▶️ Answer/Explanation
(a) Uniform circular motion [2 marks]
The velocity and acceleration both have constant magnitude.
The velocity is always perpendicular to the acceleration.
(b)(i) Speed of the ball [1 mark]
The relationship between tangential speed and angular speed is \(v=R\omega\).
Answer: \(v=R\omega\)
(b)(ii) Centripetal acceleration [2 marks]
For circular motion, \(a=\dfrac{v^2}{R}\).
Since \(v=R\omega\), \(a=\dfrac{(R\omega)^2}{R}=R\omega^2\).
Therefore, \(a=v\omega\).
Answer: \(a=v\omega\)
(c)(i) Displacement of the shadow [1 mark]
From the geometry of Fig. 1.1, \(\sin\theta=\dfrac{x}{R}\).
Hence, \(x=R\sin\theta\).
Answer: \(x=R\sin\theta\)
(c)(ii) Angular displacement [1 mark]
Angular displacement is related to angular speed by \(\theta=\omega t\), since \(\theta=0\) at \(t=0\).
Answer: \(\theta=\omega t\)
(c)(iii) Equation for the shadow [1 mark]
From (c)(i), \(x=R\sin\theta\).
Substituting \(\theta=\omega t\): \(x=R\sin(\omega t)\).
Answer: \(x=R\sin\omega t\)
(c)(iv) Simple harmonic motion [1 mark]
The equation has the form \(x=x_0\sin\omega t\), which is the standard form for simple harmonic motion.
Answer: The motion may therefore be modelled as simple harmonic motion.
(d)(i) Amplitude [1 mark]
The amplitude of the shadow is equal to the radius of the circular motion.
\(R=\dfrac{0.46}{2}=0.23\,\mathrm{m}\)
Answer: amplitude \(=0.23\,\mathrm{m}\)
(d)(ii) Period [2 marks]
Angular speed is related to period by \(\omega=\dfrac{2\pi}{T}\).
Therefore, \(T=\dfrac{2\pi}{\omega}\).
\(T=\dfrac{2\pi}{1.9}\)
\(T\approx3.3\,\mathrm{s}\)
Answer: period \(=3.3\,\mathrm{s}\)
(d)(iii) Maximum acceleration [2 marks]
For simple harmonic motion, the magnitude of acceleration is \(a=\omega^2x\).
Maximum acceleration occurs when \(x=x_0\), where \(x_0\) is the amplitude.
\(a_{\max}=\omega^2x_0\)
\(a_{\max}=(1.9)^2(0.23)\)
\(a_{\max}\approx0.83\,\mathrm{m\,s^{-2}}\)
Answer: maximum acceleration \(=0.83\,\mathrm{m\,s^{-2}}\)
(e) Position of maximum positive acceleration [1 mark]
For simple harmonic motion, acceleration is directed towards the equilibrium position.
Maximum positive acceleration therefore occurs when the shadow has maximum negative displacement, corresponding to the upper-left position of the circular path as shown in the mark scheme.
Answer: Label A at the shadow position on the screen above the left-hand edge of the circular path.
▶️ Answer/Explanation
(a)
The internal energy of a system may be changed by thermal energy being transferred to or from the system.
It may also be changed by work being done on or by the system.
Answer: thermal energy supplied to or removed from the system; work done on or by the system.
(b)(i)
When the pump is used quickly, there is little time for thermal energy to be transferred to or from the gas.
Work is done on the gas as it is compressed, decreasing its volume.
Therefore, the internal energy of the gas increases, so its temperature increases and the pump becomes hot.
Answer: The rapid compression gives little time for thermal energy transfer. Work is done on the gas, increasing its internal energy and hence its temperature.
(b)(ii)
During vaporisation, the separation between the water molecules increases.
The energy supplied is therefore used to increase the potential energy of the molecules.
The kinetic energy of the molecules remains unchanged, so the temperature remains at \(100^\circ\mathrm{C}\).
Answer: The supplied energy increases molecular separation and hence molecular potential energy, while the average kinetic energy remains unchanged. Therefore, the temperature remains constant.
▶️ Answer/Explanation
(a) Gravitational field at a point [1 mark]
Gravitational field strength at a point is the force per unit mass acting on a small test mass placed at that point.
Answer: Gravitational field strength is force per unit mass.
(b)(i) Gravitational field strength at P [2 marks]
From Newton’s law of gravitation, the force between the point mass \(M\) and the test mass \(m\) is \(F=\dfrac{GMm}{x^2}\).
Gravitational field strength is force per unit mass: \(g=\dfrac{F}{m}\).
Therefore, \(g=\dfrac{GMm}{x^2m}\).
Hence, \(g=\dfrac{GM}{x^2}\), where \(G\) is the gravitational constant.
Answer: \(g=\dfrac{GM}{x^2}\), where \(G\) is the gravitational constant.
(b)(ii) Direction of gravitational field [1 mark]
A gravitational field is directed towards the mass producing the field.
Answer: The arrow at P points directly towards the point mass \(M\).
(b)(iii) Comparison of fields at Q and P [2 marks]
The magnitude of gravitational field strength due to a point mass is \(g=\dfrac{GM}{r^2}\).
At P, the distance is \(x\), so \(g_P=\dfrac{GM}{x^2}\).
At Q, the distance is \(\dfrac{x}{2}\), so \(g_Q=\dfrac{GM}{(x/2)^2}=\dfrac{4GM}{x^2}=4g_P\).
The fields at Q and P are in opposite directions because Q and P are on opposite sides of the mass.
Answer: The gravitational field strength at Q is four times that at P, and the two fields are in opposite directions.
(c) Variation of gravitational field strength [3 marks]
At \(x=R\), point P is at the surface of sphere X. The field due to X is directed towards X, so \(g=-g_0\).
At \(x=L-R\), point P is at the surface of sphere Y. The field due to Y is directed towards Y, so \(g=+g_0\).
At the midpoint \(x=\dfrac{L}{2}\), the gravitational fields due to the identical spheres are equal in magnitude and opposite in direction.
Therefore, \(g=0\) at \(x=\dfrac{L}{2}\).
The curve becomes progressively shallower from \(x=R\) to \(x=\dfrac{L}{2}\), then becomes progressively steeper from \(x=\dfrac{L}{2}\) to \(x=L-R\).
Answer: The graph starts at \((R,-g_0)\), passes through \(\left(\dfrac{L}{2},0\right)\), and ends at \((L-R,+g_0)\), with the curve becoming shallower towards the midpoint and steeper after the midpoint.
▶️ Answer/Explanation
(a)(i) Absolute zero on the Celsius scale [1 mark]
Absolute zero corresponds to \(0\,\mathrm{K}\), which is approximately \(-273.15^\circ\mathrm{C}\).
Answer: \(-273.15^\circ\mathrm{C}\)
(a)(ii) Absolute zero on the thermodynamic scale [1 mark]
On the thermodynamic temperature scale, absolute zero is defined as \(0\,\mathrm{K}\).
Answer: \(0\,\mathrm{K}\)
(b)(i) Nature of the gas [1 mark]
The graph shows a straight-line relationship between \(pV\) and \(kT\) passing through the origin.
This indicates that the gas behaves as an ideal gas.
Answer: The gas is ideal.
(b)(ii) Number of molecules [2 marks]
For an ideal gas, \(pV=NkT\).
From the graph, when \(kT=8.0\times10^{-21}\,\mathrm{J}\), \(pV=270\,\mathrm{J}\).
Therefore, \(N=\dfrac{pV}{kT}\).
\(N=\dfrac{270}{8.0\times10^{-21}}\)
\(N=3.4\times10^{22}\)
Answer: \(N=3.4\times10^{22}\) molecules
(b)(iii) Amount of gas [1 mark]
The number of molecules is related to the amount of gas by \(N=nN_{\mathrm A}\).
Hence, \(n=\dfrac{N}{N_{\mathrm A}}\).
\(n=\dfrac{3.4\times10^{22}}{6.02\times10^{23}}\)
\(n=0.056\,\mathrm{mol}\)
Answer: \(n=0.056\,\mathrm{mol}\)
(c) Mass of one molecule [4 marks]
From kinetic theory, \(pV=\dfrac{2}{3}N\left(\dfrac{1}{2}mv_{\mathrm{rms}}^2\right)\).
Therefore, \(\dfrac{1}{2}mv_{\mathrm{rms}}^2=\dfrac{3}{2}kT\).
Since \(pV=NkT\), \(\dfrac{1}{2}mv_{\mathrm{rms}}^2=\dfrac{3pV}{2N}\).
Using \(pV=270\,\mathrm{J}\), \(N=3.4\times10^{22}\) and \(v_{\mathrm{rms}}=1900\,\mathrm{m\,s^{-1}}\):
\(\dfrac{1}{2}m(1900)^2=\dfrac{3}{2}\left(\dfrac{270}{3.4\times10^{22}}\right)\)
This gives \(m\approx6.65\times10^{-27}\,\mathrm{kg}\).
Using \(1\,\mathrm{u}=1.66\times10^{-27}\,\mathrm{kg}\):
\(m=\dfrac{6.65\times10^{-27}}{1.66\times10^{-27}}\)
\(m=4.0\,\mathrm{u}\)
Answer: \(m=4.0\,\mathrm{u}\)
▶️ Answer/Explanation
(a) Electric potential at a point [2 marks]
Electric potential is the work done per unit positive charge.
It is the work done in moving a positive test charge from infinity to the point.
Answer: Work done per unit charge in moving a positive charge from infinity to the point.
(b)(i) Show that \(V=130\,\mathrm{V}\) [2 marks]
The electric potential due to a point charge is \(V=\dfrac{Q}{4\pi\varepsilon_0r}\).
Potential due to the proton:
\(V_{\mathrm p}=\dfrac{1.60\times10^{-19}}{4\pi(8.85\times10^{-12})(10\times10^{-12})}\)
Potential due to the electron:
\(V_{\mathrm e}=\dfrac{-1.60\times10^{-19}}{4\pi(8.85\times10^{-12})(110\times10^{-12})}\)
Total potential:
\(V=\dfrac{1.60\times10^{-19}}{4\pi(8.85\times10^{-12})}\left(\dfrac{1}{10\times10^{-12}}-\dfrac{1}{110\times10^{-12}}\right)\)
\(V=130\,\mathrm{V}\)
Answer: \(130\,\mathrm{V}\)
(b)(ii) Potential when \(x=30\,\mathrm{pm}\) [2 marks]
Distance from proton \(=30\,\mathrm{pm}\).
Distance from electron \(=120-30=90\,\mathrm{pm}\).
\(V=\dfrac{1.60\times10^{-19}}{4\pi(8.85\times10^{-12})}\left(\dfrac{1}{30\times10^{-12}}-\dfrac{1}{90\times10^{-12}}\right)\)
\(V=32\,\mathrm{V}\)
Answer: \(V=+32\,\mathrm{V}\)
(b)(iii) Position where electric potential is zero [1 mark]
The electric potential is zero at the point midway between the proton and the electron because the positive and negative potentials are equal in magnitude.
Answer: Draw the cross at the midpoint, \(x=60\,\mathrm{pm}\).
(b)(iv) Sketch of variation of \(V\) with \(x\) [3 marks]
The graph begins at \((10,+130)\) and ends at \((110,-130)\).
It becomes progressively shallower until \(x=60\,\mathrm{pm}\), where it crosses the axis at \(V=0\).
After \(x=60\,\mathrm{pm}\), the curve becomes progressively steeper and passes through approximately \((30,+32)\) and \((90,-32)\).
Answer: Smooth decreasing curve through \((10,+130)\), \((30,+32)\), \((60,0)\), \((90,-32)\) and \((110,-130)\), becoming shallower before \(60\,\mathrm{pm}\) and steeper afterwards.
▶️ Answer/Explanation
(a) Series and parallel capacitors [4 marks]
| relationship between charges | relationship between p.d.s | |
|---|---|---|
| series | \(Q_{\mathrm S}=Q_1=Q_2\) | \(V_{\mathrm S}=V_1+V_2\) |
| parallel | \(Q_{\mathrm S}=Q_1+Q_2\) | \(V_{\mathrm S}=V_1=V_2\) |
(b)(i) P.d. across the capacitor [2 marks]
The energy stored in a capacitor is \(E=\dfrac{1}{2}CV^2\).
Rearranging: \(V=\sqrt{\dfrac{2E}{C}}\).
\(E=19\times10^{-3}\,\mathrm{J}\) and \(C=470\times10^{-6}\,\mathrm{F}\).
\(V=\sqrt{\dfrac{2(19\times10^{-3})}{470\times10^{-6}}}\)
\(V=9.0\,\mathrm{V}\)
Answer: \(V=9.0\,\mathrm{V}\)
(b)(ii) Charge on the capacitor [2 marks]
Using \(Q=CV\):
\(Q=(470\times10^{-6})(9.0)\)
\(Q=4.23\times10^{-3}\,\mathrm{C}\)
Answer: \(Q=4.2\times10^{-3}\,\mathrm{C}\)
(b)(iii) Total energy after connection in parallel [3 marks]
The second capacitor is initially uncharged, so when the two capacitors are connected in parallel, the total charge is conserved.
The total capacitance is \(C_{\mathrm{total}}=470+180=650\,\mu\mathrm{F}\).
The common final p.d. is \(V_{\mathrm f}=\dfrac{Q_{\mathrm{total}}}{C_{\mathrm{total}}}\).
\(V_{\mathrm f}=\dfrac{4.23\times10^{-3}}{650\times10^{-6}}\)
\(V_{\mathrm f}\approx6.51\,\mathrm{V}\)
The total stored energy is \(E=\dfrac{Q^2}{2C}\).
\(E=\dfrac{(4.23\times10^{-3})^2}{2(650\times10^{-6})}\)
\(E\approx0.014\,\mathrm{J}\)
\(E=14\,\mathrm{mJ}\)
Answer: \(14\,\mathrm{mJ}\)
▶️ Answer/Explanation
(a) Faraday’s law [2 marks]
Faraday’s law states that the induced e.m.f. is directly proportional to the rate of change of magnetic flux linkage.
For a single conductor or single-turn loop, \(\mathcal{E}=\dfrac{\Delta\Phi}{\Delta t}\).
Answer: The induced e.m.f. is directly proportional to the rate of change of magnetic flux linkage.
(b)(i) Magnetic flux cut by the wings [2 marks]
From Faraday’s law, \(\mathcal{E}=\dfrac{\Delta\Phi}{\Delta t}\).
Therefore, \(\Delta\Phi=\mathcal{E}\Delta t\).
\(\Delta\Phi=0.54\times15\)
\(\Delta\Phi=8.1\,\mathrm{Wb}\)
Answer: magnetic flux \(=8.1\,\mathrm{Wb}\)
(b)(ii) Area of flux cut [2 marks]
Magnetic flux is given by \(\Phi=BA\), since the magnetic field is perpendicular to the area swept out by the wings.
Hence, \(A=\dfrac{\Phi}{B}\).
\(B=38\times10^{-6}\,\mathrm{T}\).
\(A=\dfrac{8.1}{38\times10^{-6}}\)
\(A=2.1\times10^5\,\mathrm{m^2}\)
Answer: \(A=2.1\times10^5\,\mathrm{m^2}\)
(b)(iii) Speed of the aircraft [2 marks]
The area swept out by the wings is \(A=vtw\), where \(w=68\,\mathrm{m}\) is the distance between the wingtips.
Therefore, \(v=\dfrac{A}{tw}\).
\(v=\dfrac{2.1\times10^5}{15\times68}\)
\(v=210\,\mathrm{m\,s^{-1}}\)
Answer: \(v=210\,\mathrm{m\,s^{-1}}\)
(b)(iv) Lenz’s law and higher potential [3 marks]
The aircraft moves through a magnetic field directed downwards. The induced current must produce a magnetic effect that opposes the change causing the induction.
Using Fleming’s left-hand rule, the force on a conventional current in the wings must act backwards, opposing the forward motion of the aircraft.
For the force to act backwards, the conventional current in the wings is from Q to P.
Inside an e.m.f. source, conventional current flows from the negative terminal towards the positive terminal. Therefore, P is at the higher potential.
Answer: P is at the higher potential because the induced current is from Q to P, making P the positive terminal of the induced e.m.f.
▶️ Answer/Explanation
(a) Photon [2 marks]
A photon is a packet, or quantum, of energy of electromagnetic radiation.
Answer: A photon is a quantum of energy of electromagnetic radiation.
(b)(i) Mass of uranium-238 [3 marks]
The total energy released is related to the mass defect by \(E=\Delta mc^2\).
Hence, \(\Delta m=\dfrac{E}{c^2}\).
Convert \(4.274\,\mathrm{MeV}\) into joules: \(E=4.274\times10^6\times1.60\times10^{-19}\,\mathrm{J}\).
Therefore, \[ \Delta m =\dfrac{4.274\times10^6\times1.60\times10^{-19}} {(3.00\times10^8)^2} \]
\(\Delta m=0.00458\,\mathrm{u}\).
The uranium mass is the sum of the thorium mass, alpha-particle mass and mass defect:
\(m_{\mathrm U}=233.915174+4.000407+0.00458\)
\(m_{\mathrm U}=237.92016\,\mathrm{u}\)
Answer: \(237.92016\,\mathrm{u}\)
(b)(ii) Wavelength of gamma radiation [3 marks]
The total energy released is \(4.274\,\mathrm{MeV}\), while the alpha particle has kinetic energy \(4.200\,\mathrm{MeV}\).
With the rebound kinetic energy of the thorium nucleus assumed negligible, the remaining energy is carried by the gamma photon:
\(E_\gamma=(4.274-4.200)\,\mathrm{MeV}\)
\(E_\gamma=0.074\,\mathrm{MeV}=0.074\times1.60\times10^{-13}\,\mathrm{J}\)
For a photon, \(E=\dfrac{hc}{\lambda}\).
Hence, \(\lambda=\dfrac{hc}{E}\).
\(\lambda=\dfrac{(6.63\times10^{-34})(3.00\times10^8)} {(0.074)(1.60\times10^{-13})}\)
\(\lambda=1.7\times10^{-11}\,\mathrm{m}\)
Answer: \(\lambda=1.7\times10^{-11}\,\mathrm{m}\)
(b)(iii) Effect of thorium recoil [1 mark]
If the thorium nucleus has non-negligible recoil kinetic energy, the gamma photon has less energy than assumed in (b)(ii).
Since \(E=\dfrac{hc}{\lambda}\), a smaller photon energy corresponds to a larger wavelength.
Answer: The true wavelength is larger than the value calculated in (b)(ii).
(c) Gamma radiation from cobalt-60 [2 marks]
Neutrinos are emitted during beta decay.
The particles emitted during beta decay can carry varying amounts of energy. Therefore, the energy available for the gamma photon can also vary between decays.
Answer: The beta decay involves emission of a neutrino, and the emitted particles can carry varying amounts of energy. Consequently, the energy of the gamma photon can vary between decays, giving more than one wavelength.
▶️ Answer/Explanation
(a) Wien’s displacement law [2 marks]
Wien’s displacement law states that the temperature of a hot object is inversely proportional to the wavelength at which its emission rate is a maximum.
The relationship is \(\lambda_{\max}T=b\), where \(b\) is Wien’s displacement constant.
Answer: \(\lambda_{\max}T=\mathrm{constant}\)
(b) Conclusions about star X [3 marks]
1. Surface temperature: The peak wavelength for star X is shorter than that of the Sun. Since temperature is inversely proportional to peak wavelength, star X has a higher surface temperature than the Sun.
From the graph, the peak wavelength for star X is approximately \(4.1\times10^{-7}\,\mathrm{m}\). Therefore, \(T_{\mathrm X}\approx7000\,\mathrm{K}\).
2. Luminosity: The radiant flux intensity follows an inverse-square relationship with distance, so the graph can be used to determine the luminosity of star X.
The luminosity of star X is approximately \(2.7\times10^{27}\,\mathrm{W}\), which is greater than that of the Sun.
3. Radius: Using the luminosity and surface temperature, the radius of star X is approximately \(1.3\times10^9\,\mathrm{m}\).
Answer: Any three valid conclusions, such as \(T_{\mathrm X}\approx7000\,\mathrm{K}\), star X has a higher luminosity than the Sun, \(L_{\mathrm X}\approx2.7\times10^{27}\,\mathrm{W}\), or \(R_{\mathrm X}\approx1.3\times10^9\,\mathrm{m}\).
(c) Effect of the galaxy’s motion [2 marks]
Since the galaxy containing star X is moving away from the Earth, the radiation observed from Earth is redshifted.
Therefore, the whole emission curve would be shifted towards longer wavelengths. In particular, the wavelength of maximum emission would be greater in the observed data.
Answer: The line would be shifted towards longer wavelengths because the light from star X is redshifted as the galaxy moves away from the Earth.
▶️ Answer/Explanation
(a) Specific acoustic impedance [2 marks]
Specific acoustic impedance is the product of the density of a medium and the speed of sound in the medium.
\(Z=\rho c\)
where \(\rho\) is the density of the medium and \(c\) is the speed of sound in the medium.
(b) Detection of ultrasound by a piezoelectric crystal [2 marks]
The ultrasound waves cause the piezoelectric crystal to vibrate.
The vibrations of the crystal produce an induced e.m.f. across the crystal, which can be detected as an electrical signal.
(c)(i) Intensity reflection coefficient [2 marks]
The intensity reflection coefficient at a boundary is
\(R=\left(\dfrac{Z_2-Z_1}{Z_2+Z_1}\right)^2\)
For water, \(Z_1=1.48\times10^6\,\mathrm{kg\,m^{-2}\,s^{-1}}\), and for steel, \(Z_2=4.04\times10^7\,\mathrm{kg\,m^{-2}\,s^{-1}}\).
\(R=\left(\dfrac{40.4-1.48}{40.4+1.48}\right)^2\)
\(R=0.86\)
Answer: \(R=0.86\)
(c)(ii) Body tissue-water boundary [2 marks]
The specific acoustic impedance values of body tissue and water are very similar: \(1.38\times10^6\) and \(1.48\times10^6\,\mathrm{kg\,m^{-2}\,s^{-1}}\), respectively.
Therefore, there is very little reflection at the boundary and almost all of the ultrasound will be transmitted.
Answer: Almost all the ultrasound is transmitted and very little is reflected because the acoustic impedances are very similar.