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100+ Free HSC Physics Practice Questions
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2026 Statistics
Key Facts: HSC Physics Exam
3 hours
External examination duration (+ 5 mins reading time)
NESA
Stage 6
NESA Year 12 Higher School Certificate course
NESA
HSC Physics is the Year 12 examination set by NESA. Our 100 practice questions provide a free MCQ study adaptation covering all key syllabus modules and Working Scientifically skills.
Sample HSC Physics Practice Questions
Try these sample questions to test your HSC Physics exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.
1A projectile is launched from flat ground with an initial velocity vector at an angle to the horizontal. Neglecting air resistance, which statement correctly describes its horizontal and vertical motion?
A.Horizontal acceleration is constant and non-zero, while vertical acceleration decreases to zero at the peak.
B.Horizontal velocity remains constant, while vertical acceleration remains constant and directed downwards.
C.Horizontal velocity decreases linearly, while vertical velocity remains constant.
D.Both horizontal and vertical accelerations change continuously throughout the trajectory.
Explanation: In ideal projectile motion, no horizontal forces act on the object (ax = 0 m/s²), making horizontal velocity constant. The only force acting is gravity, which causes a constant downward vertical acceleration of g = 9.8 m/s² throughout the entire flight.
2An object moves in a horizontal circle of radius r at a constant speed v. Which option correctly identifies the directions of its velocity vector and centripetal acceleration vector?
A.Velocity is directed tangentially to the path; centripetal acceleration is directed towards the centre of the circle.
B.Velocity is directed towards the centre of the circle; centripetal acceleration is directed tangentially.
C.Both velocity and centripetal acceleration are directed radially outward.
D.Velocity is directed tangentially; centripetal acceleration is directed radially outward.
Explanation: Uniform circular motion requires a centripetal force directed towards the centre of the circle, producing a centripetal acceleration aimed towards the centre. The velocity vector is always perpendicular to acceleration, pointing tangentially along the instantaneous direction of motion.
3According to Kepler's Second Law of Planetary Motion (the Law of Areas), what happens to a planet's speed as it orbits the Sun?
A.The planet moves at a constant speed throughout its entire elliptical orbit.
B.The planet moves fastest when it is farthest from the Sun (aphelion).
C.The planet moves fastest when it is closest to the Sun (perihelion).
D.The planet's speed depends solely on its mass and is independent of its distance from the Sun.
Explanation: Kepler's Second Law states that a line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time. To sweep out an equal area when the radial distance is smaller at perihelion, the planet must cover a larger arc distance, meaning its orbital speed is highest at perihelion.
4Two spherical masses m1 and m2 are separated by a distance r. If the distance between their centres is doubled to 2r, how does the gravitational force between them change?
A.It decreases to one-quarter of its original value.
B.It decreases to one-half of its original value.
C.It doubles.
D.It quadruples.
Explanation: Newton's Law of Universal Gravitation states F = G(m1*m2)/r². Because gravitational force obeys an inverse-square relationship with distance, doubling the separation distance r reduces the force by a factor of 2² = 4.
5Which statement correctly describes the escape velocity from the surface of a celestial body?
A.It depends on the mass of the launching projectile.
B.It is the minimum velocity needed for an object to escape the gravitational field without further propulsion.
C.It is equal to the orbital velocity of a low-Earth orbit satellite.
D.It increases when the radius of the planet increases for a constant planetary mass.
Explanation: Escape velocity (v_esc = √(2GM/r)) is the minimum initial speed required for an unpropelled body to escape the gravitational pull of a primary mass M. It depends only on the mass M and radius r of the central body, independent of the projectile's mass.
6What key property defines a geostationary Earth orbit (GEO)?
A.An orbital period of 90 minutes in a polar plane.
B.An orbital period of 24 hours in the equatorial plane, matching Earth's rotational period.
C.An elliptical orbit passing over both the North and South poles.
D.An altitude of 400 km directly above the prime meridian.
Explanation: A geostationary orbit has a period of approximately 24 hours (1 sidereal day) and is situated directly above Earth's equator. This allows the satellite to remain stationary relative to a fixed point on Earth's surface.
7A cannonball is launched horizontally from a cliff height of 44.1 m above a flat plain with a horizontal speed of 20 m/s. Taking g = 9.8 m/s² and ignoring air resistance, what is the horizontal range of the cannonball?
A.30 m
B.60 m
C.80 m
D.120 m
Explanation: First calculate time of flight from vertical motion: y = 1/2 g t² => 44.1 = 0.5 * 9.8 * t² => 4.9 t² = 44.1 => t² = 9 => t = 3.0 s. Then calculate horizontal range: R = u_x * t = 20 m/s * 3.0 s = 60 m.
8A ball is launched from ground level with an initial vertical velocity component of 29.4 m/s. Taking g = 9.8 m/s², what maximum height above ground level does the ball reach?
A.44.1 m
B.88.2 m
C.29.4 m
D.132.3 m
Explanation: At maximum height, vertical velocity v_y = 0 m/s. Using v_y² = u_y² - 2g h: 0 = (29.4)² - 2(9.8)h => 19.6 h = 864.36 => h = 44.1 m.
9A car of mass 1200 kg travels around a flat circular track of radius 50 m at a constant speed of 20 m/s. What centripetal force is required to keep the car on its circular path?
A.4800 N
B.9600 N
C.2400 N
D.480 N
Explanation: Centripetal force is given by F_c = m v² / r. Substituting values: F_c = (1200 kg) * (20 m/s)² / (50 m) = 1200 * 400 / 50 = 480,000 / 50 = 9600 N.
10A frictionless track is banked at an angle θ to the horizontal for vehicles traveling around a curve of radius 80 m at a design speed of 28 m/s. Taking g = 9.8 m/s², what is tan(θ)?
A.1.0
B.0.5
C.2.0
D.0.35
Explanation: For a banked turn without friction, the horizontal component of the normal force provides centripetal force: tan(θ) = v² / (r g). Substituting v = 28 m/s, r = 80 m, g = 9.8 m/s²: tan(θ) = (28)² / (80 * 9.8) = 784 / 784 = 1.0.
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