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IB Physics SL Paper 2 Mock Paper (10 questions, 50 marks, 75 minutes, calculator allowed)
IB Physics
IB Physics SL Paper 2 mock paper: 10 questions, 50 marks, 75 minutes, calculator allowed, covering projectile motion, energy and friction on a slope, specific heat capacity and heater efficiency, double-slit fringe spacing, ideal gases and rms speed, emf and internal resistance, electromagnetic induction, satellite orbits, the photoelectric effect, and radioactive decay with binding energy, each with worked answers and marking notes.
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Paper details Level: IB Diploma SL | Subject: Physics Paper 2 | Time: 75 minutes | Marks: 50 | Questions: 10 | Calculator allowed
Section A (Questions 1 to 5, 25 marks)
Answer ALL questions. A calculator may be used. Show your working in calculations, give numerical answers to three significant figures unless told otherwise, and state units in every final answer. All data are examples and this is not a past paper of any examination board.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | A small ball leaves the edge of a table 1.8 m above the floor with a horizontal speed of 6.0 m s-1. Air resistance is negligible; take g = 9.81 m s-2. (a) Show that the ball takes about 0.61 s to reach the floor. (b) Calculate the horizontal distance travelled before landing. (c) Calculate the vertical velocity component and the magnitude of the velocity on landing. (d) Determine the angle between the velocity on landing and the horizontal. |
(a) From h = 0.5gt^2, t = sqrt(2h / g) = sqrt(2 x 1.8 / 9.81) = 0.606 s, which is about 0.61 s. (b) Horizontal distance = ut = 6.0 x 0.606 = 3.63 m. (c) Vertical component = gt = 9.81 x 0.606 = 5.94 m s-1; magnitude = sqrt(u^2 + vy^2) = sqrt(6.0^2 + 5.94^2) = 8.44 m s-1. (d) tan(theta) = 5.94 / 6.0 = 0.990, so theta = 44.7 degrees below the horizontal. Marking: (a) 1 mark, (b) 1 mark, (c) 2 marks, (d) 1 mark. |
| 2 | A block of mass 2.5 kg slides from rest down a slope inclined at 25 degrees, travelling 3.0 m along the slope. The coefficient of kinetic friction is 0.30; take g = 9.81 m s-2. (a) Calculate the normal reaction force on the block. (b) Calculate the friction force acting on the block. (c) Calculate the work done by friction. (d) Hence determine the speed of the block after sliding 3.0 m. |
(a) N = mg cos(theta) = 2.5 x 9.81 x cos 25 = 22.23 N. (b) f = mu x N = 0.30 x 22.23 = 6.67 N. (c) Wf = f x d = 6.67 x 3.0 = 20.0 J, negative work because it opposes the motion. (d) Loss of potential energy = mgh = 2.5 x 9.81 x 1.268 = 31.1 J; kinetic energy = 31.1 – 20.0 = 11.1 J; v = sqrt(2 x 11.1 / 2.5) = 2.98 m s-1. Marking: (a) 1 mark, (b) 1 mark, (c) 1 mark, (d) 2 marks. |
| 3 | An electric heater raises the temperature of 0.50 kg of water from 20 °C to 60 °C. The specific heat capacity of water is 4180 J kg-1 K-1 and the heater operates at 500 W. (a) Calculate the energy absorbed by the water. (b) Calculate the time needed if no energy is lost. (c) The actual heating time is 1.282 times the ideal time. Calculate the efficiency of the heater and the actual time taken. (d) Explain why the actual time is longer. |
(a) E = mc dT = 0.50 x 4180 x 40 = 83600 J. (b) t = E / P = 83600 / 500 = 167.2 s. (c) Efficiency = 1 / 1.282 = 0.78; actual time = 167.2 / 0.78 = 214 s (about 3.6 minutes), so the energy supplied is 107179 J. (d) Energy is transferred to the surrounding air and to the container by convection, conduction and radiation, so the energy supplied exceeds the energy absorbed by the water. Marking: (a) 1 mark, (b) 1 mark, (c) 2 marks, (d) 1 mark. |
| 4 | Monochromatic light of wavelength 630 nm is incident on a double slit of separation 0.25 mm. The screen is 2.0 m from the slits. (a) Calculate the spacing between adjacent bright fringes. (b) Find the distance between the first and the sixth bright fringe. (c) Calculate the frequency of the light. (d) The whole arrangement is immersed in a liquid of refractive index 1.33. Determine the new fringe spacing. |
(a) x = lambda D / d = 6.30e-07 x 2.0 / 2.50e-04 = 5.04 mm. (b) Five spacings = 5 x 5.04 = 25.2 mm. (c) f = c / lambda = 3.00e8 / 6.30e-07 = 4.76e+14 Hz. (d) In the liquid the wavelength becomes lambda / n = 474 nm, so the new spacing is 3.79 mm, a factor of 1/1.33 smaller. Marking: (a) 2 marks, (b) 1 mark, (c) 1 mark, (d) 1 mark. |
| 5 | 2.0 mol of oxygen is contained in a vessel of volume 0.030 m3 at 300 K. Take R = 8.31 J mol-1 K-1 and the molar mass of oxygen as 0.032 kg mol-1. (a) Calculate the pressure of the gas. (b) The gas is compressed isothermally to 0.012 m3. Determine the new pressure. (c) Calculate the root mean square speed of the oxygen molecules. (d) State whether the mean kinetic energy of the molecules changes during the compression, and why. |
(a) p = nRT / V = 2.0 x 8.31 x 300 / 0.030 = 166200 Pa (about 166.20 kPa). (b) For constant temperature pV is constant, so p2 = 166200 x 0.030 / 0.012 = 415500 Pa (about 415.5 kPa). (c) v rms = sqrt(3RT / M) = sqrt(3 x 8.31 x 300 / 0.032) = 483 m s-1. (d) The mean kinetic energy depends only on temperature, which is constant, so it does not change; the pressure rises because molecules strike the walls more frequently. Marking: (a) 2 marks, (b) 1 mark, (c) 1 mark, (d) 1 mark. |
Section B (Questions 6 to 10, 25 marks)
Answer ALL questions. A calculator may be used. Show your working in calculations, give numerical answers to three significant figures unless told otherwise, and state units in every final answer. All data are examples and this is not a past paper of any examination board.
| # | Question | Answer and marking notes |
|---|---|---|
| 6 | A battery of emf 12.0 V and internal resistance 0.50 ohms is connected to two resistors of 4.0 ohms and 6.0 ohms connected in parallel. (a) Calculate the equivalent resistance of the parallel combination. (b) Calculate the current supplied by the battery. (c) Calculate the terminal potential difference across the battery. (d) Calculate the power dissipated in the 6.0 ohm resistor. |
(a) Rp = (4.0 x 6.0) / (4.0 + 6.0) = 2.4 ohms. (b) Total resistance = 2.4 + 0.50 = 2.90 ohms; I = emf / R = 12.0 / 2.90 = 4.14 A. (c) V = emf – Ir = 12.0 – 4.14 x 0.50 = 9.93 V. (d) P = V^2 / R = 9.93^2 / 6.0 = 16.4 W (the current in this resistor is 2.48 A). Marking: (a) 1 mark, (b) 2 marks, (c) 1 mark, (d) 1 mark. |
| 7 | A coil of 40 turns and area 0.025 m2 lies in a uniform magnetic field. The flux density increases uniformly from zero to 0.80 T in 0.20 s. The resistance of the coil is 8.0 ohms. (a) Calculate the induced emf in the coil. (b) Calculate the induced current. (c) Calculate the charge that passes a given cross-section of the wire during this time. (d) One side of the coil is 0.15 m long and perpendicular to the field. Calculate the magnetic force on it. |
(a) emf = N A dB / dt = 40 x 0.025 x 0.80 / 0.20 = 4.0 V. (b) I = emf / R = 4.0 / 8.0 = 0.50 A. (c) Q = I t = 0.50 x 0.20 = 0.10 C. (d) F = B I L = 0.80 x 0.50 x 0.15 = 0.060 N, directed so as to oppose the motion (Lenz law). Marking: (a) 2 marks, (b) 1 mark, (c) 1 mark, (d) 1 mark. |
| 8 | A satellite moves in a circular orbit 500 km above the Earth surface. The Earth radius is 6.37 x 10^6 m and GM for the Earth is 3.986e+14 m3 s-2. (a) Calculate the orbital speed of the satellite. (b) Calculate the orbital period in minutes. (c) Calculate the gravitational field strength at that height. (d) Explain why the satellite does not need a continuously firing engine. |
(a) v = sqrt(GM / r) = sqrt(3.986e+14 / 6.870e+06) = 7617 m s-1. (b) T = 2 pi r / v = 2 pi x 6.870e+06 / 7617 = 5667 s, about 94.4 minutes. (c) g = GM / r^2 = 3.986e+14 / (6.870e+06)^2 = 8.45 m s-2. (d) Gravity provides the centripetal force, so only the direction of the velocity changes; in vacuum there is no drag to remove energy, so no thrust is required. Marking: (a) 2 marks, (b) 2 marks, (c) 1 mark, (d) 1 mark. |
| 9 | The work function of a metal is 2.20 eV. Light of wavelength 400 nm is incident on it (take hc = 1240 eV nm and 1 eV = 1.60e-19 J). (a) Calculate the energy of one incident photon. (b) Calculate the maximum kinetic energy of the emitted photoelectrons in eV and in J. (c) Calculate the threshold wavelength of the metal. (d) State the stopping potential, and state how the maximum kinetic energy changes if the intensity is increased. |
(a) E = hc / lambda = 1240 / 400 = 3.10 eV. (b) KE max = E – phi = 3.10 – 2.20 = 0.90 eV = 1.4e-19 J. (c) Threshold wavelength = hc / phi = 1240 / 2.20 = 564 nm. (d) Stopping potential = 0.90 V; increasing the intensity only increases the number of photons, so the maximum kinetic energy is unchanged. Marking: (a) 1 mark, (b) 2 marks, (c) 1 mark, (d) 1 mark. |
| 10 | A radioactive isotope has a half-life of 8.0 days and a sample has an initial activity of 640 Bq. The mass defect of a helium-4 nucleus is 0.03038 u; take 1 u = 931.5 MeV c-2. (a) Calculate the activity of the sample after 24 days. (b) Calculate the time taken for the activity to fall to 20 Bq. (c) Calculate the binding energy of a helium-4 nucleus. (d) Calculate the binding energy per nucleon. |
(a) 24 days is 3 half-lives, so A = 640 x (0.5)^3 = 80 Bq. (b) Falling from 640 Bq to 20 Bq takes 5 half-lives, that is 5 x 8.0 = 40 days. (c) E = dm c^2 = 0.03038 x 931.5 = 28.30 MeV. (d) Per nucleon = 28.30 / 4 = 7.07 MeV. Marking: (a) 1 mark, (b) 2 marks, (c) 1 mark, (d) 1 mark. |
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