and $15\ \Omega$ respectively, the p.d. measured by the voltmeter is $1.38\ \text{V}$ .

Calculate the internal resistance $r$ of the cell.

$r$ = ______ $\Omega$

3M Similar

DifficultyMedium

Worked solution

Working

Equivalent external resistance:

R_T = \frac{R_1R_2}{R_1+R_2} = \frac{(10)(15)}{10+15} = 6.0\ \Omega

Current in cell:

I = \frac{V}{R_T} = \frac{1.38}{6.0} = 0.230\ \text{A}

Using $\mathcal{E} = V + Ir$ :

r = \frac{\mathcal{E}-V}{I} = \frac{1.50-1.38}{0.230} = 0.52\ \Omega

Answer

r = 0.52\ \Omega

Final answer

0.52 Ω

Detailed explanation

Background Concept

A real cell can be modelled as an ideal source of e.m.f. $\mathcal{E}$ in series with an internal resistance $r$ .

The terminal p.d. across the external circuit is

V = \mathcal{E} - Ir

where $I$ is the current supplied by the cell.

The external circuit here is the parallel combination of $R_1$ and $R_2$ , with equivalent resistance $R_T$ given by

\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2}

The voltmeter reads the terminal p.d. $V$ across this external resistance.

Understanding the Question

Given:

$\mathcal{E} = 1.50\ \text{V}$
$R_1 = 10\ \Omega$ , $R_2 = 15\ \Omega$
voltmeter reading across the external parallel network: $V = 1.38\ \text{V}$
Find the internal resistance $r$ .

Approach

Find the equivalent external resistance $R_T$ of the two parallel resistors.
Use $I = V/R_T$ to find the current delivered by the cell (same current through the internal resistance, since $r$ is in series with the external network).
Use $V = \mathcal{E} - Ir$ (or $\mathcal{E} = V + Ir$ ) to solve for $r$ .

Step-by-Step Reasoning

Combine the parallel resistors:

R_T = \frac{R_1R_2}{R_1+R_2} = \frac{(10)(15)}{25} = 6.0\ \Omega

The voltmeter reading $V=1.38\ \text{V}$ is the p.d. across $R_T$ , so the current through the external circuit is

I = \frac{V}{R_T} = \frac{1.38}{6.0} = 0.230\ \text{A}

This is also the current in the cell and through $r$ .
3. The “lost volts” across internal resistance is

\mathcal{E} - V = 1.50 - 1.38 = 0.12\ \text{V}

So

Ir = 0.12\ \text{V} \Rightarrow r = \frac{0.12}{0.230} = 0.52\ \Omega

Key Takeaways

Terminal p.d. is smaller than e.m.f. when current flows because of the internal drop $Ir$ .
In a series path, the same current flows through internal resistance and the external equivalent resistance.
Parallel resistors reduce the external resistance.

Common Mistakes

Using $I = \mathcal{E}/R_T$ instead of $I = V/R_T$ (the external circuit does not have p.d. equal to $\mathcal{E}$ when there is internal resistance).
Forgetting to combine $R_1$ and $R_2$ in parallel.
Using $V = \mathcal{E} + Ir$ (wrong sign).

Things to Be Careful About

Check that $V < \mathcal{E}$ ; otherwise the arithmetic/signs are wrong.
Keep sufficient significant figures during working; round $r$ at the end (here $r \approx 0.52\ \Omega$ ).

Techniques used

calculate the equivalent resistance of parallel resistorsuse the terminal p.d. relation V = E - Irapply Ohm’s law to relate p.d., current, and resistancerearrange to solve for internal resistance

(c)

A third resistor is added in parallel with $R_1$ and $R_2$ in the circuit in Fig. 5.1.

State and explain the effect, if any, of this change on:

4M

(i)

the current in the cell

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Adding a third resistor in parallel decreases the external equivalent resistance.

Total circuit resistance $(r + R_{\text{ext}})$ decreases, so the current in the cell increases.

Final answer

The current in the cell increases.

Detailed explanation

Background Concept

For resistors in parallel, adding an extra branch provides an additional path for charge, so the equivalent resistance decreases:

\frac{1}{R_{\text{ext}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}

A cell with internal resistance $r$ supplies a current

I = \frac{\mathcal{E}}{r + R_{\text{ext}}}

So if $R_{\text{ext}}$ decreases, the denominator decreases and $I$ increases.

Understanding the Question

A third resistor is added in parallel with $R_1$ and $R_2$ . You must state the effect on the current in the cell and explain why.

Approach

Decide how the external equivalent resistance changes when adding a parallel resistor.
Use the idea that the cell current depends on the total resistance (external plus internal).

Step-by-Step Reasoning

Adding a resistor in parallel increases total conductance $\left(\frac{1}{R}\right)$ , so the equivalent external resistance $R_{\text{ext}}$ becomes smaller.
The internal resistance $r$ is in series with the external equivalent, so the total resistance $r + R_{\text{ext}}$ decreases.
Using

I = \frac{\mathcal{E}}{r + R_{\text{ext}}}

a smaller denominator gives a larger current.

Key Takeaways

More parallel branches (\to) smaller equivalent resistance.
Smaller total resistance (\to) larger supply current from the cell.

Common Mistakes

Saying the current decreases because “resistance is added” (parallel addition actually reduces the equivalent).
Confusing the current in one branch with the current in the cell (cell current is the sum of branch currents).

Things to Be Careful About

Be explicit that it is the equivalent external resistance that changes, not the internal resistance $r$ .

Techniques used

reason qualitatively about how adding a parallel branch changes equivalent resistancerelate total circuit resistance to current using I = E / (R + r)

(ii)

the p.d. measured by the voltmeter.

2M Similar

DifficultyMedium

Worked solution

Answer

The voltmeter reads the terminal p.d. $V$ across the external parallel network.

Adding a parallel resistor reduces external resistance, so current $I$ increases.

Since

V = \mathcal{E} - Ir,

$Ir$ increases, so $V$ decreases.

Final answer

The voltmeter reading (terminal p.d.) decreases.

Detailed explanation

Background Concept

The voltmeter is across the external circuit, so it measures the terminal p.d. $V$ . For a cell with internal resistance $r$ :

V = \mathcal{E} - Ir

where $Ir$ is the p.d. lost inside the cell.

An equivalent form (often useful for checking trends) is

V = I R_{\text{ext}} = \mathcal{E}\,\frac{R_{\text{ext}}}{r + R_{\text{ext}}}

This shows directly how $V$ depends on the external equivalent resistance.

Understanding the Question

A third resistor is added in parallel with $R_1$ and $R_2$ , making the external equivalent resistance smaller. You must state and explain how that affects the voltmeter reading across the external network.

Approach

Adding a parallel resistor decreases $R_{\text{ext}}$ and therefore increases the current $I$ .
Use $V = \mathcal{E} - Ir$ to see how increasing $I$ changes $V$ .

Step-by-Step Reasoning

With an extra parallel resistor, $R_{\text{ext}}$ decreases.
The cell current becomes

I = \frac{\mathcal{E}}{r + R_{\text{ext}}}

so $I$ increases when $R_{\text{ext}}$ decreases.
3. The internal p.d. drop is $Ir$ . Since $r$ is constant and $I$ increases, $Ir$ increases.
4. Therefore the terminal p.d.

V = \mathcal{E} - Ir

decreases.

(Consistency check: using $V = \mathcal{E}\,\frac{R_{\text{ext}}}{r + R_{\text{ext}}}$ , decreasing $R_{\text{ext}}$ makes the fraction smaller, so $V$ decreases.)

Key Takeaways

The voltmeter measures terminal p.d., not the e.m.f.
Increasing current increases the lost volts $Ir$ , reducing terminal p.d.

Common Mistakes

Saying the voltmeter reading increases because “more current flows” (more current causes a larger internal drop, which reduces the terminal p.d.).
Treating the voltmeter reading as fixed at $\mathcal{E}$ .

Things to Be Careful About

The p.d. across the external parallel combination is the same for all branches, but that common p.d. itself changes when the total current changes (because of internal resistance).

Techniques used

use V = E - Ir to relate terminal p.d. to currentcombine qualitative changes in current and internal voltage drop

Q6Medium-EasyParticle Physics

Nuclei of an isotope of samarium (Sm) each contain 62 protons and 85 neutrons.

(a)

State the nuclide notation in the form ${}^A_Z\text{X}$ for this isotope of samarium.

1M Similar

(b)

This isotope of samarium is radioactive and decays by emitting particles. Gamma-radiation is not emitted. The energy spectrum of the emitted particles is shown in Fig. 6.1.

Fig. 6.1

4M

(i)

Explain how Fig. 6.1 shows that this isotope of samarium emits $\alpha$ -particles and does not emit $\beta$ -particles.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Fig. 6.1 shows all emitted particles have the same kinetic energy (a single discrete line), consistent with $\alpha$ -decay.

$\beta$ -decay would give a continuous range of kinetic energies (shared with a neutrino), so $\beta$ -particles are not emitted.

Final answer

Discrete single energy line implies alpha emission; beta would give a continuous spectrum.

Detailed explanation

Background Concept

In nuclear decay, the kinetic energy distribution of emitted particles depends on how many particles share the decay energy.

$\alpha$ -decay (typically two-body):

{}^{A}_{Z}\text{X} \rightarrow {}^{A-4}_{Z-2}\text{Y} + {}^{4}_{2}\text{He}

If no $\gamma$ is emitted and the daughter nucleus is left in a single energy state, the decay energy is shared in a fixed way between just the $\alpha$ particle and the recoiling daughter nucleus. This produces (approximately) a single kinetic energy for the $\alpha$ particle (a line spectrum).

$\beta$ -decay involves three bodies because a neutrino (or antineutrino) is emitted:

\beta^-: \quad n \rightarrow p + e^- + \bar{\nu}

The available energy is shared variably between the $\beta$ particle and the neutrino, so $\beta$ particles have a continuous spectrum of kinetic energies from near $0$ up to a maximum.

Understanding the Question

You are given an energy spectrum for the emitted particles from samarium, and told no gamma radiation is emitted. You must use the spectrum shape to decide whether the emitted particles are $\alpha$ or $\beta$ .

The key visual feature is that the spectrum is a single vertical line: all particles come out with the same kinetic energy.

Approach

Identify whether the spectrum is discrete (line) or continuous (broad distribution).
Recall: $\alpha$ emission gives discrete energies (line), whereas $\beta$ emission gives a continuous distribution due to a neutrino.
State the conclusion with a brief justification.

Step-by-Step Reasoning

The graph shows one sharp line at one kinetic energy value. That means every emitted particle has essentially the same kinetic energy.
In $\alpha$ -decay (with no $\gamma$ ), the decay is effectively two-body ( $\alpha$ + daughter). Conservation of energy and momentum then fixes the kinetic energies, producing a discrete line.
In $\beta$ -decay, a neutrino is also produced, so the energy is shared in different proportions each time. This leads to a continuous range of $\beta$ kinetic energies, not a single line.
Therefore, Fig. 6.1 is evidence for $\alpha$ emission and evidence against $\beta$ emission.

Key Takeaways

A discrete line in kinetic energy suggests a two-body decay like $\alpha$ emission.
A continuous $\beta$ spectrum occurs because a neutrino/antineutrino carries away varying energy.

Common Mistakes

Saying $\beta$ has discrete energy because it is a single particle: forgetting the neutrino.
Claiming the line means $\gamma$ emission: $\gamma$ is electromagnetic radiation and is not what this particle kinetic energy spectrum represents.

Things to Be Careful About

The question states $\gamma$ is not emitted, supporting the idea of a simple two-body decay with fixed kinetic energies.
Don’t confuse a line spectrum in emitted particle energy (alpha) with line spectra in photon emission (gamma) from electronic transitions.

Techniques used

interpret an energy spectrum shapelink discrete energies to two-body decaylink continuous beta spectrum to neutrino emission

(ii)

This isotope of samarium decays to an isotope of neodymium (Nd).

Give the radioactive decay equation for this decay. Include the nucleon and proton numbers of all the particles involved.

2M Similar

DifficultyMedium-Easy

Worked solution

Working

$\alpha$ -decay: $A \rightarrow A-4$ , $Z \rightarrow Z-2$ .

{}^{147}_{62}\text{Sm} \rightarrow {}^{143}_{60}\text{Nd} + {}^{4}_{2}\text{He}

Answer

{}^{147}_{62}\text{Sm} \rightarrow {}^{143}_{60}\text{Nd} + {}^{4}_{2}\text{He}

Final answer

${}^{147}_{62}\text{Sm} \rightarrow {}^{143}_{60}\text{Nd} + {}^{4}_{2}\text{He}$

Detailed explanation

Background Concept

A nuclear (radioactive decay) equation must conserve:

Nucleon number $A$ (total protons + neutrons)
Proton number $Z$ (charge number)

For $\alpha$ -decay, the emitted $\alpha$ particle is a helium nucleus:

{}^{4}_{2}\text{He}

So the parent nucleus changes as:

A \rightarrow A - 4, \qquad Z \rightarrow Z - 2.

Understanding the Question

The samarium isotope has $Z=62$ and (from part a) $A=147$ . It decays into an isotope of neodymium ( $\text{Nd}$ ) and emits particles. From part (b)(i) we conclude the emission is $\alpha$ .

You must write the full decay equation including $A$ and $Z$ for every particle.

Approach

Start with ${}^{147}_{62}\text{Sm}$ .
Subtract $4$ from $A$ and $2$ from $Z$ to get the daughter nucleus.
Identify the daughter element from its new $Z$ .
Add the $\alpha$ particle ${}^{4}_{2}\text{He}$ on the right.

Step-by-Step Reasoning

Parent nuclide:

{}^{147}_{62}\text{Sm}

After $\alpha$ emission:

A_{\text{daughter}} = 147 - 4 = 143

Z_{\text{daughter}} = 62 - 2 = 60

Element with $Z=60$ is neodymium ( $\text{Nd}$ ), so the daughter nucleus is:

{}^{143}_{60}\text{Nd}

The decay equation is:

{}^{147}_{62}\text{Sm} \rightarrow {}^{143}_{60}\text{Nd} + {}^{4}_{2}\text{He}

Check:

Nucleon numbers: $147 = 143 + 4$
Proton numbers: $62 = 60 + 2$

Key Takeaways

In $\alpha$ -decay, $A$ decreases by $4$ and $Z$ decreases by $2$ .
Always check conservation of both $A$ and $Z$ .

Common Mistakes

Writing the alpha particle as ${}^{0}_{-1}e$ (that is $\beta^-$ , not $\alpha$ ).
Getting the daughter proton number wrong (using $Z-4$ instead of $Z-2$ ).
Changing the element symbol incorrectly after changing $Z$ .

Things to Be Careful About

The question states the daughter is neodymium; that must match your calculated $Z$ .
Include nucleon and proton numbers on all nuclides, including the emitted $\alpha$ particle.

Techniques used

apply nucleon number conservation in nuclear equationsapply proton number conservation in nuclear equationswrite a balanced alpha-decay equation

(c)

A baryon is composed of three quarks which all have different flavours. The baryon has a charge of 0.

Two of the quarks in the baryon are an up quark and a bottom quark.

3M

(i)

Determine, in terms of the elementary charge $e$ , the charge on the third quark in the baryon.

charge = ______ $e$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Charges: up quark $= +\frac{2}{3}e$ , bottom quark $= -\frac{1}{3}e$ .

Total charge $0$ :

+\frac{2}{3}e - \frac{1}{3}e + q = 0

\frac{1}{3}e + q = 0 \Rightarrow q = -\frac{1}{3}e

Answer

charge $= -\dfrac{1}{3}\,e$

Final answer

$-\dfrac{1}{3}e$

Detailed explanation

Background Concept

A baryon is a hadron made from three quarks. The total electric charge of the baryon is the sum of the charges of its three quarks.

Quark charges (in units of elementary charge $e$ ):

up ( $u$ ):

+\frac{2}{3}e

down ( $d$ ), strange ( $s$ ), bottom ( $b$ ):

-\frac{1}{3}e

charm ( $c$ ), top ( $t$ ):

+\frac{2}{3}e

Understanding the Question

You have a baryon consisting of three different-flavour quarks. Two are specified: an up quark and a bottom quark. The baryon’s total charge is $0$ . You must find the charge of the third quark in terms of $e$ .

Approach

Write down the known charges of the up and bottom quarks.
Let the third quark’s charge be $q$ .
Use charge conservation: (sum of three quark charges) $= 0$ .
Solve for $q$ .

Step-by-Step Reasoning

Known quark charges:

q_u = +\frac{2}{3}e

q_b = -\frac{1}{3}e

Let the third quark have charge $q$ .
Total charge is zero:

q_u + q_b + q = 0

Substitute:

+\frac{2}{3}e - \frac{1}{3}e + q = 0

Combine the fractions:

\left(\frac{1}{3}\right)e + q = 0

So:

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

Key Takeaways

Baryon charge is the sum of the three quark charges.
Remember: $u,c,t$ have $+\frac{2}{3}e$ ; $d,s,b$ have $-\frac{1}{3}e$ .

Common Mistakes

Using the wrong sign for the bottom quark (it is $-\frac{1}{3}e$ ).
Adding charges incorrectly: $\frac{2}{3}-\frac{1}{3}=\frac{1}{3}$ , not $\frac{3}{3}$ .
Forgetting the question asks for the answer in terms of $e$ .

Things to Be Careful About

The baryon has charge $0$ , so the quark charges must sum exactly to zero.
Don’t assume the third quark is a particular flavour in this part; only its charge is required here.

Techniques used

use quark charges to form a total baryon chargeset up and solve a simple charge-balance equationapply the condition that total charge is zero

(ii)

State a possible flavour for the third quark in the baryon.

1M Similar

Physics 9702/23 — May/June 2024

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