Free practice · VCE Units 3 & 4
These 10 free VCE Physics practice questions cover Units 3 and 4: gravitational, electric and magnetic fields, motion, electromagnetic induction, waves and light, and the quantum and relativity ideas. Each comes with a mark-by-mark scheme showing the formula, working and units examiners look for. They are original and VCAA-style, not official VCAA questions, and ATARMAxxing is not affiliated with or endorsed by the VCAA.
Work each one on paper before checking the scheme. For more, see our guide to practice exams and the best way to study for exams. You can also start free on the free plan. Always confirm exam-relevant detail against the current VCAA study design.
A satellite orbits Earth in a circular orbit at an altitude of 2.0 x 10^7 m above the surface. Take Earth's mass as 5.97 x 10^24 kg, Earth's radius as 6.37 x 10^6 m, and G = 6.67 x 10^-11 N m^2 kg^-2. (a) Calculate the orbital radius. (b) Calculate the satellite's orbital speed. (c) Calculate the orbital period in hours.
(a) 1 mark: r = R + h = 6.37 x 10^6 + 2.0 x 10^7 = 2.637 x 10^7 m. (b) 2 marks: 1 mark for equating gravitational force to centripetal force, GMm/r^2 = mv^2/r, giving v = sqrt(GM/r); 1 mark for v = sqrt(6.67 x 10^-11 x 5.97 x 10^24 / 2.637 x 10^7) = 3.89 x 10^3 m s^-1 (accept 3.9 x 10^3). (c) 2 marks: 1 mark for T = 2(pi)r/v = 2(pi)(2.637 x 10^7)/3886; 1 mark for T = 4.26 x 10^4 s = 11.8 hours (accept 11.8 to 11.9). Total: 5 marks.
A ball is launched from ground level at 25 m s^-1 at 30 degrees above the horizontal. Ignore air resistance and take g = 9.8 m s^-2. (a) Find the horizontal and vertical components of the initial velocity. (b) Find the maximum height reached. (c) Find the horizontal range.
(a) 2 marks: 1 mark for ux = 25 cos30 = 21.7 m s^-1; 1 mark for uy = 25 sin30 = 12.5 m s^-1. (b) 2 marks: 1 mark for using v^2 = uy^2 - 2gH with v = 0, so H = uy^2/(2g); 1 mark for H = 12.5^2/(2 x 9.8) = 7.97 m (accept 7.9 to 8.0 m). (c) 2 marks: 1 mark for time of flight t = 2uy/g = 2 x 12.5/9.8 = 2.55 s; 1 mark for range = ux x t = 21.7 x 2.55 = 55.2 m (accept 55 to 55.3 m). Total: 6 marks.
A coil of 50 turns has a cross-sectional area of 0.020 m^2 and sits in a uniform magnetic field of 0.40 T directed perpendicular to the plane of the coil. The field is reduced steadily to zero over 0.10 s. (a) Calculate the magnitude of the average induced EMF. (b) State the direction of the induced current relative to the original field (i.e. whether it acts to maintain or oppose the original field) and name the law that justifies it.
(a) 3 marks: 1 mark for EMF = N x (change in flux)/(change in time) with flux = BA; 1 mark for substitution EMF = 50 x (0.40 x 0.020)/0.10; 1 mark for EMF = 4.0 V. (b) 2 marks: 1 mark for stating the induced current acts to oppose the decrease, so it flows in the sense that maintains flux in the same direction as the original field; 1 mark for naming Lenz's law (Faraday's law also accepted as the governing law). Total: 5 marks.
An electron (charge magnitude 1.6 x 10^-19 C) is accelerated from rest through a potential difference of 250 V. (a) Calculate the kinetic energy gained, in joules. (b) Calculate the final speed of the electron. Take the electron mass as 9.11 x 10^-31 kg.
(a) 2 marks: 1 mark for KE = qV; 1 mark for KE = 1.6 x 10^-19 x 250 = 4.0 x 10^-17 J. (b) 2 marks: 1 mark for setting KE = (1/2)mv^2 so v = sqrt(2KE/m); 1 mark for v = sqrt(2 x 4.0 x 10^-17 / 9.11 x 10^-31) = 9.4 x 10^6 m s^-1 (accept 9.3 to 9.4 x 10^6). Total: 4 marks.
An ideal transformer has 240 turns on the primary and 1200 turns on the secondary. The primary is connected to a 12 V AC supply. (a) Calculate the secondary voltage. (b) If the secondary delivers a current of 0.50 A, calculate the primary current. (c) State one reason why a real transformer is less than 100 percent efficient.
(a) 2 marks: 1 mark for Vs/Vp = Ns/Np; 1 mark for Vs = 12 x 1200/240 = 60 V. (b) 2 marks: 1 mark for using Vp Ip = Vs Is (or Ns/Np = Ip/Is); 1 mark for Ip = 0.50 x 1200/240 = 2.5 A. (c) 1 mark: any one valid reason, e.g. resistive (I^2 R) heating in the windings, eddy currents in the core, or hysteresis losses in the core. Total: 5 marks.
In a photoelectric experiment a metal has a work function of 2.30 eV. It is illuminated with light of wavelength 400 nm. Take h = 6.63 x 10^-34 J s, c = 3.0 x 10^8 m s^-1 and 1 eV = 1.6 x 10^-19 J. (a) Calculate the energy of one photon in eV. (b) Calculate the maximum kinetic energy of the emitted photoelectrons in eV. (c) Explain why increasing the intensity of this light does not change the maximum kinetic energy.
(a) 2 marks: 1 mark for E = hc/(lambda) = 6.63 x 10^-34 x 3.0 x 10^8 / 400 x 10^-9 = 4.97 x 10^-19 J; 1 mark for converting to eV: 4.97 x 10^-19 / 1.6 x 10^-19 = 3.11 eV. (b) 2 marks: 1 mark for KEmax = E(photon) - work function = 3.11 - 2.30; 1 mark for KEmax = 0.81 eV (accept 0.80 to 0.81). (c) 1 mark: increasing intensity increases the number of photons per second (so more electrons are emitted), but each photon still carries the same energy, so the maximum kinetic energy per electron is unchanged. Total: 5 marks.
A car of mass 1200 kg travels around a flat, unbanked circular bend of radius 50 m. The coefficient of friction between the tyres and road is 0.40. Take g = 9.8 m s^-2. (a) Identify the force that provides the centripetal force. (b) Calculate the maximum speed at which the car can take the bend without skidding.
(a) 1 mark: friction between the tyres and the road provides the (net) centripetal force directed toward the centre of the circle. (b) 3 marks: 1 mark for setting maximum friction equal to required centripetal force, (mu)mg = mv^2/r; 1 mark for rearranging to v = sqrt((mu)gr); 1 mark for v = sqrt(0.40 x 9.8 x 50) = 14 m s^-1. Total: 4 marks.
A straight horizontal wire of length 0.40 m carries a current of 3.0 A. It lies perpendicular to a uniform magnetic field of magnitude 0.25 T. (a) Calculate the magnitude of the force on the wire. (b) State what happens to this force if the wire is rotated so it lies parallel to the field, and justify your answer.
(a) 2 marks: 1 mark for F = nBIL with the conductor perpendicular to the field (n = 1 here); 1 mark for F = 0.25 x 3.0 x 0.40 = 0.30 N. (b) 2 marks: 1 mark for stating the force becomes zero; 1 mark for the justification that the force depends on the component of current perpendicular to the field, and when the wire is parallel to B there is no perpendicular component, so F = 0. Total: 4 marks.
A spacecraft travels past Earth at a constant speed of 0.80c. An astronaut on board measures a journey as taking 4.0 years of their own (proper) time. (a) Calculate the Lorentz factor at 0.80c. (b) Calculate the time for the journey as measured by an observer on Earth. (c) State which observer measures the proper time and why.
(a) 2 marks: 1 mark for gamma = 1/sqrt(1 - (v/c)^2) = 1/sqrt(1 - 0.80^2); 1 mark for gamma = 1/sqrt(0.36) = 1.67 (accept 1.66 to 1.67). (b) 2 marks: 1 mark for t = gamma x t0 with t0 = 4.0 years; 1 mark for t = 1.67 x 4.0 = 6.7 years (accept 6.6 to 6.7 years). (c) 1 mark: the astronaut measures the proper time, because the two events (start and end of the journey) occur at the same location in the astronaut's frame. Total: 5 marks.
Light of a single wavelength passes through two narrow slits and forms an interference pattern on a distant screen. (a) Explain, in terms of path difference, why a bright fringe forms at a particular point on the screen. (b) Explain why this two-slit result is evidence for the wave model of light rather than a purely particle model. (c) State one experimental observation (in a different experiment) that instead supports a particle model of light.
(a) 2 marks: 1 mark for stating a bright fringe forms where light from the two slits arrives in phase, undergoing constructive interference; 1 mark for the condition that the path difference equals a whole number of wavelengths (path difference = m(lambda), m = 0, 1, 2, ...). (b) 2 marks: 1 mark for stating that interference (alternating bright and dark fringes) requires waves to superpose and cancel or reinforce; 1 mark for explaining that particles travelling independently could not produce dark fringes (regions of no light) by cancellation, so the pattern is explained only by the wave model. (c) 1 mark: any one valid observation, e.g. the photoelectric effect (existence of a threshold frequency and the independence of maximum kinetic energy from intensity). Total: 5 marks.
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