Physics 9702 / 23 — October/November 2024

(b)

A small aircraft is flying horizontally at a speed of $42\ \text{m s}^{-1}$ at a height of $63\ \text{m}$ above horizontal ground, as shown in Fig. 1.1.

Fig. 1.1

The aircraft drops a small parcel. The parcel is released from the aircraft at the instant shown in Fig. 1.1. Air resistance is negligible.

6M

(i)

On Fig. 1.1, draw a line to show the path of the parcel as it falls from the aircraft to the ground.

DifficultyMedium-Easy

Worked solution

Answer

diagram-b-i

Final answer

A forward-curving parabolic path to the right.

Detailed explanation

Background Concept

When air resistance is negligible, a dropped object keeps its horizontal velocity (no horizontal acceleration) while accelerating vertically downward due to gravity ( $g$ ). The two motions are independent:

Horizontal: constant velocity
Vertical: uniformly accelerated motion with acceleration $g$

This combination produces a parabolic path (projectile motion).

Understanding the Question

The aircraft is moving horizontally at $42\ \text{m s}^{-1}$ and releases the parcel. At the instant of release, the parcel has the same initial horizontal velocity as the aircraft, and zero initial vertical velocity. You must draw the path from release point to ground.

Approach

Sketch a projectile trajectory:

Starts at the release point.
Initially tangent is horizontal (because initial vertical velocity is zero).
Curves downward more steeply as time increases.
Moves to the right while falling.

Step-by-Step Reasoning

diagram-b-i

At release: parcel has horizontal speed $42\ \text{m s}^{-1}$ .
With no air resistance, horizontal speed stays constant.
Gravity causes increasing downward vertical speed.
So the path curves downward (parabola), landing ahead of the point directly below the release.

Key Takeaways

“Dropped from a moving object” does not mean it falls straight down.
Horizontal motion continues at constant velocity while vertical motion accelerates.

Common Mistakes

Drawing a vertical straight line down (ignores initial horizontal velocity).
Drawing a straight diagonal line (should curve because vertical speed increases).
Drawing it curving upward (gravity acts downward).

Things to Be Careful About

The curve should start horizontal at the release point (since $u_y = 0$ ) and become steeper downward.
The aircraft continues forward, so the parcel’s path is not directly under the aircraft after release.

Techniques used

identify projectile motion with independent horizontal and vertical componentssketch a parabolic trajectory from initial horizontal velocity

(ii)

Calculate the time taken from the instant of release to the instant the parcel reaches the ground.

time = ______ $\text{s}$

DifficultyMedium-Easy

Worked solution

Working

Vertical motion: $u_y = 0$ , $s = 63\ \text{m}$ , $a = g = 9.81\ \text{m s}^{-2}$ .

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

63 = \frac{1}{2}(9.81)t^2 \Rightarrow t = \sqrt{\frac{2 \times 63}{9.81}} = 3.58\ \text{s}

Answer

$3.6\ \text{s}$

Final answer

3.6 s

Detailed explanation

Background Concept

In projectile motion with negligible air resistance, horizontal and vertical motions are independent. Time of flight is determined by the vertical motion only.

For vertical motion under gravity (constant acceleration $g$ ), the displacement equation is

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

where $s$ is vertical displacement, $u$ is initial vertical velocity, $a$ is vertical acceleration.

Understanding the Question

The parcel is released from height $63\ \text{m}$ . At release, it has:

initial vertical velocity $u_y = 0$ (it is dropped, not thrown up or down),
vertical acceleration $a_y = g$ downward,
vertical displacement to the ground $s = 63\ \text{m}$ .

We want the time $t$ to reach the ground.

Approach

Use the vertical displacement equation with $u_y = 0$ . The horizontal speed $42\ \text{m s}^{-1}$ does not affect the time to fall.

Step-by-Step Reasoning

Take downward as positive (any consistent sign choice is fine):

$u_y = 0$
$a = g = 9.81\ \text{m s}^{-2}$
$s = 63\ \text{m}$

Substitute into

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

so

63 = 0 + \frac{1}{2}(9.81)t^2

Rearrange:

t^2 = \frac{2 \times 63}{9.81}

t = \sqrt{\frac{126}{9.81}} = 3.58\ \text{s} \approx 3.6\ \text{s}

Key Takeaways

Time to fall from height depends only on vertical motion.
With $u_y = 0$ , the equation simplifies to $s = \tfrac{1}{2}gt^2$ .

Common Mistakes

Using the aircraft’s $42\ \text{m s}^{-1}$ as $u$ in the vertical equation.
Using $s = gt^2$ (missing the factor $\tfrac{1}{2}$ ).
Mixing sign conventions (e.g. taking $a = -g$ but $s = +63$ without care).

Things to Be Careful About

Use $u_y = 0$ because the parcel is dropped.
Quote time to a sensible number of significant figures (here $3.6\ \text{s}$ ).

Techniques used

treat vertical motion independently of horizontal motionapply constant-acceleration equation for displacementsolve for time using a rearranged kinematics equation

(iii)

Calculate the vertical component of the velocity of the parcel immediately before it reaches the ground.

vertical component of velocity = ______ $\text{m s}^{-1}$

DifficultyMedium-Easy

Worked solution

Working

v_y = u_y + gt = 0 + 9.81(3.58) = 35.1\ \text{m s}^{-1}

Answer

$35\ \text{m s}^{-1}$ downward

Final answer

35 m s⁻¹ downward

Detailed explanation

Background Concept

For motion with constant acceleration, velocity changes linearly with time:

v = u + at

In free fall (ignoring air resistance), the vertical acceleration is constant and equal to $g$ .

Understanding the Question

We want the vertical component of velocity just before impact. The parcel starts with no vertical component ( $u_y = 0$ ) and accelerates downward at $g$ for the time found in part (ii).

Approach

Use the vertical kinematics equation $v_y = u_y + gt$ with $u_y = 0$ and $t$ from (ii).

Step-by-Step Reasoning

Initial vertical velocity: $u_y = 0$ .
Acceleration: $a_y = g = 9.81\ \text{m s}^{-2}$ .
Time of flight: $t = 3.58\ \text{s}$ .

So

v_y = 0 + 9.81 \times 3.58 = 35.1\ \text{m s}^{-1}

Direction is downward (since gravity accelerates downward), so the vertical component is about $35\ \text{m s}^{-1}$ downward.

Key Takeaways

Vertical speed increases by $g$ every second in free fall.
The vertical component at impact depends on the time spent falling.

Common Mistakes

Giving $v_y = 35\ \text{m s}^{-1}$ but not stating direction when asked for a component.
Using $42\ \text{m s}^{-1}$ (horizontal speed) instead of the vertical speed.
Using $v^2 = u^2 + 2as$ but mixing signs incorrectly.

Things to Be Careful About

State “downward” (or use a negative sign if you defined upward as positive).
Keep consistent significant figures; $35.1\ \text{m s}^{-1}$ rounds to $35\ \text{m s}^{-1}$ or $35.1\ \text{m s}^{-1}$ depending on expected precision.

Techniques used

use v = u + at for vertical motionsubstitute time of fall into a kinematics equation

(iv)

Determine the speed at which the parcel reaches the ground.

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

DifficultyMedium

Worked solution

Working

Horizontal component $v_x = 42\ \text{m s}^{-1}$ (constant).

Vertical component $v_y = 35.1\ \text{m s}^{-1}$ (downward).

Resultant speed:

v = \sqrt{v_x^2 + v_y^2} = \sqrt{42^2 + 35.1^2} = 54.7\ \text{m s}^{-1}

Answer

$55\ \text{m s}^{-1}$

Final answer

55 m s⁻¹

Detailed explanation

Background Concept

In projectile motion (no air resistance):

Horizontal velocity component is constant.
Vertical velocity component changes due to gravity.

The speed is the magnitude of the velocity vector. If the components are perpendicular,

v = \sqrt{v_x^2 + v_y^2}

This is Pythagoras because $v_x$ and $v_y$ are at right angles.

Understanding the Question

At impact, the parcel has:

horizontal component $v_x$ equal to its initial horizontal speed $42\ \text{m s}^{-1}$ ,
vertical component $v_y$ found in (iii) (downward).

You are asked for the speed (a scalar), so you need the magnitude of the combined velocity.

Approach

Take $v_x = 42\ \text{m s}^{-1}$ .
Take $v_y \approx 35\ \text{m s}^{-1}$ downward.
Combine using $v = \sqrt{v_x^2 + v_y^2}$ .

Step-by-Step Reasoning

Horizontal component:

No horizontal force (air resistance negligible), so horizontal acceleration is zero.
Therefore,

v_x = 42\ \text{m s}^{-1}

Vertical component:
From (iii),

v_y = 35.1\ \text{m s}^{-1}\ \text{(downward)}

Resultant speed:

v = \sqrt{42^2 + 35.1^2}

v = \sqrt{1764 + 1232} = \sqrt{2996} = 54.7\ \text{m s}^{-1} \approx 55\ \text{m s}^{-1}

Key Takeaways

The horizontal speed stays the same throughout (if no air resistance).
“Speed” means magnitude of velocity: combine perpendicular components with Pythagoras.

Common Mistakes

Adding components directly: $v = v_x + v_y$ (incorrect because they are perpendicular).
Using only the vertical component as the speed.
Changing $v_x$ due to gravity (gravity only affects vertical motion).

Things to Be Careful About

Use magnitudes in the Pythagoras calculation; direction (“downward”) matters for describing components but speed is always positive.
Keep consistent units ( $\text{m s}^{-1}$ throughout).
Round appropriately (typically to 2 s.f. here: $55\ \text{m s}^{-1}$ ).

Techniques used

identify horizontal velocity as constantcombine perpendicular velocity components using Pythagorascalculate resultant speed from components

Q2MediumDynamicsWork, Energy and Power

(a)

State the principle of conservation of momentum.

DifficultyEasy

Worked solution

Answer

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

Final answer

Total momentum is constant if no resultant external force acts on the system.

Detailed explanation

Background Concept

Momentum is defined as

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

The principle of conservation of momentum states that for a system of interacting bodies, internal forces cannot change the total momentum of the system. Only a resultant external force can change the total momentum.

Mathematically,

\vec{F}_{\text{external}} = \frac{\Delta \vec{p}_{\text{total}}}{\Delta t}

So if $\vec{F}_{\text{external}} = \vec{0}$ , then $\Delta \vec{p}_{\text{total}} = \vec{0}$ and total momentum is constant.

Understanding the Question

You are asked to state the principle. That means you should give a clear sentence saying what is conserved (total momentum) and under what condition (no resultant external force / isolated system).

Approach

Write the conservation statement (total momentum before = total momentum after) and include the condition for it to be valid.

Step-by-Step Reasoning

In any interaction (such as a collision), the forces the bodies exert on each other are internal forces.
Internal forces occur in equal and opposite pairs and therefore cannot change the total momentum of the system.
If there is no resultant external force on the system, then the total momentum cannot change, so it is conserved.

Key Takeaways

Conservation of momentum is about the total momentum of a system.
It is valid when the resultant external force is zero (or negligible).

Common Mistakes

Forgetting to mention the condition (no external resultant force).
Saying “momentum is always conserved” without specifying a system.
Confusing momentum with kinetic energy (kinetic energy is not necessarily conserved).

Things to Be Careful About

Momentum is a vector: direction matters.
The “system” must be defined (e.g. both balls together), and external forces must be negligible (here the surface is frictionless, so horizontal external force is negligible).

Techniques used

state the conservation law for total momentumspecify the condition of no resultant external force on the system

(b)

A ball X has mass $240\ \text{g}$ and moves in a straight line on a horizontal frictionless surface with an initial speed of $16\ \text{m s}^{-1}$ . The ball collides with a stationary ball Y that has mass $480\ \text{g}$ . After the collision, ball X is stationary, as shown in Fig. 2.1.

Fig. 2.1

4M

(i)

Show that the speed $v$ of ball Y after the collision is $8.0\ \text{m s}^{-1}$ .

DifficultyMedium-Easy

Worked solution

Working

$m_X = 0.240\ \text{kg}$ , $m_Y = 0.480\ \text{kg}$ .

Conservation of momentum:

(0.240)(16) + (0.480)(0) = (0.240)(0) + (0.480)v

v = \frac{0.240 \times 16}{0.480} = 8.0\ \text{m s}^{-1}

Answer

$8.0\ \text{m s}^{-1}$

Final answer

8.0 m s^-1

Detailed explanation

Background Concept

In a collision on a frictionless horizontal surface, the resultant external horizontal force is negligible, so horizontal momentum is conserved.

For motion in a straight line, you can treat momentum as a signed scalar (choose right as positive):

\text{total momentum} = \sum mv

Conservation means

\sum mv\,(\text{before}) = \sum mv\,(\text{after})

Understanding the Question

Ball X (mass $240\ \text{g}$ ) moves right at $16\ \text{m s}^{-1}$ and hits ball Y (mass $480\ \text{g}$ ) which is initially at rest. After collision, X is stationary and Y moves right with speed $v$ . You must show $v = 8.0\ \text{m s}^{-1}$ .

Approach

Convert grams to kilograms.
Write momentum before and after, taking right as positive.
Use conservation of momentum and solve for $v$ .

Step-by-Step Reasoning

Convert masses:

m_X = 240\ \text{g} = 0.240\ \text{kg}, \quad m_Y = 480\ \text{g} = 0.480\ \text{kg}

Write momentum before collision:

X is moving right at $16\ \text{m s}^{-1}$ , so momentum is $0.240 \times 16$ .
Y is stationary, so its momentum is $0$ .

p_{\text{before}} = (0.240)(16) + (0.480)(0)

Write momentum after collision:

X is stationary, so its momentum is $0$ .
Y moves right with speed $v$ , so momentum is $(0.480)v$ .

p_{\text{after}} = (0.240)(0) + (0.480)v

Set them equal:

(0.240)(16) = (0.480)v

Solve:

v = \frac{0.240\times 16}{0.480} = 8.0\ \text{m s}^{-1}

The speed is positive, matching the diagram (Y moves to the right).

Key Takeaways

Use momentum conservation when external forces are negligible.
Stationary means $v=0$ so momentum contribution is zero.
Always convert to SI units.

Common Mistakes

Forgetting to convert $\text{g}$ to $\text{kg}$ .
Putting masses the wrong way round when solving for $v$ .
Treating “speed” as always positive but then mishandling direction signs in momentum.

Things to Be Careful About

Choose a sign convention (e.g. right positive) and stick to it.
The question asks for speed (magnitude), but conservation of momentum is directional; here the direction is clearly to the right from the diagram.

Techniques used

apply conservation of momentum in one dimensionconvert masses to SI unitssolve a linear equation for the unknown speed

(ii)

Calculate the change in the total kinetic energy $\Delta E_K$ of the balls due to the collision.

$\Delta E_K$ = ______ $\text{J}$

DifficultyMedium-Easy

Worked solution

Working

Initial total kinetic energy:

E_{k,i} = \frac{1}{2}(0.240)(16^2) = 30.72\ \text{J}

Final total kinetic energy ( $X$ stationary, $v_Y = 8.0\ \text{m s}^{-1}$ ):

E_{k,f} = \frac{1}{2}(0.480)(8.0^2) = 15.36\ \text{J}

\Delta E_k = E_{k,f} - E_{k,i} = 15.36 - 30.72 = -15.36\ \text{J}

Answer

$\Delta E_k = -15.4\ \text{J}$

Final answer

-15.4 J

Detailed explanation

Background Concept

Kinetic energy of an object of mass $m$ moving at speed $v$ is

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

In collisions:

Momentum is conserved (if no resultant external force).
Kinetic energy is only conserved in elastic collisions.
If kinetic energy decreases, it has been transformed into other forms (internal energy, sound, deformation, etc.).

The change in total kinetic energy is defined as

\Delta E_k = E_{k,\text{final}} - E_{k,\text{initial}}

Understanding the Question

You need the total kinetic energy of both balls before the collision and after the collision, then find the change due to the collision. From part (b)(i), after collision ball Y has speed $8.0\ \text{m s}^{-1}$ and ball X is stationary.

Approach

Compute initial total $E_k$ (only X is moving initially).
Compute final total $E_k$ (only Y is moving finally).
Subtract: final minus initial.

Step-by-Step Reasoning

Initial kinetic energy:

Ball X: $m=0.240\ \text{kg}$ , $v=16\ \text{m s}^{-1}$ .

E_{k,i} = \frac{1}{2}(0.240)(16^2)

16^2 = 256 \Rightarrow E_{k,i} = 0.120 \times 256 = 30.72\ \text{J}

Ball Y is stationary initially, so its kinetic energy is $0$ .
Thus total initial $E_k = 30.72\ \text{J}$ .

Final kinetic energy:

Ball X is stationary after, so its kinetic energy is $0$ .
Ball Y: $m=0.480\ \text{kg}$ , $v=8.0\ \text{m s}^{-1}$ .

E_{k,f} = \frac{1}{2}(0.480)(8.0^2)

8.0^2 = 64 \Rightarrow E_{k,f} = 0.240 \times 64 = 15.36\ \text{J}

Change in kinetic energy:

\Delta E_k = 15.36 - 30.72 = -15.36\ \text{J}

The negative sign means the kinetic energy decreased by $15.36\ \text{J}$ (energy was converted to other forms). A suitable rounded answer is $-15.4\ \text{J}$ .

Key Takeaways

Always calculate total kinetic energy of the system (sum over all objects).
Use $\Delta E = E_f - E_i$ ; a negative value indicates a loss of kinetic energy.

Common Mistakes

Using momentum formula $mv$ instead of kinetic energy formula $\tfrac{1}{2}mv^2$ .
Forgetting that ball Y initially has zero kinetic energy.
Giving $+15.4\ \text{J}$ without stating it is a loss (sign matters because $\Delta$ is defined as final minus initial).

Things to Be Careful About

Use masses in $\text{kg}$ .
Keep enough significant figures through the working; round at the end.
The question explicitly asks for $\Delta E_k$ (not “energy lost”), so the negative sign is appropriate.

Techniques used

calculate kinetic energy before and after using $E_k = \frac{1}{2}mv^2$use the collision result to obtain final speedscompute a change in a quantity as final minus initial

(c)

The collision in (b) lasts for a time of $2.0\ \text{ms}$ . Assume that the contact force between the balls is constant during this time.

5M

(i)

Determine the magnitude and direction of the force exerted on ball X by ball Y during the collision.

magnitude = ______ $\text{N}$
direction ______

DifficultyMedium

Worked solution

Working

For ball X: $m = 0.240\ \text{kg}$ , $u = 16\ \text{m s}^{-1}$ (to the right), $v = 0$ .

\Delta p = m(v-u) = 0.240(0-16) = -3.84\ \text{kg m s}^{-1}

Collision time $\Delta t = 2.0\ \text{ms} = 2.0 \times 10^{-3}\ \text{s}$ .

F = \frac{\Delta p}{\Delta t} = \frac{-3.84}{2.0 \times 10^{-3}} = -1.92 \times 10^{3}\ \text{N}

Answer

Magnitude $= 1.9 \times 10^{3}\ \text{N}$

Direction: to the left (opposite to the initial motion of X).

Final answer

1.9 × 10^3 N to the left

Detailed explanation

Background Concept

Force is related to momentum by

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

The quantity $\Delta \vec{p}$ is the change in momentum (also called the impulse). If the force is constant during the interaction time, then

\vec{F} = \frac{m(\vec{v}-\vec{u})}{\Delta t}

Because momentum is a vector, the direction of the force is the direction of $\Delta \vec{p}$ .

Understanding the Question

The collision lasts $2.0\ \text{ms}$ and the contact force is assumed constant. Ball X (mass $0.240\ \text{kg}$ ) initially moves to the right at $16\ \text{m s}^{-1}$ and ends stationary. You must find the magnitude and direction of the force on X due to Y.

Approach

Choose right as the positive direction.
Find the change in momentum of ball X: $\Delta p = m(v-u)$ .
Divide by the collision time (converted from ms to s) to get the force.
The sign tells the direction.

Step-by-Step Reasoning

Set sign convention: right is positive.
Write initial and final velocities for X:

u_X = +16\ \text{m s}^{-1}, \quad v_X = 0

Compute change in momentum:

\Delta p_X = m_X(v_X-u_X) = 0.240(0-16) = -3.84\ \text{kg m s}^{-1}

The negative sign means X's momentum changed in the leftward direction.

Convert the time:

\Delta t = 2.0\ \text{ms} = 2.0\times 10^{-3}\ \text{s}

Compute the force:

F_X = \frac{\Delta p_X}{\Delta t} = \frac{-3.84}{2.0\times 10^{-3}} = -1.92\times 10^{3}\ \text{N}

So the magnitude is $1.92\times 10^{3}\ \text{N}$ (often rounded to $1.9\times 10^{3}\ \text{N}$ ) and the direction is to the left.

Key Takeaways

Impulse method: $F = \Delta p / \Delta t$ .
Always convert milliseconds to seconds.
Use the sign of $\Delta p$ (or $F$ ) to state direction.

Common Mistakes

Using $F = ma$ without having (or finding) the acceleration correctly.
Forgetting to convert $2.0\ \text{ms}$ to seconds.
Giving only a magnitude with no direction.
Using $\Delta p = mu - mv$ but then forgetting this changes the sign.

Things to Be Careful About

The force asked is “on ball X by ball Y”, so you must use ball X's change in momentum.
Significant figures: the time is given as $2.0\ \text{ms}$ (2 s.f.), so $F$ to 2 s.f. is reasonable.
Direction wording: “to the left” or “opposite to the motion of X”.

Techniques used

use impulse as change in momentumapply $F = \Delta p / \Delta t$ with a constant force assumptionuse a sign convention to determine force direction

(ii)

Compare the magnitude and direction of the force exerted on ball Y by ball X during the collision with the answers in (c)(i). No further calculations are required.

DifficultyMedium-Easy

Worked solution

Answer

By Newton's third law, the force exerted on ball Y by ball X has the same magnitude as in (c)(i) but is in the opposite direction.

So it is $1.9 \times 10^{3}\ \text{N}$ to the right.

Final answer

Same magnitude, opposite direction (1.9 × 10^3 N to the right on Y).

Detailed explanation

Background Concept

Newton's third law states that when two bodies interact, they exert forces on each other that are:

equal in magnitude,
opposite in direction,
of the same type,
acting on different bodies.

Symbolically, for bodies X and Y:

\vec{F}_{X\text{ on }Y} = -\vec{F}_{Y\text{ on }X}

Understanding the Question

In (c)(i) you found the force on ball X due to ball Y during contact. Now you must compare the force on ball Y due to ball X with that result, without doing new calculations.

Approach

Use Newton's third law for the contact interaction between the balls.

Step-by-Step Reasoning

The contact force is an interaction pair: “force on X by Y” and “force on Y by X”.
Newton's third law guarantees these two forces are equal in magnitude and opposite in direction at every instant (so also for their constant value in this question).
Since in (c)(i) the force on X by Y was to the left, the force on Y by X must be to the right, with the same magnitude.

Key Takeaways

Third-law pairs are equal and opposite and act on different bodies.
You do not need any numbers to compare them once you know one of the forces.

Common Mistakes

Saying the forces are equal and opposite but both acting on the same ball.
Confusing Newton's third law with equilibrium (equal and opposite forces on the same object do not generally occur in this collision).

Things to Be Careful About

State both the magnitude comparison and the direction comparison.
Make sure you label the forces correctly: “on Y by X” is the opposite of “on X by Y”.

Techniques used

apply Newton's third law to an interaction paircompare forces by stating equal magnitude and opposite direction

Q3MediumForces, Density and Pressure

(a)

State the principle of moments.

DifficultyEasy

Worked solution

Answer

For a body in equilibrium, the sum of clockwise moments about any point equals the sum of anticlockwise moments about that point (net moment is zero).

Final answer

For a body in equilibrium, the sum of clockwise moments about any point equals the sum of anticlockwise moments about that point (net moment is zero).

Detailed explanation

Background Concept

The moment of a force about a point (pivot) measures its turning effect:

\text{moment} = F \times d

where $F$ is the force and $d$ is the perpendicular distance from the pivot to the line of action of the force.

A body is in rotational equilibrium when its angular acceleration is zero. This happens when the resultant (net) moment about the pivot is zero.

Understanding the Question

You are asked to state the principle of moments. This is the standard condition used for a rigid body balanced on a pivot.

Approach

Give the equilibrium condition in words (or symbols): clockwise moments balance anticlockwise moments about the same point.

Step-by-Step Reasoning

In equilibrium, there is no tendency to rotate.
Therefore the total turning effect in one direction must equal the total turning effect in the opposite direction.
So, about any chosen pivot:

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

(Equivalently, $\sum M = 0$ with a sign convention.)

Key Takeaways

Moment is $F \times$ perpendicular distance.
Rotational equilibrium requires zero net moment.

Common Mistakes

Saying “forces are equal” instead of “moments are equal”.
Forgetting that the distance must be perpendicular.

Things to Be Careful About

The principle refers to moments about a point/pivot, not just any distances.
Use consistent clockwise/anticlockwise convention if writing it as $\sum M = 0$ .

Techniques used

state the condition for rotational equilibriumuse the definition of moment as force times perpendicular distance

(b)

A rigid uniform beam rests on a pivot at its centre, as shown in Fig. 3.1.

Fig. 3.1

A load of weight $2.6\ \text{N}$ is suspended from the beam at distance $x$ from the pivot.

A wooden cylinder of weight $4.0\ \text{N}$ is suspended from the beam at a distance of $0.40\ \text{m}$ from the pivot on the opposite side of the pivot to the load. The cylinder rests in a container of water. The lower part of the cylinder is immersed in the water to depth $h$ .

Initially, $h$ is equal to $0.10\ \text{m}$ and $x$ is equal to $0.40\ \text{m}$ . The system is in equilibrium.

5M

(i)

Use the principle of moments to show that the upthrust $U$ exerted by the water on the cylinder is $1.4\ \text{N}$ .

DifficultyMedium-Easy

Worked solution

Working

About the pivot (beam weight has no moment):

Clockwise moment = anticlockwise moment

2.6x = (4.0 - U)(0.40)

With $x = 0.40\ \text{m}$ :

2.6(0.40) = (4.0 - U)(0.40)

2.6 = 4.0 - U \Rightarrow U = 1.4\ \text{N}

Answer

$U = 1.4\ \text{N}$

Final answer

1.4 N

Detailed explanation

Background Concept

A rigid body on a pivot is in equilibrium when the net moment about the pivot is zero:

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

When an object is immersed in water, an upthrust (buoyant force) acts upward. If the object is supported by a string, the force transmitted to the support is the apparent weight:

\text{apparent weight} = W - U

This is the downward force that produces a turning effect on the beam at the point of suspension.

Understanding the Question

The beam is pivoted at its centre and is uniform, so the beam’s own weight acts through the pivot and produces no moment.

Given initially:

load weight $= 2.6\ \text{N}$ at distance $x = 0.40\ \text{m}$ on one side,
cylinder weight $= 4.0\ \text{N}$ at distance $0.40\ \text{m}$ on the other side,
cylinder experiences upthrust $U$ .

You must use moments to show $U = 1.4\ \text{N}$ .

Approach

Identify the force the cylinder applies to the beam: it is not $4.0\ \text{N}$ but $4.0 - U$ .
Take moments about the pivot (so the pivot reaction is eliminated).
Equate clockwise and anticlockwise moments and solve for $U$ .

Step-by-Step Reasoning

diagram-bi-fbd

The load produces a moment about the pivot of magnitude $2.6 \times x$ .
The cylinder side produces a moment from the tension in the string. For the cylinder:

T + U = W \Rightarrow T = W - U = 4.0 - U

This same tension acts downward on the beam at the suspension point, so the moment from that side is $(4.0 - U) \times 0.40$ .

Set moments equal (system in equilibrium):

2.6x = (4.0 - U)(0.40)

Substitute $x = 0.40\ \text{m}$ :

2.6(0.40) = (4.0 - U)(0.40)

Cancel $0.40$ and solve:

2.6 = 4.0 - U \Rightarrow U = 1.4\ \text{N}

Key Takeaways

Upthrust reduces the force transmitted to the support: $T = W - U$ .
For a uniform beam pivoted at its centre, the beam’s weight causes no moment about the pivot.

Common Mistakes

Using $4.0\ \text{N}$ as the force on the beam instead of $4.0 - U$ .
Forgetting to take moments about the pivot (and incorrectly including the pivot reaction).
Using the immersion depth $h$ in part (i) (it is not needed here).

Things to Be Careful About

Moments use perpendicular distances (here they are given directly along the beam).
Keep units consistent (distances in $\text{m}$ , moments in $\text{N m}$ ).

Techniques used

identify the effective downward force on the beam due to upthrusttake moments about the pivot and equate clockwise and anticlockwise momentssolve a linear equation for the unknown force

(ii)

The density of the water is $1.0 \times 10^3\ \text{kg m}^{-3}$ .

Calculate the area $A$ of the circular cross-section of the cylinder.

$A$ = ______ $\text{m}^2$

DifficultyMedium

Worked solution

Working

Upthrust equals weight of displaced water:

U = \rho gV = \rho gAh

So

A = \frac{U}{\rho g h}

A = \frac{1.4}{(1.0 \times 10^3)(9.81)(0.10)} = 1.43 \times 10^{-3}\ \text{m}^2

Answer

$A = 1.4 \times 10^{-3}\ \text{m}^2$

Final answer

1.4 × 10^-3 m^2

Detailed explanation

Background Concept

Archimedes’ principle states that the upthrust on a body immersed in a fluid equals the weight of the fluid displaced:

U = \rho g V

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

For a uniform cylinder with cross-sectional area $A$ immersed to a depth $h$ (and not fully submerged), the displaced volume is:

V = Ah

So:

U = \rho gAh

Understanding the Question

You are given:

upthrust from (b)(i): $U = 1.4\ \text{N}$ ,
water density: $\rho = 1.0 \times 10^3\ \text{kg m}^{-3}$ ,
immersed depth initially: $h = 0.10\ \text{m}$ .

You must find the cylinder’s cross-sectional area $A$ .

Approach

Use $U = \rho gAh$ , rearrange for $A$ , and substitute the values in SI units.

Step-by-Step Reasoning

Start with Archimedes’ principle in the useful form:

U = \rho gAh

Rearrange to isolate $A$ :

A = \frac{U}{\rho g h}

Substitute the given values:

A = \frac{1.4}{(1.0 \times 10^3)(9.81)(0.10)}

Compute the denominator:

(1.0 \times 10^3)(9.81)(0.10) = 981

So:

A = \frac{1.4}{981} = 1.43 \times 10^{-3}\ \text{m}^2

To appropriate significant figures:

A \approx 1.4 \times 10^{-3}\ \text{m}^2

Key Takeaways

Buoyant force: $U = \rho gV$ .
For a cylinder immersed to depth $h$ : $V = Ah$ .
Always keep SI units throughout.

Common Mistakes

Using $V = \pi r^2 h$ without realising $A = \pi r^2$ already represents cross-sectional area.
Using $h = 0.10$ as $10\ \text{cm}$ without converting to $0.10\ \text{m}$ .
Omitting $g$ (writing $U = \rho Ah$ ).

Things to Be Careful About

The formula $V = Ah$ assumes the cylinder is not fully submerged and has uniform cross-section.
Quote area in $\text{m}^2$ and use a sensible number of significant figures.

Techniques used

apply Archimedes' principle to relate upthrust to weight of displaced fluidmodel displaced volume as cross-sectional area times immersion depthrearrange an equation to solve for area and substitute values

(c)

More water is gradually added to the container in (b), so that depth $h$ in Fig. 3.1 gradually increases. The length $x$ is continuously adjusted so that the system remains in equilibrium.

On Fig. 3.2, sketch the variation of $x$ with $h$ . Use the space below for any working.

Fig. 3.2

DifficultyMedium

Worked solution

Working

For equilibrium about the pivot:

2.6x = (4.0 - U)(0.40)

and

U = \rho gAh \propto h

So $x$ decreases linearly with $h$ until $U = 4.0\ \text{N}$ , when $4.0-U=0$ and hence $x=0$ . For larger $h$ , the string is slack and the cylinder provides no moment, so $x$ remains $0$ .

Answer

Sketch: straight line from $(h,x)=(0.10,0.40)$ falling to $x=0$ at $h \approx 0.29\ \text{m}$ , then a horizontal line at $x=0$ up to $h=0.40\ \text{m}$ .

Final answer

A straight-line decrease of x with h to x = 0 at h ≈ 0.29 m, then x stays at 0.

Detailed explanation

Background Concept

The beam stays balanced when moments about the pivot balance:

2.6x = (4.0 - U)(0.40)

The key physics change as water level rises is that upthrust increases with displaced volume:

U = \rho gAh

So, while the cylinder is being immersed more deeply (and not yet floating freely), $U$ is directly proportional to $h$ , meaning the apparent weight $4.0-U$ decreases linearly with $h$ .

Understanding the Question

As more water is added, the immersed depth $h$ increases. You then adjust the load position $x$ so that the beam stays in equilibrium.

You must sketch how $x$ depends on $h$ on the given axes (from $h=0.10\ \text{m}$ to $0.40\ \text{m}$ ).

Approach

Write the general moments equation relating $x$ and $U$ .
Use $U = \rho gAh$ so that $U \propto h$ .
Conclude $x$ is a linear decreasing function of $h$ while the string is taut.
Identify the limit when the cylinder no longer pulls down on the beam: this happens when $U$ reaches $4.0\ \text{N}$ (apparent weight becomes zero), after which the cylinder cannot provide an upward push through a string, so the moment from that side is zero and $x$ must be zero.

Step-by-Step Reasoning

Start from the equilibrium condition about the pivot (true for all $h$ while the cylinder exerts a downward force on the beam):

2.6x = (4.0 - U)(0.40)

Rearrange for $x$ :

x = \frac{0.40}{2.6}(4.0 - U)

Use buoyancy:

U = \rho gAh

Therefore:

x = \frac{0.40}{2.6}\left(4.0 - \rho gAh\right)

This has the form $x = c - kh$ , so the graph is a straight line with negative gradient.

Now find where the straight line reaches $x=0$ (this is where the cylinder’s apparent weight becomes zero):

0 = \frac{0.40}{2.6}(4.0 - U) \Rightarrow U = 4.0\ \text{N}

Using proportionality from the initial condition ( $U=1.4\ \text{N}$ at $h=0.10\ \text{m}$ ):

\frac{U}{h} = \frac{1.4}{0.10} = 14\ \text{N m}^{-1}

So when $U=4.0\ \text{N}$ :

h = \frac{4.0}{14} \approx 0.29\ \text{m}

For $h$ greater than this, $U$ would exceed $4.0\ \text{N}$ , meaning the tension $T = 4.0-U$ would be negative. A string cannot provide a pushing force, so it goes slack and the cylinder provides no downward force on the beam. Hence the right-hand moment becomes zero, so the left-hand moment must also be zero, which requires:

2.6x = 0 \Rightarrow x = 0

So the sketch is:

a straight line from $(0.10, 0.40)$ down to $(0.29, 0)$ ,
then a horizontal line along $x=0$ until $h=0.40$ .

diagram-c-graph

Key Takeaways

As immersion depth increases, $U = \rho gAh$ increases linearly.
The cylinder’s apparent weight $4.0-U$ decreases linearly, so $x$ decreases linearly.
Physical constraints matter: when $U \ge 4.0\ \text{N}$ the string cannot push, so the cylinder stops contributing to moments.

Common Mistakes

Sketching $x$ increasing with $h$ (it must decrease because buoyancy reduces the turning effect of the cylinder side).
Extending the straight line into negative $x$ values (not meaningful here).
Not recognising the change after $U=4.0\ \text{N}$ (graph should flatten at $x=0$ ).

Things to Be Careful About

The given axis range up to $h=0.40\ \text{m}$ is larger than the $h$ value where $U$ reaches $4.0\ \text{N}$ , so you should show the ‘cut-off’ behaviour.
Your sketch should pass through the known initial point $(0.10,0.40).$

Techniques used

express upthrust as proportional to immersion depthsubstitute into a moments balance to obtain a linear relationshipidentify the physical limit where the tension becomes zero and sketch a piecewise graph

Q4MediumDeformation of SolidsForces, Density and Pressure

(a)

Define:

2M

(i)

stress

(ii)

strain.

(b)

Two wires X and Y, with equal unstretched lengths of $0.84\ \text{m}$ , are suspended from fixed points that are at the same horizontal level. The lower ends of the wires are attached to a beam of negligible mass. The beam is horizontal and in equilibrium, as shown in Fig. 4.1.

Fig. 4.1

Wire X is made from a metal that has a Young modulus of $1.9 \times 10^9\ \text{Pa}$ .
Wire Y is made from a different metal.

A load of weight $18\ \text{N}$ is suspended from the beam at a point that is equidistant from the two wires. This load causes both wires to extend by $0.47\ \text{mm}$ .

5M

(i)

Determine the cross-sectional area of wire X.

cross-sectional area = ______ $\text{m}^2$

DifficultyMedium

Worked solution

Working

Load is at the midpoint, so tensions are equal:

T_X = T_Y = \frac{18}{2} = 9.0\ \text{N}

Young modulus:

E = \frac{\text{stress}}{\text{strain}} = \frac{(F/A)}{(\Delta L/L)} = \frac{FL}{A\Delta L}

So

A = \frac{FL}{E\Delta L}

With $F = 9.0\ \text{N}$ , $L = 0.84\ \text{m}$ , $E = 1.9 \times 10^9\ \text{Pa}$ , $\Delta L = 0.47\ \text{mm} = 4.7 \times 10^{-4}\ \text{m}$ :

A = \frac{9.0 \times 0.84}{(1.9 \times 10^9)(4.7 \times 10^{-4})} = 8.5 \times 10^{-6}\ \text{m}^2

Answer

$8.5 \times 10^{-6}\ \text{m}^2$

Final answer

8.5 × 10^-6 m^2

Detailed explanation

Background Concept

For a wire under tension:

stress $\sigma = F/A$
strain $\varepsilon = \Delta L/L$
Young modulus:

E = \frac{\sigma}{\varepsilon}

Combining these gives a very useful working form:

E = \frac{(F/A)}{(\Delta L/L)} = \frac{FL}{A\Delta L}

So if you know $E$ , $F$ , $L$ and $\Delta L$ , you can find the cross-sectional area $A$ .

In addition, for a beam in equilibrium, the net force is zero and the net moment about any point is zero. Here the load is placed halfway between the two supporting wires, so the support forces must be equal by symmetry.

Understanding the Question

Two equal-length wires support a light horizontal beam. A load of weight $18\ \text{N}$ is hung at the midpoint between the wires. Both wires extend by the same amount $0.47\ \text{mm}$ .

You are told wire X has Young modulus $E = 1.9 \times 10^9\ \text{Pa}$ . You must determine the cross-sectional area $A$ of wire X.

Key ideas:

Because the load is in the middle, each wire takes half the load.
Use the Young modulus equation to relate force and extension to area.

Approach

Use equilibrium/symmetry to find the tension in wire X.
Convert the extension from mm to m.
Rearrange $E = FL/(A\Delta L)$ to solve for $A$ .
Substitute values carefully with SI units.

Step-by-Step Reasoning

diagram-b-i-1

Find the force in wire X.
The beam is symmetrical and the $18\ \text{N}$ load acts at the midpoint. For rotational equilibrium about the midpoint, the two upward tensions must be equal. Also, for vertical force equilibrium:

T_X + T_Y = 18\ \text{N}

With $T_X = T_Y$ :

T_X = 9.0\ \text{N}

Convert the extension.

\Delta L = 0.47\ \text{mm} = 0.47 \times 10^{-3}\ \text{m} = 4.7 \times 10^{-4}\ \text{m}

Use Young modulus.
Start from

E = \frac{FL}{A\Delta L}

Rearrange for $A$ :

A = \frac{FL}{E\Delta L}

Substitute.

A = \frac{(9.0)(0.84)}{(1.9 \times 10^9)(4.7 \times 10^{-4})}

Compute numerator: $9.0 \times 0.84 = 7.56$ .

Compute denominator: $(1.9 \times 10^9)(4.7 \times 10^{-4}) = 8.93 \times 10^5$ .

So

A = \frac{7.56}{8.93 \times 10^5} = 8.47 \times 10^{-6}\ \text{m}^2 \approx 8.5 \times 10^{-6}\ \text{m}^2

Key Takeaways

With a central load on a symmetric support, each support force is half the load.
For stretching:

E = \frac{FL}{A\Delta L}

Always convert mm to m before substituting into SI equations.

Common Mistakes

Using $F = 18\ \text{N}$ for a single wire instead of $9\ \text{N}$ .
Forgetting to convert $0.47\ \text{mm}$ to metres.
Rearranging wrongly (e.g. putting $A$ in the numerator).
Giving area in $\text{mm}^2$ or missing the unit.

Things to Be Careful About

The equality of tensions comes from symmetry and equilibrium (load exactly midway).
Young modulus is in pascals ( $\text{Pa} = \text{N m}^{-2}$ ), so lengths must be in metres and area in $\text{m}^2$ .
Quote the final answer to a sensible number of significant figures (typically 2–3).

Techniques used

use symmetry to determine the tension in each wireapply equilibrium of forces on the beamuse the Young modulus relation to link force, extension and arearearrange an equation to solve for cross-sectional areaconvert millimetres to metres

(ii)

Wire Y has a greater diameter than wire X.

Explain, without calculation, whether the Young modulus of the metal from which wire Y is made is less than, the same as or greater than $1.9 \times 10^9\ \text{Pa}$ .

DifficultyMedium-Easy

Worked solution

Answer

The load is at the midpoint so wire Y has the same tension as wire X. Both wires have the same extension and original length, so they have the same strain.

Wire Y has a larger area, so for the same force its stress is smaller. Since

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

a smaller stress with the same strain means $E$ for wire Y is less than $1.9 \times 10^9\ \text{Pa}$ .

Final answer

Less than 1.9 × 10^9 Pa

Detailed explanation

Background Concept

Young modulus is defined as

E = \frac{\text{stress}}{\text{strain}} = \frac{\sigma}{\varepsilon}.

For a stretched wire:

stress $\sigma = F/A$ depends on the force and the cross-sectional area.
strain $\varepsilon = \Delta L/L$ depends on how much the wire extends compared with its original length.

So if two wires have the same strain but different stress, the one with the smaller stress must have the smaller Young modulus.

Understanding the Question

Both wires extend by the same amount $0.47\ \text{mm}$ and have the same original length $0.84\ \text{m}$ , so they undergo the same strain.

The load is placed midway between them, so each wire supports the same force.

Wire Y has a greater diameter, so it has a larger cross-sectional area than wire X.

You must decide whether $E_Y$ is less than, equal to, or greater than $1.9 \times 10^9\ \text{Pa}$ , and justify it without doing any arithmetic.

Approach

Argue that the tensions (forces) in X and Y are equal.
Use diameter information to compare areas.
Use $\sigma = F/A$ to compare stresses.
Use $E = \sigma/\varepsilon$ with equal strain to compare Young moduli.

Step-by-Step Reasoning

Equal force: the beam is symmetric with the load at the midpoint, so the upward forces from the two wires must be equal for rotational equilibrium. Hence $F_Y = F_X$ .
Equal strain: both have the same extension $\Delta L$ and the same original length $L$ , so

\varepsilon_Y = \frac{\Delta L}{L} = \varepsilon_X.

Compare stresses: wire Y has larger diameter $\Rightarrow$ larger area $A_Y > A_X$ . With the same force,

\sigma_Y = \frac{F}{A_Y} < \frac{F}{A_X} = \sigma_X.

Compare Young moduli: since $E = \sigma/\varepsilon$ and $\varepsilon_Y = \varepsilon_X$ but $\sigma_Y < \sigma_X$ , it follows that

E_Y < E_X = 1.9 \times 10^9\ \text{Pa}.

Key Takeaways

Same $\Delta L$ and same $L$ means same strain.
Larger diameter means larger $A$ and therefore smaller stress for the same force.
With equal strain, smaller stress implies smaller Young modulus.

Common Mistakes

Thinking “thicker wire means larger Young modulus” (Young modulus is a material property, independent of dimensions).
Assuming the forces in the two wires are different even though the load is at the midpoint.
Mixing up stress and strain (or stating they are both smaller).

Things to Be Careful About

The conclusion depends on same force and same strain. Here, both are true because the load is central and both extensions are stated equal.
Young modulus compares materials; diameter changes stress for a given force but does not directly change $E$ unless you infer it from the observed extension.

Techniques used

use the definition of Young modulus as stress divided by straincompare stresses using force and cross-sectional areause proportional reasoning without numerical calculation

Q5MediumSuperpositionWaves

(a)

A stationary wave is formed on a string XY that has a length of $0.48\ \text{m}$ . Fig. 5.1 shows the string at one instant in time.

Fig. 5.1

The speed of the wave on the string is $1400\ \text{m s}^{-1}$ .

4M

(i)

On Fig. 5.1, draw a cross ( $\times$ ) at one position that is a node and another cross at one position that is an antinode. Label the node N and the antinode A.

DifficultyEasy

Worked solution

Answer

Node $N$ : at a point of zero displacement (e.g. at end $X$ or $Y$ ).

Antinode $A$ : at the centre of any loop (maximum displacement).

Final answer

Node at a zero-displacement point (e.g. X or Y); antinode at centre of a loop.

Detailed explanation

Background Concept

In a stationary wave on a string, two waves of the same frequency and amplitude travel in opposite directions and superpose.

A node is a point that is always at zero displacement (no oscillation).
An antinode is a point that oscillates with maximum displacement.

For a string fixed at both ends, the ends must be nodes.

Understanding the Question

You are shown the shape of a stationary wave on a string at one instant. You must mark one node and one antinode on the diagram and label them $N$ and $A$ .

Approach

Use the defining features:

Nodes occur where the string crosses its equilibrium position and stays there at all times (in the diagram these are the points separating adjacent loops; the fixed ends are also nodes).
Antinodes occur at the centres of loops, where the displacement is largest.

Step-by-Step Reasoning

Look for points that must always be stationary. Since the string is between $X$ and $Y$ and fixed, $X$ and $Y$ are nodes.
Choose any loop (any “bulge”). The midpoint of that loop is an antinode because it is furthest from the equilibrium position.
Place a cross at one node (e.g. at $X$ ) and label it $N$ .
Place a cross at the centre of any loop and label it $A$ .

Key Takeaways

Nodes: zero displacement, including fixed ends.
Antinodes: maximum displacement at the centre of each loop.

Common Mistakes

Marking a point on the string that is simply at zero displacement at that instant (a node is zero at all times, not just in the snapshot).
Labelling the end of a loop as an antinode (loop ends are nodes).

Things to Be Careful About

Any correct node/antinode pair earns the mark; you do not need to choose a specific one.
Ensure labels $N$ and $A$ are next to the correct crosses.

Techniques used

identify nodes as points of zero displacementidentify antinodes as points of maximum displacement

(ii)

Show that the wavelength of the wave produced is $0.32\ \text{m}$ . Explain your reasoning.

DifficultyMedium-Easy

Worked solution

Working

There are $3$ loops on the string, and each loop corresponds to $\frac{\lambda}{2}$ .

So

0.48 = 3\left(\frac{\lambda}{2}\right)

\lambda = \frac{2 \times 0.48}{3} = 0.32\ \text{m}

Answer

$\lambda = 0.32\ \text{m}$

Final answer

0.32 m

Detailed explanation

Background Concept

For a stationary wave on a string fixed at both ends:

Adjacent nodes are separated by $\frac{\lambda}{2}$ .
Each “loop” (one bulge between two neighbouring nodes) has length $\frac{\lambda}{2}$ .

So if you can count the number of loops along the string, you can relate the total length $L$ to $\lambda$ .

Understanding the Question

The string length is $L = 0.48\ \text{m}$ . From the diagram, you can count how many loops (segments between nodes) fit into this length. You must use that to show the wavelength is $0.32\ \text{m}$ .

Approach

Count the number of loops between $X$ and $Y$ .
Use:

L = (\text{number of loops}) \times \frac{\lambda}{2}

Rearrange to find $\lambda$ .

Step-by-Step Reasoning

From the snapshot in Fig. 5.1, the standing wave has three loops between $X$ and $Y$ .
Each loop is between two nodes, so its length is $\frac{\lambda}{2}$ .
Therefore the total length is

0.48 = 3 \times \frac{\lambda}{2}

Rearranging:

\lambda = \frac{2 \times 0.48}{3} = 0.32\ \text{m}

This matches the value you were asked to show.

Key Takeaways

In a standing wave on a string, one loop corresponds to $\lambda/2$ .
Total length = (number of loops) $\times \lambda/2$ .

Common Mistakes

Using $L = n\lambda$ instead of $L = n(\lambda/2)$ .
Counting “waves” incorrectly: the diagram shows loops, not full wavelengths.

Things to Be Careful About

Make sure you count loops between nodes, not peaks.
Keep units consistent; $L$ is already in metres so $\lambda$ comes out in metres.

Techniques used

relate string length to number of half-wavelengthsuse stationary-wave loop counting to infer wavelength

(iii)

Calculate the frequency of the wave.

frequency = ______ $\text{Hz}$

(b)

A source of sound waves of frequency $780\ \text{Hz}$ is on a rotating platform. The speed of the source is $39\ \text{m s}^{-1}$ .

The sound is detected by an observer that is a large distance from the rotating platform, as shown in Fig. 5.2.

Fig. 5.2

4M

(i)

The speed of sound in air is $320\ \text{m s}^{-1}$ .

Calculate the maximum frequency of the sound detected by the observer.

maximum frequency = ______ $\text{Hz}$

DifficultyMedium-Easy

Worked solution

Working

Maximum detected frequency occurs when the source moves directly towards the observer.

f_{\max} = f\frac{v}{v-v_s}

f_{\max} = 780\times\frac{320}{320-39} = 8.88\times 10^{2}\ \text{Hz}

Answer

$8.88\times 10^{2}\ \text{Hz}$

Final answer

8.88 × 10^2 Hz

Detailed explanation

Background Concept

The Doppler effect is the change in observed frequency due to relative motion between source and observer.

For a moving source and a stationary observer (sound speed $v$ ):

When the source moves towards the observer,

f' = f\frac{v}{v-v_s}

When the source moves away from the observer,

f' = f\frac{v}{v+v_s}

where $v_s$ is the component of the source velocity along the line joining source to observer.

Understanding the Question

Given:

emitted frequency $f = 780\ \text{Hz}$
speed of sound $v = 320\ \text{m s}^{-1}$
source speed $39\ \text{m s}^{-1}$ on a rotating platform

You need the maximum frequency detected by a distant observer. “Maximum” occurs when the velocity component towards the observer is greatest (i.e. the full speed is directed towards the observer).

Approach

Use the Doppler formula for a moving source towards the observer, taking $v_s = 39\ \text{m s}^{-1}$ for the maximum case.

Step-by-Step Reasoning

Maximum frequency corresponds to maximum approach speed along the line of sight, so take $v_s = 39\ \text{m s}^{-1}$ .
Use:

f_{\max} = f\frac{v}{v-v_s}

Substitute values:

f_{\max} = 780\times\frac{320}{320-39}

Evaluate the denominator: $320-39=281$ .
Compute:

\frac{320}{281} \approx 1.139

so

f_{\max} \approx 780\times 1.139 \approx 8.88\times 10^{2}\ \text{Hz}

Key Takeaways

For a moving source, approaching motion increases observed frequency.
Use $v-v_s$ in the denominator for approach; $v+v_s$ for recession.

Common Mistakes

Using the wrong sign (using $v+v_s$ for approach).
Using $v_s = 0$ (forgetting you need the component along the observer line).

Things to Be Careful About

$v_s$ is the component towards the observer; “maximum” means that component equals the given source speed.
Keep $v$ and $v_s$ in the same units ( $\text{m s}^{-1}$ ).

Techniques used

apply the Doppler shift formula for a moving source and stationary observeridentify the maximum shift when the source moves directly towards the observer

(ii)

At time $t = 0$ , the observer detects the sound emitted by the source when it was in the position shown in Fig. 5.2.

On Fig. 5.3, sketch the variation with $t$ of the frequency $f$ of the sound detected by the observer for one complete rotation of the platform. Calculations are not required.

Fig. 5.3

DifficultyMedium

Worked solution

graph-b-ii

Answer

Sketch a smooth periodic curve about $780\ \text{Hz}$ :

$f$ is maximum at $t=0$ ,
decreases to $780\ \text{Hz}$ (when motion is perpendicular to the line to the observer),
reaches a minimum at half a rotation,
returns to the maximum after one complete rotation.

Final answer

See working (periodic f–t sketch: max at t=0, min at half rotation, repeats).

Detailed explanation

Background Concept

For Doppler shift with a stationary observer, the detected frequency depends on the component of the source velocity along the line of sight.

If $v_s(t)$ is the velocity component towards the observer at time $t$ , then for approach:

f(t) = f\frac{v}{v-v_s(t)}

When $v_s(t)=0$ , there is no Doppler shift so $f(t)=f$ .

In uniform circular motion, the line-of-sight component of velocity varies sinusoidally with time over each rotation.

Understanding the Question

At $t=0$ , the observer detects sound that was emitted when the source was in the position shown. You must sketch how the detected frequency varies over one full rotation. No numbers are required, but the shape and key points (max/min/mean and where they occur) must be correct.

Approach

Decide when the source is moving towards, away from, or perpendicular to the observer line.
Use that to mark where the detected frequency is above $780\ \text{Hz}$ , below it, or equal to it.
Because the source repeats its motion every rotation, the frequency variation must be periodic.

Step-by-Step Reasoning

When the source moves most directly towards the observer, $v_s$ is maximum and the detected frequency is maximum (as in part (b)(i)).
When the source moves perpendicular to the observer direction, $v_s=0$ so there is no Doppler shift and the detected frequency equals the emitted frequency $780\ \text{Hz}$ .
When the source moves most directly away from the observer, $v_s$ is negative (receding), giving the minimum detected frequency.
Over one full rotation, the towards/away component changes smoothly, so the detected frequency changes smoothly, producing a sinusoidal-like curve about $780\ \text{Hz}$ .
Since the question states that at $t=0$ the observer detects sound from the shown position, you start your sketch at the frequency corresponding to that position. For the shown position (where the source is moving towards the observer most strongly), the sketch starts at a maximum.

graph-b-ii

Key Takeaways

Doppler shift depends on the line-of-sight velocity component.
Circular motion gives a periodic (approximately sinusoidal) frequency variation.
Max at strongest approach; min at strongest recession; unchanged at perpendicular motion.

Common Mistakes

Drawing a triangular or step-like graph (the velocity component changes smoothly, so $f$ should also change smoothly).
Starting at the wrong phase (e.g. starting at the mean when the diagram implies a maximum or minimum).
Making the curve not repeat after one rotation.

Things to Be Careful About

The emitted frequency $780\ \text{Hz}$ is the central value: your curve should oscillate above and below it.
Label axes correctly: $f$ on the vertical axis and $t$ on the horizontal axis, with one full period corresponding to one rotation.

Techniques used

relate observed frequency to the source velocity component along the line of sightsketch a periodic variation over one rotationidentify times of maximum, minimum, and zero Doppler shift

Q6MediumElectricityD.C. Circuits

(a)

Define resistance.

(b)

A cylindrical metal wire of length $2.4\ \text{m}$ and cross-sectional area $8.0 \times 10^{-6}\ \text{m}^2$ has a resistance of $0.33\ \Omega$ . There is a current in the wire of $4.7\ \text{A}$ .

6M

(i)

Determine the resistivity of the metal from which the wire is made.

resistivity = ______ $\Omega\text{m}$

DifficultyMedium-Easy

Worked solution

Working

For a uniform wire,

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

\rho = \frac{0.33 \times 8.0 \times 10^{-6}}{2.4} = 1.1 \times 10^{-6}\ \Omega\,\text{m}

Answer

$1.1 \times 10^{-6}\ \Omega\,\text{m}$

Final answer

1.1 × 10^-6 Ω m

Detailed explanation

Background Concept

The resistance of a uniform cylindrical conductor depends on:

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

They are related by

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

where $R$ is in $\Omega$ , $L$ in $\text{m}$ , $A$ in $\text{m}^2$ , so $\rho$ is in $\Omega\,\text{m}$ .

Understanding the Question

You are given $L = 2.4\ \text{m}$ , $A = 8.0 \times 10^{-6}\ \text{m}^2$ , and $R = 0.33\ \Omega$ for a cylindrical wire. You must determine the resistivity $\rho$ of the metal.

Approach

Use $R = \rho L/A$ , rearrange to $\rho = RA/L$ , then substitute the given values carefully.

Step-by-Step Reasoning

Start with

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

Rearrange:

\rho = \frac{RA}{L}

Substitute:

\rho = \frac{(0.33\ \Omega)(8.0 \times 10^{-6}\ \text{m}^2)}{2.4\ \text{m}}

Calculate:

0.33 \times 8.0 \times 10^{-6} = 2.64 \times 10^{-6}

\rho = \frac{2.64 \times 10^{-6}}{2.4} = 1.10 \times 10^{-6}\ \Omega\,\text{m}

Rounded appropriately:

\rho \approx 1.1 \times 10^{-6}\ \Omega\,\text{m}

Key Takeaways

Use $R = \rho L/A$ for a uniform wire.
Resistivity has unit $\Omega\,\text{m}$ .

Common Mistakes

Using $\rho = RL/A$ (wrong rearrangement).
Forgetting the $10^{-6}$ in the area.
Giving units as $\Omega/\text{m}$ instead of $\Omega\,\text{m}$ .

Things to Be Careful About

Ensure $L$ is in metres and $A$ is in $\text{m}^2$ .
Quote the final answer to 2 s.f. to match the data given.

Techniques used

apply the resistivity relation for a uniform wirerearrange an equation to make the required quantity the subjectsubstitute values with correct powers of ten and units

(ii)

Calculate the charge that passes through the wire in a time of $5.0\ \text{minutes}$ .

charge = ______ $\text{C}$

(iii)

The free electrons (charge carriers) in the wire have an average drift speed of $0.16\ \text{mm s}^{-1}$ .

Determine the number density of charge carriers in the metal.

number density = ______ $\text{m}^{-3}$

DifficultyMedium

Worked solution

Working

v = 0.16\ \text{mm s}^{-1} = 1.6 \times 10^{-4}\ \text{m s}^{-1}

I = Anvq \Rightarrow n = \frac{I}{Avq}

n = \frac{4.7}{(8.0 \times 10^{-6})(1.6 \times 10^{-4})(1.6 \times 10^{-19})} = 2.3 \times 10^{28}\ \text{m}^{-3}

Answer

$2.3 \times 10^{28}\ \text{m}^{-3}$

Final answer

2.3 × 10^28 m^-3

Detailed explanation

Background Concept

In a metal, many free electrons move randomly, but when a potential difference is applied they acquire a small drift velocity $v$ along the wire. The current is related to the drift motion by

I = Anvq

where:

$I$ is current (A)
$A$ is cross-sectional area ( $\text{m}^2$ )
$n$ is number density of charge carriers ( $\text{m}^{-3}$ )
$v$ is drift speed ( $\text{m s}^{-1}$ )
$q$ is charge per carrier (for electrons, $q = e = 1.60 \times 10^{-19}\ \text{C}$ )

Understanding the Question

You are given $I = 4.7\ \text{A}$ , $A = 8.0 \times 10^{-6}\ \text{m}^2$ , and drift speed $v = 0.16\ \text{mm s}^{-1}$ . You must determine $n$ , the number of charge carriers per unit volume.

The main hurdles are:

converting $0.16\ \text{mm s}^{-1}$ to $\text{m s}^{-1}$ ,
using the correct value of the electron charge.

Approach

Convert $v$ into SI units.
Use $I = Anvq$ and rearrange to $n = I/(Avq)$ .
Substitute values with powers of ten handled carefully.

Step-by-Step Reasoning

Convert drift speed:

0.16\ \text{mm s}^{-1} = 0.16 \times 10^{-3}\ \text{m s}^{-1} = 1.6 \times 10^{-4}\ \text{m s}^{-1}

Rearrange the drift current equation:

I = Anvq \Rightarrow n = \frac{I}{Avq}

Substitute ( $q = 1.6 \times 10^{-19}\ \text{C}$ ):

n = \frac{4.7}{(8.0 \times 10^{-6})(1.6 \times 10^{-4})(1.6 \times 10^{-19})}

Combine the denominator stepwise:

(8.0 \times 10^{-6})(1.6 \times 10^{-4}) = 12.8 \times 10^{-10} = 1.28 \times 10^{-9}

(1.28 \times 10^{-9})(1.6 \times 10^{-19}) = 2.048 \times 10^{-28}

So

n = \frac{4.7}{2.048 \times 10^{-28}} \approx 2.29 \times 10^{28}\ \text{m}^{-3}

Rounded suitably:

n \approx 2.3 \times 10^{28}\ \text{m}^{-3}

Key Takeaways

Drift current relationship: $I = Anvq$ .
Always convert drift speeds to $\text{m s}^{-1}$ .
Metals typically have very large $n$ values (order $10^{28}\ \text{m}^{-3}$ ), which is a useful check.

Common Mistakes

Using $v = 0.16\ \text{m s}^{-1}$ (missing the mm to m conversion) giving an answer $10^3$ too small.
Forgetting $q = 1.60 \times 10^{-19}\ \text{C}$ .
Using diameter instead of cross-sectional area.

Things to Be Careful About

Powers of ten: multiplying three small numbers in the denominator easily leads to exponent slips.
Units: $n$ must be in $\text{m}^{-3}$ .
Use $q$ as the magnitude of charge; the sign of the electron is not needed for current magnitude.

Techniques used

convert drift speed to SI unitsapply the charge-carrier current equation I = Anvqrearrange to find number density

(c)

The wire in (b) may be considered to be a fixed resistor. It is connected in series with a thermistor to a battery that has negligible internal resistance.

3M

(i)

Use circuit symbols to complete Fig. 6.1 to show the circuit diagram of this arrangement.

Fig. 6.1

(ii)

Explain, without calculation, how the power dissipated in the wire changes as the temperature of the thermistor is increased.

DifficultyMedium-Easy

Worked solution

Answer

As temperature increases, the thermistor resistance decreases, so the total series resistance decreases and the current increases.

Power in the wire is

P = I^2R

and $R$ for the wire is constant, so the power dissipated in the wire increases.

Final answer

Power in the wire increases as the thermistor temperature increases.

Detailed explanation

Background Concept

For a fixed resistor, the power dissipated can be written as

P = I^2R

In a series circuit, the same current $I$ flows through all components, and the current is determined by the total resistance:

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

A typical thermistor used at this level is an NTC thermistor (negative temperature coefficient), meaning its resistance decreases as its temperature increases.

Understanding the Question

You have a battery (negligible internal resistance) connected in series with:

the metal wire (fixed resistor with constant resistance), and
a thermistor.

You are asked without calculation what happens to the power dissipated in the wire when the thermistor’s temperature increases.

Approach

Link the cause-and-effect chain:

Increase temperature $\Rightarrow$ thermistor resistance changes (NTC: decreases).
In series, decreasing total resistance increases the current.
Wire power depends on $I^2$ (since wire $R$ is fixed), so power increases.

Step-by-Step Reasoning

As the thermistor temperature increases, its resistance $R_T$ decreases (NTC behavior).
Total series resistance is

R_{\text{total}} = R_{\text{wire}} + R_T

So $R_{\text{total}}$ decreases.

With a fixed supply voltage $V$ , the series current increases because

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

The wire has constant resistance $R_{\text{wire}}$ , so its power is

P_{\text{wire}} = I^2 R_{\text{wire}}

Since $I$ increases and $R_{\text{wire}}$ is constant, $P_{\text{wire}}$ increases.

Key Takeaways

NTC thermistor: higher temperature $\Rightarrow$ lower resistance.
Series circuit: lower total resistance $\Rightarrow$ higher current.
Fixed resistor power increases strongly with current because $P \propto I^2$ .

Common Mistakes

Saying power decreases because “resistance decreases” (that would only apply if the component whose resistance changes is the one you are finding power for using $P = V^2/R$ at constant $V$ across that component).
Treating the wire’s resistance as changing with thermistor temperature (the question says the wire is a fixed resistor).

Things to Be Careful About

Specify that it is the thermistor resistance that changes with temperature.
Use the correct power expression for the wire in a series circuit: $P_{\text{wire}} = I^2R_{\text{wire}}$ is the clearest qualitative route because the wire’s $R$ is constant.

Techniques used

use the qualitative temperature dependence of an NTC thermistordeduce how current changes in a series circuit when resistance changesrelate power in a fixed resistor to current using P = I^2R

Q7Medium-EasyParticle Physics

(a)

Complete Table 7.1 to show the charges, in terms of the elementary charge $e$ , on each of the flavours of quark and antiquark shown.

Table 7.1

flavour	charge / $e$
	quark	antiquark
up
down
strange

DifficultyEasy

Worked solution

Answer

flavour	quark charge / $e$	antiquark charge / $e$
up	$+\dfrac{2}{3}$	$-\dfrac{2}{3}$
down	$-\dfrac{1}{3}$	$+\dfrac{1}{3}$
strange	$-\dfrac{1}{3}$	$+\dfrac{1}{3}$

Final answer

up: +2/3, −2/3; down: −1/3, +1/3; strange: −1/3, +1/3 (in units of e)

Detailed explanation

Background Concept

Quarks are fundamental particles that carry fractional electric charge in units of the elementary charge $e$ . The common light quark flavours used at AS level are:

up ( $u$ ) with charge $+\dfrac{2}{3}e$
down ( $d$ ) with charge $-\dfrac{1}{3}e$
strange ( $s$ ) with charge $-\dfrac{1}{3}e$

For every quark there is a corresponding antiquark (e.g. $\bar{u}$ , $\bar{d}$ , $\bar{s}$ ). An antiquark has the same mass as the quark but opposite charge.

Understanding the Question

You are given a table with the flavours up, down and strange, and you must fill in the charge of each quark and each corresponding antiquark, expressed as a multiple of $e$ .

Approach

Recall the standard quark charges for $u$ , $d$ , and $s$ .
For each antiquark, change the sign of the quark charge (same magnitude, opposite sign).

Step-by-Step Reasoning

Up quark has charge $+\dfrac{2}{3}e$ , so the up antiquark has charge $-\dfrac{2}{3}e$ .
Down quark has charge $-\dfrac{1}{3}e$ , so the down antiquark has charge $+\dfrac{1}{3}e$ .
Strange quark has charge $-\dfrac{1}{3}e$ , so the strange antiquark has charge $+\dfrac{1}{3}e$ .

These are entered in the table in the form “charge / $e$ ”.

Key Takeaways

$u$ has charge $+2/3$ (in units of $e$ ); $d$ and $s$ have charge $-1/3$ .
Antiquarks have opposite charge to the corresponding quarks.

Common Mistakes

Swapping the up and down charges.
Forgetting to change the sign for the antiquark.
Writing charges as $\pm 1/3\,e$ in the “charge/ $e$ ” column (the $e$ should not be included if the heading is already “/ $e$ ”).

Things to Be Careful About

Keep the fraction format clear: $\dfrac{2}{3}$ and $\dfrac{1}{3}$ .
Ensure the sign is correct for each antiquark (exactly opposite to the quark).

Techniques used

recall quark charges in units of the elementary chargeuse the rule that antiquarks have equal magnitude and opposite sign of charge

(b)

3M

(i)

State the name of the class (group) of fundamental particles to which baryons and mesons belong.

(ii)

Compare baryons and mesons in terms of their constituent particles.

(c)

Describe $\beta^+$ decay in terms of the fundamental particles involved.