▶️ Answer/Explanation
(a) Newton’s law of gravitation [2 marks]
The gravitational force between two point masses is directly proportional to the product of their masses.
The force is inversely proportional to the square of their separation.
\(F=\dfrac{Gm_1m_2}{r^2}\)
(b)(i) Gravitational force [2 marks]
Using \(F=\dfrac{Gm_Am_B}{r^2}\):
\(F=\dfrac{(6.67\times10^{-11})(4.0\times10^{30})(2.0\times10^{30})}{(3.3\times10^{12})^2}\)
\(F=4.9\times10^{25}\,\mathrm{N}\)
Answer: \(F=4.9\times10^{25}\,\mathrm{N}\)
(b)(ii) Centripetal acceleration of star A [1 mark]
The gravitational force provides the centripetal force:
\(a=\dfrac{F}{m_A}\)
\(a=\dfrac{4.9\times10^{25}}{4.0\times10^{30}}\)
\(a=1.2\times10^{-5}\,\mathrm{m\,s^{-2}}\)
Answer: \(a=1.2\times10^{-5}\,\mathrm{m\,s^{-2}}\)
(b)(iii) Period of star A [3 marks]
For uniform circular motion, \(a=r\omega^2\) and \(\omega=\dfrac{2\pi}{T}\).
Since \(R_B=2R_A\) and \(R_A+R_B=3.3\times10^{12}\,\mathrm{m}\):
\(R_A=\dfrac{3.3\times10^{12}}{3}=1.1\times10^{12}\,\mathrm{m}\)
Therefore, \(a=R_A\left(\dfrac{2\pi}{T}\right)^2\).
Rearranging: \(T=2\pi\sqrt{\dfrac{R_A}{a}}\)
\(T=2\pi\sqrt{\dfrac{1.1\times10^{12}}{1.2\times10^{-5}}}\)
\(T=1.9\times10^9\,\mathrm{s}\)
Answer: \(T=1.9\times10^9\,\mathrm{s}\)
(b)(iv) Comparison of quantities [3 marks]
| B less than A | B equal to A | B greater than A | |
|---|---|---|---|
| centripetal acceleration | ✓ | ||
| linear speed | ✓ | ||
| period | ✓ |
Both stars orbit their common centre of gravity with the same angular speed, so their periods are equal. Since star B has the larger orbital radius, its linear speed and centripetal acceleration are greater.
▶️ Answer/Explanation
(a)(i) Meaning of symbols [3 marks]
\(p\) = pressure of the gas.
\(V\) = volume of the gas.
\(N\) = number of molecules in the gas.
\(k\) = Boltzmann constant.
\(T\) = thermodynamic temperature of the gas.
(a)(ii) Average translational kinetic energy [2 marks]
From kinetic theory, \(pV=\dfrac{1}{3}Nm\langle c^2\rangle\).
From the equation of state, \(pV=NkT\).
Therefore, \(NkT=\dfrac{1}{3}Nm\langle c^2\rangle\).
Hence, \[ \dfrac{1}{2}m\langle c^2\rangle=\dfrac{3}{2}kT \]
Answer: \(E_k=\dfrac{3}{2}kT\)
(b) Ratio of r.m.s. speeds [2 marks]
At the same temperature, the average translational kinetic energy of a molecule is the same for both gases:
\(\dfrac{1}{2}m_{\mathrm{H}}\langle c_{\mathrm{H}}^2\rangle = \dfrac{1}{2}m_{\mathrm{O}}\langle c_{\mathrm{O}}^2\rangle\)
Therefore, \(\dfrac{c_{\mathrm{H}}}{c_{\mathrm{O}}} = \sqrt{\dfrac{m_{\mathrm{O}}}{m_{\mathrm{H}}}}\).
A hydrogen molecule has nucleon number \(2\), while an oxygen molecule has nucleon number \(32\).
\[ \dfrac{c_{\mathrm{H}}}{c_{\mathrm{O}}} = \sqrt{\dfrac{32}{2}} = \sqrt{16} = 4.0 \]
Answer: ratio \(=4.0\)
▶️ Answer/Explanation
(a) Internal energy [2 marks]
The internal energy of an ideal gas is the total kinetic energy associated with the random motion of its molecules.
The molecular potential energy is taken to be zero because there are no intermolecular forces in an ideal gas.
(b)(i) Internal energy of the gas in state A [2 marks]
For an ideal gas, \(pV=NkT\) and \(U=\dfrac{3}{2}NkT\).
Therefore, \(U=\dfrac{3}{2}pV\).
\(U=\dfrac{3}{2}\times2.0\times10^5\times0.016\)
\(U=4800\,\mathrm{J}\)
Answer: \(U=4.8\times10^3\,\mathrm{J}\)
(b)(ii) Temperature in state B [1 mark]
At constant volume, \(p\propto T\).
\(\dfrac{p_A}{T_A}=\dfrac{p_B}{T_B}\)
\(T_B=400\times\dfrac{4.0\times10^5}{2.0\times10^5}\)
\(T_B=800\,\mathrm{K}\)
Answer: \(T_B=800\,\mathrm{K}\)
(b)(iii) Volume in state C [1 mark]
Change Y occurs at constant temperature, so \(pV\) is constant.
\(p_BV_B=p_CV_C\)
\((4.0\times10^5)(0.016)=(2.0\times10^5)V_C\)
\(V_C=0.032\,\mathrm{m^3}\)
Answer: \(V_C=0.032\,\mathrm{m^3}\)
(b)(iv) Pressure-volume graph [3 marks]
Change X is heating at constant volume, so it is represented by a straight vertical line from A at \((0.016,2.0)\) to B at \((0.016,4.0)\).
Change Y is an isothermal expansion, so \(pV=\text{constant}\). The graph is a curve with continuously decreasing negative gradient from B to C.
The curve ends at C at \((0.032,2.0)\).
Answer: vertical line \(X\) from A to B, followed by an isothermal curve \(Y\) from B to C.
▶️ Answer/Explanation
(a) Frequency [1 mark]
Frequency is the number of complete oscillations per unit time.
\(f=\dfrac{1}{T}\)
(b)(i) Period from Fig. 4.2 [1 mark]
The kinetic energy reaches a maximum twice during each complete oscillation. From the graph, successive maxima occur at \(t=0.20\,\mathrm{s}\) and \(t=0.60\,\mathrm{s}\).
Therefore, \(T=0.60-0.20=0.40\,\mathrm{s}\) between successive kinetic-energy maxima, giving an oscillation period of \(2\times0.40=0.80\,\mathrm{s}\).
(b)(ii) Angular frequency [2 marks]
Using \(\omega=\dfrac{2\pi}{T}\):
\(\omega=\dfrac{2\pi}{0.80}\)
\(\omega=7.85\,\mathrm{rad\,s^{-1}}\)
Answer: \(\omega=7.9\,\mathrm{rad\,s^{-1}}\)
(b)(iii) Conclusions from the graphs [3 marks]
Any three points:
• The oscillations are simple harmonic.
• The amplitude is \(0.016\,\mathrm{m}\).
• The maximum speed is approximately \(0.13\,\mathrm{m\,s^{-1}}\).
• The total energy of the oscillations is approximately \(7.0\times10^{-4}\,\mathrm{J}\).
• The mass of the object is approximately \(0.087\,\mathrm{kg}\).
• The maximum momentum is approximately \(0.011\,\mathrm{kg\,m\,s^{-1}}\) or \(0.011\,\mathrm{N\,s}\).
(b)(iv) Interchange between kinetic and potential energy [3 marks]
As the object moves from an extreme position towards the equilibrium position, potential energy decreases and is converted into kinetic energy.
At the equilibrium position, the kinetic energy is maximum and the potential energy is minimum.
As the object moves from the equilibrium position towards an extreme position, kinetic energy is converted back into potential energy. At the extreme position, kinetic energy is zero and potential energy is maximum.
The total energy remains constant for an undamped oscillation.
▶️ Answer/Explanation
(a) Electric potential near an isolated proton [3 marks]
Electric potential is defined to be zero at infinity.
A proton has a positive charge, so it repels another positive test charge.
Work must be done to move a positive test charge towards the proton, so the electric potential is positive.
(b)(i) Direction of electric field at Q [1 mark]
Since the sphere is positively charged, the electric field is directed radially away from the centre of the sphere.
Therefore, the arrow at Q should point away from the centre of the sphere, in a WSW direction.
(b)(ii) Variation of electric field with \(x\) [3 marks]
Outside the charged sphere, the electric field follows \(E=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{r^2}\).
At \(x=R\), \(E=E_0\), and at \(x=2R\), \(E=\dfrac{1}{4}E_0\).
For \(x>R\), the field is positive and decreases in magnitude as \(x\) increases. For \(x<-R\), the field is negative and its magnitude decreases as \(x\) becomes more negative.
The sketch should therefore pass through \((R,E_0)\), \((2R,\dfrac{1}{4}E_0)\), \((-R,-E_0)\) and \((-2R,-\dfrac{1}{4}E_0)\), with the curves becoming shallower further from the sphere.
(c) Electric potential energy [2 marks]
For two point charges, \(E_{\mathrm{p}}=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r}\).
For a proton and an electron, \(q_1q_2\) is negative:
\(E_{\mathrm{p}}=-\dfrac{(1.60\times10^{-19})^2}{4\pi(8.85\times10^{-12})(5.3\times10^{-11})}\)
\(E_{\mathrm{p}}=-4.3\times10^{-18}\,\mathrm{J}\)
Answer: \(E_{\mathrm{p}}=-4.3\times10^{-18}\,\mathrm{J}\)
▶️ Answer/Explanation
(a)(i) Rectification [1 mark]
Rectification is the conversion of a.c. to d.c.
(a)(ii) Type of rectification [1 mark]
The bridge rectifier produces full-wave rectification.
(a)(iii) Completing the bridge rectifier [2 marks]
The three missing diodes must be connected with the correct diode symbols and orientations.
The lower-left diode points towards the right, while the upper-left and lower-right diodes both point towards the left.
(b)(i) Period of the input voltage [2 marks]
Comparing \(V_{\mathrm{IN}}=34\sin18t\) with \(V=V_0\sin\omega t\):
\(\omega=18\,\mathrm{rad\,s^{-1}}\)
Using \(\omega=\dfrac{2\pi}{T}\):
\(T=\dfrac{2\pi}{18}\)
\(T=0.35\,\mathrm{s}\)
Answer: \(T=0.35\,\mathrm{s}\)
(b)(ii) r.m.s. input voltage [1 mark]
For a sinusoidal alternating voltage:
\(V_{\mathrm{r.m.s.}}=\dfrac{V_0}{\sqrt{2}}\)
\(V_{\mathrm{r.m.s.}}=\dfrac{34}{\sqrt{2}}\)
\(V_{\mathrm{r.m.s.}}=24\,\mathrm{V}\)
Answer: \(V_{\mathrm{r.m.s.}}=24\,\mathrm{V}\)
(b)(iii) Rectified output waveform [3 marks]
Full-wave rectification makes both halves of the input sinusoidal waveform appear on the same side of the time axis.
The waveform has a peak value of \(34\,\mathrm{V}\), a minimum value of \(0\,\mathrm{V}\), and two identical positive humps during one input period.
Therefore, \(V_{\mathrm{OUT}}=0\) at \(t=0\), \(t=0.175\,\mathrm{s}\) and \(t=0.350\,\mathrm{s}\), with non-zero \(V_{\mathrm{OUT}}\) between these points.
(c)(i) Capacitor connection [1 mark]
The capacitor is connected in parallel with resistor \(R\), across the output terminals.
(c)(ii) Time constant [2 marks]
The time constant is
\(\tau=RC\)
\(\tau=(56\times10^3)(12\times10^{-6})\)
\(\tau=0.67\,\mathrm{s}\)
Answer: \(\tau=0.67\,\mathrm{s}\)
(c)(iii) Minimum smoothed output voltage [2 marks]
During discharge, the capacitor voltage follows \(V=V_0\exp\left(-\dfrac{t}{\tau}\right)\).
Here, \(V_0=34\,\mathrm{V}\), \(t=0.14\,\mathrm{s}\), and \(\tau=0.67\,\mathrm{s}\).
\(V_{\min}=34\exp\left(-\dfrac{0.14}{0.67}\right)\)
\(V_{\min}=28\,\mathrm{V}\)
Answer: \(V_{\min}=28\,\mathrm{V}\)
▶️ Answer/Explanation
(a) Lenz’s law [2 marks]
The direction of the induced e.m.f. is such that it produces effects that oppose the change that caused it.
(b)(i) Magnetic flux cut by rotor OX [3 marks]
The area swept out by rotor OX during one complete rotation is the area of a circle of radius \(12\,\mathrm{m}\):
\(A=\pi r^2\)
The magnetic flux is \(\Phi=BA\).
\(B=0.047\,\mathrm{mT}=0.047\times10^{-3}\,\mathrm{T}\)
\(\Phi=(0.047\times10^{-3})\times\pi\times12^2\)
\(\Phi=0.021\,\mathrm{Wb}\)
Answer: \(\Phi=0.021\,\mathrm{Wb}\)
(b)(ii) Induced e.m.f. [2 marks]
The magnitude of the induced e.m.f. is the magnetic flux cut per unit time:
\(\mathrm{e.m.f.}=\dfrac{\text{flux cut}}{\text{time}}\)
The frequency is \(85\,\mathrm{Hz}\), so the time for one complete rotation is \(T=\dfrac{1}{85}\,\mathrm{s}\).
\(\mathrm{e.m.f.}=\dfrac{0.021}{1/85}\)
\(\mathrm{e.m.f.}=1.8\,\mathrm{V}\)
Answer: \(\mathrm{e.m.f.}=1.8\,\mathrm{V}\)
(b)(iii) Higher potential end [2 marks]
The induced current must produce an anticlockwise moment to oppose the rotation, according to Lenz’s law.
Using Fleming’s left-hand rule, the induced current is from O to X.
Conventional current inside an e.m.f. source flows from the lower potential to the higher potential.
Therefore, X is at the higher potential.
▶️ Answer/Explanation
(a) Tracer [2 marks]
A tracer is a radioactive substance introduced into the body.
The substance is absorbed by the tissues being studied.
(b)(i) Values of \(P\), \(Q\), \(R\) and \(S\) [2 marks]
In positron emission, the mass number is unchanged and the proton number decreases by one.
Therefore, \(P=15\), \(Q=7\), \(R=0\) and \(S=+1\).
Answer: \(P=15,\ Q=7,\ R=0,\ S=+1\)
(b)(ii) Particle Z [1 mark]
Particle Z is an electron neutrino.
(c)(i) Activity [1 mark]
Activity is the number of nuclear disintegrations per unit time.
(c)(ii) Decay constant [2 marks]
Using \(\lambda=\dfrac{\ln 2}{T_{\frac12}}\):
\(T_{\frac12}=2.04\times60=122.4\,\mathrm{s}\)
\(\lambda=\dfrac{\ln 2}{2.04\times60}\)
\(\lambda=5.66\times10^{-3}\,\mathrm{s^{-1}}\)
Answer: \(\lambda=5.66\times10^{-3}\,\mathrm{s^{-1}}\)
(c)(iii) Rate of positron production [4 marks]
The mass of one oxygen-15 nucleus is approximately \(15\times1.66\times10^{-27}\,\mathrm{kg}\).
Number of nuclei:
\(N=\dfrac{2.85\times10^{-6}}{15\times1.66\times10^{-27}}\)
Activity is given by \(A=\lambda N\).
\(A=(5.66\times10^{-3})\dfrac{2.85\times10^{-6}}{15\times1.66\times10^{-27}}\)
\(A=6.48\times10^{17}\,\mathrm{s^{-1}}\)
Answer: \(6.48\times10^{17}\,\mathrm{s^{-1}}\)
(d)(i) Detected particles [1 mark]
The detected particles are gamma photons.
(d)(ii) Formation of the detected particles [2 marks]
The positron collides with an electron inside the body.
The positron and electron annihilate, resulting in their masses being converted into photon energy.
▶️ Answer/Explanation
(a) Photoelectric effect [2 marks]
The photoelectric effect is the emission of electrons from a metal surface when electromagnetic radiation is incident on the surface.
(b)(i) Threshold frequency of metal Y [1 mark]
For a metal, the work function is related to threshold frequency by \(\phi=hf_0\).
Metal X has \(\phi=hF\).
Metal Y has work function \(2\phi\), so its threshold frequency is
\(f_{\mathrm{0,Y}}=\dfrac{2\phi}{h}=2F\)
Answer: \(2F\)
(b)(ii) Planck constant [1 mark]
For metal X, \(\phi=hF\).
Therefore, \(h=\dfrac{\phi}{F}\).
Answer: \(h=\dfrac{\phi}{F}\)
(c) Graph of maximum kinetic energy against frequency [4 marks]
The photoelectric equation is \(E_{\mathrm{K,max}}=hf-\phi\).
Therefore, both graphs are straight lines with the same positive gradient \(h\).
For metal X, the threshold frequency is \(F\), so the X line crosses the \(f\)-axis at \((F,0)\).
Its extrapolation to \(f=0\) gives \(E_{\mathrm{K}}=-\phi\).
For metal Y, the work function is \(2\phi\), so its threshold frequency is \(2F\). Therefore, the Y line crosses the \(f\)-axis at \((2F,0)\).
Its extrapolation to \(f=0\) gives \(E_{\mathrm{K}}=-2\phi\).
Both lines have equal gradients because the gradient is \(h\).
The lines should therefore be two diagonal straight lines of equal positive gradient, with X passing through \((F,0)\) and Y passing through \((2F,0)\).
The X line extrapolates back to \((0,-\phi)\), while the Y line extrapolates back to \((0,-2\phi)\).
▶️ Answer/Explanation
(a) Redshift [2 marks]
Redshift occurs when the recession of a galaxy from the observer causes its emitted light to have an increased observed wavelength.
Equivalently, the observed frequency decreases.
(b) Redshift and expansion of the Universe [2 marks]
Distant galaxies show redshift, indicating that the galaxies are moving away from us.
Since galaxies are moving apart, this provides evidence that the Universe is expanding.
(c) Hubble’s law and the Big Bang theory [3 marks]
Hubble’s law states that the speed of recession of a galaxy is proportional to its distance from us: \(v=H_0d\).
Therefore, more distant galaxies are receding faster.
Looking at more distant galaxies means observing further back in time. A long time ago, the galaxies and matter in the Universe must have been much closer together.
This leads to the idea that the Universe began in a very small, dense state and has been expanding, which forms the basis of the Big Bang theory.