3.6 Further mechanics and thermal physics

3.6.1.1

Year 2

Circular motion (A-level only)

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

An object moving at constant speed in a circle accelerates because its velocity changes direction. Its instantaneous velocity is tangential, while acceleration and resultant force point towards the centre.
Angular speed is $\omega=v/r=2\pi f$ in $\text{rad s}^{-1}$ , with one complete revolution equal to $2\pi$ radians.
Therefore $a=v^2/r=\omega^2r$ and $F=mv^2/r=m\omega^2r$ .
“Centripetal force” is not a new force: it is the inward resultant supplied by forces such as tension, friction or gravity.
The specification does not require a direction for the angular-velocity vector.

Velocity is tangential while centripetal acceleration and resultant force point towards the centre.

Worked example

A $0.50\,\text{kg}$ object travels at $6.0\,\text{m s}^{-1}$ in a circle of radius $2.0\,\text{m}$ . Find its acceleration and resultant force.

1. $a=v^2/r=6.0^2/2.0=18\,\text{m s}^{-2}$ .
2. $F=ma=0.50\times18=9.0\,\text{N}$ .
3.Both vectors point radially towards the centre.

Answer: Acceleration is 18 m s⁻² and resultant force is 9.0 N, both inwards.

Common mistakes

Don't draw the resultant force tangentially in the direction of motion.
Don't add a separate centripetal force as well as the real inward forces.
Don't use revolutions per second directly as angular speed without converting each revolution to 2π radians.

Exam tip

Identify the real forces first, then equate their inward resultant to the required centripetal force.

Tier 1 · Easy

ORIGINAL

A turntable rotates at

120\,\text{rev min}^{-1}

. Calculate its angular speed.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

A

0.45\,\text{kg}

object moves at constant speed

6.0\,\text{m s}^{-1}

in a horizontal circle of radius

0.80\,\text{m}

. Determine the resultant force and state its direction.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A

0.240\,\text{kg}

coin rests

0.350\,\text{m}

from the centre of a horizontal rotating disc. The disc can exert a maximum frictional force of

0.990\,\text{N}

on the coin. Determine the greatest rotation frequency for which the coin does not slide. Explain which force supplies the centripetal force.

[5 marks]

Total for this question: 5

3.6.1.2

Year 2

Simple harmonic motion (SHM) (A-level only)

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

Simple harmonic motion has acceleration proportional to displacement and directed towards equilibrium: $a=-\omega^2x$ .
A suitable displacement equation is $x=A\cos\omega t$ , and speed at displacement $x$ satisfies $v=\pm\omega\sqrt{A^2-x^2}$ ; the sign follows the direction of motion.
Speed is greatest at equilibrium, $v_{\max}=\omega A$ , whereas acceleration magnitude is greatest at the extremes, $a_{\max}=\omega^2A$ .
On time graphs, velocity is the gradient of displacement and acceleration is the gradient of velocity, so the three sinusoidal curves have fixed quarter-cycle phase differences.
Define the positive direction before assigning signs.

Displacement, velocity and acceleration in SHM are sinusoidal with fixed phase relationships.

Worked example

An oscillator has $A=0.040\,\text{m}$ and $\omega=5.0\,\text{rad s}^{-1}$ . At $x=+0.024\,\text{m}$ it moves towards equilibrium. Find acceleration and velocity.

1. $a=-\omega^2x=-(5.0)^2(0.024)=-0.60\,\text{m s}^{-2}$ .
2. $|v|=5.0\sqrt{0.040^2-0.024^2}=0.16\,\text{m s}^{-1}$ .
3.Motion towards equilibrium from positive $x$ gives negative velocity.

Answer: Acceleration is −0.60 m s⁻² and velocity is −0.16 m s⁻¹.

Common mistakes

Don't drop the minus sign in $a=-\omega^2x$ .
Don't assume velocity is positive because displacement is positive.
Don't place maximum speed at an extreme rather than at equilibrium.

Exam tip

Use the stated direction of motion to choose the sign after calculating speed.

Tier 1 · Easy

ORIGINAL

For an oscillator, displacement to the right is positive. At one instant

x=+0.030\,\text{m}

and

\omega=4.0\,\text{rad s}^{-1}

. Calculate its acceleration, including its sign.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

An oscillator has amplitude

0.080\,\text{m}

and angular frequency

5.0\,\text{rad s}^{-1}

. At an instant when

x=+0.048\,\text{m}

, it is moving in the negative direction. Determine its velocity and acceleration. Take displacement in the positive direction as positive.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

An oscillator starts at its positive extreme at

t=0

and has

x=0.120\cos(4.00t)

, where

x

is in metres and rightwards is positive. Determine its displacement, velocity and acceleration at

t=0.350\,\text{s}

.

[5 marks]

Total for this question: 5

3.6.1.3

Year 2

Simple harmonic systems (A-level only)

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

A mass-spring oscillator has period $T=2\pi\sqrt{m/k}$ , while a simple pendulum at small angle has $T=2\pi\sqrt{l/g}$ . Other oscillators may be analysed from supplied information.
In ideal spring SHM, total energy is $E=\tfrac12kA^2$ ; elastic potential energy is $E_{\text{p}}=\tfrac12kx^2$ and the remainder is kinetic.
Thus kinetic energy is greatest at equilibrium, potential energy is greatest at the extremes and total energy is constant.
Damping transfers energy to the surroundings, reducing amplitude.
Required practical 7 measures how period depends on mass for a spring and on length for a pendulum.

Kinetic and potential energy exchange during ideal SHM while total energy remains constant.

Worked example

A $0.20\,\text{kg}$ mass oscillates on a spring of stiffness $80\,\text{N m}^{-1}$ . Calculate its period.

1.Use $T=2\pi\sqrt{m/k}$ .
2. $T=2\pi\sqrt{0.20/80}$ .
3. $\sqrt{0.0025}=0.050$ , so multiply by $2\pi$ .

Answer: The period is 0.314 s.

Common mistakes

Don't use amplitude in the mass-spring period equation.
Don't apply the pendulum formula when the angle is not small.
Don't claim damping changes the equilibrium position rather than the amplitude.

Exam tip

For practical graphs, square the period to obtain a linear relation with mass or pendulum length.

Tier 1 · Easy

ORIGINAL

A

0.200\,\text{kg}

mass is attached to a spring of stiffness

80.0\,\text{N m}^{-1}

. Calculate the period of small vertical oscillations.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

A simple pendulum has period

1.60\,\text{s}

. Determine its length. Use

g=9.81\,\text{N kg}^{-1}

and assume the oscillation angle is small.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A

0.300\,\text{kg}

mass oscillates on a spring of stiffness

120\,\text{N m}^{-1}

with amplitude

0.0500\,\text{m}

. Determine its total energy and its speed at displacement

0.0300\,\text{m}

. Damping later reduces the amplitude to

0.0400\,\text{m}

. Calculate the percentage of the original energy that has been dissipated.

[6 marks]

Total for this question: 6

3.6.1.4

Year 2

Forced vibrations and resonance (A-level only)

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

A free vibration follows an initial disturbance and occurs at the system’s natural frequency. A forced vibration is maintained by a periodic driving force and settles at the driving frequency.
Resonance occurs when driving frequency equals natural frequency, causing maximum energy transfer and the greatest steady amplitude.
Increasing damping dissipates more energy per cycle, so the resonance peak becomes lower and broader: the resonance is less sharp.
Resonance is useful in selected mechanical systems and in producing stationary waves, but it can damage structures or machinery driven close to a natural frequency.
A complete explanation links frequency matching, energy transfer and amplitude.

Damping lowers and broadens the amplitude peak and shifts the resonant frequency slightly below the natural frequency.

Worked example

Two identical oscillators are driven through the same range of frequencies; one is more heavily damped. Compare their response curves.

1.Both show maximum response near the same natural frequency.
2.The more heavily damped oscillator loses more energy per cycle.
3.Its maximum amplitude is smaller and its peak is broader.

Answer: Greater damping produces a lower, broader and less sharp resonance peak.

Common mistakes

Don't define resonance only as a large amplitude without matching frequencies.
Don't say a forced oscillator always vibrates at its natural frequency.
Don't claim damping makes the resonance peak taller.

Exam tip

When explaining a resonance hazard, name the driver, the natural frequency and how damping changes energy loss.

Tier 1 · Easy

ORIGINAL

State what is meant by resonance in a forced oscillator.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

Describe how increased damping changes an amplitude-against-driving-frequency resonance curve.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A platform of effective oscillating mass

800\,\text{kg}

behaves like a spring of stiffness

3.20\times10^5\,\text{N m}^{-1}

. A machine drives it at

3.20\,\text{Hz}

. Determine the natural frequency and explain why adding a damper reduces the risk of a large vibration amplitude.

[5 marks]

Total for this question: 5

3.6.2.1

Year 2

Thermal energy transfer (A-level only)

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Definition

The internal energy of a body is the sum of the randomly distributed kinetic energies and potential energies of its particles. It is increased when energy is transferred to the body by heating or when work is done on it (the first law of thermodynamics, qualitatively). For a change of temperature, $Q=mc\Delta\theta$ , where $c$ is the specific heat capacity — the energy needed to raise the temperature of $1\,\text{kg}$ by $1\,\text{K}$ . For a change of state, $Q=ml$ , where $l$ is the specific latent heat: during the change the particles' potential energies change but their kinetic energies do not, so the temperature is constant.

Explanation

Internal energy is the sum of randomly distributed particle kinetic and potential energies. Heating a system or doing work on it increases internal energy; energy transferred from the system or work done by it decreases internal energy.
For a temperature change, $Q=mc\Delta\theta$ .
During a change of state, $Q=ml$ and temperature stays constant while particle potential energy changes rather than average kinetic energy.
Continuous-flow heating often uses $P=\dot{m}c\Delta\theta$ when losses are negligible.
Practical determinations of specific heat capacity or specific latent heat must account for energy transferred to the apparatus and surroundings, electrical measurements and rate or mass uncertainties.

During a change of state, energy increases while temperature remains constant.

Worked example

A

250\,\text{W}

immersion heater is used to warm

0.80\,\text{kg}

of a liquid from

20\,^{\circ}\text{C}

to

65\,^{\circ}\text{C}

. This takes

6.0

minutes. Assuming no energy is transferred to the surroundings, determine the specific heat capacity of the liquid. State an appropriate SI unit for your answer.

[4 marks]

Total for this question: 4

1.Energy supplied: $Q=Pt=250\times(6.0\times60)=90\,000\,\text{J}$ .
2.Equate to $mc\Delta\theta$ with $\Delta\theta=65-20=45\,\text{K}$ : $90\,000=0.80\times c\times45$ .
3. $c=\dfrac{90\,000}{0.80\times45}=2500$ .
4.Unit: $\text{J kg}^{-1}\,\text{K}^{-1}$ (or $\text{J kg}^{-1}\,^{\circ}\text{C}^{-1}$ ). In practice some energy heats the container and surroundings, so this value is an overestimate.

Answer: $c=2500\,\text{J kg}^{-1}\,\text{K}^{-1}$

Common mistakes

Don't use the latent-heat equation for a temperature change.
Don't say particle kinetic energy increases during a constant-temperature change of state.
Don't treat electrical input energy as entirely transferred to the sample in a practical.

Mark-scheme wording to use

energy per kg solid-to-liquid … no temperature change/at melting point — paraphrase of AQA 7408, June 2018 Paper 2, Q01.1
total/sum kinetic and potential particle energies — paraphrase of AQA 7408, June 2019 Paper 2, Q02.1
potential energy increasing … kinetic energy remaining constant — paraphrase of AQA 7408, November 2020 Paper 2, Q01.3
J kg⁻¹ K⁻¹ or J kg⁻¹ degrees C⁻¹ — paraphrase of AQA 7408, November 2020 Paper 2, Q01.5
pump does work … work raises internal energy/random motion/temperature — paraphrase of AQA 7408, June 2018 Paper 2, Q01.3

What the examiner wants

Define internal energy as the SUM of the particles' kinetic and potential energies (for an ideal gas, kinetic only — AQA rejects any implied potential energy there). During melting say potential energy increases and kinetic energy stays constant; 'all the bonds break' or 'kinetic energy increases' is not credited. Split a multi-stage process into $mc\Delta\theta$ and $ml$ terms, quote the unit with the answer, and in a practical name the energy transferred to the container and surroundings as the systematic error.

Exam tip

Split a multi-stage heating process into temperature-change and state-change energy terms before adding them.

Tier 1 · Easy

ORIGINAL

Calculate the energy required to raise the temperature of a

0.250\,\text{kg}

block of specific heat capacity

900\,\text{J kg}^{-1}\,\text{K}^{-1}

by

20.0\,\text{K}

.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

A

120\,\text{W}

heater melts

0.0800\,\text{kg}

of ice already at its melting temperature. Only

80.0\%

of the electrical energy reaches the ice. Calculate the melting time. The specific latent heat of fusion is

3.34\times10^5\,\text{J kg}^{-1}

.

[4 marks]

Total for this question: 4

Tier 3 · Hard

ORIGINAL

Water flows through an electric heater at

0.0180\,\text{kg s}^{-1}

. Its temperature rises from

18.0\,^{\circ}\text{C}

to

42.0\,^{\circ}\text{C}

. The electrical input power is

2.20\,\text{kW}

. Determine the rate of increase of the water's internal energy and the heater efficiency. Explain the energy transfer at particle level. Use

c=4200\,\text{J kg}^{-1}\,\text{K}^{-1}

.

[6 marks]

Total for this question: 6

3.6.2.2

Year 2

Ideal gases (A-level only)

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Definition

The gas laws are experimental relationships between pressure, volume, temperature and mass of gas: at constant temperature $pV$ is constant (Boyle's law); at constant pressure $V\propto T$ (Charles's law); extrapolating $V$ or $p$ to zero gives absolute zero, $-273.15\,^{\circ}\text{C}$ . The ideal gas equation is $pV=nRT$ for $n$ moles or $pV=NkT$ for $N$ molecules, with $R=N_Ak$ ; the work done by a gas expanding at constant pressure is $W=p\Delta V$ . Temperatures must be in kelvin.

Explanation

For a fixed gas mass, empirical gas laws relate pressure, volume and absolute temperature: Boyle’s law applies at constant temperature and Charles’s law at constant pressure.
Absolute zero motivates the kelvin scale, so convert using $T/\text{K}=\theta/^{\circ}\text{C}+273.15$ .
The ideal-gas equation is $pV=nRT$ for $n$ moles or $pV=NkT$ for $N$ molecules; distinguish molar mass from one molecule’s mass.
At constant pressure, work done by an expanding gas is $W=p\Delta V$ .
Required practical 8 investigates Boyle’s and Charles’s laws, with pressure, volume and temperature uncertainties and control variables stated explicitly.

At constant temperature, pressure varies inversely with volume for a fixed ideal-gas mass.

Worked example

A sealed cylinder of volume

0.024\,\text{m}^3

contains an ideal gas at a pressure of

150\,\text{kPa}

and a temperature of

27\,^{\circ}\text{C}

. Calculate the amount of gas in the cylinder, in moles, and the number of molecules of gas. (

R=8.31\,\text{J K}^{-1}\,\text{mol}^{-1}

,

N_A=6.02\times10^{23}\,\text{mol}^{-1}

)

[4 marks]

Total for this question: 4

1.Convert to kelvin: $T=27+273=300\,\text{K}$ ; pressure $p=1.50\times10^{5}\,\text{Pa}$ .
2. $n=\dfrac{pV}{RT}=\dfrac{1.50\times10^{5}\times0.024}{8.31\times300}$ .
3. $n=1.44\,\text{mol}$ .
4. $N=nN_A=1.44\times6.02\times10^{23}=8.7\times10^{23}$ molecules (or directly from $pV=NkT$ with $k=1.38\times10^{-23}\,\text{J K}^{-1}$ ).

Answer: $n=1.44\,\text{mol}$ ; $N=8.7\times10^{23}$

Required practical 8

Investigation of Boyle's law (constant temperature) and Charles's law (constant pressure) for a gas

Apparatus

Boyle's law: a sealed column of air trapped above oil in a vertical glass tube, with a pressure gauge and a foot pump (or a sealed syringe with a pressure sensor)
Charles's law: a capillary tube sealed at one end with air trapped by a bead of concentrated sulfuric acid or oil, a beaker of water, heater, thermometer and ruler
Safety screen and eye protection for the pressurised apparatus; a clamp stand to hold the tubes vertical

Method

Boyle's law: read the length (volume) of the trapped air column and the pressure; pump to raise the pressure in steps, waiting each time for the oil level to settle so the temperature returns to that of the room.
Record $p$ and $V$ for at least six pressures; plot $p$ against $1/V$ — a straight line through the origin — or $\log V$ against $\log p$ , gradient $-1$ .
Charles's law: stand the capillary tube in the water bath with the thermometer beside it; heat slowly, stirring, and record the length of the air column at intervals of about $10\,^{\circ}\text{C}$ as the water cools or warms.
Plot length (volume) against temperature in $^{\circ}\text{C}$ : a straight line whose extrapolation cuts the temperature axis near $-273\,^{\circ}\text{C}$ — absolute zero.

Variables

Independent:: pressure (Boyle) or temperature (Charles)
Dependent:: volume of the trapped air, measured as the length of the column
Control:: temperature (Boyle) — change pressure slowly and wait between readings; pressure (Charles) — the open tube stays at atmospheric pressure; mass of trapped gas (sealed apparatus, no leaks)

Hazards and precautions

Pressurised glass can shatter: use a safety screen and eye protection and do not exceed the marked maximum pressure.
Hot water and a concentrated sulfuric acid bead: heat gently, keep the tube clamped and upright, and wear goggles.

Common mistakes

Don't substitute a Celsius temperature into the ideal-gas equation.
Don't pair molecule number with the molar gas constant, or amount in moles with the Boltzmann constant.
Don't apply the constant-pressure work equation when pressure is changing.

Mark-scheme wording to use

“The volume/size of the gas molecules is negligible/point mass or point molecule … Zero volume is wrong.” — AQA 7408, June 2017 Paper 2, Q01.1
using pV=nRT with absolute temperature — paraphrase of AQA 7408, June 2024 Paper 2, Q01.1
pressure/volume/pV extrapolates to zero … random motion/kinetic energy is zero — paraphrase of AQA 7408, June 2019 Paper 2, Q02.2
same pressure … same mass/moles of air — paraphrase of AQA 7408, June 2018 Paper 3, Q03.2
“change the pressure of the gas slowly” — AQA 7408, June 2022 Paper 3, Q04.7
link gradient -1 to log V=-log p+constant/Boyle's law — paraphrase of AQA 7408, June 2022 Paper 3, Q04.5

What the examiner wants

Kelvin, always: a Celsius temperature in $pV=nRT$ loses the answer. State what is constant when you quote a gas law (Boyle's law needs constant temperature AND mass), and pair $n$ with $R$ or $N$ with $k$ — never mix them. In the Boyle's-law practical AQA credits changing the pressure slowly and waiting for the oil level to settle so the temperature stays constant; 'keep the mass of air constant' is neutral because the apparatus is sealed. A log–log gradient of $-1$ is the evidence for $pV=$ constant.

Exam tip

Write the fixed quantity beside each gas law and convert temperature to kelvin before calculating.

Tier 1 · Easy

ORIGINAL

Convert

25.0\,^{\circ}\text{C}

to kelvin.

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

A sealed container holds

0.120\,\text{mol}

of an ideal gas at

35.0\,^{\circ}\text{C}

in a volume of

2.50\times10^{-3}\,\text{m}^3

. Calculate the pressure. Use

R=8.31\,\text{J mol}^{-1}\,\text{K}^{-1}

.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

An ideal gas initially has pressure

1.00\times10^5\,\text{Pa}

, volume

2.40\times10^{-3}\,\text{m}^3

and temperature

17.0\,^{\circ}\text{C}

. It is compressed to

1.50\times10^{-3}\,\text{m}^3

and heated to

77.0\,^{\circ}\text{C}

. Determine the number of molecules and the final pressure. Use

k=1.38\times10^{-23}\,\text{J K}^{-1}

.

[5 marks]

Total for this question: 5

3.6.2.3

Year 2

Molecular kinetic theory model (A-level only)

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Definition

The kinetic theory model treats a gas as a very large number of identical molecules in continuous random motion — a range of speeds, no preferred direction — whose volume is negligible compared with the container, which exert no forces on each other except during collisions, and whose collisions (with each other and the walls) are perfectly elastic and of negligible duration. Momentum transferred to the walls gives $pV=\dfrac{1}{3}Nm(c_{\text{rms}})^2$ ; combined with $pV=NkT$ this makes the mean kinetic energy of a molecule $\dfrac{1}{2}m(c_{\text{rms}})^2=\dfrac{3}{2}kT=\dfrac{3RT}{2N_A}$ .

Explanation

Brownian motion is random motion of visible particles caused by unequal molecular impacts and supports the molecular model. Kinetic theory assumes many identical point molecules in random motion, negligible intermolecular forces except during perfectly elastic collisions, and negligible collision duration.
Momentum transfer to container walls gives $pV=\tfrac13Nm(c_{\text{rms}})^2$ ; random isotropic motion produces the factor $1/3$ .
Combining this with $pV=NkT$ gives $\tfrac12m(c_{\text{rms}})^2=\tfrac32kT=3RT/(2N_{\text{A}})$ .
For an ideal gas, internal energy is molecular kinetic energy because intermolecular potential energy is neglected.
This theory explains empirical gas laws and illustrates how evidence changes scientific models.

Random molecular collisions transfer momentum to the container walls and produce gas pressure.

Worked example

Air at a pressure of

1.01\times10^{5}\,\text{Pa}

has a density of

1.29\,\text{kg m}^{-3}

. Calculate the root mean square speed of the air molecules. The temperature of the air is then doubled at constant volume. State and explain what happens to the root mean square speed.

[4 marks]

Total for this question: 4

1.Use $p=\dfrac{1}{3}\rho(c_{\text{rms}})^2$ (from $pV=\dfrac{1}{3}Nm(c_{\text{rms}})^2$ with $\rho=Nm/V$ ).
2. $(c_{\text{rms}})^2=\dfrac{3p}{\rho}=\dfrac{3\times1.01\times10^{5}}{1.29}=2.35\times10^{5}\,\text{m}^2\,\text{s}^{-2}$ .
3. $c_{\text{rms}}=485\,\text{m s}^{-1}$ (3 significant figures, matching the data).
4.Since $\dfrac{1}{2}m(c_{\text{rms}})^2=\dfrac{3}{2}kT$ , $(c_{\text{rms}})^2\propto T$ : doubling $T$ multiplies $c_{\text{rms}}$ by $\sqrt{2}$ , to $686\,\text{m s}^{-1}$ .

Answer: $c_{\text{rms}}=485\,\text{m s}^{-1}$ ; doubling the kelvin temperature raises it by a factor of $\sqrt{2}$ to about $690\,\text{m s}^{-1}$

Common mistakes

Don't describe Brownian particles as individual gas molecules.
Don't confuse rms speed with mean speed.
Don't use the average energy per molecule as the total energy of every molecule in the gas.

Mark-scheme wording to use

molecular volume is negligible compared with container volume; collision time is negligible compared with time between collisions; collisions are perfectly elastic — paraphrase of AQA 7408, June 2022 Paper 2, Q02.1
a range of speeds or kinetic energies … no preferred direction/all directions equally likely — paraphrase of AQA 7408, June 2025 Paper 2, Q02.1
“total kinetic energy of the particles” — AQA 7408, June 2023 Paper 2, Q01.1
using p=(1/3)rho(c_rms)² — paraphrase of AQA 7408, June 2024 Paper 2, Q01.2
temperature must increase by a factor of four because (c_rms)² is proportional to T — paraphrase of AQA 7408, June 2024 Paper 2, Q01.3
collision/change momentum; force as rate of change of momentum/impulse; pressure=force/area — paraphrase of AQA 7408, June 2019 Paper 2, Q02.5

What the examiner wants

State assumptions precisely — 'molecules are small' scores nothing; 'volume negligible compared with the container' does — and give the kinetic-theory explanation of pressure as a chain: molecules collide with the wall and change momentum, force is the rate of change of momentum (Newton's second law, with the third law transferring the force to the wall), pressure is force per area. The internal energy of an IDEAL gas is the total kinetic energy of the particles; mean speed increases with temperature, never 'mean velocity'.

Exam tip

In the pressure derivation, show the momentum change, collision interval and isotropic $1/3$ step explicitly.

Tier 1 · Easy

ORIGINAL

Calculate the average translational kinetic energy of one ideal-gas molecule at

300\,\text{K}

. Use

k=1.38\times10^{-23}\,\text{J K}^{-1}

.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

A gas occupies

1.50\times10^{-2}\,\text{m}^3

at pressure

1.20\times10^5\,\text{Pa}

. It contains

3.00\times10^{23}

molecules, each of mass

4.65\times10^{-26}\,\text{kg}

. Use kinetic theory to determine

c_{\text{rms}}

.

[4 marks]

Total for this question: 4

Tier 3 · Hard

ORIGINAL

A cubical container of side

L

holds

N

identical ideal-gas molecules of mass

m

. Derive

pV=\dfrac13Nm(c_{\text{rms}})^2

from molecular collisions with a wall. Hence show that the average translational kinetic energy of one molecule is

\dfrac32kT

.

[6 marks]

Total for this question: 6

Further mechanics and thermal physics (A-level only)

Circular motion (A-level only)

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Simple harmonic motion (SHM) (A-level only)

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Simple harmonic systems (A-level only)

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Forced vibrations and resonance (A-level only)

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Thermal energy transfer (A-level only)

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Ideal gases (A-level only)

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Molecular kinetic theory model (A-level only)

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

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