Questions 1
Topic – 5.2
(a) Explain why the gravitational potential near to a point mass is negative.
(b) A planet may be assumed to be a uniform sphere. It has gravitational potential φ at distance r from the centre of the planet. The variation with \(\frac {1}{ r}\) of φ is shown in Fig. 1.1.
(i) Show that the mass of the planet is \(8.8 × 10^{25}\) kg.
(ii) The period of rotation of the planet is 0.72 Earth days. A satellite in orbit around the planet remains above the same point on the surface of the planet. Use the mass of the planet in (b)(i) to determine the radius R of the orbit of the satellite.
(iii) The speed of the satellite in (b)(ii) is \(8400ms^{–1}\). The mass of the satellite is 1200kg. Determine the additional energy required to move the satellite from its orbit to infinity.
▶️Answer/Explanation
Ans
(a) (gravitational) potential is zero at infinity (gravitational force between two masses is attractive so)
either work is done on a mass to move it away from another mass or work is done on a mass to move it to infinity
(b)(i) M = (–) gradient / G
e.g. M = \((1.76 × 10^8) / (3.0 × 10^{–8} × 6.67 × 10^{–11}\)) = \(8.8 × 10^{25}\) kg
(ii) either GMm / \(r^2\) = \(mr \omega ^2\) and \( \omega \) = \(2\pi / T\)
or GMm / \(r^2\) = \(mv^2/ r\) and v = \(2\pi r / T\)
or GMm / \(r^2\) = \(4\pi ^2 mr\) / \(T^2\)
\(R^3 = 6.67 × 10^{–11} ×8.8 × 10^{25} × (0.72 × 24 × 60 ×60)^2/ 4\pi ^2\)
R = \(8.3 × 10^7\) m
(iii) \(\Delta E\) = (GMm / r) – ½mv2
kinetic energy = \((½ × 1200 × 8400^2)\)
potential energy = \((–)[(6.67 × 10^{–11} × 8.8 × 10^{25} × 1200) / (8.3 × 10^7)]\)
\(\Delta E\) = \([(6.67 × 10^{–11} × 8.8 × 10^{25} × 1200) / (8.3 × 10^7)]– (½ × 1200 × 8400^2)\)
= \(4.3 × 10^{10}\) J
Questions 2
Topic – 4.3
(a) By referring to both kinetic energy and potential energy, explain what is meant by the internal energy of an ideal gas.
(b) A fixed mass of an ideal gas at a temperature of 20°C is sealed in a cylinder by a piston, as shown in Fig. 2.1.
The initial volume of the gas is \(1.24 × 10^{–4}m^3\). Thermal energy is supplied to the gas and its volume increases by \(5.20 × 10^{–5}m^3\).
(i) The piston is freely moving so that the gas is always at atmospheric pressure. Atmospheric pressure is \(1.01 × 10^5Pa\). Calculate the work done by the gas.
(ii) Calculate the final thermodynamic temperature T of the gas.
(iii) The mass of the gas is 16g. For this expansion, there is a net transfer of 960J of thermal energy to the gas. Calculate the specific heat capacity c of the gas at this pressure.
(c) The gas in (b) is allowed to return to its starting temperature. The piston is now fixed in position. Thermal energy is supplied to increase the temperature to the same final temperature as in (b). Use the first law of thermodynamics to suggest and explain how the specific heat capacity of the gas for this situation compares with the value in (b)(iii).
▶️Answer/Explanation
Ans
(a) total kinetic energy associated with random motion of molecules
plus total potential energy (of molecules) but potential energy is zero
(b)(i) W =\( p\Delta V \)
= \(1.01 \times 10^5 \times 5.20 \times 10^{–5}\)
= (+)5.25 J
(ii)\( V \propto T\) or V / T = constant
1.24 / (273 + 20) = (1.24 + 0.520) / T
T = 416 K
(iii) \(c = Q / m\Delta T \)
= \(960 / (0.016 (416 – 293)\)
= \(490 J kg^{–1} K^{–1}\)
(c) no change in volume so no work is done (by the gas)
(same temperature change so) same change in internal energy
less thermal energy needs to be supplied so c is less
Questions 3
Topic – 17.1
A small object of mass 24g rests on a platform. The platform is attached to an oscillator, as shown in Fig. 3.1.
The oscillator moves the platform up and down.
(a) The total energy of the oscillations of the object is \(2.2 × 10^{–4}\) J. In one oscillation the object travels a total distance of 14mm. Calculate the angular frequency ω of the oscillations.
(b) The frequency of the oscillator is fixed, and the amplitude of the oscillations is gradually increased.
(i) Calculate the maximum amplitude of the oscillations so the object does not lose contact with the platform.
(ii) The amplitude of the oscillations is increased so it is greater than the value in (b)(i). State and explain the position in an oscillation where the object first loses contact with the platform.
▶️Answer/Explanation
Ans
(a) \(E = ½ m \omega ^2 X_o^2\)
\(2.2 10^{–4} = ½ \times 24 \times10^{–3} \times (14 \times 10^{–3}/ 4)^2 \times \omega ^2 \)
\(\omega = 39 rad s^{–1}\)
(b)(i) use of acceleration = \(9.81 m s^{–2}\)
\(X_o = 9.81 / 39^2\)
= \(6.4 \times 10^{–3}\) m
(ii) at top of oscillation
any one point from:
where the downward acceleration first exceeds free-fall acceleration
where the greatest downwards acceleration occurs
where the resultant force is the maximum downwards
where the contact force is a minimum
Questions 4
Topic – 19.1
(a) Three capacitors are connected as shown in Fig. 4.1.
Determine the total capacitance, in μF, of the network of three capacitors.
(b) A capacitor of capacitance 45μF is connected to a variable power supply initially set at 8.0V. The output of the power supply increases so that the potential difference (p.d.) across the capacitor increases to 9.6V. Calculate the increase in energy ΔE stored in the capacitor.
(c) A sinusoidal a.c. power supply is connected to the input of a bridge rectifier. The output of the rectifier is connected to a load resistor.
(i) Complete the circuit in Fig. 4.2 by adding a capacitor to smooth the p.d. across the load resistor
(ii) The variation with time t of the p.d. V of the smoothed output is shown in Fig. 4.3. 
Determine the time constant, in ms, of the smoothing circuit.
(d) A sinusoidal a.c. power supply has a maximum power of 16W. State the value of the mean power when the output of the power supply is:
(i) full‑wave rectified
mean power = ……………….
(ii) half‑wave rectified……….
▶️Answer/Explanation
Ans
(a) combined capacitance of parallel capacitors = 30 \((\mu F)\)
total capacitance = \((1 / 45 + 1 / 30)^{–1}\)
= \(18 \mu F\)
(b) \(E = ½CV^2\)
\(\Delta E = ½ \times 45 \times 10^{–6}(9.6^2 – 8.0^2)\)
= \(6.3 \times 10^{–4} J\)
(c)(i) gaps in circuit closed and correct symbol for capacitor shown in parallel with load resistor
4(c)(ii) two correct pairs of values of t and V read off from within same discharge cycle, e.g. (5.0, 4.0) and (13.0, 3.2)
correct substitution of values of V, V0 and \(\Delta t\) into V = V0 exp \((–\Delta t / \tau)\)
e.g. \(3.2 = 4.0 exp (–8.0 / \tau)\)
\(\tau = 36 ms\)
(d)(i) 8.0 W
(ii) 4.0 W
Questions 5
Topic – 18.1
(a) An object travels in a circle at constant speed. State the names of two quantities that vary during the motion of the object.
(b) A charged particle of mass m and with charge q enters a region of uniform magnetic field, perpendicular to the field lines. The magnetic flux density is B. The particle travels in a circle with period T and radius r.
(i) By considering the magnetic force acting on the particle, show that \( B=\frac{2\pi m}{qT}\).
(ii) The particle is an alpha particle. The period of the circular motion is 2.5μs. Calculate B.
(iii) A second alpha particle is in the same uniform field. It travels in a circle of radius 2r. State and explain how the periods of the motion of the two particles compare.
(iv) The speed of the alpha particle in (b)(ii) is \(1.1 × 10^6ms^{–1}\). An electric field is applied so that this particle now moves with constant velocity. Use your answer in (b)(ii) to calculate the electric field strength E. Give the unit with your answer.
▶️Answer/Explanation
Ans
(a) any 2 points from:
• (angular) displacement
• velocity
• momentum
• (centripetal) acceleration
• (resultant) force
(b)(i) \(Bqv = mv^2/ r\)
\(v = 2\pi r / T\)
completion of algebra leading to \(B = 2\pi m / qT\)
(ii) B = \((2\pi × 4 × 1.66 × 10^{–27}) / (2 × 1.60 × 10^{–19} × 2.5 × 10^{–6})\)
= 0.052 T
(iii) either the same because T is independent of r
or the same because B, q and m are unchanged
or the same because both radius and speed have doubled
(iv) qE = Bqv
E = Bv = \(0.052 × 1.1 × 10^6\)
= \(5.7 × 10^4 N C^{–1}\)
Questions 6
Topic – 20.4
(a) A small coil C has 64 turns and cross‑sectional area \(0.71cm^2\). The coil is placed inside a solenoid as shown in Fig. 6.1.
The centre of coil C is on the central axis of the solenoid.
(i) There is a constant current in the solenoid. Coil C is moved through the solenoid from position X to position Y. On Fig. 6.2, sketch a line to show the variation of the magnetic flux linkage in coil C with position as it moves from X to Y. 
(ii) Explain the shape of your line in (a)(i).
(iii) Coil C is now held stationary at X. The current in the solenoid varies so that the magnetic flux density B at X varies from time 0 to time 4t as shown in Fig. 6.3. 
Calculate the maximum magnetic flux linkage in coil C.
(iv) On Fig. 6.4, sketch a line to show the induced electromotive force (e.m.f.) E in coil C from time 0 to time 4t.
(b) A metal spring rests on a smooth table. The turns of the spring are equally spaced. The ends of the spring are connected to a d.c. power supply, as shown in Fig. 6.5. 
The spring is connected to the d.c. power supply using flexible leads. The spring is not under tension. With reference to magnetic fields, describe and explain the change in the distance between the turns of the spring when the power supply is first switched on.
▶️Answer/Explanation
Ans
(a)(i) non-zero horizontal straight line from X to Y
(ii) constant flux density (inside coil)
either (magnetic) flux linkage proportional to flux density
or \(\phi \)= BAN and B, A and N are all constant
(iii) \(\phi \)= BAN
= \(0.080 \times 0.71 \times 10^{–4} \times 64\)
= \(3.6 \times 10^{–4} Wb\)
(iv) sketch showing:
E is zero from time 0 to time t and non-zero after time t
E has constant non-zero magnitude between time t and time 4t
E has non-zero value of one sign between time t and time 2t, and non-zero value of the opposite sign between time 2t and time 4t
(b) current in spring creates a magnetic field around the spring
either (magnetic) fields around adjacent turns interact to cause a force to be exerted (between the turns) or current in one turn interacts with (magnetic) field due to adjacent turns to cause force to be exerted (between the turns)
(magnetic force) is attractive so distance (between turns) decreases
Questions 7
Topic – 22.1
(a) A photon has an energy of \(3.11 × 10^{–19}\) J. Calculate the momentum of the photon.
(b) A laser beam has a power of 350mW. The light from the laser has a wavelength of 640nm.
(i) Determine the number of photons emitted by the laser in a time of 1.0s.
(ii) The laser beam is incident normally on a surface that absorbs all of the photons. Show that the force F exerted on the surface by the laser beam is given by \(F=\frac{P}{c}\) where P is the power of the laser beam and c is the speed of light.
(c) Light of a single wavelength is incident on the surface of different metals. The work function energy of the metals is given in Table 7.1.
(i) Explain the term threshold wavelength.
(ii) For the metals in Table 7.1, calculate the value of the largest threshold wavelength.
▶️Answer/Explanation
Ans
(a) p= E / c
= \((3.11 \times 10^{–19}) / (3.00 \times 10^8)\)
= \(1.04 \times 10^{–27} N s\)
(b)(i) E = hf and c = \(f\lambda \) so
energy of one photon = \(hc / \lambda \)
\(350 \times 10^{–3} = N \times (6.63 \times 10^{–34} \times 3.00 \times 10^8) / (640 \times 10^{–9})\)
\(N = 1.1 \times 10^{18}\)
(ii) F = (change in) momentum / time
Clear use of p = E / c and t = E / P to complete the algebra and arrive at the final equation:
e.g. F = [E / c] / [E / P] = P / c
(c)(i) maximum wavelength (of electromagnetic radiation) that causes electrons to be emitted (from surface of metal)
(ii) work function = \(2.26 \times 1.60 \times 10^{–19}\) (J)
\(E = hc / \lambda \)so
\((2.26 \times 1.60 \times 10^{–19}) = (6.63 \times 10^{–34} \times 3.00 \times 10^8) / \lambda_0\)
\(\lambda _0 = 5.50 \times 10^{–7 m}\)
Questions 8
Topic – 23.1
(a) State what is meant by the binding energy of a nucleus.
(b) A nucleus of uranium‑235 absorbs a neutron and becomes unstable. It then undergoes a fission reaction. One possible reaction is
(i) Determine the number of neutrons produced in this fission reaction.
(ii) Data for the binding energies per nucleon for this fission reaction are given in Table 8.1. 
Calculate the energy released, in MeV, from the fission of one nucleus of uranium‑235.
(iii) The isotope xenon‑142 is unstable. The isotope xenon‑132 is stable. Suggest a reason why xenon‑142 is unstable.
(iv) Xenon‑142 decays into the isotope caesium‑142. A sample initially contains only nuclei of xenon‑142. After a time equal to 6.0s, the ratio \(\frac{number of decayed nuclei of xenon‑142}{number of undecayed nuclei of xenon‑142}\) is equal to 31. Calculate the half‑life of xenon‑142. Show your working
▶️Answer/Explanation
Ans
(a) (minimum) energy required to separate the nucleons (of a nucleus) to infinity
(b)(i) 4
(ii) energy= \((142 \times 8.37) + (90 \times 8.72) – (235 \times 7.59)\)
= 190 MeV
(iii) either it has too many neutrons (for the number of protons) or its neutron to proton ratio is too high
(iv) (when t = 6.0 s), N / No = 1 / 32
either
\((1 / 32) = exp (– ln2 \times 6.0 / t½)\)
t½ = 1.2 s
or
\(32 / 2^n\) = 1 so n = 5 (half-lives)
t1/2 = 6.0 / 5
= 1.2 s
Questions 9
Topic -24.1
(a) Electrons in a vacuum are accelerated through a potential difference of 84kV. The electrons then strike a metal target and X‑rays are produced.
(i) Calculate the minimum wavelength of the X‑rays that are produced.
(ii) The melting points of two metals are given in Table 9.1.
Suggest why the metal target is made from tungsten rather than copper.
(b) An X‑ray beam is incident normally on a sample of soft tissue and bone as shown in Fig. 9.1. 
Data for the two materials are given in Table 9.2. 
The total thickness of soft tissue is x. The total thickness of bone is also x. The incident intensity of the X‑ray beam is \(I_0\). The transmitted intensity of the X‑ray beam is 13% of the incident intensity. Determine x, in cm.
(c) (i) Define the specific acoustic impedance of a medium.
(ii) Use data from Table 9.2 to calculate the percentage of the intensity of ultrasound that is transmitted at a boundary between soft tissue and bone.
(iii) The ultrasound is now incident on the sample of soft tissue and bone shown in Fig. 9.1. Suggest two reasons why the transmitted intensity through the sample is less than the answer in (c)(ii).
▶️Answer/Explanation
Ans
(a)(i) \(eV = hc /\lambda \)
\(\lambda = (6.63 \times 10^{–34} \times 3.00 \times 10^8) / (84 \times 10^3 \times 1.60 \times 10^{–19})\)
= \(1.5 \times 10^{–11} m\)
(ii) either (some) kinetic energy (of electrons) is converted to thermal energy at target
or some X-rays are absorbed by the target so its temperature increases
(tungsten) has higher melting point so does not melt quickly / easily
(b) \(I = I_0 exp (–\mu t)\)
\(0.13 = [exp (–3.0x)] \times [exp (–0.22x)]\)
= exp (–3.22x)
x = 0.63 cm
(c)(i) product of density and speed
speed of ultrasound in medium
(ii) \(IR / I0 = (7.8 – 1.7)^2/ (7.8 + 1.7)^2\)
= 0.41
fraction transmitted = 1.00 – 0.41 = 0.59
percentage transmitted = 59%
(iii) more than one boundary so more reflections
some ultrasound is attenuated in matter
Questions 10
Topic – 25.1
(a) The Sun has a surface temperature of 5780K. The luminosity of the Sun is \(3.85 × 10^{26}\)W.
(i) Calculate the radius of the Sun.
(ii) The Earth is a distance of \(1.50 × 10^{11}\)m from the Sun. Calculate the radiant flux intensity F of the radiation from the Sun at a distance of \(1.50 × 10^{11}\)m. Give a unit with your answer.
(iii) The variation with wavelength of the intensity of radiation emitted from the Sun is shown in Fig. 10.1.
Another star has the same radius as the Sun but has a lower surface temperature. On Fig. 10.1, sketch a line to show the variation with wavelength of the intensity of the radiation emitted for this star
(b) A galaxy in the constellation Corona Borealis is moving away from the Earth.
(i) The visible emission spectrum for the Sun is shown in Fig. 10.2.
The lines are at wavelengths of 397nm, 410nm, 434nm, 486nm and 656nm. The compositions of the Sun and a star in the Corona Borealis galaxy are similar. On Fig. 10.3, sketch the emission spectrum for the star in the Corona Borealis galaxy as observed from the Earth. No calculations are required.
(ii) The galaxy in Corona Borealis is moving away from the Earth at a speed of \(21400kms^{–1}\). Use information from (b)(i) to calculate, in nm, the observed wavelength of the lowest visible energy emission for the star in the Corona Borealis galaxy.
(iii) The wavelength in (b)(ii) is used to calculate a value for the surface temperature of the star in the Corona Borealis galaxy. The calculation does not give an accurate value. State and explain whether this value of temperature is too high or too low.
▶️Answer/Explanation
Ans
(a)(i) \(L = 4\pi σr^2T^4\)
\(3.85 \times 10^{26} = 4\pi \times 5.67 \times 10^{–8} \times r^2 \times 5780^4\)
\(r = 6.96 \times 10^8 m\)
(ii) \(F = L / 4\pi d^2\)
= \((3.85 \times 10^{26}) / (4\pi \times (1.50 \times 10^{11})^2)\)
= \(1.36 \times 10^3 W m^{–2}\)
(iii) line of same shape showing peak intensity at greater wavelength
line of same shape showing lower peak intensity
(b)(i) 5 lines in same pattern shifted to longer wavelengths
(ii) \(\Delta \lambda /\lambda \)= v / c
\(\Delta \lambda\)= \((21400 / 300000) \times 656\)
= 46.8 nm
wavelength = 656 + 46.8
= 703 nm
(iii) (peak) wavelength too high so temperature too low