Further mechanics and thermal physics (A-level only)

3.6 Further mechanics and thermal physics (A-level only) — revision question pack

7 specification points · notes, questions, answers and worked methods

Checked against AQA 7408 section 3.6. Review basis: the qualification registry sourced from the AQA A-level Physics (7408) specification; registry verification recorded 11 July 2026.

How this checking works

Answer all questions in the spaces provided.

3.6.1.1 · Circular motion

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

A turntable rotates at $120\,\text{rev min}^{-1}$ . Calculate its angular speed.
[2 marks]
Total for this question: 2
A rotating arm turns through $135^{\circ}$ . Calculate its angular displacement in radians.
[1 mark]
Total for this question: 1

Tier 2 · Standard

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
A point on a centrifuge rotor is $0.25\,\text{m}$ from the axis and rotates with angular speed $18\,\text{rad s}^{-1}$ . Determine its linear speed and centripetal acceleration.
[3 marks]
Total for this question: 3
A pulley of radius $0.0800\,\text{m}$ rotates at $360\,\text{rev min}^{-1}$ and drives a second pulley of radius $0.240\,\text{m}$ using a belt that does not slip. Determine the angular speed of the second pulley and the centripetal acceleration of a point $0.180\,\text{m}$ from its axis.
[4 marks]
Total for this question: 4

Tier 3 · Hard

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
An aircraft of mass $1.20\times10^3\,\text{kg}$ flies at $65.0\,\text{m s}^{-1}$ in a level circular path of radius $4.50\times10^2\,\text{m}$ . The lift force is perpendicular to the plane of the wings. Determine the centripetal acceleration, the banking angle to the horizontal and the lift force. Use $g=9.81\,\text{N kg}^{-1}$ .
[5 marks]
Total for this question: 5
A $0.350\,\text{kg}$ mass moves as a conical pendulum on a string of length $1.20\,\text{m}$ , with the string at $28.0^{\circ}$ to the vertical. Determine the angular speed, the period and the tension in the string. The maximum safe tension is $4.00\,\text{N}$ ; state whether the string is safe and justify your answer. Use $g=9.81\,\text{N kg}^{-1}$ .
[6 marks]
Total for this question: 6
A $0.180\,\text{kg}$ mass on a light string moves in a vertical circle of radius $0.650\,\text{m}$ . Its speed at the bottom is $5.80\,\text{m s}^{-1}$ . Neglect resistive forces. Determine its speed and the string tension at the top. Determine the minimum speed at the bottom for the string to remain taut throughout the circle, and hence assess the given motion. Use $g=9.81\,\text{N kg}^{-1}$ .
[6 marks]
Total for this question: 6
A cylindrical rotating habitat has an outer floor $42.0\,\text{m}$ from its axis. Its angular speed is chosen so that the centripetal acceleration at this floor is $9.20\,\text{m s}^{-2}$ . Determine the angular speed, rotation period and rotation rate in revolutions per minute. Calculate the contact force on a $68.0\,\text{kg}$ person at the outer floor and at a platform $30.0\,\text{m}$ from the axis, and determine the percentage reduction in contact force at the platform.
[6 marks]
Total for this question: 6

3.6.1.2 · Simple harmonic motion (SHM)

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

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
An oscillator has amplitude $0.075\,\text{m}$ and angular frequency $8.0\,\text{rad s}^{-1}$ . Calculate its maximum speed.
[1 mark]
Total for this question: 1

Tier 2 · Standard

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
An acceleration-against-displacement graph for an oscillator is a straight line through the origin with gradient $-25\,\text{s}^{-2}$ . Determine the angular frequency and period of the motion.
[3 marks]
Total for this question: 3
Using $x=A\sin(2\pi t/T)$ , an oscillator of amplitude $0.0600\,\text{m}$ and period $0.800\,\text{s}$ starts at equilibrium moving in the positive direction. Determine the first time at which its displacement is $+0.0300\,\text{m}$ and its acceleration at that instant.
[4 marks]
Total for this question: 4

Tier 3 · Hard

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
At one instant an oscillator has acceleration $-1.80\,\text{m s}^{-2}$ and velocity $+0.400\,\text{m s}^{-1}$ . Its angular frequency is $6.00\,\text{rad s}^{-1}$ . Determine its displacement, amplitude, maximum speed and the time until it next reaches maximum positive displacement.
[5 marks]
Total for this question: 5
An oscillator has speed $0.500\,\text{m s}^{-1}$ at displacement $0.0300\,\text{m}$ and speed $0.300\,\text{m s}^{-1}$ at displacement $0.0500\,\text{m}$ . Determine its angular frequency, amplitude and period. Determine the maximum acceleration and the fraction of one period spent within $0.0200\,\text{m}$ of equilibrium.
[6 marks]
Total for this question: 6
An oscillator in SHM has maximum speed $0.720\,\text{m s}^{-1}$ and maximum acceleration $5.40\,\text{m s}^{-2}$ . Determine its angular frequency, amplitude and period. At one instant its displacement is $+0.0576\,\text{m}$ and it is moving towards equilibrium. Determine its velocity and acceleration, including their signs when positive displacement is to the right.
[6 marks]
Total for this question: 6
An oscillator of amplitude $0.0800\,\text{m}$ starts at equilibrium moving in the positive direction. It first passes $x=+0.0400\,\text{m}$ and passes the same position again $0.200\,\text{s}$ later. Determine the angular frequency and period. Determine the velocity at each of these two crossings, the acceleration at either crossing and the maximum speed.
[6 marks]
Total for this question: 6

3.6.1.3 · Simple harmonic systems

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

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
A $0.450\,\text{kg}$ trolley attached between two springs oscillates at $1.50\,\text{Hz}$ . Calculate the effective spring constant.
[2 marks]
Total for this question: 2

Tier 2 · Standard

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
A graph of $T^2$ against mass for a mass-spring oscillator has gradient $0.0800\,\text{s}^2\,\text{kg}^{-1}$ . Determine the spring constant.
[3 marks]
Total for this question: 3
A mass suspended from a vertical spring produces a static extension of $0.0850\,\text{m}$ . The mass is displaced slightly and released. Determine the period of its vertical oscillations without first determining either the mass or the spring constant. Use $g=9.81\,\text{N kg}^{-1}$ .
[3 marks]
Total for this question: 3

Tier 3 · Hard

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
A liquid column of total length $0.420\,\text{m}$ oscillates in a U-tube. Its period is given by $T=2\pi\sqrt{L/(2g)}$ . The measured time for 25 oscillations is $(23.5\pm0.1)\,\text{s}$ . Determine the experimental value of $g$ , its percentage difference from $9.81\,\text{m s}^{-2}$ and the percentage uncertainty in $g$ due to the timing uncertainty.
[5 marks]
Total for this question: 5
A platform of unknown mass oscillates on a spring with period $0.800\,\text{s}$ . Adding a $0.200\,\text{kg}$ mass to the platform increases the period to $1.00\,\text{s}$ . Determine the spring constant and the platform mass. The original platform then oscillates with amplitude $0.0600\,\text{m}$ ; determine its total oscillation energy.
[6 marks]
Total for this question: 6
Two light springs are joined end to end on a horizontal surface. One end of the pair is fixed and a $0.300\,\text{kg}$ mass is attached to the other end, so the same force acts through both springs. Their spring constants are $72.0\,\text{N m}^{-1}$ and $120\,\text{N m}^{-1}$ . Show that their effective spring constant is $45.0\,\text{N m}^{-1}$ , then determine the oscillation period and total energy for amplitude $0.0800\,\text{m}$ . At maximum displacement, determine the extension and elastic energy of each spring.
[6 marks]
Total for this question: 6
A $0.600\,\text{kg}$ mass on a spring of stiffness $24.0\,\text{N m}^{-1}$ oscillates with amplitude $0.0900\,\text{m}$ . A small-angle pendulum is adjusted to have the same period. Determine the common period and the pendulum length. The pendulum bob is released from a position $2.50\times10^{-3}\,\text{m}$ above its equilibrium position. Determine the maximum speed of each oscillator and the ratio of the spring oscillator's maximum speed to that of the pendulum. Use $g=9.81\,\text{N kg}^{-1}$ .
[5 marks]
Total for this question: 5

3.6.1.4 · Forced vibrations and resonance

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

State what is meant by resonance in a forced oscillator.
[2 marks]
Total for this question: 2
Give two differences between a free vibration and a forced vibration.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Describe how increased damping changes an amplitude-against-driving-frequency resonance curve.
[3 marks]
Total for this question: 3
An undriven oscillator completes 22 cycles in $4.80\,\text{s}$ . Determine its period. The oscillator is then driven. Calculate the driving frequency for resonance and the corresponding angular frequency.
[3 marks]
Total for this question: 3
The driving frequency of a lightly damped oscillator is increased slowly from well below its natural frequency to well above it. Explain how the steady oscillation amplitude changes and how this is related to energy transfer.
[4 marks]
Total for this question: 4

Tier 3 · Hard

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
A loudspeaker drives the air in a pipe while the driving frequency is varied. Explain why large-amplitude stationary waves occur only near particular frequencies, how increased damping changes the response and how a larger driving-force amplitude affects a resonance peak.
[5 marks]
Total for this question: 5
A vehicle suspension has natural frequency $1.50\,\text{Hz}$ . The vehicle resonates when travelling at $13.0\,\text{km h}^{-1}$ over equally spaced road ridges. Determine the ridge spacing and the frequency at which the vehicle crosses the ridges at $30.0\,\text{km h}^{-1}$ . Explain why the vibration amplitude is large near $13.0\,\text{km h}^{-1}$ and why the ride is smoother at $30.0\,\text{km h}^{-1}$ .
[5 marks]
Total for this question: 5
The steady amplitudes of a forced oscillator are measured at driving frequencies $4.60$ , $4.80$ , $5.00$ , $5.20$ and $5.40\,\text{Hz}$ . Before a damper is fitted, the corresponding amplitudes are $3.2$ , $8.6$ , $15.2$ , $7.9$ and $3.0\,\text{mm}$ . With the damper fitted, they are $4.1$ , $5.4$ , $6.0$ , $5.3$ and $4.0\,\text{mm}$ . Determine the best estimate of the natural frequency and the factor by which the measured peak amplitude is reduced. For each set, use the sampled frequencies to estimate the span over which the amplitude is at least half its own maximum. Hence identify the more strongly damped response. Evaluate a student's claim that fitting the damper changed the natural frequency to $4.80\,\text{Hz}$ .
[6 marks]
Total for this question: 6
A forced oscillator has mass $0.400\,\text{kg}$ and spring constant $100\,\text{N m}^{-1}$ . It is driven at resonance, and the driver supplies $3.84\times10^{-2}\,\text{J}$ per cycle. At its steady amplitude, damping removes $12.0\%$ of the oscillator's total energy each cycle. Determine the resonant driving frequency, the steady amplitude and the mean power supplied by the driver. A stronger damper then removes $20.0\%$ of the total energy per cycle while the driving energy per cycle is unchanged. Determine the new steady amplitude. State the driving frequency at which the steady amplitude is greatest and explain why driving well above this frequency gives a much smaller amplitude.
[6 marks]
Total for this question: 6

3.6.2.1 · Thermal energy transfer

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

Find the energy needed to heat $0.50\,\text{kg}$ of aluminium through $20\,\text{K}$ when $c=900\,\text{J kg}^{-1}\text{K}^{-1}$ .

1.This is a temperature change, so use $Q=mc\Delta\theta$ .
2. $Q=0.50\times900\times20$ .
3.Keep mass in kilograms and temperature change in kelvin.

Answer: The energy required is 9.0 × 10³ J.

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.

Exam tip

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

Tier 1 · Easy

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
A $0.0180\,\text{kg}$ sample of water at its boiling temperature is vaporised. Calculate the energy transferred. The specific latent heat of vaporisation of water is $2.26\times10^6\,\text{J kg}^{-1}$ .
[2 marks]
Total for this question: 2

Tier 2 · Standard

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
A $0.150\,\text{kg}$ metal sample at $95.0\,^{\circ}\text{C}$ is placed in $0.250\,\text{kg}$ of water at $18.0\,^{\circ}\text{C}$ . The final temperature is $22.5\,^{\circ}\text{C}$ . Determine the specific heat capacity of the metal, assuming no energy is transferred to the surroundings. Use $c_{\text{water}}=4200\,\text{J kg}^{-1}\,\text{K}^{-1}$ .
[3 marks]
Total for this question: 3
A $12.0\,\text{V}$ heater carrying $1.50\,\text{A}$ heats water in a calorimeter for $300\,\text{s}$ . The temperature of the water and calorimeter rises by $15.0\,\text{K}$ . The calorimeter has heat capacity $120\,\text{J K}^{-1}$ . Determine the mass of water, assuming no energy is transferred to the surroundings. Use $c_{\text{water}}=4200\,\text{J kg}^{-1}\,\text{K}^{-1}$ .
[4 marks]
Total for this question: 4

Tier 3 · Hard

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
A $0.800\,\text{kg}$ metal block is heated for $2.40\times10^2\,\text{s}$ by a heater operating at $12.0\,\text{V}$ and $3.20\,\text{A}$ . Its temperature rises by $10.5\,\text{K}$ . Determine the measured specific heat capacity and explain the effect of energy transfer to the surroundings on this result. If the accepted specific heat capacity is $900\,\text{J kg}^{-1}\,\text{K}^{-1}$ , calculate the mean rate of energy transfer to the surroundings.
[5 marks]
Total for this question: 5
A $0.500\,\text{kg}$ solid block of specific heat capacity $800\,\text{J kg}^{-1}\,\text{K}^{-1}$ is heated by a constant $32.0\,\text{W}$ heater in a room at $20.0\,^{\circ}\text{C}$ . When the block is at $50.0\,^{\circ}\text{C}$ , its temperature is rising at $0.0500\,\text{K s}^{-1}$ . Assume the rate of energy transfer to the surroundings is proportional to the block's temperature difference from the room. Determine the rate of energy transfer to the surroundings at $50.0\,^{\circ}\text{C}$ and the predicted equilibrium temperature.
[6 marks]
Total for this question: 6
A $1.80\,\text{kg}$ thermal-storage material is initially at $22.0\,^{\circ}\text{C}$ and melts at $48.0\,^{\circ}\text{C}$ . Its specific latent heat of fusion is $1.65\times10^5\,\text{J kg}^{-1}$ ; its specific heat capacity while solid is $2.10\times10^3\,\text{J kg}^{-1}\,\text{K}^{-1}$ . A $240\,\text{W}$ heater operates for $36.0\,\text{min}$ , and $72.0\%$ of its energy reaches the material. Determine whether all the material melts and, if not, the mass and percentage that melt.
[6 marks]
Total for this question: 6
A $0.950\,\text{kg}$ alloy block of specific heat capacity $520\,\text{J kg}^{-1}\,\text{K}^{-1}$ is initially at $165\,^{\circ}\text{C}$ . It is placed in an insulated container holding $0.300\,\text{kg}$ of a phase-change material initially at $18.0\,^{\circ}\text{C}$ . The material melts at $38.0\,^{\circ}\text{C}$ ; its specific heat capacity is $1.80\times10^3\,\text{J kg}^{-1}\,\text{K}^{-1}$ when solid, its specific latent heat of fusion is $1.40\times10^5\,\text{J kg}^{-1}$ and its liquid specific heat capacity is $2.40\times10^3\,\text{J kg}^{-1}\,\text{K}^{-1}$ . Determine whether it melts completely and calculate the final equilibrium temperature.
[6 marks]
Total for this question: 6

3.6.2.2 · Ideal gases

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

Find the pressure of $0.20\,\text{mol}$ of ideal gas at $300\,\text{K}$ in $5.0\times10^{-3}\,\text{m}^3$ using $R=8.31\,\text{J mol}^{-1}\text{K}^{-1}$ .

1.Rearrange $pV=nRT$ to $p=nRT/V$ .
2. $p=(0.20)(8.31)(300)/(5.0\times10^{-3})$ .
3.All quantities are already in SI units.

Answer: The pressure is 1.0 × 10⁵ Pa to two significant figures.

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.

Exam tip

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

Tier 1 · Easy

Convert $25.0\,^{\circ}\text{C}$ to kelvin.
[1 mark]
Total for this question: 1
A fixed mass of gas at constant temperature has pressure $1.20\times10^5\,\text{Pa}$ and volume $3.50\times10^{-4}\,\text{m}^3$ . Calculate its volume when the pressure is $2.80\times10^5\,\text{Pa}$ .
[2 marks]
Total for this question: 2

Tier 2 · Standard

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
An ideal gas expands at constant pressure $1.40\times10^5\,\text{Pa}$ from $4.50\times10^{-3}\,\text{m}^3$ to $7.20\times10^{-3}\,\text{m}^3$ . Calculate the work done by the gas and state, with a reason, whether the gas is heated during the expansion.
[3 marks]
Total for this question: 3
In a Boyle's-law experiment, one reading is $p=(1.02\times10^5\,\text{Pa})\pm2\%$ and $V=(60.0\,\text{cm}^3)\pm1\%$ . A second is $p=(1.50\times10^5\,\text{Pa})\pm2\%$ and $V=(40.5\,\text{cm}^3)\pm1\%$ . Determine whether the readings are consistent with constant $pV$ . For this calculation, add percentage uncertainties in multiplied quantities.
[4 marks]
Total for this question: 4

Tier 3 · Hard

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
A $0.860\,\text{g}$ gas sample occupies $150\,\text{cm}^3$ at $1.01\times10^5\,\text{Pa}$ and $298\,\text{K}$ . Determine the amount of gas, its molar mass and the mass of one molecule. Use $R=8.31\,\text{J mol}^{-1}\,\text{K}^{-1}$ and $N_{\text{A}}=6.02\times10^{23}\,\text{mol}^{-1}$ .
[5 marks]
Total for this question: 5
A rigid vessel of volume $1.20\times10^{-2}\,\text{m}^3$ contains nitrogen gas at pressure $4.80\times10^5\,\text{Pa}$ and temperature $320\,\text{K}$ . After some gas escapes, the remaining gas is at $2.60\times10^5\,\text{Pa}$ and $290\,\text{K}$ . Determine the fraction of the molecules that escaped and the mass of nitrogen lost. Use $R=8.31\,\text{J mol}^{-1}\,\text{K}^{-1}$ and nitrogen molar mass $2.80\times10^{-2}\,\text{kg mol}^{-1}$ .
[6 marks]
Total for this question: 6
A constant-pressure gas apparatus has an unmarked connecting tube whose volume must be included. The graduated cylinder reads $42.0\,\text{cm}^3$ at $290\,\text{K}$ and $58.0\,\text{cm}^3$ at $370\,\text{K}$ . The gas pressure is $1.05\times10^5\,\text{Pa}$ . Determine the unmarked volume, the cylinder reading predicted at $273.15\,\text{K}$ , the amount of trapped gas and the work done by the gas between the two measured states.
[6 marks]
Total for this question: 6
A cylinder contains $0.350\,\text{mol}$ of ideal gas at $285\,\text{K}$ and $1.10\times10^5\,\text{Pa}$ . The gas is heated at constant pressure until its volume is $1.40$ times its initial value. Determine the initial and final volumes, the temperature after this expansion and the work done by the gas. The piston is then locked in position and the gas is heated further to $510\,\text{K}$ . Determine the final pressure.
[6 marks]
Total for this question: 6

3.6.2.3 · Molecular kinetic theory model

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

Calculate the average translational kinetic energy of one ideal-gas molecule at $300\,\text{K}$ using $k=1.38\times10^{-23}\,\text{J K}^{-1}$ .

1.Use $\overline{E_{\text{k}}}=\tfrac32kT$ .
2. $\overline{E_{\text{k}}}=1.5(1.38\times10^{-23})(300)$ .
3.This is the average for one molecule, not the whole gas.

Answer: The average translational kinetic energy is 6.21 × 10⁻²¹ J.

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.

Exam tip

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

Tier 1 · Easy

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
Explain how Brownian motion provides evidence for the existence of atoms or molecules.
[2 marks]
Total for this question: 2

Tier 2 · Standard

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
Explain, using the molecular model, why the pressure of a fixed volume of ideal gas increases when its temperature rises.
[3 marks]
Total for this question: 3
Helium and argon gases are at the same temperature. Their molar masses are $4.00\times10^{-3}\,\text{kg mol}^{-1}$ and $4.00\times10^{-2}\,\text{kg mol}^{-1}$ respectively. Determine the ratio $c_{\text{rms, helium}}/c_{\text{rms, argon}}$ and explain the difference.
[4 marks]
Total for this question: 4

Tier 3 · Hard

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
A sample contains $0.240\,\text{mol}$ of monatomic helium at $350\,\text{K}$ . Determine its internal energy and the root-mean-square speed of its atoms. Use $R=8.31\,\text{J mol}^{-1}\,\text{K}^{-1}$ and helium molar mass $4.00\times10^{-3}\,\text{kg mol}^{-1}$ . Explain why compressing this ideal gas at constant temperature does not change its internal energy.
[5 marks]
Total for this question: 5
An ideal gas at $310\,\text{K}$ has pressure $1.05\times10^5\,\text{Pa}$ and density $0.164\,\text{kg m}^{-3}$ . Determine the root-mean-square molecular speed, the mass of one molecule and the number of molecules per unit volume. Use $k=1.38\times10^{-23}\,\text{J K}^{-1}$ .
[6 marks]
Total for this question: 6
Nitrogen at $420\,\text{K}$ and helium at an unknown temperature have the same root-mean-square molecular speed. Their molar masses are $2.80\times10^{-2}\,\text{kg mol}^{-1}$ and $4.00\times10^{-3}\,\text{kg mol}^{-1}$ respectively. Determine the common rms speed and the helium temperature. If the gases also have the same pressure, determine the ratio of their number densities and the ratio of their average translational kinetic energies per molecule.
[6 marks]
Total for this question: 6
An ideal gas initially has pressure $1.60\times10^5\,\text{Pa}$ , volume $4.50\times10^{-3}\,\text{m}^3$ and temperature $300\,\text{K}$ . After heating and allowing some gas to escape, the number of molecules remaining is $0.800$ times the initial number and their root-mean-square speed is $1.25$ times its initial value. The volume is unchanged. Determine the final temperature and pressure, the initial and final internal energies and the percentage change in internal energy. Explain, using molecular impacts, why the pressure rises even though fewer molecules remain.
[6 marks]
Total for this question: 6

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

3.6.1.1 · Circular motion

Tier 1 · Easy

Mark scheme for 3.6.1.1 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$12.6\,\text{rad s}^{-1}$	$120\,\text{rev min}^{-1}=120/60=2.00\,\text{Hz}$ . Therefore $\omega=2\pi f=2\pi\times2.00=12.6\,\text{rad s}^{-1}$ .	2
02.1	$2.36\,\text{rad}$	Use $\theta_{\text{rad}}=\theta_{\text{deg}}\pi/180$ . Thus $\theta=135\pi/180=3\pi/4=2.36\,\text{rad}$ to three significant figures.	1

Tier 2 · Standard

Mark scheme for 3.6.1.1 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$20\,\text{N}$ towards the centre of the circle	Use $F=\dfrac{mv^2}{r}=\dfrac{0.45\times6.0^2}{0.80}=20.25\,\text{N}$ . To two significant figures this is $20\,\text{N}$ , directed radially inwards towards the centre.	3
02.1	$4.5\,\text{m s}^{-1}$ $81\,\text{m s}^{-2}$	The linear speed is $v=\omega r=18\times0.25=4.5\,\text{m s}^{-1}$ . The centripetal acceleration is $a=\omega^2r=18^2\times0.25=81\,\text{m s}^{-2}$ , both to two significant figures.	3
03.1	$12.6\,\text{rad s}^{-1}$ $28.4\,\text{m s}^{-2}$	The first pulley rotates at $360/60=6.00\,\text{Hz}$ , so its angular speed is $\omega_1=2\pi f=12\pi\,\text{rad s}^{-1}$ . The belt speed is $v=\omega_1r_1=12\pi(0.0800)=0.96\pi\,\text{m s}^{-1}$ . With no slipping, this is the rim speed of the second pulley, so $\omega_2=v/r_2=0.96\pi/0.240=4\pi=12.6\,\text{rad s}^{-1}$ . Hence $a=\omega_2^2r=(4\pi)^2(0.180)=28.4\,\text{m s}^{-2}$ .	4

Tier 3 · Hard

Mark scheme for 3.6.1.1 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$0.546\,\text{Hz}$ ; static friction acts towards the centre	At the limiting frequency, the maximum static friction supplies the required centripetal force, so $0.990=m\omega^2r$ . Hence $\omega=\sqrt{0.990/(0.240\times0.350)}=3.43\,\text{rad s}^{-1}$ . Therefore $f=\omega/(2\pi)=3.43/(2\pi)=0.546\,\text{Hz}$ . Static friction is the real horizontal force and acts radially inwards.	5
02.1	$9.39\,\text{m s}^{-2}$ $43.7^{\circ}$ $1.63\times10^4\,\text{N}$	The centripetal acceleration is $a=v^2/r=65.0^2/450=9.39\,\text{m s}^{-2}$ . For level flight, $L\cos\theta=mg$ , while the horizontal component supplies the centripetal force: $L\sin\theta=ma$ . Hence $\tan\theta=a/g=9.39/9.81$ , giving $\theta=43.7^{\circ}$ . Also $L=m\sqrt{g^2+a^2}=1200\sqrt{9.81^2+9.3889^2}=1.63\times10^4\,\text{N}$ to three significant figures.	5
03.1	$3.04\,\text{rad s}^{-1}$ $2.06\,\text{s}$ $3.89\,\text{N}$ The string is safe because $3.89\,\text{N}<4.00\,\text{N}$ .	The circular-path radius is $r=l\sin\theta$ . Resolving the tension vertically gives $T\cos\theta=mg$ , while its horizontal component supplies the centripetal force: $T\sin\theta=m\omega^2r$ . Dividing the equations and using $r=l\sin\theta$ gives $\omega=\sqrt{g/(l\cos\theta)}=\sqrt{9.81/[1.20\cos(28.0^{\circ})]}=3.04\,\text{rad s}^{-1}$ . Hence the period is $2\pi/\omega=2.06\,\text{s}$ . The tension is $T=mg/\cos\theta=0.350(9.81)/\cos(28.0^{\circ})=3.89\,\text{N}$ , below the $4.00\,\text{N}$ safe limit.	6
04.1	$2.85\,\text{m s}^{-1}$ at the top $0.487\,\text{N}$ $5.65\,\text{m s}^{-1}$ minimum at the bottom The string remains taut because $5.80\,\text{m s}^{-1}>5.65\,\text{m s}^{-1}$ .	Energy conservation between bottom and top gives $\tfrac12mv_b^2=\tfrac12mv_t^2+mg(2r)$ , so $v_t=\sqrt{5.80^2-4(9.81)(0.650)}=2.852\,\text{m s}^{-1}$ . At the top, both tension and weight act towards the centre: $T+mg=mv_t^2/r$ . Hence $T=m(v_t^2/r-g)=0.180[(2.852)^2/0.650-9.81]=0.487\,\text{N}$ . At the limiting condition the top tension is zero, so $v_t^2=gr$ . Combining this with energy conservation gives $v_b^2=5gr$ , and $v_{b,\min}=\sqrt{5(9.81)(0.650)}=5.65\,\text{m s}^{-1}$ . The stated bottom speed exceeds this value, so the string remains taut.	6
05.1	$0.468\,\text{rad s}^{-1}$ $13.4\,\text{s}$ $4.47\,\text{rev min}^{-1}$ $626\,\text{N}$ at the outer floor and $447\,\text{N}$ at the platform $28.6\%$ reduction	Using $a=\omega^2r$ , $\omega=\sqrt{9.20/42.0}=0.4680\,\text{rad s}^{-1}$ . The period is $T=2\pi/\omega=13.4249\,\text{s}=13.4\,\text{s}$ and the rotation rate is $(\omega/2\pi)60=4.469\,\text{rev min}^{-1}=4.47\,\text{rev min}^{-1}$ . The floor contact force supplies the centripetal force, so at the outer floor $F=ma=68.0(9.20)=625.6\,\text{N}=626\,\text{N}$ . At $30.0\,\text{m}$ , $F=m\omega^2r=68.0(0.4680)^2(30.0)=446.9\,\text{N}=447\,\text{N}$ . The reduction is $(625.6-446.9)/625.6\times100=28.6\%$ .	6

3.6.1.2 · Simple harmonic motion (SHM)

Tier 1 · Easy

Mark scheme for 3.6.1.2 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$-0.48\,\text{m s}^{-2}$	Use $a=-\omega^2x=-(4.0)^2(0.030)=-0.48\,\text{m s}^{-2}$ . The negative sign shows that the acceleration is to the left, towards equilibrium.	2
02.1	$0.60\,\text{m s}^{-1}$	The maximum speed is $v_{\max}=\omega A=8.0\times0.075=0.60\,\text{m s}^{-1}$ .	1

Tier 2 · Standard

Mark scheme for 3.6.1.2 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$-0.32\,\text{m s}^{-1}$ $-1.2\,\text{m s}^{-2}$	The speed is $\omega\sqrt{A^2-x^2}=5.0\sqrt{0.080^2-0.048^2}=0.32\,\text{m s}^{-1}$ . It is moving in the negative direction, so $v=-0.32\,\text{m s}^{-1}$ . Also $a=-\omega^2x=-(5.0)^2(0.048)=-1.2\,\text{m s}^{-2}$ .	3
02.1	$5.0\,\text{rad s}^{-1}$ $1.26\,\text{s}$	For SHM, $a=-\omega^2x$ , so the graph gradient is $-\omega^2$ . Hence $\omega=\sqrt{25}=5.0\,\text{rad s}^{-1}$ and $T=2\pi/\omega=2\pi/5.0=1.26\,\text{s}$ .	3
03.1	$0.0667\,\text{s}$ $-1.85\,\text{m s}^{-2}$	For this initial condition, $x=A\sin(2\pi t/T)$ . Since $x/A=0.0300/0.0600=0.500$ , the first phase angle is $\pi/6$ . Hence $t=(\pi/6)T/(2\pi)=T/12=0.0667\,\text{s}$ . The angular frequency is $\omega=2\pi/T=7.85\,\text{rad s}^{-1}$ , so $a=-\omega^2x=-(7.85)^2(0.0300)=-1.85\,\text{m s}^{-2}$ .	4

Tier 3 · Hard

Mark scheme for 3.6.1.2 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$+0.0204\,\text{m}$ $-0.473\,\text{m s}^{-1}$ $-0.326\,\text{m s}^{-2}$	At $t=0.350\,\text{s}$ , $\omega t=4.00\times0.350=1.40\,\text{rad}$ . Thus $x=0.120\cos1.40=+0.0204\,\text{m}$ . Differentiating gives $v=-A\omega\sin\omega t$ , so $v=-0.120\times4.00\sin1.40=-0.473\,\text{m s}^{-1}$ . Finally $a=-\omega^2x=-(4.00)^2(0.0204)=-0.326\,\text{m s}^{-2}$ .	5
02.1	$+0.0500\,\text{m}$ $0.0833\,\text{m}$ $0.500\,\text{m s}^{-1}$ $0.155\,\text{s}$	From $a=-\omega^2x$ , $x=-a/\omega^2=1.80/6.00^2=+0.0500\,\text{m}$ . Using $v^2=\omega^2(A^2-x^2)$ gives $A=\sqrt{x^2+(v/\omega)^2}=\sqrt{0.0500^2+(0.400/6.00)^2}=0.0833\,\text{m}$ . Therefore $v_{\max}=\omega A=0.500\,\text{m s}^{-1}$ . Since the positive velocity shows that the oscillator is moving towards $+A$ , the remaining phase angle is $\cos^{-1}(x/A)=\cos^{-1}(0.600)=0.927\,\text{rad}$ , so the time is $0.927/6.00=0.155\,\text{s}$ .	5
03.1	$10.0\,\text{rad s}^{-1}$ $0.0583\,\text{m}$ $0.628\,\text{s}$ $5.83\,\text{m s}^{-2}$ $0.223$ of one period	Using $v^2=\omega^2(A^2-x^2)$ at both positions and subtracting gives $0.500^2-0.300^2=\omega^2(0.0500^2-0.0300^2)$ . Thus $0.160=0.00160\omega^2$ , so $\omega=10.0\,\text{rad s}^{-1}$ . Then $A=\sqrt{x^2+v^2/\omega^2}=\sqrt{0.0300^2+0.500^2/10.0^2}=0.0583\,\text{m}$ and $T=2\pi/\omega=0.628\,\text{s}$ . The maximum acceleration is $a_{\max}=\omega^2A=10.0^2(0.0583095)=5.83\,\text{m s}^{-2}$ . Writing $x=A\sin\phi$ , the motion is within $0.0200\,\text{m}$ of equilibrium for a fraction $4\sin^{-1}(0.0200/A)/(2\pi)=0.222884$ , or $0.223$ , of each period.	6
04.1	$7.50\,\text{rad s}^{-1}$ $0.0960\,\text{m}$ $0.838\,\text{s}$ $-0.576\,\text{m s}^{-1}$ $-3.24\,\text{m s}^{-2}$	For SHM, $v_{\max}=\omega A$ and $a_{\max}=\omega^2A$ , so $\omega=a_{\max}/v_{\max}=5.40/0.720=7.50\,\text{rad s}^{-1}$ . Then $A=v_{\max}/\omega=0.720/7.50=0.0960\,\text{m}$ and $T=2\pi/\omega=0.83776\,\text{s}=0.838\,\text{s}$ . The speed at $x=0.0576\,\text{m}$ is $\omega\sqrt{A^2-x^2}=7.50\sqrt{0.0960^2-0.0576^2}=0.576\,\text{m s}^{-1}$ . Motion towards equilibrium from positive displacement gives $v=-0.576\,\text{m s}^{-1}$ . Also $a=-\omega^2x=-(7.50)^2(0.0576)=-3.24\,\text{m s}^{-2}$ .	6
05.1	$10.5\,\text{rad s}^{-1}$ $0.600\,\text{s}$ $+0.726\,\text{m s}^{-1}$ then $-0.726\,\text{m s}^{-1}$ $-4.39\,\text{m s}^{-2}$ $0.838\,\text{m s}^{-1}$	Write $x=A\sin\omega t$ . Since $x/A=0.0400/0.0800=0.500$ , the first crossing has phase $\pi/6$ and the next has phase $5\pi/6$ . The phase change is therefore $2\pi/3$ in $0.200\,\text{s}$ , so $\omega=(2\pi/3)/0.200=10.472\,\text{rad s}^{-1}=10.5\,\text{rad s}^{-1}$ . Equivalently, using the booklet form $x=A\cos(\omega t)$ measured from a displacement maximum, the two crossings are symmetric about that maximum, so $\cos(0.100\omega)=0.500$ and $\omega=\pi/(3\times0.100)$ , the same value. The period is $2\pi/\omega=0.600\,\text{s}$ . At the crossings, $\|v\|=\omega\sqrt{A^2-x^2}=10.472\sqrt{0.0800^2-0.0400^2}=0.7255\,\text{m s}^{-1}$ , positive at the first crossing and negative at the second. The acceleration is $a=-\omega^2x=-(10.472)^2(0.0400)=-4.386\,\text{m s}^{-2}$ . The maximum speed is $\omega A=10.472(0.0800)=0.8378\,\text{m s}^{-1}$ .	6

3.6.1.3 · Simple harmonic systems

Tier 1 · Easy

Mark scheme for 3.6.1.3 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$0.314\,\text{s}$	$T=2\pi\sqrt{m/k}=2\pi\sqrt{0.200/80.0}=2\pi(0.0500)=0.314\,\text{s}$ .	2
02.1	$40.0\,\text{N m}^{-1}$	For a mass-spring oscillator, $f=(1/2\pi)\sqrt{k/m}$ , so $k=4\pi^2mf^2=4\pi^2(0.450)(1.50)^2=40.0\,\text{N m}^{-1}$ to three significant figures.	2

Tier 2 · Standard

Mark scheme for 3.6.1.3 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$0.636\,\text{m}$	From $T=2\pi\sqrt{l/g}$ , $l=g\left(\dfrac{T}{2\pi}\right)^2$ . Therefore $l=9.81\left(\dfrac{1.60}{2\pi}\right)^2=0.636\,\text{m}$ .	3
02.1	$493\,\text{N m}^{-1}$	Squaring $T=2\pi\sqrt{m/k}$ gives $T^2=(4\pi^2/k)m$ , so the gradient is $4\pi^2/k$ . Hence $k=4\pi^2/0.0800=493\,\text{N m}^{-1}$ to three significant figures.	3
03.1	$0.585\,\text{s}$	At static equilibrium, $kx=mg$ , so $m/k=x/g$ . Substituting this into $T=2\pi\sqrt{m/k}$ gives $T=2\pi\sqrt{x/g}$ . Therefore $T=2\pi\sqrt{0.0850/9.81}=0.585\,\text{s}$ to three significant figures.	3

Tier 3 · Hard

Mark scheme for 3.6.1.3 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$0.150\,\text{J}$ $0.800\,\text{m s}^{-1}$ $36.0\%$	Initially $E=\tfrac12kA^2=\tfrac12(120)(0.0500)^2=0.150\,\text{J}$ . At $x=0.0300\,\text{m}$ , $E_{\text{p}}=\tfrac12kx^2=0.0540\,\text{J}$ , so $E_{\text{k}}=0.150-0.0540=0.0960\,\text{J}$ . From $E_{\text{k}}=\tfrac12mv^2$ , $v=\sqrt{2(0.0960)/0.300}=0.800\,\text{m s}^{-1}$ . The later energy is $\tfrac12(120)(0.0400)^2=0.0960\,\text{J}$ . The dissipated fraction is $(0.150-0.0960)/0.150=0.360$ , or $36.0\%$ .	6
02.1	$9.38\,\text{m s}^{-2}$ $4.4\%$ $0.85\%$	The period is $T=23.5/25=0.940\,\text{s}$ . Rearranging the supplied equation gives $g=2\pi^2L/T^2=2\pi^2(0.420)/(0.940)^2=9.38\,\text{m s}^{-2}$ . The percentage difference is $\|9.3826-9.81\|/9.81\times100=4.4\%$ . Since $g\propto T^{-2}$ and dividing by 25 does not change percentage uncertainty, the percentage uncertainty in $g$ is $2(0.1/23.5)\times100=0.85\%$ .	5
03.1	$21.9\,\text{N m}^{-1}$ $0.356\,\text{kg}$ $3.95\times10^{-2}\,\text{J}$	For the original platform, $T_1^2=4\pi^2m/k$ . With the added mass, $T_2^2=4\pi^2(m+0.200)/k$ . Subtracting gives $T_2^2-T_1^2=4\pi^2(0.200)/k$ , so $k=4\pi^2(0.200)/(1.00^2-0.800^2)=21.932\ldots\,\text{N m}^{-1}$ , or $21.9\,\text{N m}^{-1}$ . Using the unrounded $k$ , $m=kT_1^2/(4\pi^2)=0.35556\ldots\,\text{kg}=0.356\,\text{kg}$ . The total energy is $E=\tfrac12kA^2=\tfrac12(21.932\ldots)(0.0600)^2=3.9478\times10^{-2}\,\text{J}=3.95\times10^{-2}\,\text{J}$ .	6
04.1	$45.0\,\text{N m}^{-1}$ $0.513\,\text{s}$ $0.144\,\text{J}$ $0.0500\,\text{m}$ and $0.0900\,\text{J}$ for the $72.0\,\text{N m}^{-1}$ spring $0.0300\,\text{m}$ and $0.0540\,\text{J}$ for the $120\,\text{N m}^{-1}$ spring	Series springs carry the same force and their extensions add, so $1/k_{\text{eff}}=1/72.0+1/120$ , giving $k_{\text{eff}}=45.0\,\text{N m}^{-1}$ . The period is $T=2\pi\sqrt{m/k_{\text{eff}}}=2\pi\sqrt{0.300/45.0}=0.513\,\text{s}$ . The total energy is $E=\tfrac12k_{\text{eff}}A^2=\tfrac12(45.0)(0.0800)^2=0.144\,\text{J}$ . At maximum displacement the force is $F=k_{\text{eff}}A=45.0(0.0800)=3.60\,\text{N}$ . Thus $x_1=F/k_1=3.60/72.0=0.0500\,\text{m}$ and $x_2=3.60/120=0.0300\,\text{m}$ . Their energies are $\tfrac12(72.0)(0.0500)^2=0.0900\,\text{J}$ and $\tfrac12(120)(0.0300)^2=0.0540\,\text{J}$ , which sum to $0.144\,\text{J}$ .	6
05.1	$0.993\,\text{s}$ $0.245\,\text{m}$ $0.569\,\text{m s}^{-1}$ for the spring oscillator $0.221\,\text{m s}^{-1}$ for the pendulum $2.57$ (accept $2.57$ to $2.58$ )	For the spring oscillator, $T=2\pi\sqrt{m/k}=2\pi\sqrt{0.600/24.0}=0.99346\,\text{s}=0.993\,\text{s}$ . Equal periods require $2\pi\sqrt{l/g}=2\pi\sqrt{m/k}$ , so $l=gm/k=9.81(0.600)/24.0=0.24525\,\text{m}=0.245\,\text{m}$ . The spring oscillator has $v_{\max}=A\sqrt{k/m}=0.0900\sqrt{24.0/0.600}=0.56921\,\text{m s}^{-1}=0.569\,\text{m s}^{-1}$ . For the pendulum, conservation of energy gives $mgh=\tfrac12mv^2$ , so $v_{\max}=\sqrt{2gh}=\sqrt{2(9.81)(2.50\times10^{-3})}=0.22147\,\text{m s}^{-1}=0.221\,\text{m s}^{-1}$ . Their speed ratio is $0.56921/0.22147=2.57$ .	5

3.6.1.4 · Forced vibrations and resonance

Tier 1 · Easy

Mark scheme for 3.6.1.4 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	The driving frequency equals the natural frequency of the oscillator, producing maximum amplitude.	State both marking points: the periodic driving frequency equals the system's natural frequency, and the forced-oscillation amplitude is then maximum.	2
02.1	A free vibration follows an initial displacement and occurs at the natural frequency, whereas a forced vibration is maintained by a periodic driver and occurs at the driving frequency.	Compare the cause and the frequency: free motion follows an initial disturbance and occurs at the natural frequency; forced motion is maintained by a periodic force and occurs at the driving frequency.	2

Tier 2 · Standard

Mark scheme for 3.6.1.4 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	The maximum amplitude is smaller and the peak is broader and less sharp, with appreciable response over a wider range of frequencies.	Increased damping removes more energy during each cycle. The resonant maximum therefore has a lower amplitude. The peak also becomes wider, so resonance is less sharp and the oscillator responds over a broader range of driving frequencies.	3
02.1	$0.218\,\text{s}$ $4.58\,\text{Hz}$ $28.8\,\text{rad s}^{-1}$	The period is $T=4.80/22=0.218\,\text{s}$ . At resonance the driving frequency equals the natural frequency, so $f=1/T=22/4.80=4.58\,\text{Hz}$ . The angular frequency is $\omega=2\pi f=2\pi(22/4.80)=28.8\,\text{rad s}^{-1}$ , with each result given to three significant figures.	3
03.1	The amplitude rises to a maximum when the driving frequency equals the natural frequency, then falls; at resonance the driver transfers energy most effectively, while away from resonance the phase relationship makes the energy transferred per cycle smaller.	Well below the natural frequency the steady amplitude is relatively small. As the driving frequency approaches the natural frequency, energy is transferred more effectively and the amplitude increases. At equality, resonance gives maximum energy transfer per cycle and the greatest steady amplitude. Above the natural frequency the frequency mismatch reduces the net energy transfer per cycle, so the amplitude falls again.	4

Tier 3 · Hard

Mark scheme for 3.6.1.4 Tier 3 · Hard
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01.1	$3.18\,\text{Hz}$ ; the driver is close to resonance, and extra damping dissipates more energy per cycle and lowers the resonance peak.	For the equivalent mass-spring system, $\omega_0=\sqrt{k/m}=\sqrt{3.20\times10^5/800}=20.0\,\text{rad s}^{-1}$ . Hence $f_0=\omega_0/(2\pi)=20.0/(2\pi)=3.18\,\text{Hz}$ . The $3.20\,\text{Hz}$ drive is very close to this natural frequency, so energy transfer is resonantly enhanced. A damper transfers more mechanical energy to the surroundings each cycle, lowering and broadening the resonance peak and therefore limiting the amplitude.	5
02.1	At an allowed natural frequency the driver resonates with an air-column mode, giving maximum energy transfer and a large stationary-wave amplitude. Increased damping removes more energy per cycle, lowering and broadening each peak; a larger driving force transfers more energy per cycle and raises the peak.	The pipe supports modes with discrete natural frequencies fixed by its length and boundary conditions. When the loudspeaker frequency matches one of them, resonance produces maximum energy transfer to that mode and a large amplitude. Greater damping dissipates more energy per cycle, so each maximum is lower and spread over a wider frequency range. Increasing the driving-force amplitude increases the energy supplied per cycle and therefore raises the resonance peak.	5
03.1	$2.41\,\text{m}$ $3.46\,\text{Hz}$ Near $13.0\,\text{km h}^{-1}$ , the ridge-crossing frequency matches the natural frequency, so resonance gives maximum energy transfer and a large amplitude. At $30.0\,\text{km h}^{-1}$ , the ridge-crossing frequency is well above the natural frequency, so energy transfer and the steady amplitude are smaller.	The resonance speed is $13.0/3.60=3.61\,\text{m s}^{-1}$ . The driving frequency is the number of ridges crossed per second, $f=v/d$ . At resonance $f=f_0$ , so $d=v/f_0=3.61/1.50=2.41\,\text{m}$ . At $30.0\,\text{km h}^{-1}$ , $v=30.0/3.60=8.33\,\text{m s}^{-1}$ and $f=8.33/2.41=3.46\,\text{Hz}$ . Near $13.0\,\text{km h}^{-1}$ , the driving frequency matches the suspension's natural frequency, giving resonance, maximum energy transfer and a large amplitude. At $30.0\,\text{km h}^{-1}$ , the driving frequency is well above the natural frequency, so the energy transferred per cycle and the steady vibration amplitude are smaller, making the ride smoother.	5
04.1	$5.00\,\text{Hz}$ The measured peak is reduced by a factor of $2.53$ . At least $0.40\,\text{Hz}$ but less than $0.80\,\text{Hz}$ before fitting, and at least $0.80\,\text{Hz}$ after fitting. The response with the damper is more strongly damped because its peak is lower and broader. The claim is not supported: both sampled peaks occur at $5.00\,\text{Hz}$ , and measurements spaced by $0.20\,\text{Hz}$ could not establish a smaller shift.	Both data sets have their greatest measured amplitude at $5.00\,\text{Hz}$ , so this is the best estimate of the natural frequency. The peak reduction factor is $15.2/6.0=2.533=2.53$ . Before fitting, half the maximum is $7.6\,\text{mm}$ ; the sampled amplitudes meet or exceed this from $4.80$ to $5.20\,\text{Hz}$ , a span of about $0.40\,\text{Hz}$ . After fitting, half the maximum is $3.0\,\text{mm}$ ; every sampled amplitude from $4.60$ to $5.40\,\text{Hz}$ exceeds this, so the measured span is at least $0.80\,\text{Hz}$ . The lower, broader peak identifies the more strongly damped response. The data do not support a move to $4.80\,\text{Hz}$ because both sampled maxima remain at $5.00\,\text{Hz}$ . With $0.20\,\text{Hz}$ sampling, only a shift on that scale could be resolved reliably.	6
05.1	$2.52\,\text{Hz}$ $0.0800\,\text{m}$ $9.66\times10^{-2}\,\text{W}$ $0.0620\,\text{m}$ with the stronger damper The steady amplitude is greatest when the driving frequency equals the natural frequency, about $2.52\,\text{Hz}$ ; well above this frequency the driving force reverses before the oscillator can respond, so the steady amplitude is much smaller.	The resonant frequency is the natural frequency, $f=(1/2\pi)\sqrt{k/m}=(1/2\pi)\sqrt{100/0.400}=2.5165\,\text{Hz}=2.52\,\text{Hz}$ . At steady amplitude, energy supplied per cycle equals energy dissipated per cycle. Since $E=\tfrac12kA^2$ , $3.84\times10^{-2}=0.120[\tfrac12(100)A^2]$ , giving $A^2=6.40\times10^{-3}\,\text{m}^2$ and $A=0.0800\,\text{m}$ . The mean input power is $(3.84\times10^{-2})(2.5165)=9.663\times10^{-2}\,\text{W}=9.66\times10^{-2}\,\text{W}$ . With the stronger damper, $3.84\times10^{-2}=0.200[\tfrac12(100)A^2]$ , so $A=\sqrt{3.84\times10^{-3}}=0.06197\,\text{m}=0.0620\,\text{m}$ . Stronger damping removes more energy at any given amplitude, so energy balance is reached at a lower amplitude. The amplitude peaks when the driving frequency matches the natural frequency; driven well above it, the driving force reverses direction before the oscillator can follow, so the steady amplitude is far below the resonant value.	6

3.6.2.1 · Thermal energy transfer

Tier 1 · Easy

Mark scheme for 3.6.2.1 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$4.50\times10^3\,\text{J}$	$Q=mc\Delta\theta=0.250\times900\times20.0=4.50\times10^3\,\text{J}$ .	2
02.1	$4.07\times10^4\,\text{J}$	Use $Q=ml$ , so $Q=0.0180(2.26\times10^6)=4.07\times10^4\,\text{J}$ to three significant figures.	2

Tier 2 · Standard

Mark scheme for 3.6.2.1 Tier 2 · Standard
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01.1	$278\,\text{s}$	The energy received by the ice is $Q=ml=0.0800\times3.34\times10^5=2.672\times10^4\,\text{J}$ . The useful heating power is $0.800\times120=96.0\,\text{W}$ . Hence $t=Q/P=2.672\times10^4/96.0=278\,\text{s}$ .	4
02.1	$434\,\text{J kg}^{-1}\,\text{K}^{-1}$	Energy lost by the metal equals energy gained by the water: $0.150c(95.0-22.5)=0.250(4200)(22.5-18.0)$ . Therefore $c=4725/10.875=434\,\text{J kg}^{-1}\,\text{K}^{-1}$ to three significant figures.	3
03.1	$5.71\times10^{-2}\,\text{kg}$	The electrical energy supplied is $E=VIt=12.0(1.50)(300)=5400\,\text{J}$ . The calorimeter receives $Q_{\text{cal}}=C\Delta\theta=120(15.0)=1800\,\text{J}$ , leaving $5400-1800=3600\,\text{J}$ for the water. Hence $m=Q/(c\Delta\theta)=3600/[4200(15.0)]=5.71\times10^{-2}\,\text{kg}$ .	4

Tier 3 · Hard

Mark scheme for 3.6.2.1 Tier 3 · Hard
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01.1	$1.81\,\text{kW}$ $82.5\%$ Heating increases the random kinetic energy of the water molecules; the unused input is transferred to the heater and surroundings.	The temperature rise is $\Delta\theta=42.0-18.0=24.0\,\text{K}$ . For continuous flow, $P_{\text{water}}=\dot{m}c\Delta\theta=0.0180\times4200\times24.0=1814\,\text{W}=1.81\,\text{kW}$ . The efficiency is $1814/2200=0.825$ , or $82.5\%$ . Heating transfers energy to the water, increasing the random kinetic energy of its molecules and hence its internal energy; the remaining input is transferred to the heater and surroundings.	6
02.1	$1.10\times10^3\,\text{J kg}^{-1}\,\text{K}^{-1}$ Energy loss makes the measured value too large. $6.90\,\text{W}$	The electrical input is $E=VIt=12.0\times3.20\times240=9216\,\text{J}$ . Assuming it all heats the block gives $c=E/(m\Delta\theta)=9216/(0.800\times10.5)=1.10\times10^3\,\text{J kg}^{-1}\,\text{K}^{-1}$ . If some input heats the apparatus or surroundings, the calculation assigns too much energy to the block, so the measured $c$ is too large. The input power is $VI=38.4\,\text{W}$ , while the mean rate received by the block is $mc\Delta\theta/t=0.800(900)(10.5)/240=31.5\,\text{W}$ . The mean loss rate is therefore $38.4-31.5=6.90\,\text{W}$ .	5
03.1	$12.0\,\text{W}$ $100\,^{\circ}\text{C}$	The block's heat capacity is $mc=0.500(800)=400\,\text{J K}^{-1}$ . At $50.0\,^{\circ}\text{C}$ , its internal energy is increasing at $mc(dT/dt)=400(0.0500)=20.0\,\text{W}$ . Therefore the loss rate is $32.0-20.0=12.0\,\text{W}$ . The temperature difference from the room is then $30.0\,\text{K}$ , so the proportionality constant is $12.0/30.0=0.400\,\text{W K}^{-1}$ . At equilibrium the temperature is constant, so the loss rate equals the $32.0\,\text{W}$ heater power. The required temperature difference is $32.0/0.400=80.0\,\text{K}$ , giving an equilibrium temperature of $20.0+80.0=100\,^{\circ}\text{C}$ .	6
04.1	The material does not all melt. $1.67\,\text{kg}$ melts $92.6\%$ melts	The useful heater energy is $E=0.720(240)(36.0\times60)=373248\,\text{J}$ . Heating the solid to its melting temperature requires $Q_1=mc\Delta\theta=1.80(2.10\times10^3)(48.0-22.0)=98280\,\text{J}$ . This leaves $373248-98280=274968\,\text{J}$ for melting. Complete melting would require $Q_2=ml=1.80(1.65\times10^5)=297000\,\text{J}$ , so the material does not all melt. The melted mass is $274968/(1.65\times10^5)=1.6665\,\text{kg}=1.67\,\text{kg}$ , which is $(1.6665/1.80)\times100=92.6\%$ .	6
05.1	The phase-change material melts completely. $46.2\,^{\circ}\text{C}$	If the alloy cools to the melting temperature, it releases $0.950(520)(165-38.0)=62738\,\text{J}$ . Heating the solid material to $38.0\,^{\circ}\text{C}$ requires $0.300(1.80\times10^3)(38.0-18.0)=10800\,\text{J}$ and melting all of it requires $0.300(1.40\times10^5)=42000\,\text{J}$ . Their sum is $52800\,\text{J}$ , less than $62738\,\text{J}$ , so complete melting occurs and the final temperature is above $38.0\,^{\circ}\text{C}$ . Let that temperature be $T$ . Energy conservation gives $0.950(520)(165-T)=10800+42000+0.300(2.40\times10^3)(T-38.0)$ . Thus $81510-494T=25440+720T$ , so $T=56070/1214=46.186\,^{\circ}\text{C}=46.2\,^{\circ}\text{C}$ .	6

3.6.2.2 · Ideal gases

Tier 1 · Easy

Mark scheme for 3.6.2.2 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$298\,\text{K}$	$T=25.0+273.15=298.15\,\text{K}$ , which is $298\,\text{K}$ to the nearest kelvin.	1
02.1	$1.50\times10^{-4}\,\text{m}^3$	At constant temperature, $p_1V_1=p_2V_2$ . Thus $V_2=(1.20\times10^5)(3.50\times10^{-4})/(2.80\times10^5)=1.50\times10^{-4}\,\text{m}^3$ .	2

Tier 2 · Standard

Mark scheme for 3.6.2.2 Tier 2 · Standard
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01.1	$1.23\times10^5\,\text{Pa}$	Convert the temperature first: $T=35.0+273.15=308.15\,\text{K}$ . Then $p=\dfrac{nRT}{V}=\dfrac{0.120\times8.31\times308.15}{2.50\times10^{-3}}=1.23\times10^5\,\text{Pa}$ .	3
02.1	$378\,\text{J}$ The gas is heated: at constant pressure the temperature rises during expansion, so its internal energy increases while it also does work, and both must be supplied by heating.	The volume change is $\Delta V=(7.20-4.50)\times10^{-3}=2.70\times10^{-3}\,\text{m}^3$ . Hence $W=p\Delta V=(1.40\times10^5)(2.70\times10^{-3})=378\,\text{J}$ to three significant figures. At constant pressure, $V\propto T$ , so the expansion raises the temperature: $\Delta U>0$ while $W>0$ , and the first law gives $Q=\Delta U+W>0$ — the gas must be heated.	3
03.1	$p_1V_1=(6.12\pm0.18)\,\text{J}$ and $p_2V_2=(6.08\pm0.18)\,\text{J}$ ; the uncertainty intervals overlap, so the readings are consistent with constant $pV$ .	Convert the volumes to SI units. The products are $p_1V_1=(1.02\times10^5)(60.0\times10^{-6})=6.12\,\text{J}$ and $p_2V_2=(1.50\times10^5)(40.5\times10^{-6})=6.075\,\text{J}$ , reported as $6.08\,\text{J}$ . Each product has percentage uncertainty $2\%+1\%=3\%$ , giving absolute uncertainties of about $0.18\,\text{J}$ . The intervals overlap, so any difference between the products is smaller than the measurement uncertainty and the readings are consistent with Boyle's law.	4

Tier 3 · Hard

Mark scheme for 3.6.2.2 Tier 3 · Hard
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01.1	$5.99\times10^{22}$ molecules $1.93\times10^5\,\text{Pa}$	Convert both temperatures: $T_1=17.0+273.15=290.15\,\text{K}$ and $T_2=77.0+273.15=350.15\,\text{K}$ . Initially, $N=\dfrac{p_1V_1}{kT_1}=\dfrac{1.00\times10^5\times2.40\times10^{-3}}{1.38\times10^{-23}\times290.15}=5.99\times10^{22}$ . For the fixed number of molecules, $\dfrac{pV}{T}$ is constant, so $p_2=p_1\dfrac{V_1T_2}{T_1V_2}=1.00\times10^5\dfrac{2.40\times10^{-3}\times350.15}{290.15\times1.50\times10^{-3}}=1.93\times10^5\,\text{Pa}$ .	5
02.1	$6.12\times10^{-3}\,\text{mol}$ $0.141\,\text{kg mol}^{-1}$ $2.34\times10^{-25}\,\text{kg}$	Convert the data: $m=8.60\times10^{-4}\,\text{kg}$ and $V=1.50\times10^{-4}\,\text{m}^3$ . The amount is $n=pV/(RT)=(1.01\times10^5)(1.50\times10^{-4})/[8.31(298)]=6.12\times10^{-3}\,\text{mol}$ . Hence $M=m/n=0.141\,\text{kg mol}^{-1}$ . One molecule has mass $M/N_{\text{A}}=0.1406/(6.02\times10^{23})=2.34\times10^{-25}\,\text{kg}$ , to three significant figures.	5
03.1	$0.402$ (or $40.2\%$ ) $2.44\times10^{-2}\,\text{kg}$	Initially, $n_1=p_1V/(RT_1)=(4.80\times10^5)(1.20\times10^{-2})/[8.31(320)]=2.17\,\text{mol}$ . Finally, $n_2=(2.60\times10^5)(1.20\times10^{-2})/[8.31(290)]=1.29\,\text{mol}$ . The escaped amount is $n_1-n_2=0.871\,\text{mol}$ , so the escaped fraction is $(n_1-n_2)/n_1=0.402$ , or $40.2\%$ . The mass lost is $0.871(2.80\times10^{-2})=2.44\times10^{-2}\,\text{kg}$ .	6
04.1	$16.0\,\text{cm}^3$ $38.6\,\text{cm}^3$ $2.53\times10^{-3}\,\text{mol}$ $1.68\,\text{J}$	Let the unmarked volume be $D$ in $\text{cm}^3$ . Charles's law gives $(42.0+D)/290=(58.0+D)/370$ . Hence $370(42.0+D)=290(58.0+D)$ , so $80D=1280$ and $D=16.0\,\text{cm}^3$ . The total volume per kelvin is $(42.0+16.0)/290=0.200\,\text{cm}^3\,\text{K}^{-1}$ . At $273.15\,\text{K}$ the total volume is $54.63\,\text{cm}^3$ , so the cylinder reading is $54.63-16.0=38.63\,\text{cm}^3=38.6\,\text{cm}^3$ . At $290\,\text{K}$ , $n=pV/(RT)=(1.05\times10^5)(58.0\times10^{-6})/[8.31(290)]=2.527\times10^{-3}\,\text{mol}$ . The measured volume change is $16.0\,\text{cm}^3=16.0\times10^{-6}\,\text{m}^3$ , so $W=p\Delta V=(1.05\times10^5)(16.0\times10^{-6})=1.68\,\text{J}$ .	6
05.1	$7.54\times10^{-3}\,\text{m}^3$ initially $1.05\times10^{-2}\,\text{m}^3$ after expansion $399\,\text{K}$ $332\,\text{J}$ $1.41\times10^5\,\text{Pa}$	Initially, $V_1=nRT_1/p=0.350(8.31)(285)/(1.10\times10^5)=7.5357\times10^{-3}\,\text{m}^3$ . Hence $V_2=1.40V_1=1.05499\times10^{-2}\,\text{m}^3$ . At constant pressure, $V/T$ is constant, so $T_2=1.40(285)=399\,\text{K}$ . The work is $W=p(V_2-V_1)=(1.10\times10^5)(1.05499\times10^{-2}-7.5357\times10^{-3})=331.569\,\text{J}=332\,\text{J}$ . Once the piston is locked, volume and amount are fixed, so $p_3/p_2=T_3/T_2$ . Thus $p_3=(1.10\times10^5)(510/399)=1.406\times10^5\,\text{Pa}=1.41\times10^5\,\text{Pa}$ .	6

3.6.2.3 · Molecular kinetic theory model

Tier 1 · Easy

Mark scheme for 3.6.2.3 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$6.21\times10^{-21}\,\text{J}$	The average energy per molecule is $\overline{E_{\text{k}}}=\tfrac32kT=\tfrac32(1.38\times10^{-23})(300)=6.21\times10^{-21}\,\text{J}$ .	2
02.1	The observed particle moves randomly because it receives unequal, rapidly changing impacts from unseen molecules, showing that matter contains moving microscopic particles.	Link the irregular motion of a visible suspended particle to unbalanced collisions by much smaller, invisible molecules. These impacts provide indirect evidence that the molecules exist and are in continual random motion.	2

Tier 2 · Standard

Mark scheme for 3.6.2.3 Tier 2 · Standard
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01.1	$6.22\times10^2\,\text{m s}^{-1}$	From $pV=\tfrac13Nm(c_{\text{rms}})^2$ , $(c_{\text{rms}})^2=\dfrac{3pV}{Nm}$ . Therefore $c_{\text{rms}}=\sqrt{\dfrac{3(1.20\times10^5)(1.50\times10^{-2})}{(3.00\times10^{23})(4.65\times10^{-26})}}=6.22\times10^2\,\text{m s}^{-1}$ .	4
02.1	Higher temperature means greater average molecular kinetic energy and speed, so wall collisions are more frequent and each has a larger momentum change; the resulting force per unit area increases.	A temperature rise increases the molecules' average translational kinetic energy, so their mean speeds rise. They strike the container walls more often and transfer more momentum per collision. The greater rate of momentum transfer increases the force on the walls and therefore the pressure.	3
03.1	$3.16$ At the same temperature the average translational kinetic energy $\tfrac12mc_{\text{rms}}^2=\tfrac32kT$ is equal, so the lighter helium atoms have the greater rms speed.	At the same temperature, each gas has the same average translational kinetic energy per molecule. Since $\tfrac12m(c_{\text{rms}})^2=\tfrac32kT$ , $c_{\text{rms}}\propto1/\sqrt{m}$ and the same relation holds using molar mass. Therefore $c_{\text{rms, helium}}/c_{\text{rms, argon}}=\sqrt{M_{\text{argon}}/M_{\text{helium}}}=\sqrt{(4.00\times10^{-2})/(4.00\times10^{-3})}=\sqrt{10}=3.16$ . Helium atoms are lighter, so they must have the greater rms speed for the same average kinetic energy.	4

Tier 3 · Hard

Mark scheme for 3.6.2.3 Tier 3 · Hard
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01.1	$pV=\dfrac13Nm(c_{\text{rms}})^2$ and $\overline{E_{\text{k}}}=\dfrac32kT$	Take $c_x$ to be the magnitude of one molecule's velocity component normal to the wall. An elastic collision changes its momentum by $2mc_x$ . The time between successive collisions with the same wall is $2L/c_x$ , so its mean force on that wall is $2mc_x/(2L/c_x)=mc_x^2/L$ . Summing over all molecules gives $F=\dfrac{m}{L}\sum c_x^2$ . Since the wall area is $L^2$ and $V=L^3$ , $p=F/L^2=\dfrac{m}{V}\sum c_x^2=\dfrac{Nm}{V}\overline{c_x^2}$ . Random isotropic motion gives $\overline{c_x^2}=\tfrac13\overline{c^2}=\tfrac13(c_{\text{rms}})^2$ . Hence $pV=\tfrac13Nm(c_{\text{rms}})^2$ . Equating this with $pV=NkT$ gives $\tfrac13m(c_{\text{rms}})^2=kT$ , so $\overline{E_{\text{k}}}=\tfrac12m(c_{\text{rms}})^2=\tfrac32kT$ .	6
02.1	$1.05\times10^3\,\text{J}$ $1.48\times10^3\,\text{m s}^{-1}$ An ideal gas has negligible intermolecular potential energy, and constant temperature keeps the molecules' average kinetic energy unchanged.	For a monatomic ideal gas, $U=\tfrac32nRT=\tfrac32(0.240)(8.31)(350)=1.05\times10^3\,\text{J}$ . Also $\tfrac12M(c_{\text{rms}})^2=\tfrac32RT$ , so $c_{\text{rms}}=\sqrt{3RT/M}=1.48\times10^3\,\text{m s}^{-1}$ . In the ideal-gas model, intermolecular potential energy is negligible and internal energy is the molecules' kinetic energy. Constant temperature means unchanged average kinetic energy, so compression alone does not change $U$ .	5
03.1	$1.39\times10^3\,\text{m s}^{-1}$ $6.68\times10^{-27}\,\text{kg}$ $2.45\times10^{25}\,\text{m}^{-3}$	Since $Nm/V$ is the gas density $\rho$ , kinetic theory gives $p=\tfrac13\rho(c_{\text{rms}})^2$ . Hence $c_{\text{rms}}=\sqrt{3p/\rho}=\sqrt{3(1.05\times10^5)/0.164}=1.39\times10^3\,\text{m s}^{-1}$ . Using $\tfrac12m(c_{\text{rms}})^2=\tfrac32kT$ , the molecular mass is $m=3kT/(c_{\text{rms}})^2=3(1.38\times10^{-23})(310)/(1.3859\times10^3)^2=6.68\times10^{-27}\,\text{kg}$ . The number density is $N/V=\rho/m=0.164/(6.68\times10^{-27})=2.45\times10^{25}\,\text{m}^{-3}$ .	6
04.1	$6.12\times10^2\,\text{m s}^{-1}$ $60.0\,\text{K}$ $(N/V)_{\text{helium}}/(N/V)_{\text{nitrogen}}=7.00$ for number density $\overline{E}_{\text{k,helium}}/\overline{E}_{\text{k,nitrogen}}=1/7$	For a gas, $c_{\text{rms}}=\sqrt{3RT/M}$ . Nitrogen therefore has $c_{\text{rms}}=\sqrt{3(8.31)(420)/(2.80\times10^{-2})}=611.5\,\text{m s}^{-1}=6.12\times10^2\,\text{m s}^{-1}$ . Equal rms speeds require $T_{\text{He}}/M_{\text{He}}=T_{\text{N}}/M_{\text{N}}$ , so $T_{\text{He}}=420(4.00\times10^{-3})/(2.80\times10^{-2})=60.0\,\text{K}$ . Number density is $N/V=p/(kT)$ , so at equal pressure its helium-to-nitrogen ratio is $420/60.0=7.00$ . Average translational kinetic energy is $\tfrac32kT$ , so the helium-to-nitrogen energy ratio is $60.0/420=1/7$ .	6
05.1	$469\,\text{K}$ $2.00\times10^5\,\text{Pa}$ $1.08\times10^3\,\text{J}$ initially and $1.35\times10^3\,\text{J}$ finally $25.0\%$ increase The lower collision rate caused by fewer molecules is outweighed by the greater momentum transfer and collision frequency per molecule at the higher rms speed.	For identical molecules, $c_{\text{rms}}^2\propto T$ , so $T_2/T_1=(1.25)^2=1.5625$ and $T_2=1.5625(300)=468.75\,\text{K}=469\,\text{K}$ . From $pV=\tfrac13Nm(c_{\text{rms}})^2$ at fixed volume, $p_2/p_1=(N_2/N_1)(c_2/c_1)^2=0.800(1.25)^2=1.250$ . Hence $p_2=1.250(1.60\times10^5)=2.00\times10^5\,\text{Pa}$ . In this kinetic model, $U=\tfrac32pV$ . Therefore $U_1=\tfrac32(1.60\times10^5)(4.50\times10^{-3})=1080\,\text{J}=1.08\times10^3\,\text{J}$ and $U_2=\tfrac32(2.00\times10^5)(4.50\times10^{-3})=1350\,\text{J}=1.35\times10^3\,\text{J}$ . The increase is $(1350-1080)/1080\times100=25.0\%$ . Although there are fewer molecules striking the walls, each faster molecule collides more often and has a greater momentum change per collision; the speed-squared increase outweighs the $20.0\%$ fall in molecule number.	6