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Key Facts: SACE Stage 2 Physics Exam
30% weighting for the official SACE Stage 2 Physics external written examination
130-minute (2 hours 10 minutes) duration for the end-of-year external examination
70% school-based assessment weighting split between Skills & Applications Tasks and Investigation Folio
Three core topics: Motion & Relativity, Electricity & Magnetism, and Light & Atoms
Calculators and SACE Physics Formula and Data Sheet are permitted in the exam
Achievement reported on a 15-point scale from A+ to E-, with C- required for satisfactory completion
SACE Stage 2 Physics is the senior secondary physics curriculum in South Australia, evaluated through 70% school-based tasks and a 30% 130-minute external written examination. Covering Motion and Relativity, Electricity and Magnetism, and Light and Atoms, this 100-question practice module offers multiple-choice preparation with step-by-step mathematical problem solving.
Sample SACE Stage 2 Physics Practice Questions
Try these sample questions to review concepts for the SACE Stage 2 Physics exam. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.
1A projectile is launched horizontally from a cliff height of 19.6 m at an initial speed of 15.0 m/s. Assuming g = 9.80 m/s² and negligible air resistance, what is the horizontal distance travelled by the projectile before hitting the ground?
A.15.0 m
B.30.0 m
C.45.0 m
D.60.0 m
Explanation: The time of flight t is determined solely by the vertical motion: t = sqrt(2h / g) = sqrt(2 * 19.6 / 9.80) = sqrt(4.00) = 2.00 s. Because the horizontal velocity remains constant at 15.0 m/s, the horizontal range is x = v_x * t = 15.0 m/s * 2.00 s = 30.0 m.
2A projectile is launched at an angle of 30.0° above the horizontal with an initial speed of 25.0 m/s over level ground. What is the maximum height reached above the launch point?
A.3.98 m
B.7.97 m
C.15.9 m
D.31.9 m
Explanation: The initial vertical velocity component is v_0y = v_0 * sin(30.0°) = 25.0 * 0.500 = 12.5 m/s. At maximum height, vertical velocity is zero. Using v_y² = v_0y² - 2gh gives h_max = v_0y² / (2g) = (12.5)² / (2 * 9.80) = 156.25 / 19.6 = 7.97 m.
3For an angled projectile launch over flat terrain, which statement correctly describes the velocity and acceleration vectors at the trajectory's highest point?
A.Velocity is zero and acceleration is zero.
B.Velocity is purely horizontal and acceleration is directed vertically downwards.
C.Velocity is purely vertical and acceleration is directed horizontally.
D.Velocity and acceleration are both directed horizontally in the direction of motion.
Explanation: At the apex of a parabolic trajectory, the vertical component of velocity decreases to zero, leaving only the constant horizontal component v_x = v_0 * cos(theta). Gravitational acceleration remains constant at 9.80 m/s² directed vertically downwards throughout the entire flight.
4A soccer ball is kicked at 20.0 m/s at an angle of 45.0° to the horizontal over level ground. What is the total time of flight before it hits the ground?
A.1.44 s
B.2.04 s
C.2.89 s
D.4.08 s
Explanation: The initial vertical velocity component is v_0y = 20.0 * sin(45.0°) = 14.14 m/s. The time of flight to return to ground level is t = 2 * v_0y / g = (2 * 14.14) / 9.80 = 2.89 s.
5In ideal projectile motion without air resistance, why does the horizontal component of velocity remain constant throughout the flight?
A.No horizontal force acts on the projectile after launch.
B.Gravity acts equally in both horizontal and vertical directions.
C.The momentum of the projectile continuously increases.
D.Centripetal force balances the gravitational force.
Explanation: According to Newton's first law of motion, an object maintains constant velocity along a straight line unless acted upon by a net external force. In ideal projectile motion, gravity is the only force acting and acts purely in the vertical direction, leaving net horizontal force equal to zero.
6How does air resistance modify the parabolic trajectory of a high-speed projectile compared to ideal vacuum motion?
A.It increases both the maximum height and the horizontal range.
B.It increases maximum height but decreases total flight time.
C.It reduces maximum height and horizontal range, creating a steeper angle of descent.
D.It maintains symmetrical ascent and descent paths while reducing horizontal velocity.
Explanation: Air resistance exerts a drag force opposing the velocity vector. This reduces both the vertical component of velocity (reducing peak height) and the horizontal component (reducing range), resulting in an asymmetric trajectory with a steeper descent.
7A car of mass 1200 kg travels at a constant speed of 20.0 m/s around a flat circular track of radius 50.0 m. What is the magnitude of the centripetal acceleration experienced by the car?
A.0.40 m/s²
B.2.50 m/s²
C.8.00 m/s²
D.16.0 m/s²
Explanation: Centripetal acceleration for uniform circular motion is given by a_c = v² / r. Substituting v = 20.0 m/s and r = 50.0 m yields a_c = (20.0)² / 50.0 = 400 / 50.0 = 8.00 m/s².
8What magnitude of net horizontal force must static friction provide between tyres and road for a 1200 kg car to navigate a flat 50.0 m radius curve at 20.0 m/s without slipping?
A.480 N
B.2400 N
C.9600 N
D.19200 N
Explanation: The required centripetal force is provided by static friction: F_c = m * v² / r. Using m = 1200 kg, v = 20.0 m/s, and r = 50.0 m gives F_c = 1200 * (400 / 50.0) = 1200 * 8.00 = 9600 N.
9A racetrack curve of radius 80.0 m is banked at an angle theta so that a car travelling at 28.0 m/s can negotiate the turn without relying on friction. What is the design banking angle theta? (g = 9.80 m/s²)
A.25.0°
B.35.0°
C.45.0°
D.55.0°
Explanation: For a friction-free banked turn, the horizontal component of the normal force provides the centripetal force (F_N * sin(theta) = m*v²/r) while the vertical component balances gravity (F_N * cos(theta) = m*g). Dividing these yields tan(theta) = v² / (r*g). Here, tan(theta) = (28.0)² / (80.0 * 9.80) = 784 / 784 = 1.00, giving theta = arctan(1.00) = 45.0°.
10A satellite of mass m orbits Earth (mass M = 5.97 x 10²⁴ kg) in a circular orbit of radius r = 7.00 x 10⁶ m from Earth's centre. Using G = 6.67 x 10⁻¹¹ N m²/kg², what is the orbital speed of the satellite?
A.5.74 km/s
B.7.54 km/s
C.9.80 km/s
D.11.2 km/s
Explanation: Gravitational force provides the centripetal force: G*M*m / r² = m*v² / r, which simplifies to orbital velocity v = sqrt(G*M / r). Substituting values: v = sqrt((6.67 x 10⁻¹¹ * 5.97 x 10²⁴) / (7.00 x 10⁶)) = sqrt(3.982 x 10¹⁴ / 7.00 x 10⁶) = sqrt(5.688 x 10⁷) = 7542 m/s = 7.54 km/s.
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