IB Physics SL Paper 2 Mock Paper (second set, 10 questions, 50 marks, 75 minutes, calculator allowed)

2026-11-01

IB Physics

IB Physics SL Paper 2 second mock paper: 10 original questions, 50 marks in 75 minutes, calculator allowed, covering projectiles, Newton laws and friction, momentum, circular motion and gravitation, ideal gases, waves and resonance, circuits, electromagnetic induction, the photoelectric effect and radioactive decay.

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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 ball is thrown horizontally at 12.0 m s-1 from a height of 45.0 m above the ground. Air resistance is negligible and g = 9.81 m s-2. (a) Calculate the time taken for the ball to reach the ground, in seconds. (b) Calculate the horizontal distance travelled before landing, in metres. (c) Calculate the vertical component of the velocity just before landing, in m s-1. (d) Calculate the speed of the ball just before landing, in m s-1. (e) Calculate the angle between the velocity at landing and the horizontal, in degrees.	(a) Vertically the ball falls from rest, so h = 1/2 g t2 and t = square root of (2h / g) = square root of (2 x 45.0 / 9.81) = 3.03 s. (b) Horizontally the motion is uniform, so x = u t = 12.0 x 3.029 = 36.3 m. (c) The vertical velocity is v_y = g t = 9.81 x 3.029 = 29.7 m s-1, directed downwards. (d) The speed is the magnitude of the resultant velocity: square root of (12.0 squared plus 29.71 squared) = 32.0 m s-1. (e) The angle is the inverse tangent of (29.71 / 12.0) = 68.0 degrees below the horizontal. Marking notes: 1 mark each for (a) to (e).
2	A block of mass 4.0 kg rests on a rough horizontal floor. A force of 30 N is applied at 25 degrees above the horizontal. The coefficient of kinetic friction is 0.25 and g = 9.81 m s-2. (a) Calculate the normal reaction force from the floor on the block, in newtons. (b) Calculate the friction force acting on the block, in newtons. (c) Calculate the acceleration of the block, in m s-2. (d) The block moves 3.0 m. Calculate the work done by friction on the block, in joules. (e) State whether the acceleration would increase or decrease if the same force were applied horizontally, and explain why.	(a) Resolving vertically: N + F sin 25 = m g, so N = 4.0 x 9.81 – 30 x 0.4226 = 39.24 – 12.68 = 26.6 N. (b) Kinetic friction f = mu N = 0.25 x 26.56 = 6.64 N, opposing the motion. (c) Horizontally F cos 25 – f = m a, so 30 x 0.9063 – 6.64 = 20.5 N and a = 20.5 / 4.0 = 5.14 m s-2. (d) Friction acts opposite to the displacement, so the work is -6.64 x 3.0 = -19.9 J, a magnitude of 19.9 J. (e) The acceleration increases. Applied horizontally, the whole 30 N acts along the floor instead of only 30 cos 25, so the driving force is larger. The vertical component no longer lifts the block, so the normal force rises to mg and friction increases slightly, but the gain in driving force is bigger, so the resultant force and the acceleration both increase. Marking notes: 1 mark each for (a) to (e).
3	A toy car A of mass 1.2 kg moving at 2.5 m s-1 on a smooth horizontal track collides with a stationary toy car B of mass 1.8 kg. After the collision the two cars move together. (a) Calculate the momentum of A before the collision, in kg m s-1. (b) Calculate the common speed of the two cars after the collision, in m s-1. (c) Calculate the kinetic energy lost in the collision, in joules. (d) Calculate the impulse on B during the collision, in N s. (e) State whether the collision is elastic or inelastic and justify your answer with your calculated values.	(a) p = m_A u_A = 1.2 x 2.5 = 3.0 kg m s-1. (b) Momentum is conserved: 1.2 x 2.5 = (1.2 + 1.8) x v, so v = 3.0 / 3.0 = 1.0 m s-1. (c) The kinetic energy before is 0.5 x 1.2 x 2.5 squared = 3.75 J and after is 0.5 x 3.0 x 1.0 squared = 1.50 J, so 2.25 J is lost. (d) The impulse equals the change in momentum of B: 1.8 x 1.0 = 1.8 N s in the original direction of motion of A. (e) The collision is inelastic: the kinetic energy falls by 2.25 J, from 3.75 J to 1.50 J, and is transferred to thermal and sound energy while momentum is conserved. The cars sticking together is also characteristic of an inelastic collision. Marking notes: 1 mark each for (a) to (e).
4	A satellite moves in a circular orbit of radius 7.0 x 106 m about the centre of the Earth. The mass of the Earth is 5.97 x 1024 kg and G = 6.67 x 10-11 N m2 kg-2. (a) Calculate the orbital speed of the satellite, in m s-1. (b) Calculate the orbital period of the satellite, in minutes. (c) Calculate the gravitational field strength at the satellite, in N kg-1. (d) Calculate the centripetal acceleration of the satellite, in m s-2, and compare it with your answer to (c). (e) Explain whether the speed needed for a circular orbit of larger radius is greater or smaller, and why.	(a) Gravity provides the centripetal force, so v = square root of (G M / r) = square root of (6.67 x 10-11 x 5.97 x 1024 / 7.0 x 106) = 7.542e+03 m s-1, about 7.54 x 103 m s-1. (b) T = 2 pi r / v = 2 pi x 7.0 x 106 / 7542 = 5831 s, about 97.2 minutes. (c) g = G M / r2 = 6.67 x 10-11 x 5.97 x 1024 / (7.0 x 106)2 = 8.13 N kg-1. (d) a = v2 / r = 7542 squared / 7.0 x 106 = 8.13 m s-2, identical to the field strength in (c), because in a circular orbit gravity is the only centripetal force. (e) A smaller speed is needed. Since v is proportional to the square root of (1 / r), the required speed falls as the radius rises. In practice a satellite first speeds up into an elliptical transfer orbit and then slows down at the higher altitude to circularise. Marking notes: 1 mark each for (a) to (e).
5	A rigid container of volume 5.0 x 10-3 m3 holds 0.30 mol of a monatomic ideal gas at 300 K. R = 8.31 J mol-1 K-1 and the Boltzmann constant k = 1.38 x 10-23 J K-1. (a) Calculate the pressure of the gas, in pascals. (b) The gas is heated to 400 K at constant volume. Calculate the new pressure, in pascals. (c) Calculate the mean kinetic energy of the gas molecules at 300 K, in joules. (d) Calculate the internal energy of the gas at 300 K, in joules. (e) State two changes in the motion of the molecules as the gas is heated.	(a) p = n R T / V = 0.30 x 8.31 x 300 / 5.0 x 10-3 = 1.496e+05 Pa, about 1.50 x 105 Pa. (b) At constant volume the pressure is proportional to the temperature, so p = 1.496e+05 x 400 / 300 = 1.994e+05 Pa, about 1.99 x 105 Pa. (c) The mean kinetic energy is 1.5 k T = 1.5 x 1.38 x 10-23 x 300 = 6.21e-21 J. (d) For a monatomic ideal gas the internal energy is 1.5 n R T = 1.5 x 0.30 x 8.31 x 300 = 1122 J. (e) The mean kinetic energy and the root-mean-square speed of the molecules increase, so each collision with the wall involves a larger momentum change and the collisions are more frequent. The whole speed distribution shifts towards higher speeds, while the number of molecules and their mean separation stay the same. Marking notes: 1 mark each for (a) to (e).

Section B (Questions 6 to 10, 25 marks)

#	Question	Answer and marking notes
6	A transverse wave on a string has frequency 250 Hz and wavelength 1.20 m. (a) Calculate the speed of the wave on the string, in m s-1. (b) Calculate the period of the wave, in milliseconds. (c) A pipe of length 0.65 m, closed at one end, resonates at its fundamental frequency. The speed of sound is 340 m s-1. Calculate the fundamental frequency, in hertz. (d) Calculate the third harmonic of the same pipe, in hertz. (e) State where the displacement of the air molecules is smallest and where it is largest when the pipe vibrates at its fundamental frequency.	(a) v = f lambda = 250 x 1.20 = 300 m s-1. (b) T = 1 / f = 1 / 250 = 4.0 ms. (c) For a pipe closed at one end the fundamental wavelength is 4L, so f1 = v / (4L) = 340 / (4 x 0.65) = 130.8 Hz. (d) A closed pipe supports only odd harmonics, so f3 = 3 x 130.8 = 392.3 Hz. (e) At the closed end, which is fixed, the displacement is smallest (a displacement node and a pressure antinode); at the open end the displacement is largest (a displacement antinode and a pressure node). Adjacent nodes and antinodes are one quarter of a wavelength apart. Marking notes: 1 mark each for (a) to (e).
7	A battery of emf 12.0 V and internal resistance 0.50 ohm is connected to a parallel combination of a 4.0 ohm and a 6.0 ohm resistor. (a) Calculate the equivalent resistance of the parallel section, in ohms. (b) Calculate the current delivered by the battery, in amperes. (c) Calculate the terminal potential difference of the battery, in volts. (d) Calculate the current in the 6.0 ohm resistor, in amperes. (e) Calculate the power dissipated in the internal resistance, in watts, and the efficiency of the circuit, in per cent.	(a) 1 / R_p = 1 / 4.0 + 1 / 6.0, so R_p = 24 / 10 = 2.4 ohm. (b) The total resistance is 2.4 + 0.50 = 2.9 ohm, so I = 12.0 / 2.9 = 4.14 A. (c) The terminal potential difference is 12.0 – 4.14 x 0.50 = 9.93 V. (d) The parallel section has the same potential difference, so the current is 9.93 / 6.0 = 1.66 A. (e) The power wasted in the internal resistance is I squared r = 4.14 squared x 0.50 = 8.56 W, and the efficiency is 9.93 / 12.0 = 82.8 per cent. Marking notes: 1 mark each for (a) to (e).
8	A coil of 200 turns and cross-sectional area 4.0 x 10-4 m2 lies with its plane perpendicular to a uniform magnetic field. The flux density increases uniformly from 0.050 T to 0.250 T in 0.20 s. (a) Calculate the change in magnetic flux through each turn, in webers. (b) Calculate the induced emf in the coil, in volts. (c) Explain how the direction of the induced current satisfies Lenz law. (d) A straight wire of length 0.40 m carrying 2.5 A lies perpendicular to a uniform field of 0.35 T. Calculate the magnetic force on the wire, in newtons. (e) State what happens to the induced emf if the coil instead moves parallel to the field without changing the flux through it.	(a) The flux change per turn is A times the change in B = 4.0 x 10-4 x (0.250 – 0.050) = 8.0e-05 Wb. (b) emf = N times flux change divided by time = 200 x 8.0e-05 / 0.20 = 0.08 V. (c) When the flux into the page increases, the induced current produces a field out of the page, opposing the increase; if the flux decreases the current reverses to maintain the original flux. This is Lenz law: the induced effect opposes the change that caused it. (d) F = B I L sin theta = 0.35 x 2.5 x 0.40 x 1 = 0.35 N, with the direction given by Fleming left-hand rule. (e) The induced emf is zero. Moving parallel to the field does not change the flux through the coil, so the rate of change of flux is zero and no emf or current is induced. Marking notes: 1 mark each for (a) to (e).
9	Light of wavelength 400 nm is incident on a sodium surface. The work function of sodium is 2.28 eV. Take hc = 1240 eV nm and 1 eV = 1.60 x 10-19 J. (a) Calculate the energy of an incident photon, in eV. (b) Calculate the maximum kinetic energy of the emitted photoelectrons, in eV and in J. (c) Calculate the threshold wavelength of sodium, in nm. (d) Calculate the stopping potential, in volts. (e) State how the maximum kinetic energy and the photocurrent change if only the intensity of the light is increased at constant frequency.	(a) E = hc / lambda = 1240 / 400 = 3.10 eV. (b) The maximum kinetic energy is 3.10 – 2.28 = 0.82 eV, which is 0.82 x 1.60 x 10-19 = 1.31e-19 J. (c) The threshold wavelength is 1240 / 2.28 = 544 nm. (d) The stopping potential is 0.82 V. (e) The maximum kinetic energy is unchanged because it depends only on the frequency of the light and the work function of the metal. The photocurrent increases, below saturation, because more photons arrive per second and more photoelectrons are released per second. Marking notes: 1 mark each for (a) to (e).
10	A radionuclide has a half-life of 8.0 days. A sample has an initial activity of 640 Bq. (a) Calculate the activity of the sample after 24 days, in becquerels. (b) Calculate the decay constant of the nuclide, in s-1. (c) Use your answer to (a) to calculate the number of undecayed nuclei present at that time. (d) A nuclear reaction has a mass defect of 0.0030 u. Taking 1 u = 931.5 MeV, calculate the energy released, in MeV. (e) Explain why after 24 days the activity has fallen greatly while the total mass of the sample has decreased only slightly.	(a) 24 days is three half-lives, so the activity is 640 / 8 = 80 Bq. (b) The decay constant is ln 2 divided by the half-life = 0.693 / (8.0 x 86400) = 1.00e-06 s-1. (c) Since A = lambda N, N = 80 / 1.00e-06 = 7.98e+07, about 8.0 x 107 nuclei. (d) The energy released is 0.0030 x 931.5 = 2.79 MeV, about 2.8 MeV. (e) Activity is proportional to the number of undecayed nuclei, which halves every half-life, so after three half-lives only one eighth remains and the activity falls from 640 Bq to 80 Bq. Each alpha or beta decay changes the mass by only a tiny fraction of an atomic mass, and most of the decay products stay in the sample, so the total mass falls only slightly. Marking notes: 1 mark each for (a) to (e).

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