Physics 5054 / 22 — October/November 2025

(ii)

the distance travelled by the car between the driver seeing the obstruction at $t = 5.7\ \text{s}$ and the car beginning to slow down

distance = ______ $\text{m}$

DifficultyMedium-Easy

Worked solution

Working

The driver sees the obstruction at $t = 5.7\ \text{s}$ . The car begins to slow down when the graph starts to slope downwards, which is at $t = 6.4\ \text{s}$ .

During this interval, the car is travelling at a constant speed of $v = 29\ \text{m/s}$ .

Time interval: $t = 6.4 - 5.7 = 0.7\ \text{s}$

\text{distance} = v \times t

\text{distance} = 29 \times 0.7

\text{distance} = 20.3\ \text{m}

Rounding to 2 significant figures (consistent with the data given):

\text{distance} \approx 20\ \text{m}

Answer

20

Final answer

20 m

Detailed explanation

Walkthrough

The question asks for the distance travelled between $t = 5.7\ \text{s}$ and the moment the car begins to slow down. From the graph, the car travels at a constant speed until $t = 6.4\ \text{s}$ , where the line starts to slope downwards. Thus, the braking begins at $t = 6.4\ \text{s}$ .

The time interval during which the car is moving at constant speed is:
$t = 6.4 - 5.7 = 0.7\ \text{s}$

During constant speed, distance is calculated using:
$s = v \times t$

Substituting the values:
$s = 29\ \text{m/s} \times 0.7\ \text{s} = 20.3\ \text{m}$

Given that the speed (29 m/s) and the time difference (0.7 s) have 2 and 1 significant figures respectively, the answer is appropriately rounded to 2 significant figures, giving 20 m.

Key Takeaways

The area under a speed-time graph represents distance. For a constant speed, this area is a rectangle, and the calculation simplifies to $s = vt$ . Identifying the correct start and end times from the problem description and the graph is crucial.

Common Mistakes

Using the wrong time interval (e.g., using 5.7 s instead of the difference $6.4 - 5.7$ ).
Forgetting to convert or calculate the time difference correctly.
Not rounding to the appropriate number of significant figures.

Things to Be Careful About

Always calculate the time interval ( $\Delta t$ ) rather than just using the end time. Pay attention to significant figures; 20.3 m rounds to 20 m to 2 s.f.

Techniques used

calculate distance from constant speed using s = vt

(iii)

the distance travelled by the car between $t = 0$ and $t = 11.2\ \text{s}$ .

distance = ______ $\text{m}$

DifficultyMedium

Worked solution

Working

The total distance travelled is the area under the speed-time graph from $t = 0$ to $t = 11.2\ \text{s}$ .

The shape under the graph is a trapezium, which can be split into a rectangle and a triangle, or calculated directly using the trapezium formula.

Method 1: Rectangle + Triangle
Area of rectangle (constant speed phase):

A_1 = \text{speed} \times \text{time} = 29 \times 6.4 = 185.6\ \text{m}

Area of triangle (braking phase):

A_2 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (11.2 - 6.4) \times 29

A_2 = \frac{1}{2} \times 4.8 \times 29 = 69.6\ \text{m}

Total distance:

\text{Total distance} = A_1 + A_2 = 185.6 + 69.6 = 255.2\ \text{m}

Method 2: Trapezium formula

\text{Area} = \frac{1}{2} \times (a + b) \times h

where $a = 6.4\ \text{s}$ , $b = 11.2\ \text{s}$ , and $h = 29\ \text{m/s}$ .

\text{Area} = \frac{1}{2} \times (6.4 + 11.2) \times 29 = \frac{1}{2} \times 17.6 \times 29 = 255.2\ \text{m}

Rounding to 2 significant figures:

\text{distance} \approx 260\ \text{m}

Answer

260

Final answer

260 m

Detailed explanation

Walkthrough

The total distance travelled by the car is equal to the total area under the speed-time graph from $t = 0$ to $t = 11.2\ \text{s}$ .

The graph consists of two sections:

A rectangle from $t = 0$ to $t = 6.4\ \text{s}$ at a constant speed of 29 m/s.
A triangle from $t = 6.4\ \text{s}$ to $t = 11.2\ \text{s}$ where the speed decreases uniformly from 29 m/s to 0.

Calculating the area of the rectangle:
$\text{Area}_1 = 29 \times 6.4 = 185.6\ \text{m}$

Calculating the area of the triangle:
$\text{Base} = 11.2 - 6.4 = 4.8\ \text{s}$
$\text{Height} = 29\ \text{m/s}$
$\text{Area}_2 = \frac{1}{2} \times 4.8 \times 29 = 69.6\ \text{m}$

Total distance = $185.6 + 69.6 = 255.2\ \text{m}$ .

Alternatively, the entire shape is a trapezium with parallel sides $a = 6.4\ \text{s}$ and $b = 11.2\ \text{s}$ , and height $h = 29\ \text{m/s}$ :
$\text{Area} = \frac{1}{2}(a + b)h = \frac{1}{2}(6.4 + 11.2) \times 29 = 255.2\ \text{m}$

Rounding to 2 significant figures (as the speed 29 m/s has 2 s.f.), the distance is 260 m.

Key Takeaways

The area under a speed-time graph always represents the distance travelled. For uniform acceleration (straight line slope), the area is a triangle; for constant speed, it is a rectangle. These can be combined or treated as a trapezium.

Common Mistakes

Calculating the area of the triangle using the full time base (11.2 s) instead of the braking time (4.8 s).
Forgetting to add the area of the constant speed section.
Not rounding to the correct number of significant figures.

Things to Be Careful About

Ensure the base of the triangle is the difference in time ( $11.2 - 6.4$ ), not the absolute time value. Check significant figures at the end; 255.2 rounds to 260 to 2 s.f.

Techniques used

calculate distance from area under a speed-time graph (trapezium)

(b)

The speed of the car affects both the thinking distance and the braking distance.

State one factor that affects only the braking distance of the car.

Explain how a change in this factor increases the braking distance.

factor ______

explanation ______

DifficultyMedium

Worked solution

Answer

factor: condition of the road (or: wear on tyres / mass of the car / condition of brakes)

explanation: a slippery or smooth road reduces the friction between the tyres and the road surface, which reduces the deceleration of the car and increases the time taken to stop, thereby increasing the braking distance.

Final answer

Factor: condition of the road. Explanation: reduces friction, leading to reduced deceleration and a longer distance to stop.

Detailed explanation

Walkthrough

The question asks for a factor that affects only the braking distance, not the thinking distance. Thinking distance is affected by the driver's reaction time (e.g., alcohol, drugs, fatigue). Braking distance is affected by the vehicle and the road conditions.

Valid factors include:

Mass of the car or load (heavier vehicle)
Condition/wear of the tyres (worn tyres)
State of the road (slippery, wet, icy, smooth)
Condition of the brakes (worn brakes)

Let's choose road condition (e.g., a slippery or icy road).

Explanation chain:

A slippery road reduces the friction between the tyres and the road surface.
Reduced friction means the braking force is smaller.
By Newton's second law ( $F = ma$ ), a smaller resultant force leads to a reduced deceleration.
With a smaller deceleration, it takes longer for the car to stop from the same initial speed, meaning the braking distance increases.

Key Takeaways

Stopping distance = thinking distance + braking distance. Thinking distance depends on the driver; braking distance depends on the vehicle and road. Factors affecting braking distance include mass, tyre condition, road surface, and brake condition. The explanation must link the factor to friction, then to deceleration, and finally to the increased distance.

Common Mistakes

Suggesting a factor that affects thinking distance (e.g., driver fatigue, alcohol, speed).
Stopping the explanation at "reduced friction" without explaining how that leads to a longer distance (i.e., failing to mention reduced deceleration or increased stopping time).
Saying "less force" without specifying it is friction or braking force.

Things to Be Careful About

The question asks for a factor that affects only the braking distance. Speed affects both, so it is not a correct answer here. Ensure the explanation is a clear causal chain: factor $\rightarrow$ effect on friction/force $\rightarrow$ effect on deceleration $\rightarrow$ effect on distance.

Techniques used

identify factors affecting braking distanceexplain the relationship between friction, deceleration and stopping distance

Q29MMediumForcesEnergy, Work and Power

A spherical container is used to carry measuring instruments to the bottom of a lake.

Fig. 2.1 shows the container held at rest just below the water surface of a lake.

Fig. 2.1

The container is released and falls through the water to the bottom of the lake.

(a)

The container reaches terminal velocity before it hits the bottom of the lake.

Explain, in terms of the forces on the container, the motion of the container through the water until it reaches terminal velocity.

DifficultyMedium

Worked solution

Answer

Initially, the weight is greater than the drag, so there is a resultant force downwards and the container accelerates (speeds up). As the speed increases, the drag (or friction / viscous force) increases. This reduces the resultant force and so the acceleration decreases. When the drag equals the weight, the resultant force is zero, the acceleration is zero, and the container falls at a constant terminal velocity.

Final answer

Container accelerates as weight > drag; drag increases with speed until it equals weight, resultant force is zero and terminal velocity is reached.

Detailed explanation

Walkthrough

To explain the motion until terminal velocity is reached, we must link the forces acting on the container to its acceleration using Newton's second law ( $F = ma$ ).

Step 1: Initial motion. When the container is released, its speed is zero, so the upward drag force is zero. The only significant force is the downward weight. The resultant force is downwards, causing the container to accelerate and gain kinetic energy.

Step 2: Increasing drag. As the container speeds up, the water resistance (drag / viscous force) acting upwards increases. Because the weight remains constant, the resultant downward force ( $F_{\text{resultant}} = \text{weight} - \text{drag}$ ) decreases.

Step 3: Terminal velocity. According to $a = F_{\text{resultant}} / m$ , a decreasing resultant force means the acceleration decreases. Eventually, the upward drag force grows large enough to exactly equal the downward weight. The forces balance, the resultant force becomes zero, and the acceleration drops to zero. The container then continues to fall at a constant maximum speed, known as terminal velocity.

Key Takeaways

Terminal velocity is reached when the resistive force (drag) equals the driving force (weight).
Acceleration is proportional to the resultant force; as drag increases, resultant force and acceleration decrease.
At terminal velocity, the object moves at a constant speed with zero acceleration.

Common Mistakes

Saying "forces become equal so it stops" — it does not stop; it continues at a constant speed.
Using the word "gravity" instead of "weight" when referring to the downward force.
Forgetting to mention that drag increases as speed increases.

Things to Be Careful About

The question specifically asks to explain the motion in terms of the forces. Simply stating "it speeds up then goes at a constant speed" is not enough; you must link the speed changes to the resultant force and acceleration.
Acceptable terms for the upward force include drag, friction, resistance, or viscous force.

Techniques used

explain terminal velocity in terms of forces and motionrelate drag to speed and resultant force

(b)

The container falls through a vertical distance of $32\ \text{m}$ before it hits the bottom of the lake.

The container and instruments have a total mass of $74\ \text{kg}$ .

6M

(i)

Calculate the energy transferred from the gravitational potential energy store of the container.

energy transferred = ______ $\text{J}$

DifficultyMedium-Easy

Worked solution

Working

\Delta E_p = mg\Delta h

\Delta E_p = 74 \times 9.8 \times 32

\Delta E_p = 23206.4 \text{ J}

Answer

$2.3 \times 10^4 \text{ J}$

Final answer

2.3 x 10^4 J

Detailed explanation

Walkthrough

The energy transferred from the gravitational potential energy store is the loss in gravitational potential energy ( $\Delta E_p$ ) as the container falls.

Step 1: Identify the equation. The change in gravitational potential energy is given by $\Delta E_p = mg\Delta h$ , where $m$ is the mass, $g$ is the gravitational field strength, and $\Delta h$ is the vertical distance fallen.

Step 2: Substitute the values. The mass $m = 74 \text{ kg}$ , the vertical distance $\Delta h = 32 \text{ m}$ , and we use $g = 9.8 \text{ N/kg}$ (or $9.81 \text{ N/kg}$ ).

\Delta E_p = 74 \times 9.8 \times 32 = 23206.4 \text{ J}

Step 3: Round to an appropriate number of significant figures. The given values (74 and 32) have two significant figures, so we round the answer to two significant figures: $2.3 \times 10^4 \text{ J}$ .

Key Takeaways

Gravitational potential energy depends on mass, gravitational field strength, and vertical height: $\Delta E_p = mg\Delta h$ .
Always ensure mass is in kg and distance is in metres to get energy in joules.
Round final answers to the same number of significant figures as the least precise given data.

Common Mistakes

Using $g = 10 \text{ N/kg}$ when the mark scheme expects $9.8$ (though $2.4 \times 10^4 \text{ J}$ would typically be accepted if $g=10$ is used consistently, the scheme explicitly shows $9.8$ ).
Forgetting to convert units (e.g., using cm instead of m, or g instead of kg).
Writing the answer as $23206.4$ without rounding to 2 significant figures.

Things to Be Careful About

The question asks for the energy transferred from the gravitational potential energy store, which is simply the loss in $E_p$ . No efficiency or other losses need to be considered for this specific calculation.
Use standard form ( $2.3 \times 10^4$ ) for large numbers to clearly show significant figures.

Techniques used

calculate change in gravitational potential energysubstitute mass, g and height into E_p = mgh

(ii)

State the principle of conservation of energy.

(iii)

Work is done to transfer energy from the gravitational potential energy store as the container falls at terminal velocity.

Describe, in terms of the work done, how the energy is transferred from the gravitational potential energy store to one other energy store in the water.

DifficultyMedium

Worked solution

Answer

The container does work against the friction (or drag / resistive force) of the water. This work transfers energy from the gravitational potential energy store to the internal energy store of the water.

Final answer

Work is done against drag, transferring energy to the internal energy store of the water.

Detailed explanation

Walkthrough

The question asks how the energy is transferred in terms of the work done and to one other energy store in the water.

Step 1: Identify the mechanism of transfer. When a force acts over a distance, work is done. As the container falls at terminal velocity, it pushes through the water, doing work against the drag (friction / resistive force) exerted by the water.

Step 2: Identify the destination energy store. The work done against the resistive force causes the water to heat up slightly. In the 5054 energy stores model, this is described as transferring energy to the internal energy store of the water. (Avoid using the word 'heat' as an energy store; 'thermal energy' or 'internal energy' is the correct terminology for the store itself, though 'heat transfer' is acceptable for the process).

Step 3: Combine the statements. Energy is transferred from the gravitational potential energy store to the internal energy store of the water by doing work against the drag force.

Key Takeaways

Work done against friction or drag results in an increase in the internal energy of the surrounding medium.
Energy transfers can be described as 'doing work against a force'.
The correct energy store for a heated substance is the 'internal energy store', not 'heat'.

Common Mistakes

Saying the energy is transferred to the 'kinetic energy store' of the water — while the water molecules do move more, the correct 5054 terminology for the store that increases when something is heated is the 'internal energy store'.
Using the word 'heat' as a noun for the energy store (e.g., 'transferred to the heat store'). 'Heat' is the process of energy transfer, not the store.
Forgetting to mention 'work done' or 'doing work against drag', which the question specifically asks for.

Things to Be Careful About

The question specifies 'in terms of the work done', so you must include the phrase 'work done against friction/drag'.
Ensure you specify the energy store is in the water, not the container (though both heat up, the question asks about the water's store).

Techniques used

describe energy transfer in terms of work doneidentify the internal energy store of the water

Q39MMediumMass, Weight and DensityMomentumForces

Fig. 3.1 shows a firefighter directing a jet of water at a wall.

Fig. 3.1

In $1.0\ \text{s}$ , a volume of $7.5 \times 10^{-3}\ \text{m}^3$ of water hits the wall horizontally, at a speed of $24\ \text{m / s}$ .

The density of water is $1000\ \text{kg / m}^3$ .

(a)

Calculate:

4M

(i)

the mass of the water that hits the wall in $1.0\ \text{s}$

mass = ______ $\text{kg}$

(ii)

the horizontal momentum of the water that hits the wall in $1.0\ \text{s}$ .

momentum = ______ $\text{kg m / s}$

(b)

The horizontal momentum of the water decreases to zero when the water hits the wall. None of the water bounces back from the wall.

Explain why the momentum calculated in (a)(ii) is equal in size to the horizontal force exerted on the water by the wall.

DifficultyMedium-Easy

Worked solution

Answer

Force is defined as the rate of change of momentum: $F = \frac{\Delta p}{t}$ . The change in momentum of the water is $180 \text{ kg m/s}$ and the time taken is $1.0 \text{ s}$ . Therefore, the force is $\frac{180}{1.0} = 180 \text{ N}$ . Since the time is exactly $1.0 \text{ s}$ , the numerical value of the force in newtons is equal in size to the change in momentum in $\text{kg m/s}$ .

Final answer

Force equals the change in momentum divided by time; since the time is 1.0 s, the force is numerically equal to the change in momentum.

Detailed explanation

Walkthrough

Newton's second law can be stated as: the force acting on an object is equal to the rate of change of its momentum. Mathematically, $F = \frac{\Delta p}{t}$ . The water hits the wall with a momentum of $180 \text{ kg m/s}$ and comes to rest, so the change in momentum $\Delta p$ is $180 \text{ kg m/s}$ . This change happens over a time $t = 1.0 \text{ s}$ . Calculating the force gives $F = \frac{180}{1.0} = 180 \text{ N}$ . The question asks why the momentum value is equal in size to the force; the answer is that dividing by $1.0 \text{ s}$ leaves the numerical value unchanged.

Key Takeaways

Force is the rate of change of momentum. When the time interval is $1 \text{ s}$ , the force in newtons is numerically equal to the change in momentum.

Common Mistakes

Stating "force is momentum" without mentioning the time factor or rate of change. The mark scheme specifically looks for the relationship $F = \frac{\text{change in momentum}}{\text{time}}$ .

Things to Be Careful About

The question asks about the force on the water by the wall. The momentum calculated in (a)(ii) is the momentum of the water, which is fully lost (change is 180). The time given is exactly 1.0 s, which is the key to the numerical equality.

Techniques used

relate force to the rate of change of momentumexplain why force equals change in momentum numerically for a 1.0 s interval

(c)

Explain, in terms of Newton's third law, why there is a horizontal force exerted on the wall by the water.

(d)

The pump that forces the water to flow through the hose is adjusted, and the speed of the water leaving the hose doubles.

Explain why the force exerted by the water on the wall increases by a factor of 4.

DifficultyMedium

Worked solution

Answer

If the speed of the water doubles, the volume of water hitting the wall in 1.0 s also doubles, so the mass hitting the wall in 1.0 s doubles. Since momentum $p = mv$ , doubling both the mass and the velocity means the momentum hitting the wall in 1.0 s increases by a factor of $2 \times 2 = 4$ . Because force is the rate of change of momentum over 1.0 s, the force also increases by a factor of 4.

Final answer

Doubling the speed doubles the mass hitting the wall per second, and quadruples the momentum transferred per second, so the force increases by a factor of 4.

Detailed explanation

Walkthrough

The force is the rate of change of momentum: $F = \frac{\Delta p}{t}$ . For a 1.0 s interval, $F = \Delta p = mv$ . If the speed $v$ doubles to $2v$ , the volume of water flowing through the hose in 1.0 s also doubles (since flow rate = area $\times$ velocity). This means the mass $m$ hitting the wall in 1.0 s also doubles to $2m$ . The new momentum transferred in 1.0 s is $(2m)(2v) = 4mv$ . Since the time is still 1.0 s, the new force is 4 times the original force.

Key Takeaways

Flow rate is proportional to velocity. Doubling velocity doubles the mass flow rate. Momentum transfer rate depends on both mass flow rate and velocity, so it scales with the square of the velocity.

Common Mistakes

Only considering the effect of velocity on momentum ( $p=mv$ ) and forgetting that doubling the speed also doubles the amount of water (mass) hitting the wall in the same time. This leads to an incorrect factor of 2 instead of 4.

Things to Be Careful About

The question asks to explain why the force increases by a factor of 4. You must explicitly state that the mass hitting the wall in 1.0 s doubles, and then combine this with the doubling of velocity to get the factor of 4 for momentum (and thus force).

Techniques used

relate flow rate to velocityshow that doubling velocity quadruples the momentum transfer rate

Q48MMedium-EasyThermal Properties of MatterKinetic Particle Model of Matter

A large test-tube contains a thermometer and some solid wax at a temperature of $21^\circ\text{C}$ .

At time $t = 0$ , the test-tube is partially immersed in a beaker of boiling water and the temperature of the wax is recorded for the next $600\ \text{s}$ .

Fig. 4.1 shows the variation of the temperature of the wax with time $t$ .

Fig. 4.1

(a)

Explain how Fig. 4.1 shows that the melting temperature of the wax is $63^\circ\text{C}$ .

(b)

As the temperature of the wax increases, the energy in the internal energy store increases.

5M

(i)

The mass of the wax in the test-tube is $0.040\ \text{kg}$ and its temperature at time $t = 0$ is $21^\circ\text{C}$ . The specific heat capacity of wax is $2100\ \text{J / (kg }^\circ\text{C)}$ .

Using Fig. 4.1, determine the increase in the energy in the internal energy store of the wax between $t = 0$ and $t = 220\ \text{s}$ . Show your working.

increase in energy in internal energy store = ______ $\text{J}$

DifficultyMedium-Easy

Worked solution

Working

From Fig. 4.1, the temperature of the wax at $t = 220\ \text{s}$ is $48\ ^\circ\text{C}$ .

The increase in temperature is:

\Delta\theta = 48 - 21 = 27\ ^\circ\text{C}

The increase in energy in the internal energy store is given by:

\Delta E = mc\Delta\theta

Substituting the values:

\Delta E = 0.040 \times 2100 \times 27

\Delta E = 2268\ \text{J}

Rounding to 2 significant figures:

\Delta E \approx 2300\ \text{J}

Answer

increase in energy in internal energy store = 2300 $\text{J}$

Final answer

2300

Detailed explanation

Walkthrough

First, read the temperature of the wax at $t = 220\ \text{s}$ from the graph. The curve passes through $48\ ^\circ\text{C}$ at this time. The initial temperature at $t = 0$ is $21\ ^\circ\text{C}$ . The temperature change is $\Delta\theta = 48 - 21 = 27\ ^\circ\text{C}$ . Next, use the specific heat capacity formula $\Delta E = mc\Delta\theta$ . Substitute the mass $m = 0.040\ \text{kg}$ , specific heat capacity $c = 2100\ \text{J/(kg}\ ^\circ\text{C)}$ , and $\Delta\theta = 27\ ^\circ\text{C}$ . The calculation gives $2268\ \text{J}$ . Since the given values ( $0.040$ , $2100$ , $27$ ) have 2 significant figures, the final answer is rounded to $2300\ \text{J}$ .

Key Takeaways

The equation $\Delta E = mc\Delta\theta$ calculates the thermal energy transferred to change the temperature of a substance. Always ensure you read the temperature change from the graph correctly and match units (kg, $^\circ$ C, J/(kg $^\circ$ C)).

Common Mistakes

Reading the wrong temperature from the graph (e.g., reading 45 $^\circ$ C instead of 48 $^\circ$ C). Forgetting to calculate the temperature change $\Delta\theta$ and using the final temperature directly in the equation. Not rounding to the correct number of significant figures.

Things to Be Careful About

Ensure the mass is in kg (it is given as 0.040 kg, so no conversion is needed). The specific heat capacity is in J/(kg $^\circ$ C), so temperature in $^\circ$ C is correct. Do not use $g = 10\ \text{N/kg}$ here; this is a thermal energy calculation, not a weight calculation.

Techniques used

read a temperature from a heating curvecalculate thermal energy using delta E = mc delta theta

(ii)

Describe what happens to the motion of the molecules of the solid wax as the internal energy of the wax increases.

(c)

When the wax reaches its melting temperature, energy continues to be transferred thermally to the wax.

Describe, in terms of the forces between particles, why energy is required to melt the wax.

Q58MMedium-EasyElectromagnetic SpectrumGeneral Properties of Waves

The microwave region is one region of the electromagnetic spectrum.

(a)

Fig. 5.1 represents the electromagnetic spectrum divided into the seven main regions.

Fig. 5.1

The regions are arranged from left to right in order of increasing wavelength.

On Fig. 5.1:

indicate the microwave region by marking it with the letter M
indicate the ultraviolet region by marking it with the letters UV.

(b)

The frequency of the microwaves used by some satellite television systems is $12\ \text{GHz}$ .

5M

(i)

The speed of microwaves in a vacuum is $3.0 \times 10^8\ \text{m / s}$ .

Calculate the wavelength of these microwaves in a vacuum.

wavelength = ______ $\text{m}$

DifficultyMedium-Easy

Worked solution

Working

The wave equation is:

\lambda = \frac{c}{f}

Given:

Speed of microwaves, $c = 3.0 \times 10^8 \text{ m/s}$
Frequency, $f = 12 \text{ GHz} = 12 \times 10^9 \text{ Hz} = 1.2 \times 10^{10} \text{ Hz}$

Substitute the values:

\lambda = \frac{3.0 \times 10^8}{1.2 \times 10^{10}}

\lambda = 2.5 \times 10^{-2} \text{ m} = 0.025 \text{ m}

Answer

wavelength = 0.025 m

Final answer

0.025 m

Detailed explanation

Walkthrough

Identify the given values: The speed of electromagnetic waves in a vacuum is $c = 3.0 \times 10^8 \text{ m/s}$ . The frequency is given as $12 \text{ GHz}$ .
Convert units: The prefix 'G' (giga) means $10^9$ . So, $12 \text{ GHz} = 12 \times 10^9 \text{ Hz}$ . In standard scientific notation, this is $1.2 \times 10^{10} \text{ Hz}$ . It is crucial to convert GHz to Hz before substituting into the equation.
Select the equation: The wave equation links speed, frequency, and wavelength: $v = f \lambda$ . For electromagnetic waves in a vacuum, $v = c$ , so $c = f \lambda$ . Rearranging for wavelength gives $\lambda = c / f$ .
Substitute and calculate:

\lambda = \frac{3.0 \times 10^8}{1.2 \times 10^{10}} = 2.5 \times 10^{-2} \text{ m}

Final answer: $2.5 \times 10^{-2} \text{ m}$ is equal to $0.025 \text{ m}$ . Both forms are acceptable, but the question blank implies a decimal or standard form value with the unit 'm'.

Key Takeaways

Always convert units to standard SI units (Hz for frequency, m/s for speed) before calculating.
The wave equation $v = f \lambda$ applies to all waves, including electromagnetic waves where $v = c$ .
$1 \text{ GHz} = 10^9 \text{ Hz}$ .

Common Mistakes

Forgetting to convert GHz to Hz (using 12 instead of $1.2 \times 10^{10}$ ).
Rearranging the equation incorrectly (e.g., $\lambda = c \times f$ ).
Arithmetic errors with powers of ten.

Things to Be Careful About

Pay close attention to metric prefixes (kilo, mega, giga, milli, micro, nano).
Ensure the final answer includes the correct unit (meters).

Techniques used

convert GHz to Hzapply the wave equation to find wavelength

(ii)

Describe how microwaves are used in satellite television systems.

(c)

All microwave ovens have a switch that turns the oven off when the door is opened.

Suggest why it is important for a microwave oven to switch off when the door is opened.

DifficultyMedium-Easy

Worked solution

Answer

If the oven is not switched off when the door opens, microwaves could escape from the oven and spread out into the surrounding environment.
Microwaves can be absorbed by living tissue (including people), causing heating, burns, or tissue damage.

Final answer

To prevent microwaves from escaping and causing burns or tissue damage to people nearby.

Detailed explanation

Walkthrough

Microwave ovens work by emitting microwaves that are absorbed by water molecules in food, causing them to vibrate and generate heat. This same heating effect can occur in human tissue.

Preventing leakage: The metal casing and the mesh in the door are designed to contain the microwaves inside the oven. If the door is opened and the magnetron (the device that generates the microwaves) continues to operate, microwaves could leak out into the kitchen.
Health hazard: If a person is exposed to leaking microwaves, the radiation can penetrate their skin and be absorbed by their body's water content. This causes rapid heating of the tissue, which can lead to severe burns, blisters, and damage to internal tissues (especially eyes and testes, which have poor blood circulation to dissipate heat).

Therefore, the interlock switch is a critical safety feature that cuts power to the magnetron the moment the door latch is released.

Key Takeaways

Electromagnetic radiation can be hazardous if not contained.
Microwaves cause heating in matter, including biological tissue.
Safety interlocks are essential in appliances that generate ionising or high-energy non-ionising radiation.

Common Mistakes

Saying microwaves are 'radioactive' (they are not; they are non-ionising radiation).
Stating that microwaves cause cancer (the primary immediate danger is thermal damage/burns, not cancer, at the power levels used in ovens).
Simply saying 'it is dangerous' without explaining why (tissue heating/burns).

Things to Be Careful About

Distinguish between the danger of microwaves (heating/burns) and the danger of ionising radiation like X-rays or gamma rays (cell damage/cancer).

Techniques used

explain the safety purpose of a microwave oven door switch

Q610MMedium-EasyCurrent, Voltage and Resistance

An electric car is driven by a direct current (d.c.) motor that is powered by a direct current power supply.

(a)

Describe how a direct current differs from an alternating current (a.c.).

(b)

Fig. 6.1 represents the motor circuit. Point P is a point in the circuit between the motor and the positive terminal of the power supply.

Point P is shown to the left of the motor in the circuit diagram.

Fig. 6.1

During a time of $120\ \text{s}$ , the current in the circuit at point P is $41\ \text{A}$ .

4M

(i)

Calculate the charge that flows through point P in the time of $120\ \text{s}$ .

charge = ______ $\text{C}$

(ii)

The current in the circuit is due to the flow of electrons. Each electron carries a charge of magnitude $1.6 \times 10^{-19}\ \text{C}$ .

State the direction which the electrons move past point P and determine the number of electrons that pass point P in $1.0\ \text{s}$ .

direction ______

number of electrons = ______

DifficultyMedium-Easy

Worked solution

Answer

direction: towards the positive terminal (or away from the motor)

Working

Conventional current flows from the positive terminal to the negative terminal. Since electrons are negatively charged, they flow in the opposite direction to the conventional current. At point P (on the wire from the positive terminal to the motor), conventional current flows towards the motor, so electrons move towards the positive terminal (away from the motor).

Number of electrons in $1.0 \text{ s}$ :

\text{charge in } 1.0 \text{ s} = I \times t = 41 \times 1.0 = 41 \text{ C}

\text{number of electrons} = \frac{\text{total charge}}{\text{charge per electron}} = \frac{41}{1.6 \times 10^{-19}}

\text{number of electrons} = 2.5625 \times 10^{20} \approx 2.6 \times 10^{20}

number of electrons = $2.6 \times 10^{20}$

Final answer

direction: towards the positive terminal; number of electrons = 2.6 x 10^20

Detailed explanation

Walkthrough

First, determine the direction of electron flow. Conventional current is defined as the flow of positive charge, from the positive terminal of the supply to the negative terminal. In the circuit, point P is on the wire connecting the positive terminal to the motor, so conventional current flows from the positive terminal towards the motor (to the right in Fig. 6.1). Electrons are negatively charged, so they flow in the opposite direction to conventional current. Therefore, electrons move past point P towards the positive terminal (or away from the motor).

Second, calculate the number of electrons passing point P in $1.0 \text{ s}$ . The current is $41 \text{ A}$ , which means $41 \text{ C}$ of charge flows past point P every second. Each electron carries a charge of $1.6 \times 10^{-19} \text{ C}$ . The number of electrons is the total charge divided by the charge per electron:

\text{number} = \frac{41}{1.6 \times 10^{-19}} = 2.5625 \times 10^{20}

Rounding to 2 significant figures gives $2.6 \times 10^{20}$ .

Key Takeaways

Conventional current flows from positive to negative; electrons flow from negative to positive (opposite direction).
The number of electrons can be found by dividing the total charge by the elementary charge $e = 1.6 \times 10^{-19} \text{ C}$ .

Common Mistakes

Stating that electrons flow in the same direction as the current.
Using the total charge from part (b)(i) ( $4920 \text{ C}$ ) and dividing by $1.6 \times 10^{-19}$ without dividing by the time (120 s) first, or forgetting to adjust for the 1.0 s time interval.

Things to Be Careful About

The direction must be stated clearly: 'towards the positive terminal' or 'away from the motor'. Vague answers like 'left' may not score if not referenced to circuit components.
Use 2 or 3 significant figures for the final number of electrons. $2.6 \times 10^{20}$ is appropriate.

Techniques used

determine electron flow direction relative to conventional currentcalculate the number of electrons using the elementary charge

(c)

The power supply for the electric car is a number of batteries, each made from a large number of identical cells.

The e.m.f. (electromotive force) of each cell is $3.7\ \text{V}$ .

5M

(i)

Define the term 'electromotive force'.

(ii)

Each battery consists of 92 cells in series.

Calculate the e.m.f. of one battery.

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

(iii)

The power supply is made by connecting 85 of these batteries in parallel.

State the e.m.f. of the power supply.

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

Q79MMediumPractical Electricity

An electric kettle has an outer casing made of metal. Fig. 7.1 shows the kettle.

Fig. 7.1

The heater in the electric kettle is connected to a $230\ \text{V}$ mains power supply and has a power rating of $2300\ \text{W}$ .

(a)

The kettle is switched on.

4M

(i)

Calculate the current in the heater.

current = ______ $\text{A}$

(ii)

The manufacturer of the kettle states that the cable for the kettle is safe for currents that are smaller than $15\ \text{A}$ . The wiring in the mains power supply in the wall is safe for currents that are smaller than $20\ \text{A}$ .

The fuses available for the kettle have the ratings shown.

3\ \text{A} \quad 5\ \text{A} \quad 7\ \text{A} \quad 10\ \text{A} \quad 13\ \text{A} \quad 15\ \text{A} \quad 18\ \text{A} \quad 20\ \text{A} \quad 25\ \text{A}

State which of these fuses is the most appropriate. Explain why it is the most appropriate.

most appropriate fuse rating = ______ $\text{A}$

explanation ______

DifficultyMedium-Easy

Worked solution

Answer

most appropriate fuse rating: $13\ \text{A}$

explanation: It is slightly larger than the normal operating current of $10\ \text{A}$ so the fuse does not blow during normal use, but it is less than $15\ \text{A}$ so it protects the cable from overheating.

Final answer

13 A; slightly larger than the operating current so it does not blow during normal use, and smaller than 15 A to protect the cable

Detailed explanation

Walkthrough

A fuse must satisfy two conditions:

Its current rating must be slightly higher than the device's normal working current ( $10\ \text{A}$ ), so it will not melt (blow) when the appliance is operating normally.
Its current rating must be lower than the maximum safe current of the cable ( $15\ \text{A}$ ) and wall wiring ( $20\ \text{A}$ ), so that if a fault occurs, the fuse blows before the cable overheats and catches fire.

From the given list ( $3\ \text{A}, 5\ \text{A}, 7\ \text{A}, 10\ \text{A}, 13\ \text{A}, 15\ \text{A}, 18\ \text{A}, 20\ \text{A}, 25\ \text{A}$ ):

$10\ \text{A}$ or lower would blow during normal operation.
$15\ \text{A}$ or higher would not protect the cable safely.
Therefore, $13\ \text{A}$ is the correct choice.

Key Takeaways

A fuse rating must be just above the normal operating current of the appliance.
The fuse prevents excessive currents that could cause wires to overheat and cause a fire.

Common Mistakes

Choosing $10\ \text{A}$ (which would blow as soon as the kettle is turned on due to normal operating current or surges).
Choosing $15\ \text{A}$ or higher, which does not provide adequate safety margin for the $15\ \text{A}$ cable.

Things to Be Careful About

Ensure both aspects of the explanation are mentioned: operating without blowing under normal current, and protecting the wiring/cable.

Techniques used

select an appropriate fuse rating based on normal operating current and cable safety limit

(b)

There are three wires in the cable that connects the kettle to the mains power supply:

the earth wire
the live (line) wire
the neutral wire.

The wires are all correctly connected.

3M

(i)

State the name of the wire that the fuse is connected into. Explain why the fuse is connected into this wire.

name of wire ______

explanation ______

(ii)

The insulation on the live wire is damaged so that the wire is exposed.

Explain what happens when the exposed live wire touches the metal outer casing.

(c)

The kettle is switched on for a total time of $34\ \text{h}$ during one year.

The cost of electricity is $0.32 per kilowatt-hour (kWh).

Calculate the cost of using the kettle during the year.

cost = $ ______

DifficultyMedium-Easy

Worked solution

Working

\text{power in kW} = \frac{2300\ \text{W}}{1000} = 2.3\ \text{kW}

\text{energy used} = \text{power} \times \text{time} = 2.3\ \text{kW} \times 34\ \text{h} = 78.2\ \text{kW h}

\text{cost} = 78.2\ \text{kW h} \times \$0.32/\text{kW h} = \$25.024 \approx \$25.02

Answer

$25.02

Final answer

$25.02

Detailed explanation

Walkthrough

Convert power from watts to kilowatts: $P = \frac{2300\ \text{W}}{1000} = 2.3\ \text{kW}$
Calculate energy consumed in kilowatt-hours ( $\text{kW h}$ ): $E = P \times t = 2.3\ \text{kW} \times 34\ \text{h} = 78.2\ \text{kW h}$
Calculate total cost: $\text{cost} = \text{energy (in kW h)} \times \text{unit price}$ $\text{cost} = 78.2\ \text{kW h} \times \$0.32/\text{kW h} = \$25.024$ Rounding to two decimal places (cents) gives $25.02.

Key Takeaways

$1\ \text{kW h}$ is the electrical energy used by a $1\ \text{kW}$ appliance operating for $1\ \text{hour}$ .
$\text{Energy in kW h} = \text{Power in kW} \times \text{Time in hours}$ .
$\text{Total cost} = \text{Energy in kW h} \times \text{Cost per kW h}$ .

Common Mistakes

Forgetting to convert power from $\text{W}$ to $\text{kW}$ (dividing by 1000).
Converting time into seconds instead of leaving it in hours when computing $\text{kW h}$ .

Things to Be Careful About

Ensure money is written to 2 decimal places (e.g., $25.02).

Techniques used

calculate electrical energy in kilowatt-hourscalculate cost of electrical energy

Q88MMedium-EasyThe Nuclear AtomRadioactivity

The element lithium has several different isotopes.

(a)

2M

(i)

State one way in which the compositions of the atoms of all lithium isotopes are the same.

(ii)

State one way in which the composition of a lithium atom of one isotope is different from the composition of a lithium atom of a different isotope.

(b)

Fig. 8.1 is a diagram that represents a neutral atom of a radioactive isotope of lithium (Li).

Fig. 8.1

There are three electrons in orbit around the nucleus.

3M

(i)

The nuclide symbols for isotopes of lithium are represented as $^{X}_{Y}\text{Li}$ where $X$ and $Y$ are numbers.

Determine the values of $X$ and $Y$ for the isotope represented by Fig. 8.1.

$X$ = ______

$Y$ = ______

(ii)

An ion of this isotope of lithium has a single positive charge and is represented as ( $^{X}_{Y}\text{Li}^+$ ).

Describe how the diagram that represents this ion differs from Fig. 8.1.

(c)

Radioactive isotopes emit nuclear radiation which can affect living things.

3M

(i)

State one damaging effect of nuclear radiation.

(ii)

The emission of radioactive radiation from an unstable isotope is described as random and spontaneous.

State what is meant by 'random' and what is meant by 'spontaneous'.

random ______

spontaneous ______

DifficultyMedium-Easy

Worked solution

Answer

random: The emission of radiation is unpredictable in time and direction; there is no fixed time interval between decays, and it follows no pattern.

spontaneous: The emission is not influenced by external factors (such as temperature, pressure, or chemical combination); it happens on its own without any trigger.

Final answer

random: unpredictable in time/direction; spontaneous: not influenced by external factors

Detailed explanation

Walkthrough

Radioactive decay is a quantum mechanical process. Two key characteristics define it:

Random: You cannot predict when a specific nucleus will decay, nor in which direction the emitted radiation will travel. If you have a large sample, you can predict the overall decay rate (half-life), but for any individual nucleus, the decay is completely unpredictable. There is no pattern or fixed time interval between individual decay events.
Spontaneous: The decay happens on its own. It is not triggered or influenced by external conditions such as changes in temperature, pressure, magnetic fields, or the chemical environment (e.g., whether the atom is in a compound or pure element). An unstable nucleus will decay regardless of what is happening to it externally.

Key Takeaways

Random refers to the unpredictability of when and where (direction) a decay event occurs for an individual nucleus.
Spontaneous refers to the independence of the decay process from external physical or chemical conditions.

Common Mistakes

For 'random': Saying "it happens at random times" without explaining that it is unpredictable or has no pattern.
For 'spontaneous': Saying "it happens quickly" or "it happens by itself" without explicitly stating that external factors (temperature, pressure, etc.) do not influence it.

Things to Be Careful About

Ensure both definitions are distinct. 'Random' is about unpredictability and lack of pattern; 'spontaneous' is about independence from external influences.
Do not confuse 'spontaneous' with 'immediate' — some isotopes have half-lives of billions of years, but the decay of each individual nucleus is still spontaneous.

Techniques used

define 'random' in the context of radioactive decaydefine 'spontaneous' in the context of radioactive decay

Q910MMediumEarth and the Solar SystemKinematicsStars and the Universe

Fig. 9.1 shows how the Sun and the rest of the Solar System orbit around the centre of the Milky Way in a circular path.

Fig. 9.1

The Sun travels around the circular path at a speed $v$ .

(a)

The speed of light in a vacuum is $9.5 \times 10^{12}\ \text{km / year}$ .

The Sun is $26\ 000\ \text{light-years}$ from the centre of the Milky Way.

4M

(i)

Determine the distance from the Sun to the centre of the Milky Way in kilometres (km).

distance = ______ $\text{km}$

(ii)

It takes the Sun $2.3 \times 10^8\ \text{years}$ to complete one orbit around the centre of the Milky Way.

Calculate the speed $v$ .

$v$ = ______ $\text{m / s}$

DifficultyMedium

Worked solution

Working

v = \frac{2\pi r}{T}

Convert radius to metres:

r = 2.5 \times 10^{17}\ \text{km} = 2.5 \times 10^{20}\ \text{m}

Convert period to seconds:

T = 2.3 \times 10^8 \times 365 \times 24 \times 3600 = 7.255 \times 10^{15}\ \text{s}

Substitute into the equation:

v = \frac{2\pi \times 2.5 \times 10^{20}}{7.255 \times 10^{15}} = 2.16 \times 10^5\ \text{m/s}

Answer

$2.1 \times 10^5\ \text{m/s}$

Final answer

2.1 x 10^5

Detailed explanation

Walkthrough

The Sun travels in a circular orbit. The speed $v$ is the circumference of the orbit divided by the orbital period $T$ . The circumference is $2\pi r$ . The radius $r$ must be in metres and the period $T$ in seconds to give the speed in m/s.
Radius $r = 2.5 \times 10^{17}\ \text{km} = 2.5 \times 10^{20}\ \text{m}$ . (Using the unrounded value $2.47 \times 10^{20}\ \text{m}$ gives $2.14 \times 10^5\ \text{m/s}$ , which is also accepted). Period $T = 2.3 \times 10^8\ \text{years}$ . Convert years to seconds: $2.3 \times 10^8 \times 365 \times 24 \times 3600 = 7.255 \times 10^{15}\ \text{s}$ . Calculate $v = \frac{2\pi \times 2.47 \times 10^{20}}{7.255 \times 10^{15}} \approx 2.1 \times 10^5\ \text{m/s}$ .

Key Takeaways

For circular motion, speed = distance / time. The distance for one complete orbit is the circumference $2\pi r$ . Always convert units to SI (metres and seconds) before calculating speed in m/s.

Common Mistakes

Forgetting to convert the radius from km to m (dividing by 1000) or the period from years to seconds (multiplying by $365 \times 24 \times 3600$ ).
Using the wrong formula for circumference (e.g., $\pi r^2$ instead of $2\pi r$ ).
Rounding intermediate values too early, which can lead to $2.2 \times 10^5\ \text{m/s}$ ; both $2.1 \times 10^5$ and $2.2 \times 10^5$ are accepted depending on rounding.

Things to Be Careful About

Be careful with the powers of ten. $10^{20} / 10^{15} = 10^5$ . Make sure your calculator handles the large numbers correctly, or break the calculation into parts: calculate the numerator $2\pi r$ and the denominator $T$ separately before dividing.

Techniques used

calculate orbital speed using v = 2*pi*r / Tconvert distance to metres and time to seconds

(b)

Astronomical observations suggest that, at the centre of the Milky Way, there is a black hole.

Black holes are produced from red supergiants.

6M

(i)

Near the end of its life, a massive star becomes a red supergiant.

Describe what happens inside the star as it becomes a red supergiant.

(ii)

Describe how a red supergiant produces a black hole.

(iii)

Describe how the heaviest elements are produced.