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A Level Physics 9702 Paper 4 (10 Questions, 100 Marks, 2 Hours)
A Level Physics 9702 Paper 4
A Level Physics 9702 Paper 4 practice paper: 10 original questions, 100 marks in 2 hours, with full answers and marking notes on mechanics, fields and quantum physics.
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Paper details Level: A Level | Subject: Physics | Paper: 4 | Time: 2 hours | Marks: 100 | Questions: 10
Section A Mechanics, Thermal Physics and Oscillations (50 marks)
Questions 1-2
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | 1. A satellite orbits Earth in a circular orbit of radius 7.0 x 10^6 m (GM = 3.99 x 10^14 m3 s-2). (a) Show that the orbital speed is 7.55 x 10^3 m s-1. (3 marks) (b) Calculate the orbital period in minutes. (4 marks) (c) A second satellite orbits at a larger radius. Explain why its period is longer. (3 marks) |
Answer: (a) GMm/r2 = mv2/r gives v = sqrt(GM/r) = sqrt(5.7e+07) = 7.55 x 10^3 m s-1. (3) (b) T = 2(pi)r/v = 2(pi) x 7.0e+06 / 7550 = 5826 s = 97 minutes. (4) (c) A larger radius weakens the gravitational field, so a smaller centripetal acceleration is needed, but the circumference is greater; the net result is a smaller speed and a longer period, T proportional to r^(3/2). (3) |
| 2 | 2. A cylinder of volume 2.0 x 10^-3 m3 contains 5.0 x 10^22 molecules of a monatomic ideal gas at 300 K (k = 1.38 x 10^-23 J K-1). (a) Calculate the gas pressure. (3 marks) (b) Calculate the mean kinetic energy of one molecule. (4 marks) (c) The gas is heated at constant volume. Explain in molecular terms why the pressure rises. (3 marks) |
Answer: (a) p = NkT/V = 5.0e+22 x 1.38e-23 x 300 / 2.0e-03 = 1.03 x 10^5 Pa. (3) (b) E = 3/2 kT = 1.5 x 1.38e-23 x 300 = 6.21e-21 J. (4) (c) Higher temperature means greater mean kinetic energy and speed; molecules strike the walls more often and each collision changes momentum more, so the force per unit area increases. (3) |
Questions 3-5
| # | Question | Answer and marking notes |
|---|---|---|
| 3 | 3. A block of mass 0.25 kg on a spring of stiffness 40 N m-1 oscillates with amplitude 40 mm. (a) Calculate the period. (4 marks) (b) Calculate the maximum speed. (3 marks) (c) Calculate the maximum acceleration and state where it occurs. (3 marks) |
Answer: (a) omega = sqrt(k/m) = sqrt(40/0.25) = 12.6 rad s-1; T = 2(pi)/omega = 0.497 s. (4) (b) v0 = omega A = 12.6 x 0.040 = 0.506 m s-1. (3) (c) a0 = omega2 A = 12.6^2 x 0.040 = 6.4 m s-2, at maximum displacement. (3) |
| 4 | 4. A 470 microfarad capacitor is charged to 12 V and then discharged through a 2.2 kohm resistor. (a) Calculate the initial energy stored. (3 marks) (b) Calculate the time constant. (4 marks) (c) Calculate the time for the voltage to fall to 3.0 V. (3 marks) |
Answer: (a) E = 1/2 C V2 = 0.5 x 4.70e-04 x 12^2 = 0.0338 J. (3) (b) tau = RC = 2200 x 4.70e-04 = 1.03 s. (4) (c) V = V0 exp(-t/tau), so t = tau ln(V0/V) = 1.03 x ln(12/3.0) = 1.43 s. (3) |
| 5 | 5. A coil of 200 turns, each of area 4.5 x 10^-4 m2, lies perpendicular to a uniform magnetic field that rises from 0.020 T to 0.060 T in 0.15 s. (a) Calculate the induced emf. (3 marks) (b) The coil resistance is 0.60 ohm; calculate the induced current. (3 marks) (c) Calculate the charge that flows in this time. (4 marks) |
Answer: (a) emf = N A dB/dt = 200 x 4.5e-04 x 0.04 / 0.15 = 24.0 x 10^-3 V. (3) (b) I = emf/R = 24.0e-3 / 0.60 = 40 mA. (3) (c) Q = I t = 40e-3 x 0.15 = 6.0 x 10^-3 C (also Q = N dPhi/R). (4) |
Section B Fields, Quantum and Nuclear Physics and Applications (50 marks)
Questions 6-7
| # | Question | Answer and marking notes |
|---|---|---|
| 6 | 6. Light of wavelength 250 nm falls on a metal of work function 2.0 eV (h = 6.63 x 10^-34 J s, c = 3.00 x 10^8 m s-1). (a) Calculate the photon energy in eV. (3 marks) (b) Calculate the maximum kinetic energy of the photoelectrons. (4 marks) (c) Calculate the threshold wavelength. (3 marks) |
Answer: (a) E = hc/lambda = 6.63e-34 x 3.00e8 / 2.5e-07 = 7.956e-19 J = 4.97 eV. (3) (b) KEmax = 4.97 – 2.0 = 2.97 eV = 4.756e-19 J. (4) (c) lambda0 = hc/phi = 6.63e-34 x 3.00e8 / 3.2e-19 = 622 nm. (3) |
| 7 | 7. A helium-4 nucleus contains 2 protons and 2 neutrons. Proton mass 1.00728 u, neutron mass 1.00867 u, helium-4 nucleus mass 4.00260 u; 1 u is equivalent to 931.5 MeV. (a) Calculate the mass defect. (4 marks) (b) Calculate the total binding energy in MeV. (3 marks) (c) Calculate the binding energy per nucleon. (3 marks) |
Answer: (a) Sum of nucleon masses = 2 x 1.00728 + 2 x 1.00867 = 4.03190 u; mass defect = 4.03190 – 4.00260 = 0.02930 u. (4) (b) E = 0.02930 x 931.5 = 27.29 MeV. (3) (c) 27.29 / 4 = 6.82 MeV per nucleon. (3) |
Questions 8-10
| # | Question | Answer and marking notes |
|---|---|---|
| 8 | 8. An ultrasound probe of frequency 2.5 MHz operates in soft tissue where the speed of sound is 1540 m s-1 and the density is 1060 kg m-3. (a) Calculate the wavelength in tissue. (3 marks) (b) An echo returns 52 microseconds after the pulse; calculate the depth of the reflecting boundary. (4 marks) (c) The acoustic impedance of air is about 400 kg m-2 s-1; calculate the intensity reflection coefficient at a tissue-air boundary and state the purpose of coupling gel. (3 marks) |
Answer: (a) lambda = c/f = 1540 / 2.5e+06 = 6.16e-04 m. (3) (b) d = ct/2 = 1540 x 5.2e-05 / 2 = 4.00 cm. (4) (c) Z1 = 1.63e+06; R = ((Z1-Z2)/(Z1+Z2))^2 = 0.9990, about 99.9 per cent reflected, so gel removes the air layer. (3) |
| 9 | 9. A point charge of +3.2 nC is in a vacuum (k = 8.99 x 10^9 N m2 C-2). (a) Calculate the field strength 12 cm from the charge. (3 marks) (b) A second charge of +5.0 nC moves from 12 cm to 30 cm; calculate the work done by the electric force. (4 marks) (c) State the relationship between equipotential surfaces and field lines. (3 marks) |
Answer: (a) E = kQ/r2 = 8.99e+09 x 3.2e-09 / 0.12^2 = 1998 V m-1. (3) (b) V = kQ/r: V1 = 239.7 V, V2 = 95.9 V; W = q(V1-V2) = 5.0e-09 x 143.8 = 7.19e-07 J. (4) (c) Equipotential surfaces are everywhere perpendicular to field lines; no work is done moving charge along one, and field lines run from high to low potential. (3) |
| 10 | 10. A gas at 1.5 x 10^5 Pa expands at constant pressure from 2.0 x 10^-3 m3 to 3.2 x 10^-3 m3 while absorbing 500 J of heat. (a) Calculate the work done by the gas. (4 marks) (b) Calculate the change in internal energy. (3 marks) (c) Explain why the heat absorbed exceeds the increase in internal energy. (3 marks) |
Answer: (a) W = p dV = 1.5e+05 x (3.2e-03 – 2.0e-03) = 180 J. (4) (b) dU = Q – W = 500 – 180 = 320 J. (3) (c) At constant pressure the gas must push back the surroundings, so part of the heat supplied leaves as work. (3) |
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