3.6 Further mechanics and thermal physics (A-level only) — printable coverage pack

3.6.1.1 · Circular motion

Constant speed in a circle still involves acceleration because velocity changes direction; the acceleration and resultant force are directed towards the centre.
Angular speed is $\omega=\dfrac{v}{r}=2\pi f$ and is measured in $\text{rad s}^{-1}$ .
Centripetal acceleration is $a=\dfrac{v^2}{r}=\omega^2r$ , giving $F=\dfrac{mv^2}{r}=m\omega^2r$ .
Centripetal force is not an additional type of force: it is the name for the inward resultant of real forces such as tension, friction or gravity.
A common error is to draw a force in the direction of motion; the instantaneous velocity is tangential and perpendicular to the centripetal acceleration.

Tier 1 · Easy

1. A turntable rotates at $120\,\text{rev min}^{-1}$ . Calculate its angular speed.[2 marks]
Answer
- $12.6\,\text{rad s}^{-1}$
Method: $120\,\text{rev min}^{-1}=120/60=2.00\,\text{Hz}$ . Therefore $\omega=2\pi f=2\pi\times2.00=12.6\,\text{rad s}^{-1}$ .

Tier 2 · Standard

1. 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]
Answer
- $20\,\text{N}$ towards the centre of the circle
Method: 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.

Tier 3 · Hard

1. A coin rests $0.350\,\text{m}$ from the centre of a horizontal rotating disc. The coefficient of limiting friction is $0.420$ . Determine the greatest rotation frequency for which the coin does not slide. Explain which force supplies the centripetal force. Use $g=9.81\,\text{N kg}^{-1}$ .[5 marks]
Answer
- $0.546\,\text{Hz}$ ; static friction acts towards the centre
Method: At the limiting frequency, the maximum static friction $\mu mg$ supplies the required centripetal force $m\omega^2r$ . Hence $\mu mg=m\omega^2r$ , so $\omega=\sqrt{\dfrac{\mu g}{r}}=\sqrt{\dfrac{0.420\times9.81}{0.350}}=3.43\,\text{rad s}^{-1}$ . Therefore $f=\dfrac{\omega}{2\pi}=\dfrac{3.43}{2\pi}=0.546\,\text{Hz}$ . Static friction is the real horizontal force and acts radially inwards.

3.6.1.2 · Simple harmonic motion (SHM)

The defining condition for SHM is $a\propto-x$ , written $a=-\omega^2x$ ; the minus sign means the acceleration is always towards equilibrium.
For $x=A\cos\omega t$ , the speed-displacement relation is $v=\pm\omega\sqrt{A^2-x^2}$ , with the sign chosen from the stated direction of motion.
The maximum speed is $v_{\max}=\omega A$ at equilibrium and the maximum acceleration magnitude is $a_{\max}=\omega^2A$ at either extreme.
The velocity-time graph is the gradient of the displacement-time graph, and the acceleration-time graph is the gradient of the velocity-time graph.
Always define the positive direction before assigning signs; a common error is to report only acceleration magnitude when the direction is required.

Tier 1 · Easy

1. 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]
Answer
- $-0.48\,\text{m s}^{-2}$
Method: 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.

Tier 2 · Standard

1. 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]
Answer
- $-0.32\,\text{m s}^{-1}$
- $-1.2\,\text{m s}^{-2}$
Method: 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 $a=-\omega^2x=-(5.0)^2(0.048)=-1.2\,\text{m s}^{-2}$ .

Tier 3 · Hard

1. 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]
Answer
- $+0.0204\,\text{m}$
- $-0.473\,\text{m s}^{-1}$
- $-0.326\,\text{m s}^{-2}$
Method: 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 $a=-\omega^2x=-(4.00)^2(0.0204)=-0.326\,\text{m s}^{-2}$ .

3.6.1.3 · Simple harmonic systems

For a mass-spring oscillator, $T=2\pi\sqrt{m/k}$ ; use the total oscillating mass if the question requires it.
For a simple pendulum at small angle, $T=2\pi\sqrt{l/g}$ . The model fails when the small-angle approximation is not valid.
For an ideal spring system, total energy is $E=\tfrac12kA^2$ ; at displacement $x$ , elastic potential energy is $E_{\text{p}}=\tfrac12kx^2$ and the remainder is kinetic.
During ideal SHM, kinetic energy is maximum at equilibrium and potential energy is maximum at the extremes; the total stays constant.
Damping transfers energy from the oscillator to the surroundings, so its amplitude decreases. A common error is to say that damping changes the equilibrium position.

Tier 1 · Easy

1. 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]
Answer
- $0.314\,\text{s}$
Method: $T=2\pi\sqrt{m/k}=2\pi\sqrt{0.200/80.0}=2\pi(0.0500)=0.314\,\text{s}$ .

Tier 2 · Standard

1. 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]
Answer
- $0.636\,\text{m}$
Method: From $T=2\pi\sqrt{l/g}$ , $l=g\left(\dfrac{T}{2\pi}\right)^2$ . Therefore $l=9.81\left(\dfrac{1.60}{2\pi}\right)^2=0.636\,\text{m}$ .

Tier 3 · Hard

1. 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]
Answer
- $0.150\,\text{J}$
- $0.800\,\text{m s}^{-1}$
- $36.0\%$
Method: 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 $36.0\%$ .

3.6.1.4 · Forced vibrations and resonance

A free vibration occurs at the system's natural frequency after an initial disturbance; a forced vibration is maintained by a periodic driving force.
Resonance occurs when the driving frequency equals the natural frequency, producing maximum energy transfer and maximum amplitude.
Increasing damping lowers and broadens the resonance peak, making the resonance less sharp.
Resonance can be useful, as in tuning a stationary-wave system, or hazardous, as in structures driven near a natural frequency.
A common error is to define resonance only as a large amplitude without comparing driving and natural frequencies.

Tier 1 · Easy

1. State what is meant by resonance in a forced oscillator.[2 marks]
Answer
- The driving frequency equals the natural frequency of the oscillator, producing maximum amplitude.
Method: State both marking points: the periodic driving frequency equals the system's natural frequency, and the forced-oscillation amplitude is then maximum.

Tier 2 · Standard

1. Describe how increased damping changes an amplitude-against-driving-frequency resonance curve.[3 marks]
Answer
- The maximum amplitude is smaller and the peak is broader and less sharp, with appreciable response over a wider range of frequencies.
Method: 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.

Tier 3 · Hard

1. 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]
Answer
- $3.18\,\text{Hz}$ ; the driver is close to resonance, and extra damping dissipates more energy per cycle and lowers the resonance peak.
Method: 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.

3.6.2.1 · Thermal energy transfer

Internal energy is the sum of the randomly distributed kinetic and potential energies of a body's particles.
Heating a system or doing work on it increases its internal energy; the reverse transfers decrease it. State the direction of energy transfer in qualitative first-law answers.
For a temperature change, $Q=mc\Delta\theta$ . For a change of state, $Q=ml$ and temperature remains constant while particle potential energy changes.
Continuous-flow calculations use energy transferred per unit time, often $P=\dot{m}c\Delta\theta$ when losses and other changes are negligible.
A common error is to use the latent-heat equation during a temperature change, or to claim that particle kinetic energy rises during a constant-temperature change of state.

Tier 1 · Easy

1. 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]
Answer
- $4.50\times10^3\,\text{J}$
Method: $Q=mc\Delta\theta=0.250\times900\times20.0=4.50\times10^3\,\text{J}$ .

Tier 2 · Standard

1. 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]
Answer
- $278\,\text{s}$
Method: 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 $t=Q/P=2.672\times10^4/96.0=278\,\text{s}$ .

Tier 3 · Hard

1. 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]
Answer
- $1.81\,\text{kW}$
- $82.5\%$
- Heating increases the random kinetic energy of the water molecules; the unused input is transferred to the heater and surroundings.
Method: 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.

3.6.2.2 · Ideal gases

The ideal-gas equations are $pV=nRT$ for $n$ moles and $pV=NkT$ for $N$ molecules.
Temperature must be absolute: convert using $T/\text{K}=\theta/^{\circ}\text{C}+273.15$ before substitution.
At fixed gas mass, the empirical gas laws relate $p$ , $V$ and $T$ ; state which quantity is constant when applying Boyle's or Charles's law.
For a constant-pressure volume change, the work done by the gas is $W=p\Delta V$ ; use the signed volume change consistently.
Distinguish amount of substance in moles from number of molecules, and distinguish molar mass from the mass of one molecule.

Tier 1 · Easy

1. Convert $25.0\,^{\circ}\text{C}$ to kelvin.[1 mark]
Answer
- $298\,\text{K}$
Method: $T=25.0+273.15=298.15\,\text{K}$ , which is $298\,\text{K}$ to the nearest kelvin.

Tier 2 · Standard

1. 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]
Answer
- $1.23\times10^5\,\text{Pa}$
Method: Convert the temperature first: $T=35.0+273.15=308.15\,\text{K}$ . Then $p=\dfrac{nRT}{V}=\dfrac{0.120\times8.31\times308.15}{2.50\times10^{-3}}=1.23\times10^5\,\text{Pa}$ .

Tier 3 · Hard

1. 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]
Answer
- $5.99\times10^{22}$ molecules
- $1.93\times10^5\,\text{Pa}$
Method: 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 $p_2=p_1\dfrac{V_1T_2}{T_1V_2}=1.00\times10^5\dfrac{2.40\times10^{-3}\times350.15}{290.15\times1.50\times10^{-3}}=1.93\times10^5\,\text{Pa}$ .

3.6.2.3 · Molecular kinetic theory model

Brownian motion provides evidence for atoms because visible particles move randomly through unequal molecular impacts.
The kinetic-theory derivation assumes many identical molecules in random motion, negligible molecular volume, negligible intermolecular forces except during elastic collisions, and collision times negligible compared with flight times.
Momentum changes at container walls lead to $pV=\tfrac13Nm(c_{\text{rms}})^2$ ; isotropy supplies the factor $\tfrac13$ .
Combining kinetic theory with $pV=NkT$ gives the average translational kinetic energy per molecule: $\tfrac12m(c_{\text{rms}})^2=\tfrac32kT$ .
Do not confuse rms speed with average speed, or the average energy of one molecule $\tfrac32kT$ with the total energy of $N$ molecules.
For a monatomic ideal gas, internal energy is the random translational kinetic energy of its atoms; an ideal gas has no intermolecular potential-energy contribution.

Tier 1 · Easy

1. 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]
Answer
- $6.21\times10^{-21}\,\text{J}$
Method: The average energy per molecule is $\overline{E_{\text{k}}}=\tfrac32kT=\tfrac32(1.38\times10^{-23})(300)=6.21\times10^{-21}\,\text{J}$ .

Tier 2 · Standard

1. 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]
Answer
- $6.22\times10^2\,\text{m s}^{-1}$
Method: From $pV=\tfrac13Nm(c_{\text{rms}})^2$ , $(c_{\text{rms}})^2=\dfrac{3pV}{Nm}$ . Therefore $c_{\text{rms}}=\sqrt{\dfrac{3(1.20\times10^5)(1.50\times10^{-2})}{(3.00\times10^{23})(4.65\times10^{-26})}}=6.22\times10^2\,\text{m s}^{-1}$ .

Tier 3 · Hard

1. 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]
Answer
- $pV=\dfrac13Nm(c_{\text{rms}})^2$ and $\overline{E_{\text{k}}}=\dfrac32kT$
Method: 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 $\overline{E_{\text{k}}}=\tfrac12m(c_{\text{rms}})^2=\tfrac32kT$ .