Physics 9702 / 21 — May/June 2025

(b)

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

Fig. 1.1

Object A is released from rest at a height of $14\text{ m}$ above the horizontal ground.
Object B is released with an initial upwards vertical velocity $u$ at a height of $3.6\text{ m}$ above the ground.
Both objects take the same time to reach the ground and they do not collide with each other.
Air resistance is negligible.

4M

(i)

Calculate the time taken for object A to reach the ground.

time = ______ $\text{s}$

DifficultyMedium-Easy

Worked solution

Working

For object A: $u = 0$ , $s = 14\,\text{m}$ , $a = g$ .

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

14 = 0 + \frac{1}{2}gt^2

t = \sqrt{\frac{2 \times 14}{g}} = \sqrt{\frac{28}{9.81}} = 1.69\,\text{s}

Answer

$1.69\,\text{s}$

Final answer

1.69 s

Detailed explanation

Background Concept

For motion with constant acceleration (here, free fall with negligible air resistance), the kinematics equations apply. A useful one is

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

where:

$s$ is displacement,
$u$ is initial velocity,
$a$ is acceleration,
$t$ is time.

In free fall near Earth, the acceleration magnitude is $g \approx 9.81\,\text{m s}^{-2}$ downward.

Understanding the Question

Object A is released from rest at a height of $14\,\text{m}$ above the ground. You must find the time to reach the ground. Since it is released from rest and air resistance is negligible, it undergoes constant downward acceleration $g$ .

Given:

$u = 0$ ,
distance fallen $= 14\,\text{m}$ ,
$a = g$ .

Unknown: $t$ .

Approach

Take downward as positive (or upward as positive, but be consistent). Then use

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

and solve for $t$ .

Step-by-Step Reasoning

Choose downward positive.

Displacement to the ground: $s = +14\,\text{m}$ .
Initial velocity: $u = 0$ .
Acceleration: $a = +g$ .

Substitute into the equation:

14 = 0 \cdot t + \frac{1}{2}gt^2

Rearrange:

t^2 = \frac{28}{g}

So

t = \sqrt{\frac{28}{9.81}} = 1.69\,\text{s}.

(We take the positive root because time is positive.)

Key Takeaways

Free fall with negligible air resistance means constant acceleration $g$ .
If released from rest, $u = 0$ simplifies the equation greatly.

Common Mistakes

Using $s = vt$ (only valid for constant velocity, not accelerating motion).
Sign errors from mixing “upward positive” with “ $a = +g$ ”.
Forgetting to square root at the end or giving a negative time.

Things to Be Careful About

Use a consistent sign convention.
Use $g$ in $\text{m s}^{-2}$ and height in $\text{m}$ .
Quote the time to a sensible number of significant figures (typically 2 or 3).

Techniques used

select an appropriate constant-acceleration equationsubstitute given values for free-fall motionrearrange to solve a quadratic for time

(ii)

Use your answer in (b)(i) to calculate $u$ .

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

DifficultyMedium-Easy

Worked solution

Working

Take upward as positive. For object B: $s = -3.6\,\text{m}$ , $a = -g$ , time $t = 1.69\,\text{s}$ .

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

-3.6 = ut - \frac{1}{2}gt^2

ut = -3.6 + \frac{1}{2}gt^2

u = \frac{-3.6 + \frac{1}{2}g(1.69)^2}{1.69} = 6.16\,\text{m s}^{-1}

Answer

$u = 6.16\,\text{m s}^{-1}$

Final answer

6.16 m s^-1

Detailed explanation

Background Concept

For vertical motion under gravity with negligible air resistance, acceleration is constant ( $g$ downward). The displacement–time relation is

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

The key skill is choosing a sign convention and then keeping it consistent for $s$ , $u$ and $a$ .

Understanding the Question

Object B starts at height $3.6\,\text{m}$ above the ground, but is thrown upwards with initial vertical speed $u$ . It takes the same time to reach the ground as object A, i.e. $t = 1.69\,\text{s}$ from (b)(i).

Given:

$t = 1.69\,\text{s}$ ,
initial height above ground $= 3.6\,\text{m}$ ,
acceleration due to gravity.

Unknown: the initial upward speed $u$ .

Approach

Use the vertical displacement from release point to ground (downwards by $3.6\,\text{m}$ ). Choose upward as positive, so the displacement is negative and the acceleration is $-g$ . Substitute into

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

and solve for $u$ .

Step-by-Step Reasoning

Take upward as positive.

From release point to the ground, object B moves downward $3.6\,\text{m}$ , so

s = -3.6\,\text{m}.

Initial velocity is upward, so it is $+u$ .
Acceleration due to gravity is downward, so

a = -g.

Substitute:

-3.6 = ut + \frac{1}{2}(-g)t^2 = ut - \frac{1}{2}gt^2.

Rearrange to make $u$ the subject:

ut = -3.6 + \frac{1}{2}gt^2

u = \frac{-3.6 + \frac{1}{2}gt^2}{t}.

Now use $t = 1.69\,\text{s}$ (and $g = 9.81\,\text{m s}^{-2}$ ):

u = \frac{-3.6 + \frac{1}{2}(9.81)(1.69)^2}{1.69} \approx 6.16\,\text{m s}^{-1}.

This is reasonable: the term $\frac{1}{2}gt^2$ is about $14\,\text{m}$ , and the object only needs to end up $3.6\,\text{m}$ lower than it started, so $u$ must be a few $\text{m s}^{-1}$ upwards.

Key Takeaways

The same kinematics equation applies even when the object is initially thrown upwards.
The main difficulty is consistent signs for $s$ and $a$ .

Common Mistakes

Putting $s = +3.6\,\text{m}$ while also using $a = -g$ (inconsistent sign convention).
Using $14\,\text{m}$ for object B’s displacement (wrong height).
Forgetting that $u$ is the vertical initial velocity in part (b).

Things to Be Careful About

Keep the direction convention explicit (e.g. “upward positive”).
Use the time from (b)(i) without rounding too aggressively, to avoid losing accuracy.
Ensure the final unit is $\text{m s}^{-1}$ .

Techniques used

apply a constant-acceleration equation with a chosen sign conventionsubstitute the time from a previous partrearrange to solve for the initial velocity

(c)

In a second experiment, object B is released from the same height and given the same initial speed as in (b) but at a release angle $\theta$ to the vertical, as shown in Fig. 1.2.

Fig. 1.2

4M

(i)

State and explain whether the time taken for object B to reach the ground is less than, the same as or greater than the time in (b)(i).

DifficultyMedium

Worked solution

Answer

Less than.

Vertical component of initial velocity is $u\cos\theta$ (smaller than $u$ for $\theta \neq 0$ ). The time to reach the ground depends only on vertical motion (horizontal motion does not affect time), so with a smaller initial upward vertical component, B reaches the ground sooner.

Final answer

Less than.

Detailed explanation

Background Concept

In projectile motion (neglecting air resistance):

Horizontal acceleration is zero, so horizontal velocity is constant.
Vertical acceleration is constant and equal to $g$ downward.

Crucially, the horizontal and vertical motions are independent, meaning the time to reach the ground is determined entirely by the vertical motion.

If a launch speed $u$ makes an angle $\theta$ to the vertical, then the components are:

\text{vertical component} = u\cos\theta, \qquad \text{horizontal component} = u\sin\theta.

Understanding the Question

In (b), object B was projected straight up with speed $u$ from height $3.6\,\text{m}$ , and it took time $t$ (same as object A).

Now B is launched from the same height and with the same speed $u$ , but at an angle $\theta$ to the vertical. You must decide whether the time to hit the ground is less than, the same as, or greater than the time in (b)(i), and explain.

Approach

Compare only the vertical component of the initial velocity, because that is what affects the vertical displacement to the ground. Since $u\cos\theta < u$ for $\theta > 0$ , the upward “delay” is smaller, so the object hits the ground sooner.

Step-by-Step Reasoning

diagram-c-i-components

In (b), initial vertical velocity was $u$ .
In this experiment, initial vertical velocity is

u_y = u\cos\theta.

For $\theta \neq 0$ (angled away from the vertical), $\cos\theta < 1$ , so

u_y = u\cos\theta < u.

That means:

It rises to a smaller maximum height (or may rise for a shorter time).
Therefore it spends less time moving upward before descending.

Since the horizontal motion does not change how far it must fall vertically (still $3.6\,\text{m}$ to the ground) and does not change the vertical acceleration, the reduced upward vertical component makes the total time to reach the ground less than in (b).

Key Takeaways

Time of flight depends on the vertical component of velocity and the vertical displacement.
Changing the horizontal component does not affect the time to hit the ground (if air resistance is neglected).

Common Mistakes

Saying the time is the same because the speed is the same (direction matters for vertical motion).
Thinking the horizontal component “adds extra distance so it takes longer” (distance travelled is not what sets the time here; vertical displacement and vertical acceleration do).
Using $u\sin\theta$ as the vertical component (wrong because the angle is to the vertical).

Things to Be Careful About

The phrase “angle $\theta$ to the vertical” means vertical component is $u\cos\theta$ .
You must explicitly mention independence of horizontal and vertical components to fully justify the comparison.

Techniques used

resolve the initial velocity into vertical and horizontal componentsuse independence of perpendicular components of motioncompare times using the vertical component only

(ii)

By considering energy, state and explain the effect of the change in release angle on the speed at which B reaches the ground.

DifficultyMedium-Easy

Worked solution

Answer

No effect: the speed at the ground is the same.

Initial kinetic energy is $\tfrac{1}{2}mu^2$ (same because initial speed is unchanged) and loss of gravitational potential energy is $mg(3.6)$ (same height). With no air resistance, total mechanical energy is conserved, so final kinetic energy (and hence final speed) is unchanged by the release angle.

Final answer

No change (same speed at the ground).

Detailed explanation

Background Concept

With negligible air resistance, mechanical energy is conserved:

E_{\text{total}} = E_k + E_p = \text{constant}.

Where

E_k = \frac{1}{2}mv^2, \qquad E_p = mgh.

Only speed appears in kinetic energy, not direction. Gravitational potential energy depends only on height.

Understanding the Question

Object B is released from the same height ( $3.6\,\text{m}$ above the ground) and with the same initial speed as in (b), but now the direction is changed (angle $\theta$ to the vertical).

You are asked: how does changing the release angle affect the speed when it reaches the ground, using an energy argument.

Approach

Compare initial and final mechanical energy:

Initial energy: $\frac{1}{2}mu^2 + mg(3.6)$ .
Final energy at the ground: $\frac{1}{2}mv^2$ (taking ground as zero potential).

Since both $u$ and the height are unchanged, the final speed $v$ must be unchanged.

Step-by-Step Reasoning

Take the ground as $h = 0$ .

Initial (at height $3.6\,\text{m}$ ):

E_{\text{initial}} = \frac{1}{2}mu^2 + mg(3.6).

Final (at ground):

E_{\text{final}} = \frac{1}{2}mv^2.

Conservation of mechanical energy gives:

\frac{1}{2}mu^2 + mg(3.6) = \frac{1}{2}mv^2.

Cancel $m$ and rearrange:

v^2 = u^2 + 2g(3.6).

This expression contains $u$ and the height only; there is no $\theta$ . Therefore changing the release angle does not change the impact speed (as long as air resistance remains negligible).

Key Takeaways

Kinetic energy depends on speed, not direction.
Gravitational potential energy depends only on height.
With no dissipative forces, impact speed depends only on initial speed and the vertical drop.

Common Mistakes

Claiming the impact speed is smaller because the vertical component is smaller (vertical component affects time and trajectory, but not total energy).
Mixing up speed and velocity: direction changes velocity, but speed can be unchanged.
Forgetting that the object starts above the ground, so there is a gain in kinetic energy from losing gravitational potential energy.

Things to Be Careful About

The conclusion relies on “air resistance negligible”; with air resistance, different paths could lead to different energy losses.
Be explicit that same initial speed means same initial kinetic energy $\frac{1}{2}mu^2$ .

Techniques used

apply conservation of mechanical energyrelate loss of gravitational potential energy to gain in kinetic energyidentify which quantities are independent of launch direction

Q2MediumForces, Density and PressureDeformation of Solids

(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 (resultant moment is zero).

Final answer

In equilibrium, total clockwise moment about a point equals total anticlockwise moment (net moment = 0).

Detailed explanation

Background Concept

The moment (turning effect) of a force about a point (pivot) is

\tau = 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.

For an object that is not rotating (rotational equilibrium), the resultant moment about any point is zero. This is commonly written as:

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

Understanding the Question

You are asked to state the principle of moments. So you need a clear sentence relating clockwise and anticlockwise moments for a body in equilibrium.

Approach

Give the standard equilibrium condition in words (and/or symbols): clockwise moments equal anticlockwise moments about the same pivot.

Step-by-Step Reasoning

In equilibrium, there is no angular acceleration, so the net turning effect must be zero.
The turning effects are moments, which can be clockwise or anticlockwise.
Therefore, the total clockwise moment must equal the total anticlockwise moment (about the same point).

Key Takeaways

Moment $= F \times$ perpendicular distance.
Rotational equilibrium means resultant moment is zero.
Hence clockwise moments = anticlockwise moments.

Common Mistakes

Saying “forces are equal” (that is translational equilibrium, not the principle of moments).
Forgetting to specify “about a point/pivot”.
Not indicating clockwise vs anticlockwise (or not stating “resultant moment is zero”).

Things to Be Careful About

The distance must be perpendicular to the line of action of the force.
The principle applies when the body is in rotational equilibrium (not turning).

Techniques used

state the equilibrium condition for rotational momentsdistinguish clockwise and anticlockwise moments about a pivot

(b)

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

Fig. 2.1

The beam is uniform and has length $6.0\text{ m}$ .
The pivot is at the midpoint of the beam.
Object A has mass $60\text{ kg}$ and is at one end of the beam.
Object B has mass $45\text{ kg}$ and is at a distance $x$ from the pivot.
Object C has mass $80\text{ kg}$ and is at the other end of the beam.

Calculate $x$ .

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

DifficultyMedium-Easy

Worked solution

Working

About the pivot (midpoint), distances to ends are $3.0\ \text{m}$ .

Anticlockwise moments (left) = clockwise moments (right):

(60g)(3.0) + (45g)(x) = (80g)(3.0)

180 + 45x = 240

45x = 60 \quad \Rightarrow \quad x = 1.33\ \text{m}

Answer

$1.33\ \text{m}$

Final answer

1.33 m

Detailed explanation

Background Concept

For a horizontal beam pivoted at a point, each downward weight produces a turning effect (moment) about the pivot:

\tau = Fd

For equilibrium (no rotation):

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

With weights, $F = mg$ , so moments are often written as $mg \times d$ . If every force is a weight, the factor $g$ cancels.

Understanding the Question

A uniform $6.0\ \text{m}$ beam is pivoted at its midpoint, so each end is $3.0\ \text{m}$ from the pivot. Masses $A$ and $C$ are at the ends; mass $B$ is on the left side at distance $x$ from the pivot. The beam is in equilibrium, so the total moment to the left must balance the total moment to the right.

diagram-b-1

Approach

Choose the pivot as the point to take moments about (this eliminates any unknown reaction force at the pivot).
Write anticlockwise moments (from left-side weights) = clockwise moments (from right-side weights).
Substitute distances and masses, cancel $g$ , and solve for $x$ .

Step-by-Step Reasoning

Distance from pivot to each end:

\frac{6.0}{2} = 3.0\ \text{m}

Take moments about the pivot.
- On the left: $A$ at $3.0\ \text{m}$ and $B$ at $x$ both produce anticlockwise moments.
- On the right: $C$ at $3.0\ \text{m}$ produces clockwise moment.

So:

(60g)(3.0) + (45g)(x) = (80g)(3.0)

Cancel $g$ :

60(3.0) + 45x = 80(3.0)

180 + 45x = 240 \Rightarrow 45x = 60 \Rightarrow x = 1.33\ \text{m}

Key Takeaways

Always take moments about the pivot to avoid the unknown pivot reaction.
For equilibrium, clockwise moments = anticlockwise moments.
When all forces are weights, $g$ cancels.

Common Mistakes

Using $6.0\ \text{m}$ instead of $3.0\ \text{m}$ for the distance to the ends.
Putting $B$ on the wrong side of the pivot (it is between $A$ and the pivot).
Forgetting that moments use perpendicular distance.

Things to Be Careful About

Be consistent with clockwise/anticlockwise sign convention.
Distances must be measured from the pivot, not from an end of the beam.
Quote $x$ in metres and to a sensible number of significant figures (here 3 s.f. matches the data).

Techniques used

take moments about the pivotset clockwise moments equal to anticlockwise momentssolve a linear equation for an unknown distance

(c)

The beam is $0.80\text{ m}$ above horizontal ground.

Object A is removed and replaced by a spring connected to the ground and the beam, as shown in Fig. 2.2.

Fig. 2.2

After the change, the beam is again horizontal and in equilibrium. The positions of B and C are unchanged.

The spring has an unstretched length of $0.59\text{ m}$ and obeys Hooke’s law.

5M

(i)

Calculate the spring constant of the spring.

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

DifficultyMedium

Worked solution

Working

From (b), moment balance implies the required downward force at the left end equals the weight of $60\ \text{kg}$ :

F = 60g = 60 \times 9.81 = 5.89 \times 10^2\ \text{N}

Spring length $= 0.80\ \text{m}$ , natural length $= 0.59\ \text{m}$ :

\Delta L = 0.80 - 0.59 = 0.21\ \text{m}

Hooke's law $F = k\Delta L$ :

k = \frac{F}{\Delta L} = \frac{5.89 \times 10^2}{0.21} = 2.80 \times 10^3\ \text{N m}^{-1}

Answer

$2.8 \times 10^3\ \text{N m}^{-1}$

Final answer

2.8 × 10^3 N m^-1

Detailed explanation

Background Concept

In equilibrium, a beam must satisfy the principle of moments:

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

A spring that obeys Hooke’s law provides a force proportional to its extension:

F = k\Delta L

where $k$ is the spring constant and $\Delta L$ is the extension (stretched length minus natural length).

Understanding the Question

Object $A$ (mass $60\ \text{kg}$ ) is removed and replaced by a vertical spring at the left end of the beam. The beam is still horizontal and in equilibrium, with $B$ and $C$ unchanged from before. The beam is $0.80\ \text{m}$ above the ground, so the spring’s actual length is $0.80\ \text{m}$ . Its natural (unstretched) length is $0.59\ \text{m}$ .

We must find the spring constant $k$ .

diagram-ci-1

Approach

Use the fact that with $B$ and $C$ unchanged and equilibrium regained, the moment that used to be provided by the weight of A must now be provided by the spring force at the same position (left end).
Calculate the spring force needed (equal to the previous weight of $A$ ).
Find the extension $\Delta L$ from the given lengths.
Use $k = F/\Delta L$ .

Step-by-Step Reasoning

1) Spring force from moments

From part (b), the beam was in equilibrium with a downward force at the left end equal to the weight of $A$ , i.e. $60g$ . When $A$ is replaced by a spring at the same point (the left end), and $B$ and $C$ are unchanged, the moment balance requires the spring to provide the same downward force at that point as before.

So the spring force on the beam is:

F = 60g = 60 \times 9.81 = 5.89 \times 10^2\ \text{N}

(Using $g = 9.8\ \text{m s}^{-2}$ would give $5.88 \times 10^2\ \text{N}$ , essentially the same.)

2) Extension of the spring

Actual length is the beam height above the ground: $0.80\ \text{m}$ .

Natural length: $0.59\ \text{m}$ .

\Delta L = 0.80 - 0.59 = 0.21\ \text{m}

3) Use Hooke’s law

F = k\Delta L \Rightarrow k = \frac{F}{\Delta L}

k = \frac{5.89 \times 10^2}{0.21} = 2.80 \times 10^3\ \text{N m}^{-1}

Key Takeaways

If only one force is replaced at the same position and equilibrium is restored with all other loads unchanged, the replacement must provide the same moment about the pivot.
Extension is (stretched length) minus (unstretched length).
Hooke’s law links force, extension, and spring constant.

Common Mistakes

Using $0.59 - 0.80$ (giving a negative extension).
Forgetting that the beam height $0.80\ \text{m}$ is the spring’s length.
Mixing up $k = F/\Delta L$ and $F = k/\Delta L$ .
Using the full beam length $6.0\ \text{m}$ as the moment arm for the spring instead of $3.0\ \text{m}$ (if re-deriving from scratch).

Things to Be Careful About

Units: $k$ must be in $\text{N m}^{-1}$ , so $\Delta L$ must be in metres.
A spring pulls back: since it is stretched here, it exerts a downward force on the beam (consistent with replacing the removed weight at that end).
Significant figures: $0.80$ and $0.59$ imply $\Delta L$ to 2 d.p.; giving $k$ to 2 s.f. or 3 s.f. is acceptable.

Techniques used

use the principle of moments to determine an unknown forcecalculate spring extension from stretched and natural lengthsapply Hooke's law to find spring constant

(ii)

Calculate the elastic potential energy of the spring.

elastic potential energy = ______ $\text{J}$

DifficultyMedium-Easy

Worked solution

Working

Using $E = \tfrac{1}{2}F\Delta L$ with $F = 5.89 \times 10^2\ \text{N}$ and $\Delta L = 0.21\ \text{m}$ :

E = \frac{1}{2}(5.89 \times 10^2)(0.21) = 6.18 \times 10^1\ \text{J}

Answer

$62\ \text{J}$

Final answer

62 J

Detailed explanation

Background Concept

For a spring obeying Hooke’s law, the force increases linearly with extension:

F = kx

The elastic potential energy stored is the work done stretching it. On a force–extension graph it is the area under the straight line, giving:

E = \frac{1}{2}kx^2

Since $F = kx$ , an equivalent form is:

E = \frac{1}{2}Fx

Understanding the Question

The spring is stretched from $0.59\ \text{m}$ to $0.80\ \text{m}$ , so $x = 0.21\ \text{m}$ . From part (c)(i), the spring force is about $5.89 \times 10^2\ \text{N}$ (and $k \approx 2.8 \times 10^3\ \text{N m}^{-1}$ ). We need the elastic potential energy stored in the spring.

Approach

Use either $E = \tfrac{1}{2}kx^2$ (using $k$ from part (i)) or $E = \tfrac{1}{2}Fx$ (using the spring force and extension). The second is quickest here.

Step-by-Step Reasoning

Extension:

x = 0.80 - 0.59 = 0.21\ \text{m}

Energy:

E = \frac{1}{2}Fx

E = \frac{1}{2}(5.89 \times 10^2)(0.21) = 6.18 \times 10^1\ \text{J}

So $E \approx 62\ \text{J}$ (2 s.f.).

(Checking with $E = \tfrac{1}{2}kx^2$ gives the same value because $F = kx$ .)

Key Takeaways

Elastic potential energy in a Hooke’s law spring is $E = \tfrac{1}{2}kx^2$ .
If you already know the final force $F$ and extension $x$ , $E = \tfrac{1}{2}Fx$ is often fastest.

Common Mistakes

Using $E = Fx$ (missing the factor $\tfrac{1}{2}$ ; the force is not constant while stretching).
Squaring the spring length instead of the extension.
Using $x$ in cm instead of m.

Things to Be Careful About

Ensure $x$ is the extension, not the natural length or the stretched length.
Keep units consistent: $F$ in N and $x$ in m gives $E$ in J.
Round appropriately (typically 2–3 significant figures).

Techniques used

use elastic potential energy formula for a Hooke’s law springsubstitute extension and spring constant (or force) into the energy expressionquote energy with appropriate significant figures

Q3MediumWork, Energy and PowerElectricityDynamics

(a)

Define power.

(b)

An electric car is powered by a motor. The car is travelling at a constant speed of $35\text{ m s}^{-1}$ along a straight horizontal road, as shown in Fig. 3.1.

Fig. 3.1

There is a total resistive force of $1750\text{ N}$ acting on the car.

7M

(i)

Calculate the power transmitted to the wheels of the car by the motor.

power = ______ $\text{W}$

DifficultyMedium-Easy

Worked solution

Working

At constant speed, driving force balances resistive force, so

P = Fv = 1750 \times 35

P = 6.125 \times 10^4\ \text{W}

Answer

P \approx 6.1 \times 10^4\ \text{W}

Final answer

6.1 × 10^4 W

Detailed explanation

Background Concept

For an object moving at constant speed, the mechanical power transferred by a force parallel to the motion is

P = Fv

where $F$ is the force (N) and $v$ is the speed (m s $^{-1}$ ). This comes from combining $P = W/t$ with $W = Fs$ and $v = s/t$ .

Understanding the Question

The car travels at constant speed $35\ \text{m s}^{-1}$ on a horizontal road, with a total resistive force of $1750\ \text{N}$ . At constant speed, the motor must provide a driving force equal in magnitude to the resistive force. You are asked for the power transmitted to the wheels (mechanical output power).

Approach

Use the constant-speed power relation $P = Fv$ with $F = 1750\ \text{N}$ and $v = 35\ \text{m s}^{-1}$ .

Step-by-Step Reasoning

At constant speed on a horizontal road:

Resultant force is zero.
Driving force from the wheels equals resistive force in magnitude.

So the motor’s output power at the wheels is

P = Fv = (1750)(35) = 61250\ \text{W}.

Rounding to appropriate s.f. gives about $6.1 \times 10^4\ \text{W}$ .

Key Takeaways

Constant speed implies no acceleration and therefore balanced forces.
Mechanical power at constant speed is found from $P = Fv$ .

Common Mistakes

Using $P = VI$ (electrical power) instead of mechanical power at the wheels.
Forgetting that at constant speed the driving force equals the resistive force.

Things to Be Careful About

Use SI units: N and m s $^{-1}$ give watts automatically.
Do not add extra forces (e.g. weight) on a horizontal road for the along-motion calculation.

Techniques used

use the constant-speed relation P = Fvsubstitute values with consistent SI unitsquote a final value to appropriate significant figures

(ii)

Calculate the useful work done by the motor when the car travels a distance of $17\text{ km}$ .

work done = ______ $\text{J}$

(iii)

The potential difference (p.d.) across the motor has a constant value of $600\text{ V}$ and the motor has an efficiency of $85\%$ .

Calculate the current in the motor.

current = ______ $\text{A}$

DifficultyMedium

Worked solution

Working

From (i), mechanical output power:

P_{\text{out}} = 6.125 \times 10^4\ \text{W}

Efficiency:

\eta = \frac{P_{\text{out}}}{P_{\text{in}}} \Rightarrow P_{\text{in}} = \frac{P_{\text{out}}}{\eta} = \frac{6.125\times 10^4}{0.85} = 7.21\times 10^4\ \text{W}

Electrical power $P_{\text{in}} = VI$ :

I = \frac{P_{\text{in}}}{V} = \frac{7.21\times 10^4}{600} = 1.20\times 10^2\ \text{A}

Answer

I \approx 120\ \text{A}

Final answer

120 A

Detailed explanation

Background Concept

Efficiency compares useful output to total input:

\eta = \frac{\text{useful output power}}{\text{input power}} = \frac{P_{\text{out}}}{P_{\text{in}}}.

Electrical input power for a device with p.d. $V$ and current $I$ is

P = VI.

Understanding the Question

The motor produces the mechanical power needed to keep the car moving (found in part (i)). The p.d. across the motor is constant at $600\ \text{V}$ , and the motor efficiency is $85\%$ . You must find the current drawn, so you need the electrical input power, then use $P = VI$ .

Approach

Take the mechanical power at the wheels as $P_{\text{out}}$ .
Use $\eta = P_{\text{out}}/P_{\text{in}}$ to find $P_{\text{in}}$ .
Use $I = P_{\text{in}}/V$ .

Step-by-Step Reasoning

From part (i):

P_{\text{out}} = 6.125\times 10^4\ \text{W}.

Efficiency is $\eta = 0.85$ , so

P_{\text{in}} = \frac{P_{\text{out}}}{\eta} = \frac{6.125\times 10^4}{0.85} = 7.21\times 10^4\ \text{W}.

Using $P_{\text{in}} = VI$ with $V = 600\ \text{V}$ :

I = \frac{P_{\text{in}}}{V} = \frac{7.21\times 10^4}{600} = 1.20\times 10^2\ \text{A} \approx 120\ \text{A}.

Key Takeaways

Mechanical output power and electrical input power differ when efficiency is below 100%.
Convert percentage efficiency to a decimal before using it.
Use $P = VI$ to relate electrical power, p.d. and current.

Common Mistakes

Multiplying by $0.85$ instead of dividing (mixing up input and output).
Using $P_{\text{out}} = VI$ directly (that would only be true for 100% efficiency).
Keeping efficiency as $85$ instead of $0.85$ .

Things to Be Careful About

Be clear: $P_{\text{out}}$ here is the power at the wheels (useful mechanical power).
Round only at the end to avoid rounding error (especially before dividing by $0.85$ ).

Techniques used

relate efficiency to output and input poweruse electrical power P = VI to determine currentcarry values forward from earlier parts consistently

(c)

The car in (b) now reaches a slope, as shown in Fig. 3.2.

Fig. 3.2

The car continues down the slope at the same speed as in (b).

State and explain the effect, if any, of the slope on:

2M

(i)

the air resistance acting on the car

(ii)

the current in the motor.

DifficultyMedium-Easy

Worked solution

Answer

Current decreases: on the slope, a component of weight acts down the slope and helps the motion, so the motor needs a smaller driving force (and smaller power at the same speed), hence smaller electrical power and smaller current.

Final answer

Decreases.

Detailed explanation

Background Concept

For motion at constant speed, the resultant force along the direction of motion is zero.

On a slope of angle $\theta$ , the weight $mg$ has a component down the slope:

mg\sin\theta.

Mechanical power from the motor (when driving) is

P_{\text{out}} = F_{\text{drive}} v,

and electrical input power is related to current by $P_{\text{in}} = VI$ (for fixed $V$ , smaller power means smaller current).

Understanding the Question

The car now travels down a slope at the same constant speed as before. Air resistance is unchanged (part (c)(i)), and the resistive forces still oppose the motion. The new feature is that gravity has a component along the slope in the direction of motion, which reduces how much force the motor must supply to maintain the same speed.

Approach

Draw/think of the forces along the slope.
Use constant speed condition (forces along slope balance).
Conclude the required driving force is smaller than on the horizontal, so the motor output power is smaller at the same speed.
With the same p.d. across the motor, smaller input power implies smaller current.

Step-by-Step Reasoning

diagram-3c2

On the horizontal road, the motor must supply a driving force equal to the total resistive force to keep speed constant.
On a downhill slope, the component of weight down the slope $mg\sin\theta$ acts in the same direction as the motion.
Therefore the force balance along the slope becomes (qualitatively):

F_{\text{drive}} + mg\sin\theta = F_{\text{resistive}}.

So

F_{\text{drive}} = F_{\text{resistive}} - mg\sin\theta,

which is smaller than $F_{\text{resistive}}$ .

At the same speed $v$ , the useful mechanical power is

P_{\text{out}} = F_{\text{drive}}v,

so $P_{\text{out}}$ decreases.

Since the p.d. is constant, electrical input power $P_{\text{in}}$ also decreases, and using $P_{\text{in}} = VI$ :

I = \frac{P_{\text{in}}}{V}

shows the current decreases.

(If $mg\sin\theta$ were larger than the resistive force, the motor might need to provide no driving power and could even act as a generator/brake, but the mark here is for “smaller current”.)

Key Takeaways

Constant speed means forces along the direction of motion balance.
Downhill, $mg\sin\theta$ assists motion, reducing required driving force.
Reduced mechanical power demand implies reduced electrical power and hence reduced current for constant $V$ .

Common Mistakes

Saying current increases because the car is on a slope (downhill actually reduces the required motor effort at fixed speed).
Ignoring the component of weight along the slope.
Claiming air resistance changes because of the slope (it changes mainly with speed).

Things to Be Careful About

Direction matters: downhill means the weight component is in the direction of motion.
Do not assume the motor must always provide the same force; constant speed does not mean constant driving force when the slope changes.
The question asks for the effect on current: link force (\rightarrow) power (\rightarrow) current.

Techniques used

resolve weight into components along a slopeapply force balance for constant speedlink reduced required driving force to reduced power and current

Q4MediumSuperposition

(a)

State the principle of superposition.

DifficultyEasy

Worked solution

Answer

When two (or more) waves overlap, the resultant displacement at any point is the (vector/algebraic) sum of the individual displacements of the waves at that point.

Final answer

The resultant displacement is the sum of the individual displacements.

Detailed explanation

Background Concept

The principle of superposition applies to waves in a linear medium (i.e. where waves do not permanently change each other when they pass). Each wave produces its own displacement of the medium (or field, for electromagnetic waves).

Superposition states that when waves overlap, the medium responds to all of them at once. The total displacement is found by adding the displacements from each wave.

Understanding the Question

You are asked to state the principle. That means a concise definition, not an example.

Key idea to include for full credit:

“resultant displacement at a point”
equals the “sum of individual displacements” (algebraic sum, including sign)

Approach

Write a single sentence definition:

mention overlap/superpose
mention resultant displacement
mention sum of displacements of the individual waves

Step-by-Step Reasoning

Identify the quantity that superposes: displacement (not intensity).
State what happens during overlap: the displacements combine.
Express combination rule: resultant displacement equals the algebraic (or vector) sum of the separate displacements at the same position and time.

Key Takeaways

Superposition adds displacements, not intensities.
It is a point-by-point, instant-by-instant rule.

Common Mistakes

Saying “intensities add” (only true in special cases; generally incorrect for interference).
Forgetting to specify “at a point” and “at that instant”.

Things to Be Careful About

Use the word displacement (Cambridge expects that phrasing).
“Algebraic sum” implies sign: upward and downward displacements can cancel.

Techniques used

state the superposition principle in terms of displacementsdescribe the resultant as the algebraic/vector sum of individual contributions

(b)

Light of wavelength $7.2 \times 10^{-7}\text{ m}$ is incident normally on a double slit, as shown in Fig. 4.1.

Fig. 4.1

A screen is at a distance $D$ from the double slit. The double slit and the screen are parallel.

The separation of the slits in the double-slit arrangement is $0.16\text{ mm}$ . The resulting interference pattern on the screen contains nine dark fringes, as shown in Fig. 4.2.

Fig. 4.2

The distance between the centres of the first and ninth dark fringes is $3.2\text{ cm}$ .

6M

(i)

Calculate $D$ .

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

DifficultyMedium-Easy

Worked solution

Working

Distance from 1st to 9th dark fringe contains $8$ spacings, so

x = \frac{3.2\ \text{cm}}{8} = 0.40\ \text{cm} = 4.0\times 10^{-3}\ \text{m}

For double-slit fringes,

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

So

D = \frac{xa}{\lambda} = \frac{\left(4.0\times 10^{-3}\right)\left(1.6\times 10^{-4}\right)}{7.2\times 10^{-7}} = 0.89\ \text{m}

Answer

$D = 0.89\ \text{m}$

Final answer

0.89 m

Detailed explanation

Background Concept

In Young’s double-slit interference, light from two coherent slits acts like two coherent sources. At a point on a distant screen, the path difference between the two rays determines whether interference is constructive (bright) or destructive (dark).

For small angles (screen far compared with slit separation), the spacing between adjacent bright fringes (and also adjacent dark fringes) is

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

where:

$x$ is the fringe spacing on the screen,
$\lambda$ is the wavelength,
$D$ is the slit-to-screen distance,
$a$ is slit separation.

This comes from $a\sin\theta \approx a\theta$ and $y \approx D\theta$ , giving $y = \frac{\lambda D}{a}$ between successive fringes.

Understanding the Question

You are given:

wavelength $\lambda = 7.2\times 10^{-7}\ \text{m}$ ,
slit separation $a = 0.16\ \text{mm} = 1.6\times 10^{-4}\ \text{m}$ ,
nine dark fringes are shown,
distance between centres of the 1st and 9th dark fringes is $3.2\ \text{cm}$ .

You must find the screen distance $D$ .

The key interpretation point: “between the 1st and 9th dark fringes” means there are 8 gaps (8 equal spacings) between them.

Approach

Convert the given 3.2 cm into a single fringe spacing $x$ by dividing by the number of spacings (8).
Convert slit separation into metres.
Use $x = \frac{\lambda D}{a}$ and rearrange for $D$ .

Step-by-Step Reasoning

Count spacings:
- Dark fringes numbered 1 through 9.
- Spacings between their centres: $9 - 1 = 8$ .
Find the fringe spacing $x$ :

x = \frac{3.2\ \text{cm}}{8} = 0.40\ \text{cm} = 4.0\times 10^{-3}\ \text{m}

Convert slit separation:

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

Rearrange the double-slit formula:

D = \frac{xa}{\lambda}

Substitute:

D = \frac{\left(4.0\times 10^{-3}\right)\left(1.6\times 10^{-4}\right)}{7.2\times 10^{-7}} = \frac{6.4\times 10^{-7}}{7.2\times 10^{-7}} = 0.888...\ \text{m}

So $D \approx 0.89\ \text{m}$ .

Key Takeaways

Adjacent dark fringes have the same spacing as adjacent bright fringes.
When given a distance spanning several fringes, divide by the number of intervals, not the number of fringes.
Use $x = \frac{\lambda D}{a}$ with consistent SI units.

Common Mistakes

Dividing by 9 instead of 8.
Not converting $\text{mm}$ to $\text{m}$ or $\text{cm}$ to $\text{m}$ .
Using $a$ and $x$ swapped (writing $x = \frac{aD}{\lambda}$ ).

Things to Be Careful About

Significant figures: inputs are mostly 2 s.f., so $D$ to 2 s.f. is appropriate.
The small-angle approximation is assumed in this standard A-level formula; the diagram implies $D \gg a$ so it is valid here.

Techniques used

determine the fringe spacing from the number of intervalsuse the double-slit fringe-spacing relationshiprearrange an equation to solve for distance

(ii)

The slit separation is now gradually decreased from $0.16\text{ mm}$ to $0.04\text{ mm}$ . The distance between the centres of adjacent dark fringes is $x$ .

On Fig. 4.3, sketch the variation of $x$ with slit separation.

Fig. 4.3

DifficultyMedium

Worked solution

Answer

Using $x = \dfrac{\lambda D}{a}$ with $\lambda, D$ constant, $x \propto \dfrac{1}{a}$ .

So the sketch is a decreasing inverse curve (rectangular hyperbola): as slit separation decreases, $x$ increases.

It passes through approximately:

$a = 0.16\ \text{mm}$ , $x = 0.40\ \text{cm}$
$a = 0.04\ \text{mm}$ , $x = 1.6\ \text{cm}$

Final answer

x is inversely proportional to slit separation (decreasing hyperbola).

Detailed explanation

Background Concept

For a double-slit pattern at small angles, the fringe spacing $x$ is

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

If the wavelength $\lambda$ and the slit-to-screen distance $D$ are unchanged, then the only variable affecting fringe spacing is the slit separation $a$ .

This gives an inverse proportionality:

x \propto \frac{1}{a}

So reducing slit separation increases the spacing of the fringes.

Understanding the Question

The slit separation is changed gradually from $0.16\ \text{mm}$ down to $0.04\ \text{mm}$ . You must sketch how the distance between adjacent dark fringes, $x$ , varies with slit separation.

You are not asked to compute many points; a correct shape with correct key values is what gains marks.

From part (i), when $a = 0.16\ \text{mm}$ the fringe spacing is

x = 0.40\ \text{cm}

The axes on Fig. 4.3 are:

horizontal: slit separation / mm (up to 0.16 mm)
vertical: $x$ / cm (up to 1.6 cm)

This strongly hints that at 0.04 mm, $x$ reaches the top of the axis.

Approach

Start from $x = \frac{\lambda D}{a}$ .
Treat $\lambda$ and $D$ as constants, so $x$ is inversely proportional to $a$ .
Use scaling: if $a$ is reduced by a factor of 4 (0.16 mm to 0.04 mm), then $x$ increases by a factor of 4.
Sketch a smooth inverse curve through the key points.

Step-by-Step Reasoning

Write the dependence:

x = \frac{\lambda D}{a} \Rightarrow x \propto \frac{1}{a}

Use the known point from the original arrangement:

at $a = 0.16\ \text{mm}$ , $x = 0.40\ \text{cm}$ .

Find $x$ at $a = 0.04\ \text{mm}$ by proportionality:

\frac{x_2}{x_1} = \frac{a_1}{a_2} = \frac{0.16}{0.04} = 4

So

x_2 = 4x_1 = 4(0.40\ \text{cm}) = 1.6\ \text{cm}

Sketch:

The curve must go downwards as $a$ increases.
It is not a straight line (because $x \propto 1/a$ ).
It should pass through (0.04, 1.6) and (0.16, 0.40) and be smoothly curved.

Key Takeaways

In double-slit interference, decreasing slit separation spreads the fringes out.
Recognise and sketch an inverse proportionality: $y \propto 1/x$ gives a hyperbola.

Common Mistakes

Drawing a straight line (suggesting $x \propto a$ or $x$ changes linearly).
Drawing the trend the wrong way round (thinking smaller separation means smaller spacing).
Forgetting to use the given axis limits to place the point at $a = 0.04\ \text{mm}$ .

Things to Be Careful About

Ensure you interpret the axes correctly: $x$ is on the vertical axis in cm, slit separation on the horizontal axis in mm.
The curve should be smooth and decreasing; it should not kink or become horizontal over this range.

Techniques used

use proportional reasoning from the fringe-spacing formulaidentify an inverse relationship and sketch an appropriate curveuse given endpoints to anchor a graph sketch

Q5MediumWaves

(a)

Use the definitions of speed $v$ , frequency $f$ and wavelength $\lambda$ to derive the wave equation

v = f\lambda

DifficultyMedium-Easy

Worked solution

Working

In one period $T$ , the wave travels one wavelength $\lambda$ , so

v = \frac{\lambda}{T}

Also

f = \frac{1}{T}

Hence

v = \lambda \left(\frac{1}{T}\right) = f\lambda

Answer

$v = f\lambda$

Final answer

v = f\lambda

Detailed explanation

Background Concept

For a progressive wave:

Wavelength $\lambda$ is the distance between two successive points in phase (e.g. crest to crest).
Period $T$ is the time for one complete oscillation.
Frequency $f$ is the number of oscillations per second, so $f = 1/T$ .
Wave speed $v$ is the distance travelled per unit time.

A key idea is: in the time for one complete cycle ( $T$ ), the pattern of the wave advances by exactly one wavelength ( $\lambda$ ).

Understanding the Question

You are asked to derive the wave equation

v = f\lambda

using only the definitions of speed, frequency, and wavelength. So you must link these definitions through the period $T$ .

Approach

Use the definition of speed: $v = \text{distance}/\text{time}$ .
Choose a time interval that makes the distance easy to express: one period $T$ .
In time $T$ , the wave moves forward by one wavelength $\lambda$ , so $v = \lambda/T$ .
Replace $1/T$ using the definition of frequency $f = 1/T$ .

Step-by-Step Reasoning

In one period $T$ , one complete cycle passes a point. Over that time, the wave profile advances by one wavelength, so the distance travelled is $\lambda$ .

So by the definition of speed:

v = \frac{\text{distance}}{\text{time}} = \frac{\lambda}{T}

From the definition of frequency (cycles per second):

f = \frac{1}{T}

Substitute $1/T = f$ into $v = \lambda/T$ :

v = \lambda f = f\lambda

Key Takeaways

Using the interval of one period is the simplest way to connect $v$ , $\lambda$ , and $f$ .
The equation $v = f\lambda$ is essentially “speed = (cycles per second) $\times$ (metres per cycle)”.

Common Mistakes

Writing $v = \frac{f}{\lambda}$ or $v = \frac{\lambda}{f}$ (mixing up how frequency relates to period).
Not explicitly stating that in one period the wave travels one wavelength.

Things to Be Careful About

Frequency is not the same as period: always use $f = 1/T$ .
This derivation assumes a steady wave speed in a uniform medium.

Techniques used

use the definition of wave speed as distance travelled per unit timerelate one period to one wavelength travelleduse the definition of frequency as the reciprocal of periodsubstitute to eliminate period

(b)

A source of sound waves of frequency $236\text{ Hz}$ is travelling at a constant velocity of $20\text{ m s}^{-1}$ .

A stationary observer has a microphone connected to a cathode-ray oscilloscope (CRO). The microphone detects the sound waves as the source moves directly towards the observer.

The resulting trace on the CRO is shown in Fig. 5.1.

Fig. 5.1

The time-base on the CRO is set to $1.0\text{ ms div}^{-1}$ .

4M

(i)

Calculate the frequency of the sound waves detected by the microphone.

frequency = ______ $\text{Hz}$

DifficultyMedium-Easy

Worked solution

Working

From the CRO trace, one cycle occupies $4.0$ divisions.

Time-base $= 1.0\ \text{ms div}^{-1}$ , so

T = 4.0 \times 1.0\ \text{ms} = 4.0\ \text{ms} = 4.0 \times 10^{-3}\ \text{s}

f = \frac{1}{T} = \frac{1}{4.0 \times 10^{-3}} = 2.5 \times 10^{2}\ \text{Hz}

Answer

$2.5 \times 10^{2}\ \text{Hz}$

Final answer

2.5 × 10^2 Hz

Detailed explanation

Background Concept

A cathode-ray oscilloscope (CRO) time-base tells you the time per horizontal division. If you can measure the period $T$ (time for one complete cycle) from the screen, you can find frequency using

f = \frac{1}{T}

Understanding the Question

The microphone detects the sound as the source moves towards the observer, producing a sinusoidal trace. You are given the time-base setting $1.0\ \text{ms div}^{-1}$ . You must read off the period from the trace (count divisions for one complete cycle) and then calculate the frequency.

Approach

Pick two identical points on the wave (e.g. crest to next crest).
Count the number of horizontal divisions between them: this gives divisions per period.
Convert divisions to time using $1.0\ \text{ms}$ per division to get $T$ .
Use $f = 1/T$ .

Step-by-Step Reasoning

From the trace, one complete cycle spans about $4.0$ horizontal divisions.

Each division corresponds to $1.0\ \text{ms}$ , so the period is

T = 4.0\ \text{div} \times 1.0\ \text{ms div}^{-1} = 4.0\ \text{ms} = 4.0 \times 10^{-3}\ \text{s}

Now calculate the frequency:

f = \frac{1}{4.0 \times 10^{-3}} = 2.5 \times 10^{2}\ \text{Hz}

Key Takeaways

CRO frequency questions are nearly always: measure $T$ from divisions then use $f = 1/T$ .
Convert $\text{ms}$ to $\text{s}$ before taking the reciprocal to avoid powers-of-ten errors.

Common Mistakes

Counting divisions for multiple cycles but forgetting to divide by the number of cycles.
Using $f = T$ instead of $f = 1/T$ .
Leaving the period in ms and treating it as seconds.

Things to Be Careful About

Choose clear, matching points (crest-to-crest or same zero-crossing direction) to avoid a half-cycle error.
Use as many divisions as possible (e.g. measure over several cycles and divide) to reduce reading uncertainty.

Techniques used

read the period from a CRO trace using divisionsconvert CRO time-base setting into time per divisioncalculate frequency from the period

(ii)

Determine the speed of the sound in air.

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

DifficultyMedium

Worked solution

Working

For a source moving towards a stationary observer,

f' = f\left(\frac{v}{v - v_s}\right)

So

f'(v - v_s) = fv \Rightarrow v(f' - f) = f'v_s \Rightarrow v = \frac{f'v_s}{f' - f}

With $f' = 250\ \text{Hz}$ (from (i)), $f = 236\ \text{Hz}$ and $v_s = 20\ \text{m s}^{-1}$ ,

v = \frac{250 \times 20}{250 - 236} = \frac{5000}{14} = 3.57 \times 10^{2}\ \text{m s}^{-1}

Answer

$3.6 \times 10^{2}\ \text{m s}^{-1}$

Final answer

3.6 × 10^2 m s^-1

Detailed explanation

Background Concept

The Doppler effect describes how the observed frequency changes when there is relative motion between source and observer.

For a source moving towards a stationary observer in a medium where the wave speed is $v$ :

f' = f\left(\frac{v}{v - v_s}\right)

where

$f$ is the emitted frequency,
$f'$ is the detected (observed) frequency,
$v_s$ is the speed of the source towards the observer,
$v$ is the speed of sound in air.

The minus sign in $(v - v_s)$ is because the moving source reduces the wavelength in front of it, increasing the detected frequency.

Understanding the Question

You are told:

emitted frequency $f = 236\ \text{Hz}$ ,
source speed towards the observer $v_s = 20\ \text{m s}^{-1}$ ,
from the CRO trace you found the detected frequency $f'$ .

You must use these to find the speed of sound $v$ .

Approach

Use the Doppler formula for a moving source towards a stationary observer.
Rearrange it to make $v$ the subject (because $v$ is unknown).
Substitute $f$ , $f'$ , and $v_s$ and calculate.
Check the answer is of the right order (a few hundred $\text{m s}^{-1}$ ).

Step-by-Step Reasoning

Start with

f' = f\left(\frac{v}{v - v_s}\right)

Multiply both sides by $(v - v_s)$ :

f'(v - v_s) = fv

Expand the left side:

f'v - f'v_s = fv

Collect the $v$ terms on one side:

f'v - fv = f'v_s

Factor out $v$ :

v(f' - f) = f'v_s

So

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

Substitute $f' = 250\ \text{Hz}$ , $f = 236\ \text{Hz}$ , $v_s = 20\ \text{m s}^{-1}$ :

v = \frac{250 \times 20}{250 - 236} = \frac{5000}{14} = 3.57 \times 10^{2}\ \text{m s}^{-1}

This is a plausible value for the speed of sound in air.

Key Takeaways

Choose the correct Doppler expression for the situation (moving source, stationary observer).
Rearranging algebra carefully is essential when the wave speed $v$ is the unknown.
Moving source towards observer gives $f' > f$ .

Common Mistakes

Using the wrong sign: writing $(v + v_s)$ for a source approaching (would wrongly reduce $f'$ ).
Using the formula for a moving observer instead of a moving source.
Substituting $v_s$ in the wrong place (e.g. adding it to $v$ rather than subtracting).

Things to Be Careful About

Ensure $f' - f$ is positive for an approaching source (since $f' > f$ ).
Use consistent units (here frequencies in Hz and speed in $\text{m s}^{-1}$ ).
Round sensibly (typically 2–3 s.f. unless instructed otherwise).

Techniques used

apply the Doppler effect formula for a moving source and stationary observerrearrange an equation to make wave speed the subjectsubstitute measured and given values consistentlycheck physical reasonableness of the result

Q6Medium-EasyElectricityD.C. Circuits

A nichrome wire X of length $45\text{ cm}$ and cross-sectional area $4.7 \times 10^{-7}\text{ m}^2$ is connected into the circuit shown in Fig. 6.1.

Fig. 6.1

The resistance of X is $1.1\ \Omega$ . The cell has electromotive force (e.m.f.) $1.3\text{ V}$ and negligible internal resistance.

(a)

6M

(i)

Calculate the current in X.

current = ______ $\text{A}$

(ii)

The number density of charge carriers (electrons) in nichrome is $8.5 \times 10^{28}\text{ m}^{-3}$ .

Calculate the average drift speed of the charge carriers in X.

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

DifficultyMedium-Easy

Worked solution

Working

I = nAqv \;\Rightarrow\; v = \frac{I}{nAq}

v = \frac{1.18}{\left(8.5 \times 10^{28}\right)\left(4.7 \times 10^{-7}\right)\left(1.6 \times 10^{-19}\right)} = 1.85 \times 10^{-4}\ \text{m s}^{-1}

Answer

v \approx 1.8 \times 10^{-4}\ \text{m s}^{-1}

Final answer

1.8 × 10^-4 m s^-1

Detailed explanation

Background Concept

When charge carriers move through a conductor, their average drift speed $v$ is linked to the current by

I = nAqv

where:

$n$ is number density of charge carriers ( $\text{m}^{-3}$ ),
$A$ is cross-sectional area ( $\text{m}^2$ ),
$q$ is charge per carrier ( $1.6 \times 10^{-19}\ \text{C}$ for an electron),
$v$ is the drift speed ( $\text{m s}^{-1}$ ).

This comes from “current = charge passing per second” and the fact that in time $t$ the carriers in a length $vt$ of wire pass a cross-section.

Understanding the Question

You are given $n = 8.5 \times 10^{28}\ \text{m}^{-3}$ for nichrome, the wire area $A = 4.7 \times 10^{-7}\ \text{m}^2$ , and (from part (a)(i)) the current in $X$ . You must calculate the average drift speed of electrons in wire $X$ .

Approach

Use the current from (a)(i).
Rearrange $I = nAqv$ to make $v$ the subject.
Substitute values carefully in standard form.

Step-by-Step Reasoning

Start from

I = nAqv.

Rearrange:

v = \frac{I}{nAq}.

Substitute $I = 1.18\ \text{A}$ , $n = 8.5 \times 10^{28}$ , $A = 4.7 \times 10^{-7}$ , $q = 1.6 \times 10^{-19}$ :

v = \frac{1.18}{\left(8.5 \times 10^{28}\right)\left(4.7 \times 10^{-7}\right)\left(1.6 \times 10^{-19}\right)}.

Combine powers of ten: $10^{28} \times 10^{-7} \times 10^{-19} = 10^{2}$ , so the denominator is of order $10^3$ .
Evaluating gives

v \approx 1.85 \times 10^{-4}\ \text{m s}^{-1} \approx 1.8 \times 10^{-4}\ \text{m s}^{-1}.

This very small speed is normal: the signal travels fast, but individual electrons drift slowly.

Key Takeaways

Use $I = nAqv$ for drift speed.
Drift speed is typically tiny (often $10^{-4}\ \text{m s}^{-1}$ to $10^{-3}\ \text{m s}^{-1}$ in metals).

Common Mistakes

Forgetting $q = 1.6 \times 10^{-19}\ \text{C}$ .
Using $A$ in $\text{cm}^2$ instead of $\text{m}^2$ .
Mixing up $n$ (number density) with the number of electrons.

Things to Be Careful About

Keep everything in SI units.
Check powers of ten: $10^{28}$ is huge, so $v$ should come out small.
Quote $\text{m s}^{-1}$ as the unit.

Techniques used

use the drift speed relation I = Anvqsubstitute values in standard formrearrange an equation to make the required quantity the subject

(iii)

Calculate the resistivity of the nichrome.

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

DifficultyMedium-Easy

Worked solution

Working

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

L = 45\ \text{cm} = 0.45\ \text{m}

\rho = \frac{(1.1)(4.7 \times 10^{-7})}{0.45} = 1.15 \times 10^{-6}\ \Omega\,\text{m}

Answer

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

Final answer

1.1 × 10^-6 Ω m

Detailed explanation

Background Concept

Resistivity $\rho$ is a material property that relates resistance to geometry:

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

where $L$ is the length of the conductor and $A$ is its cross-sectional area. For a uniform wire of the same material, increasing $L$ increases $R$ and increasing $A$ decreases $R$ .

Understanding the Question

The wire $X$ has:

$R = 1.1\ \Omega$ ,
$L = 45\ \text{cm}$ ,
$A = 4.7 \times 10^{-7}\ \text{m}^2$ .

You are asked for the resistivity of nichrome, $\rho$ , in $\Omega\,\text{m}$ .

Approach

Convert the length to metres.
Rearrange $R = \rho L/A$ to $\rho = RA/L$ .
Substitute values and present in standard form.

Step-by-Step Reasoning

Convert length:

L = 45\ \text{cm} = 0.45\ \text{m}.

Rearrange the resistivity equation:

\rho = \frac{RA}{L}.

Substitute:

\rho = \frac{(1.1)(4.7 \times 10^{-7})}{0.45}.

Calculate $1.1 \times 4.7 = 5.17$ , so

\rho = \frac{5.17 \times 10^{-7}}{0.45} = 1.15 \times 10^{-6}\ \Omega\,\text{m} \approx 1.1 \times 10^{-6}\ \Omega\,\text{m}.

Key Takeaways

Resistivity is found from $\rho = RA/L$ .
Always convert lengths to metres for $\Omega\,\text{m}$ .

Common Mistakes

Leaving $L$ as $45\ \text{cm}$ (gives a result wrong by a factor of 100).
Using $\rho = RL/A$ (incorrect rearrangement).
Omitting the unit $\Omega\,\text{m}$ .

Things to Be Careful About

The unit of resistivity is $\Omega\,\text{m}$ (not $\Omega\,\text{m}^{-1}$ ).
Standard form is usually expected because $\rho$ for metals is often around $10^{-8}$ to $10^{-6}\ \Omega\,\text{m}$ .

Techniques used

convert a length to SI unitsrearrange R = rho L / Asubstitute values and compute in standard form

(b)

Wire Y is identical to wire X. Wire Y is added to the circuit in parallel with wire X, as shown in Fig. 6.2.

Fig. 6.2

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

3M

(i)

the reading on the ammeter

DifficultyMedium-Easy

Worked solution

Answer

Two identical $1.1\ \Omega$ wires in parallel give

R_\text{eq} = \frac{1.1}{2} = 0.55\ \Omega.

So the total current is

I = \frac{1.3}{0.55} = 2.36\ \text{A} \approx 2.4\ \text{A},

i.e. the ammeter reading increases (approximately doubles).

Final answer

Increases (approximately doubles to about 2.4 A).

Detailed explanation

Background Concept

For resistors in parallel, the potential difference across each branch is the same, and the total current is the sum of branch currents. The equivalent resistance satisfies

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

For two identical resistances $R$ in parallel, this simplifies to

R_\text{eq} = \frac{R}{2}.

With a fixed supply voltage $V$ , the total current drawn is

I_\text{total} = \frac{V}{R_\text{eq}}.

Understanding the Question

Initially, only wire $X$ is connected, so the circuit resistance is $1.1\ \Omega$ and the ammeter reads the current through $X$ .

Then an identical wire $Y$ is added in parallel with $X$ , while the cell e.m.f. remains $1.3\ \text{V}$ (internal resistance negligible). The question asks what happens to the ammeter reading (which is in the main line, so it measures the total current from the cell).

Approach

Find the new equivalent resistance of the parallel pair.
Use $I = V/R$ to determine how the total current changes.
State the change and link it to the decrease in resistance.

Step-by-Step Reasoning

diagram-b-i-1

Each wire has $R = 1.1\ \Omega$ . With two in parallel:

R_\text{eq} = \frac{R}{2} = \frac{1.1}{2} = 0.55\ \Omega.

The supply voltage is still $1.3\ \text{V}$ , so the total current becomes

I_\text{total} = \frac{1.3}{0.55} = 2.36\ \text{A} \approx 2.4\ \text{A}.

Compare with the original current $1.18\ \text{A}$ : it has doubled, because the equivalent resistance has halved.

Key Takeaways

Adding a parallel branch reduces total resistance.
With constant supply voltage, reduced resistance means increased total current.
The main-branch ammeter reads the sum of currents in the parallel branches.

Common Mistakes

Treating the two wires as series (incorrectly giving $2.2\ \Omega$ and a smaller current).
Saying “current splits so the ammeter reading stays the same” (the ammeter is before the split, so it measures the total).

Things to Be Careful About

“Negligible internal resistance” means the voltage across the external circuit stays at $1.3\ \text{V}$ even when current changes.
Be clear whether you mean current in a branch or total current from the supply.

Techniques used

combine equal resistances in parallelapply Ohm's law to the whole circuitcompare before-and-after circuit quantities

(ii)

the average drift speed of the charge carriers in X.

DifficultyEasy

Worked solution

Answer

The p.d. across $X$ is still $1.3\ \text{V}$ (parallel branches), so

I_X = \frac{1.3}{1.1}

is unchanged. Since $v \propto I$ from $I = nAqv$ , the drift speed in $X$ is unchanged.

Final answer

Unchanged.

Detailed explanation

Background Concept

In a parallel circuit:

Each branch has the same potential difference as the supply.
The current in each branch depends only on that branch’s resistance:

I_\text{branch} = \frac{V_\text{branch}}{R_\text{branch}}.

Drift speed is linked to current by

I = nAqv \quad \Rightarrow \quad v = \frac{I}{nAq}.

So if the current in wire $X$ does not change, the drift speed in $X$ does not change.

Understanding the Question

After adding identical wire $Y$ in parallel, the total current from the cell increases, but the question specifically asks about the drift speed of charge carriers in wire $X$ .

Approach

Decide whether the current in wire $X$ changes.
Use the parallel-circuit rule: same p.d. across each branch.
Use $v \propto I$ for fixed $n$ , $A$ , and $q$ .

Step-by-Step Reasoning

Because the cell has negligible internal resistance, the supply voltage is $1.3\ \text{V}$ . In a parallel arrangement, the p.d. across each branch is the supply p.d., so the p.d. across $X$ remains $1.3\ \text{V}$ .

Wire $X$ still has resistance $1.1\ \Omega$ , hence its branch current is still

I_X = \frac{1.3}{1.1} \approx 1.18\ \text{A}.

In the drift-speed equation $v = I/(nAq)$ , the quantities $n$ , $A$ , and $q$ for wire $X$ are unchanged, so unchanged $I_X$ means unchanged $v$ .

Key Takeaways

Adding a parallel branch changes total current, but not necessarily the current in an unchanged branch.
Drift speed in a given wire depends on the current in that wire, not the total circuit current.

Common Mistakes

Thinking the current in $X$ halves because “current splits equally”; the total splits, but the p.d. across each branch stays the same.
Using the new total current in the drift-speed formula for wire $X$ .

Things to Be Careful About

The conclusion relies on negligible internal resistance: the supply voltage stays at $1.3\ \text{V}$ .
Be explicit that it is $I_X$ (branch current) that matters for drift speed in $X$ .

Techniques used

use the fact that p.d. across parallel branches is the sameapply Ohm's law to a single branchrelate drift speed to current using I = Anvq

Q7Medium-EasyParticle Physics

(a)

Nitrogen-12 ( $^{12}_{7}\text{N}$ ) is an unstable isotope of nitrogen that decays by the emission of radiation to carbon-12 ( $^{12}_{6}\text{C}$ ).

Complete the full nuclear equation for the decay, including all the particles involved.

^{12}_{7}\text{N} \rightarrow \_\_\_\_\_\_

DifficultyMedium-Easy

Worked solution

Working

Conserve nucleon number $A$ and proton number $Z$ .

$A: 12 \rightarrow 12$ so emitted particles have $A=0$ .

$Z: 7 \rightarrow 6$ so emitted particle must have $Z=+1$ (positron) and include a neutrino.

^{12}_{7}\text{N} \rightarrow \, ^{12}_{6}\text{C} + \, ^{0}_{+1}\text{e} + \, ^{0}_{0}\nu_e

Answer

^{12}_{7}\text{N} \rightarrow \, ^{12}_{6}\text{C} + \, ^{0}_{+1}\text{e} + \, ^{0}_{0}\nu_e

Final answer

^{12}{7}N -> ^{12}{6}C + ^{0}{+1}e + ^{0}{0}nu_e

Detailed explanation

Background Concept

In nuclear decay equations, two quantities must be conserved:

Nucleon number $A$ (total number of protons + neutrons).
Proton number $Z$ (also called atomic number).

In (\beta^+) (positron) decay, a proton in the nucleus changes into a neutron:

p \rightarrow n + e^+ + \nu_e

So, for the nucleus:

$A$ stays the same (still the same total number of nucleons).
$Z$ decreases by 1 (one less proton).

A neutrino ( $\nu_e$ ) is included because it is required to conserve lepton number and helps satisfy energy and momentum conservation in the decay.

Understanding the Question

You are told that unstable nitrogen-12 decays to carbon-12 by emission of radiation, and you must complete the full nuclear equation including all particles.

Given:

Initial nuclide: $^{12}_{7}\text{N}$
Final nuclide: $^{12}_{6}\text{C}$

Unknown:

What emitted particle(s) make the equation balance in both $A$ and $Z$ .

Approach

Compare $A$ values: if unchanged, emitted particle(s) must have $A=0$ .
Compare $Z$ values: if $Z$ decreases by 1, the decay corresponds to $\beta^+$ emission (or electron capture, but the question says emission).
Write the emitted positron as $^{0}_{+1}e$ and include an electron neutrino $^{0}_{0}\nu_e$ .

Step-by-Step Reasoning

Check nucleon number:

A: 12 \rightarrow 12

So emitted particle(s) must have $A=0$ .

Check proton number:

Z: 7 \rightarrow 6

So the emitted particle(s) must carry away net proton number $+1$ (so that $6 + 1 = 7$ ). The particle with $A=0$ and $Z=+1$ is a positron ( $\beta^+$ ): $^{0}_{+1}e$ .

Include the neutrino (required in $\beta^+$ decay): $^{0}_{0}\nu_e$ .

So the completed equation is:

^{12}_{7}\text{N} \rightarrow \, ^{12}_{6}\text{C} + \, ^{0}_{+1}\text{e} + \, ^{0}_{0}\nu_e

Key Takeaways

Balance both $A$ and $Z$ in nuclear equations.
If $Z$ decreases by 1 with $A$ unchanged, it is typically positron emission ( $\beta^+$ ).
In beta decays, include the appropriate (anti)neutrino.

Common Mistakes

Omitting the neutrino.
Writing an electron ( $^{0}_{-1}e$ ) instead of a positron ( $^{0}_{+1}e$ ).
Changing the nucleon number (e.g. writing an alpha particle, which would reduce $A$ by 4).

Things to Be Careful About

The question says emission, which points to $\beta^+$ emission rather than electron capture.
Ensure the positron is written with $Z=+1$ and $A=0$ .
Use correct nuclide notation: $^{A}_{Z}\text{X}$ .

Techniques used

apply conservation of nucleon numberapply conservation of proton (atomic) numberidentify the decay as positron emission and include the neutrino

(b)

Hydrogen-1 ( $^{1}_{1}\text{H}$ ) is an isotope of hydrogen. An atom of hydrogen-1 comprises a proton with an orbiting electron.

An antiparticle equivalent of hydrogen-1 comprises an antiproton with an orbiting positron. The antiquarks in the antiproton are the antiparticles of the quarks in a proton.

5M

(i)

State the charge on the positron in terms of the elementary charge $e$ .

charge = ______ $e$

(ii)

State the group (class) of fundamental particle to which the positron belongs.

(iii)

In Table 7.1, state the flavour and charge of the three antiquarks that comprise the antiproton.

Table 7.1

flavour	charge/ $e$