Physics 5054 / 21 — May/June 2025

(b)

Explain how the graph shows that the acceleration decreases as the speed increases.

(c)

During the first part of the fall, there is a resultant vertical force acting downwards on the skydiver.

3M

(i)

One of the vertical forces acting on the skydiver is her weight.

State the name of the other vertical force that acts on the skydiver.

(ii)

Explain why the resultant vertical force eventually becomes zero.

(d)

At one instant, the vertical force on the skydiver is $400\ \text{N}$ downwards.

At the same instant, the wind causes an additional horizontal force of $100\ \text{N}$ to the right to act on the skydiver.

Draw a vector diagram to determine the resultant of the $400\ \text{N}$ vertical force and the $100\ \text{N}$ horizontal force. Place an arrow on all of the forces to show their directions.

Determine the magnitude (size) of this resultant force and its direction to the vertical.

magnitude of resultant = ______ $\text{N}$
direction = ______ $^\circ$ to vertical

DifficultyMedium

Worked solution

Working

Magnitude of resultant force:

R = \sqrt{400^2 + 100^2} = \sqrt{160000 + 10000} = \sqrt{170000} \approx 412 \text{ N}

Direction to the vertical:

\tan \theta = \frac{100}{400} = 0.25

\theta = \arctan(0.25) \approx 14^\circ

Answer

magnitude of resultant = 412 N (410–413 N accepted)
direction = 14° (12–16°) to vertical

Final answer

412 N, 14°

Detailed explanation

Walkthrough

The skydiver experiences a vertical force of 400 N downwards and a horizontal force of 100 N to the right. These two forces are perpendicular to each other.

To find the resultant, draw a vector diagram: draw a vertical arrow pointing downwards labelled '400 N'. From the tip of this arrow, draw a horizontal arrow pointing to the right labelled '100 N'. Draw the resultant vector from the tail of the 400 N arrow to the tip of the 100 N arrow.

Since the forces are perpendicular, we can use Pythagoras' theorem to find the magnitude of the resultant:

R = \sqrt{F_1^2 + F_2^2} = \sqrt{400^2 + 100^2} = \sqrt{170000} \approx 412 \text{ N}

To find the direction to the vertical, use trigonometry. The angle $\theta$ between the resultant and the vertical force is given by:

\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{100}{400} = 0.25

\theta = \arctan(0.25) \approx 14^\circ

Key Takeaways

Resultant of two perpendicular vectors can be found using Pythagoras' theorem.
Direction can be found using trigonometric ratios (tan, sin, cos).
Vector diagrams must show all forces with correct directions and the resultant.

Common Mistakes

Forgetting to include units in the final answer.
Calculating the angle to the horizontal instead of the vertical (would be $90^\circ - 14^\circ = 76^\circ$ ).
Drawing the vector diagram incorrectly (e.g., not placing the tail of the second vector at the tip of the first).

Things to Be Careful About

The question asks for the direction to the vertical, so use the vertical force as the adjacent side in the tangent calculation.
Graphical methods may yield slightly different results (e.g., 410 N, 12–16°), which are accepted according to the mark scheme.

Techniques used

draw a vector diagram for two perpendicular forcescalculate the magnitude of a resultant using Pythagoras' theoremcalculate the direction of a resultant using trigonometry

Q28MMediumTurning Effect of ForcesMass, Weight and Density

(a)

State the principle of moments.

(b)

An airline passenger wishes to check the weight of a suitcase that he carries as hand luggage onto an aeroplane.

He uses a uniform plank of wood, pivoted at its centre, as shown in Fig. 2.1.

Fig. 2.1

The plank is balanced with the suitcase on one side of the pivot and three bags of sugar, each of mass $2.0\ \text{kg}$ , on the other side.

The distances are shown on Fig. 2.1.

3M

(i)

Calculate the weight of one bag of sugar.

weight = ______ $\text{N}$

(ii)

The maximum weight of a suitcase that can be carried onto the aeroplane is $67\ \text{N}$ .

Determine whether the weight of the suitcase exceeds the maximum allowed.

Show a calculation in your answer.

DifficultyMedium-Easy

Worked solution

Working

Using the principle of moments (anticlockwise moment = clockwise moment):

80 \times 3 \times 19.6 = 90 \times W_{\text{suitcase}}

4704 = 90 \times W_{\text{suitcase}}

W_{\text{suitcase}} = \frac{4704}{90} = 52.3 \approx 52\ \text{N}

Answer

The weight of the suitcase is $52\ \text{N}$ . Since $52\ \text{N} < 67\ \text{N}$ , the weight of the suitcase does not exceed the maximum allowed.

Final answer

52 N; the weight does not exceed the maximum allowed

Detailed explanation

Walkthrough

The plank is balanced, so the total anticlockwise moment about the pivot equals the total clockwise moment. The sugar bags create an anticlockwise moment: $3 \times 19.6\ \text{N} \times 80\ \text{cm} = 4704\ \text{N cm}$ . The suitcase creates a clockwise moment: $W_{\text{suitcase}} \times 90\ \text{cm}$ . Equating the two gives $90 \times W_{\text{suitcase}} = 4704$ , so $W_{\text{suitcase}} = 52.3\ \text{N}$ (or $53.3\ \text{N}$ if $g=10\ \text{N/kg}$ was used in part (i)). The calculated weight ( $52\ \text{N}$ ) is less than the maximum allowed weight of $67\ \text{N}$ , so the suitcase is within the limit.

Key Takeaways

The principle of moments can be used to find an unknown force (or weight) when the system is balanced.
Distances must be measured from the pivot to the line of action of the force.

Common Mistakes

Forgetting to multiply by the number of bags (3) when calculating the total anticlockwise moment.
Using distances in metres without converting, though since both sides use the same units (cm), they cancel out.
Failing to explicitly state whether the weight exceeds the limit after calculating it.

Things to Be Careful About

Carry forward the answer from part (b)(i) (error carried forward is allowed). If a student used $20\ \text{N}$ for the weight of one bag, their moment would be $4800\ \text{N cm}$ , giving a suitcase weight of $53.3\ \text{N}$ , which still correctly leads to the conclusion that it does not exceed $67\ \text{N}$ .

Techniques used

apply the principle of moments to find the unknown weightcompare the calculated weight with the maximum allowed

(c)

A tall bus is tested for stability.

Fig. 2.2 shows the bus on a slope.

The centre of gravity of the bus is marked.

Fig. 2.2

3M

(i)

State what is meant by the 'centre of gravity' of an object.

(ii)

When the slope is made very steep, the bus falls over by rotating about point P.

Explain why the bus falls over.

Q38MMedium-EasyEnergy, Work and Power

A battery, a pulley and a motor are used to lift a load as shown in Fig. 3.1.

Fig. 3.1

(a)

Describe the transfers between energy stores that occur as the load is lifted.

DifficultyMedium-Easy

Worked solution

Answer

The chemical energy store in the battery decreases.
Electrical energy is transferred to the motor.
Mechanical energy is transferred to the load.
The gravitational energy store of the load increases.
Thermal energy increases in the motor and the surrounding air due to friction and electrical heating.

Final answer

See working

Detailed explanation

Walkthrough

The question asks for the transfers between energy stores as the load is lifted. We trace the energy from its source to its final useful and wasted destinations:

The battery provides energy from its chemical energy store, which decreases as it is used.
This chemical energy is converted into electrical energy that is transferred to the motor.
The motor converts electrical energy into mechanical energy, which is transferred via the cord to the load.
As the load is lifted, its height increases, so its gravitational energy store increases.
No system is perfectly efficient, so some energy is always wasted as thermal energy (heating the motor windings due to electrical resistance, and heating the pulley and air due to friction).

Any three of these linked statements earn the three marks.

Key Takeaways

Energy is never lost; it is only transferred between stores or dissipated.
Use the correct terminology: 'chemical energy store', 'gravitational energy store', and 'thermal energy', rather than informal terms like 'heat' or 'electricity'.

Common Mistakes

Saying 'energy is used up' or 'lost' instead of 'transferred' or 'dissipated'.
Using incorrect store names, such as 'battery energy' or 'lift energy'.
Confusing the energy store with the energy transfer (e.g., saying 'electrical energy is transferred to the battery' instead of 'from the battery').

Things to Be Careful About

The mark scheme accepts 'increase in thermal energy' or 'electrical heating'. Ensure you specify that thermal energy increases in the motor or the air, not just 'heat is produced'.

Techniques used

identify energy storesdescribe energy transfers

(b)

The efficiency of the motor, pulley and load system is less than 100%.

3M

(i)

By comparing the input energy and the useful output energy, explain why the efficiency is less than 100%.

(ii)

Explain how the principle of conservation of energy applies in lifting the load.

(c)

The input power to the motor is $15\ \text{W}$ . The motor is used for $20\ \text{s}$ . The efficiency of the motor is 60%.

Calculate the energy supplied to the load.

energy = ______ $\text{J}$

Q47MMedium-EasyThermal Properties of MatterTransfer of Thermal Energy

(a)

Evaporation of water from the surface of the skin causes cooling.

3M

(i)

Describe, using ideas about particles, how evaporation causes cooling.

(ii)

State one difference between evaporation and boiling.

(b)

When a refrigerator is switched on, cooling coils placed at the top of the space inside the refrigerator become cold. This causes a convection current which cools the air inside the refrigerator. The refrigerator is shown in Fig. 4.1.

Fig. 4.1

4M

(i)

Explain how the cooling coils cause a convection current in the air inside the refrigerator.

(ii)

The food in the refrigerator is initially at a temperature of $20.0\ ^\circ\text{C}$ . The food has a mass of $3.6\ \text{kg}$ and a specific heat capacity of $3000\ \text{J / (kg }^\circ\text{C)}$ .

Calculate the final temperature of the food after $160\,000\ \text{J}$ of thermal energy is removed from it.

final temperature = ______ $^\circ\text{C}$

Q58MMediumGeneral Properties of Waves

(a)

Fig. 5.1 is a diagram showing the arrangement of air particles as a longitudinal wave passes through them.

Fig. 5.1

2M

(i)

On Fig. 5.1, mark the centre of a compression with the letter C, and mark the centre of a rarefaction with the letter R.

(ii)

Describe the difference between a compression and a rarefaction.

(b)

In a ripple tank, a water wave is produced by a wooden bar moving up and down on the surface of water.

5M

(i)

The wooden bar makes 45 complete oscillations in $1.0\ \text{minute}$ .

Calculate the frequency of the wave produced.

frequency = ______ $\text{Hz}$

(ii)

The frequency of the water wave is increased by moving the wooden bar up and down more quickly.

State what happens to the speed and what happens to the wavelength of the wave produced.

speed ______
wavelength ______

(iii)

The crests of the water wave move into the shallow region shown in Fig. 5.2.

Fig. 5.2

On Fig. 5.2, draw the crests in the shallow region.

DifficultyMedium

Worked solution

Answer

On Fig. 5.2, draw at least 3 new wavefronts (straight lines) in the shallow region. These lines must be:

Parallel to each other.
Spanning the full width of the shallow region.
Closer together than the wavefronts in the deep water (shorter wavelength, because speed has decreased).
Bent so they are more parallel to the shallow region boundary (refracted towards the normal to the boundary).
They must join smoothly with the incoming wavefronts at the boundary.

Final answer

At least 3 parallel wavefronts in the shallow region, closer together than in deep water, and bent towards the normal (more parallel to the boundary) joining the incoming wavefronts.

Detailed explanation

Walkthrough

When water waves travel from deep water into shallow water, their speed decreases. Because $v = f\lambda$ and frequency $f$ remains constant, the wavelength $\lambda$ must decrease. This means the wavefronts (crests) in the shallow region must be drawn closer together.

Furthermore, because one end of a wavefront enters the shallow region and slows down before the other end, the wavefront pivots. This is refraction. The wavefronts bend to become more parallel to the boundary (i.e., they refract towards the normal to the boundary surface).

To draw this correctly on Fig. 5.2:

The incoming wavefronts are vertical lines travelling right.
The shallow region boundary is a diagonal line. The normal to this boundary is perpendicular to it.
Draw 3 or more straight lines in the shallow region. They should be parallel to each other, spaced closer together than the deep-water wavefronts, and angled so they are more parallel to the diagonal boundary line than the incoming vertical lines. Ensure they connect seamlessly to the incoming wavefronts at the boundary.

Key Takeaways

Water waves slow down in shallow water. This causes refraction: wavefronts bend towards the normal and wavelength decreases. Always draw wavefronts (not rays) for ripple tank refraction.

Common Mistakes

Drawing rays (arrows) instead of wavefronts (lines). The question asks for 'crests', which are wavefronts.
Drawing the wavefronts in the shallow region further apart (this would happen if speed increased, e.g., deep to shallow is wrong; shallow to deep increases speed).
Drawing the wavefronts parallel to the incoming ones (no refraction drawn).
Drawing the wavefronts bending the wrong way (away from the normal).

Things to Be Careful About

The mark scheme specifically requires 'at least 3 wavefronts', 'parallel to each other', 'on correct side of normal', and 'join wavefronts shown'. Ensure your lines actually connect to the ends of the incoming wavefronts at the boundary.

Techniques used

draw refracted wavefronts at a boundary where wave speed changes

(c)

Describe what is meant by the diffraction of a water wave.

Q610MMediumCurrent, Voltage and ResistanceElectric Circuits

A student sets up a circuit to determine the resistance of a length of wire.

The circuit contains a battery of unknown e.m.f., a length of a wire used to make a resistor X, an ammeter, a voltmeter and a variable resistor R.

(a)

Fig. 6.1 shows part of the circuit diagram.

Fig. 6.1

3M

(i)

On Fig. 6.1, complete the circuit diagram by adding one voltmeter and one ammeter in suitable places to allow the determination of the resistance of X.

(ii)

Explain how X and R act as a variable potential divider.

(b)

The student determines the resistance of the resistor X for five different lengths of the wire making it. The lengths of wire range from $20\ \text{cm}$ to $60\ \text{cm}$ . The type of wire and the cross-sectional area of the wire are kept constant.

Fig. 6.2 shows a graph of the results.

Fig. 6.2

6M

(i)

State the relationship between the resistance of the wire and the length of the wire.

(ii)

Calculate the current in a $90\ \text{cm}$ length of the wire when there is a potential difference (p.d.) of $9.0\ \text{V}$ across it.

Show your working.

current = ______ $\text{A}$

DifficultyMedium

Worked solution

Working

From the graph, at length $L = 60\ \text{cm}$ , resistance $R = 3.6\ \Omega$ .

Since resistance is directly proportional to length:

\text{resistance of } 90\ \text{cm} \text{ wire} = \frac{90}{60} \times 3.6 = 5.4\ \Omega

Using Ohm's law:

I = \frac{V}{R}

I = \frac{9.0}{5.4} = 1.667\ \text{A}

I = 1.7\ \text{A}

Answer

1.7

Final answer

1.7 A

Detailed explanation

Walkthrough

Find the resistance of the $90\ \text{cm}$ length of wire:
From the graph in Fig. 6.2, we can read off a point, e.g., at $L = 60\ \text{cm}$ , $R = 3.6\ \Omega$ (or at $L = 20\ \text{cm}$ , $R = 1.2\ \Omega$ ).
The resistance per unit length is:
$\frac{3.6\ \Omega}{60\ \text{cm}} = 0.060\ \Omega/\text{cm}$
For a $90\ \text{cm}$ length:
$R = 90\ \text{cm} \times 0.060\ \Omega/\text{cm} = 5.4\ \Omega$
Calculate the current:
We are given a potential difference $V = 9.0\ \text{V}$ .
Using $I = \frac{V}{R}$ :
$I = \frac{9.0\ \text{V}}{5.4\ \Omega} = 1.666...\ \text{A} \approx 1.7\ \text{A}$

Key Takeaways

Direct proportionality allows simple scaling: $R_2 = R_1 \times \frac{L_2}{L_1}$ .
Current is determined from potential difference and resistance via $I = \frac{V}{R}$ .

Common Mistakes

Inverting the ratio when scaling the resistance.
Forgetting to calculate resistance for $90\ \text{cm}$ and using $3.6\ \Omega$ directly.
Rounding to 1 significant figure instead of 2.

Things to Be Careful About

State the calculated resistance value ( $5.4\ \Omega$ ) clearly in working to secure intermediate marks.

Techniques used

extrapolate resistance from a graphapply Ohm's law (I = V / R)

(iii)

The p.d. across the wire making the resistor X is kept constant for all the measurements of resistance.

Describe the relationship between the current in the wire and the length of the wire.

(c)

State how the resistance of a wire depends upon the cross-sectional area of the wire.

Q79MMedium-EasyElectromagnetic Induction and TransformersUses of an Oscilloscope

Fig. 7.1 shows an alternating current (a.c.) power supply connected to a transformer.

Fig. 7.1

(a)

Explain how an alternating current in the primary coil produces an alternating output voltage.

DifficultyMedium-Easy

Worked solution

Answer

The alternating current in the primary coil produces a changing (or alternating) magnetic field. The iron core transmits this changing magnetic field to the secondary coil. The changing magnetic field induces an alternating voltage (e.m.f.) in the secondary coil.

Final answer

See working

Detailed explanation

Walkthrough

To explain how a transformer works, we follow the chain of events from the primary coil to the secondary coil. First, an alternating current in the primary coil produces a magnetic field that is constantly changing in strength and reversing in direction. Second, the soft iron core is a good conductor of magnetic flux, so it transmits this changing magnetic field from the primary coil to the secondary coil. Third, by the principle of electromagnetic induction, a changing magnetic field through a coil induces an e.m.f. across it. Because the magnetic field is itself alternating, the induced e.m.f. in the secondary coil is also alternating.

Key Takeaways

A transformer requires an alternating current to produce a changing magnetic field. The iron core links the two coils magnetically. Electromagnetic induction occurs when the magnetic flux through the secondary coil changes.

Common Mistakes

Saying 'the current flows from primary to secondary'. Current does not flow across the gap; energy is transferred magnetically.
Forgetting the word 'changing' or 'alternating'. A steady direct current would produce a constant magnetic field, which would induce zero voltage in the secondary coil.
Saying 'magnetism induces voltage'. The specific phrase required is 'changing magnetic field' or 'changing magnetic flux'.

Things to Be Careful About

Each of the three marks requires a distinct physical link in the chain: (1) AC produces changing magnetic field, (2) core transmits it, (3) changing field induces voltage. Stating only two of these will lose a mark.

Techniques used

explain electromagnetic induction in a transformerrelate changing magnetic field to induced e.m.f.

(b)

A student uses a voltmeter set on a $0\text{--}10\ \text{V}$ range to measure the input and output voltages. She obtains the values shown in Table 7.1.

Table 7.1

input voltage / $\text{V}$	output voltage / $\text{V}$
1.2	2.4
2.3	4.6
4.8	9.6
6.4	no reading

3M

(i)

Suggest why no output voltage reading is obtained with this voltmeter when the input voltage is $6.4\ \text{V}$ .

(ii)

The number of turns on the primary coil is 48.

Calculate the number of turns on the secondary coil.

number of turns = ______

DifficultyEasy

Worked solution

Working

The transformer turns-ratio equation is:

\frac{V_S}{V_P} = \frac{N_S}{N_P}

Rearranging for the number of turns on the secondary coil $N_S$ :

N_S = N_P \times \frac{V_S}{V_P}

Using the values from the third row of the table ( $V_P = 4.8\ \text{V}$ , $V_S = 9.6\ \text{V}$ ) and $N_P = 48$ :

N_S = 48 \times \frac{9.6}{4.8} = 48 \times 2 = 96

Answer

number of turns = 96

Final answer

96

Detailed explanation

Walkthrough

The transformer equation links the voltages across the coils to the number of turns on each coil: $V_S / V_P = N_S / N_P$ . We are given $N_P = 48$ . We can use any of the valid data pairs from the table to find the ratio. Using $V_P = 4.8\ \text{V}$ and $V_S = 9.6\ \text{V}$ :

\frac{9.6}{4.8} = \frac{N_S}{48}

2 = \frac{N_S}{48}

N_S = 96

Any consistent pair from the table gives the same result. The number of turns on the secondary coil is 96.

Key Takeaways

The transformer equation $V_S / V_P = N_S / N_P$ is valid for ideal transformers. The ratio of voltages equals the ratio of turns.

Common Mistakes

Inverting the ratio: writing $N_S / N_P = V_P / V_S$ . This would give $N_S = 24$ , which is wrong.
Using the invalid data point (6.4 V input, no reading) in the calculation.
Forgetting to write the unit or just writing the number. The question asks for 'number of turns', which is a count, so no unit is strictly required, but 96 is the value.

Things to Be Careful About

Ensure you match $V_S$ with $N_S$ and $V_P$ with $N_P$ . The equation can also be written as $V_P / V_S = N_P / N_S$ , which is algebraically equivalent. Make sure your substitution is consistent.

Techniques used

apply the transformer turns-ratio equation

(c)

The student uses an oscilloscope to display an alternating output voltage from the transformer.

Fig. 7.2 shows the front of the oscilloscope before it is connected to the transformer.

Fig. 7.2

When the oscilloscope is connected to the output of the transformer, a trace representing the alternating output voltage is displayed on the screen.

3M

(i)

On Fig. 7.2, draw a trace representing an alternating output voltage on the screen.

(ii)

Describe how to use the trace to measure the maximum value of the output voltage.

DifficultyEasy

Worked solution

Answer

Measure the vertical height (in divisions) of the trace from the centre line to the peak.
Multiply this height by the Y-gain setting (2.0 V/division) to find the maximum voltage.

Final answer

Measure the vertical height of the trace in divisions from the centre line to the peak, then multiply by the Y-gain setting (2.0 V/division).

Detailed explanation

Walkthrough

The maximum value of the output voltage is the peak voltage. On the oscilloscope trace, this corresponds to the maximum vertical distance from the central zero line to the highest point (peak) of the wave. First, count the number of vertical divisions from the centre line to the peak. Second, look at the Y-gain control setting, which is given as 2.0 V/division. This means each vertical division represents 2.0 volts. Multiply the number of divisions by 2.0 to calculate the maximum voltage in volts. For example, if the peak is 2.5 divisions above the centre, the maximum voltage is $2.5 \times 2.0 = 5.0\ \text{V}$ .

Key Takeaways

The vertical deflection on an oscilloscope is proportional to the voltage. The Y-gain setting tells you how many volts each vertical division represents. Maximum voltage = (peak height in divisions) × (Y-gain in V/division).

Common Mistakes

Measuring the total peak-to-peak height instead of the height from the centre line to the peak. The maximum value (amplitude) is half the peak-to-peak value.
Forgetting to multiply by the Y-gain setting. Just stating 'measure the height in divisions' is only half the answer.
Using the timebase setting. The timebase is used for measuring time or frequency, not voltage.

Things to Be Careful About

Be precise with wording: say 'height from the centre line to the peak', not just 'height of the trace'. Ensure you mention multiplying by the Y-gain value. The question asks to 'describe how', so a two-step procedure is required.

Techniques used

measure the height of the trace in divisionsmultiply by Y-gain setting

Q811MMedium-EasyThe Nuclear AtomRadioactivity

Plutonium-239 ( $^{239}_{94}\text{Pu}$ ) is an isotope that is used as the fuel in some nuclear reactors.

(a)

State the names of the types of particles found in a nucleus of plutonium-239, and state how many there are of each type.

(b)

Fig. 8.1 shows the nuclear fission process that occurs within the fuel rods of the nuclear reactor.

Fig. 8.1

4M

(i)

Explain how the fission process produces a chain reaction.

(ii)

Explain how control rods are used to increase and decrease the rate of the chain reaction in a nuclear reactor.

(c)

Plutonium-239 decays by the emission of an alpha particle ( $\alpha$ -particle).

State two differences between an alpha particle and a beta particle ( $\beta$ -particle).

difference 1 ______
difference 2 ______

(d)

Alpha particles from the radioactive source are detected in a cloud chamber or with a spark counter.

3M

(i)

Draw a labelled diagram of either a cloud chamber or a spark counter.

Label the position of the radioactive source with an S.

DifficultyMedium-Easy

Worked solution

Answer

Spark counter:
A spark counter consists of a fine metal wire suspended parallel to a flat metal plate. The radioactive source S is placed between the wire and the plate, very close to the wire. A high voltage supply is connected across the wire and the plate.

Cloud chamber (alternative):
A cloud chamber is a closed container with a volatile liquid (like alcohol) at the bottom and a cooling source (like dry ice) at the base to create a vapour. The radioactive source S is placed inside or just outside the chamber.

Final answer

See diagram

Detailed explanation

Walkthrough

Both a cloud chamber and a spark counter are used to detect alpha particles, which produce clear, straight tracks due to their strong ionising ability.

Spark counter: A fine wire and a metal plate are arranged with a small gap between them. A high voltage is applied across them, just below the threshold for spontaneous sparking. When an alpha particle passes through the gap, it ionises the air molecules. The freed electrons are accelerated by the high voltage and cause a cascade of further ionisations, resulting in a visible spark jumping between the wire and the plate along the ionised track.

Cloud chamber: A sealed container contains air supersaturated with vapour (from a volatile liquid like alcohol). The bottom is cooled, creating a temperature gradient. When an alpha particle passes through, it ionises the air. The supersaturated vapour condenses on these ions, forming tiny liquid droplets that make the particle's path visible as a white track.

Key Takeaways

Spark counter: fine wire, plate, high voltage, source nearby.
Cloud chamber: closed container, volatile liquid/vapour, cooling source, source inside/outside.
Both rely on the ionising effect of alpha particles.

Common Mistakes

Forgetting to label the radioactive source with 'S'.
Drawing a cloud chamber without the cooling mechanism or vapour.
Drawing a spark counter without the high voltage supply or the fine wire.
Confusing the components with a Geiger-Müller tube.

Things to Be Careful About

You only need to draw and label ONE of the two apparatuses. Ensure all key components are clearly labelled.
The source must be explicitly labelled 'S'.

Techniques used

draw a labelled diagram of a cloud chamber or spark counter

(ii)

State what causes the tracks in a cloud chamber or state what causes the sparks in a spark counter.

Q910MMediumStars and the Universe

(a)

The life cycle of a star begins with a large cloud of dust and gas which collapses.

Five later stages of the life cycle of a very massive star are:

$\text{black hole}\quad \text{protostar}\quad \text{red supergiant}\quad \text{stable star}\quad \text{supernova}$

Place these stages in Table 9.1 in the order in which they occur.

Table 9.1

earlier time
cloud of dust and gas
______
______
______
______
______
later time

DifficultyEasy

Worked solution

Answer

Completed Table 9.1:

earlier time
cloud of dust and gas
protostar
stable star
red supergiant
supernova
black hole
later time

Final answer

protostar, stable star, red supergiant, supernova, black hole

Detailed explanation

Walkthrough

The question asks for the order of stages after the initial cloud of dust and gas collapses. For a very massive star, the collapse first produces a protostar, which then becomes a stable star while it is fusing hydrogen. Later it swells into a red supergiant, explodes as a supernova, and the core left behind becomes a black hole.

The mark scheme gives one mark for the first two stages in the correct order (protostar, stable star) and one mark for the last three in the correct order (red supergiant, supernova, black hole). Writing all five in the correct order earns the full two marks.

Key Takeaways

A very massive star follows: protostar → stable star → red supergiant → supernova → black hole.
The stable star is the long, middle stage of a star's life.
The final remnant of a very massive star after a supernova is a black hole.

Common Mistakes

Writing black hole before supernova: the explosion happens before the black hole is formed.
Placing red supergiant before the stable star: a star becomes stable first, then evolves into a red supergiant.
Confusing the life cycle of a low-mass star (which ends as a white dwarf) with that of a very massive star.

Things to Be Careful About

The words “earlier time” and “later time” in the table show the direction of time; read from top to bottom.
The supernova is an explosion, so it must come immediately before the formation of the black hole.

Techniques used

recall the stages of a very massive star's life cycleplace the stages in chronological order in a table

(b)

The original collapse of the cloud of dust and gas that formed the Sun was caused by an inward force.

2M

(i)

State the name of the inward force.

(ii)

Further collapse is prevented by an outward force. The Sun will remain in the stable stage of its life cycle for a few billion years.

Describe what causes the outward force.

(c)

One of the first supernovas ever observed is known as SN185. It was formed from the explosion of a star in the Milky Way galaxy.

The remnants of SN185 are at a distance of 8200 light-years from Earth.

4M

(i)

State what is meant by a 'light-year'.

(ii)

State the time that passed between the explosion that formed SN185 and the electromagnetic radiation from the explosion reaching Earth.

(iii)

A recently observed supernova is SN2014J.

The remnants of SN2014J are 12 million light-years from Earth, outside the Milky Way.

There is no redshift seen in the electromagnetic radiation from the remnants of SN185 but a large redshift is seen in the electromagnetic radiation from the remnants of SN2014J.

Explain this difference.

DifficultyMedium

Worked solution

Answer

SN185 is not moving (or has very little speed) away from Earth, so little or no redshift is seen. SN2014J is much further from Earth, so the expansion of the Universe is carrying it away faster, giving a large redshift.

Final answer

SN185 is not moving / has little speed away from Earth; SN2014J is much further away and moving away faster because of the expansion of the Universe, so its redshift is large

Detailed explanation

Walkthrough

Redshift happens when a source of light is moving away from an observer: the wavelengths of light are stretched, making the light appear redder. If a source has little or no motion away from Earth, little or no redshift is seen.

SN185 is in the Milky Way, close to Earth, so its motion away from Earth is negligible and no redshift is seen.

SN2014J is outside the Milky Way and much further away. The Universe is expanding, and more distant objects are moving away from us faster. Therefore SN2014J is moving away much faster than SN185, so its light is strongly redshifted.

The mark scheme gives one mark for the idea that SN185 is not moving/little speed away, and one mark for the idea that SN2014J is moving away faster (or that the redshift is larger because of the expansion of the Universe).

Key Takeaways

Redshift indicates that a source of light is moving away from the observer.
In an expanding Universe, more distant objects generally recede faster, so they show a larger redshift.
An object in our own galaxy may show little or no redshift because its motion away from us is small.

Common Mistakes

Saying “SN2014J is farther away, so it has redshift” without explaining that it is moving away faster: distance alone does not cause redshift; it is the recession due to expansion.
Saying the supernova explosion itself caused the redshift.
Forgetting to compare the two supernovae explicitly.

Things to Be Careful About

The word “redshift” is one word in 5054 and in the mark scheme.
Use “expansion of the Universe” rather than “the Universe is moving” without explanation.
Make sure both credited points are clearly stated: little/no motion for SN185 and faster recession for SN2014J.

Techniques used

relate redshift to the motion of a source away from Earthrelate greater distance and expansion of the Universe to greater recessional speedcompare the two supernovae

(d)

Most of the atoms found in the early Universe were hydrogen and helium.

The Universe now contains atoms of heavier elements.

Explain how the heavier elements are formed.