HKDSE Physics Practice Paper: Mechanics And Circular Motion

HKDSE Physics Practice Paper: Mechanics and Circular Motion

2026-11-04

HKDSE Physics: Mechanics and Circular Motion

HKDSE Physics paper: 12 questions, 100 marks, 75 minutes on kinematics, momentum, circular motion, gravity, energy and moments, with full worked answers.

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Paper details　Level: S4-S6 | Subject: Physics | Time: 75 minutes | Marks: 100 | Questions: 12

Section A: Kinematics, Force and Momentum (Questions 1-4, 32 marks)

Each question carries 8 marks; show the formula used, the substitution and the unit.

#	Question	Answer and marking notes
1	A car starts from rest and accelerates uniformly, reaching 20 m/s in 8.0 s. (a) Find the acceleration. (b) Find the distance travelled in these 8.0 s. (c) Find the speed at 4.0 s. (d) Find the average speed for the whole journey.	(a) a = (v – u)/t = 20/8.0 = 2.5 m s⁻². [2 marks] (b) s = 0.5at² = 0.5 × 2.5 × 64 = 80 m. [3 marks] (c) v = at = 2.5 × 4.0 = 10 m/s. [1 mark] (d) Average speed = 80/8.0 = 10 m/s, equal to the mean of the initial and final speeds. [2 marks] Marking notes: use s = 0.5(u + v)t; units must be given.
2	A small ball is projected horizontally at 15 m/s from a table 20 m high. Take g = 10 m s⁻² and ignore air resistance. (a) Find the time taken to reach the ground. (b) Find the horizontal range. (c) Find the vertical velocity component on landing. (d) Find the speed on landing.	(a) From h = 0.5gt², t = √(2h/g) = √4.0 = 2.0 s. [2 marks] (b) Range = 15 × 2.0 = 30 m. [2 marks] (c) Vertical velocity = gt = 20 m/s downwards. [2 marks] (d) Speed = √(15² + 20²) = 25 m/s. [2 marks] Marking notes: horizontal and vertical motions are independent; speeds combine as vectors, not as scalars.
3	A 4.0 kg box rests on a horizontal floor. The coefficient of kinetic friction is 0.25. A horizontal push of 30 N is applied. Take g = 10 m s⁻². (a) Find the friction acting on the box. (b) Find the acceleration of the box. (c) Find the distance moved in the first 3.0 s from rest. (d) If the push is removed at 3.0 s, find the further distance the box slides.	(a) f = μmg = 0.25 × 4.0 × 10 = 10 N. [2 marks] (b) Net force = 30 – 10 = 20 N, a = 20/4.0 = 5.0 m s⁻². [2 marks] (c) s = 0.5 × 5.0 × 9.0 = 22.5 m. [2 marks] (d) v = 5.0 × 3.0 = 15 m/s; deceleration = μg = 2.5 m s⁻²; distance = v²/(2 × 2.5) = 45 m. [2 marks] Marking notes: friction opposes motion; after the push is removed only friction acts.
4	A 2.0 kg toy car moving at 3.0 m/s collides with a stationary 4.0 kg toy car. The two cars stick together after the collision. (a) Find the total momentum before the collision. (b) Find their common velocity after the collision. (c) Find the loss of kinetic energy. (d) Explain why kinetic energy is not conserved in this collision.	(a) Total momentum = 2.0 × 3.0 = 6.0 kg m s⁻¹. [2 marks] (b) By conservation of momentum, 6.0 = 6.0v, so v = 1.0 m/s. [2 marks] (c) KE before = 0.5 × 2.0 × 9.0 = 9.0 J; KE after = 0.5 × 6.0 × 1.0 = 3.0 J; loss = 6.0 J. [2 marks] (d) A sticking collision is perfectly inelastic: part of the kinetic energy becomes thermal and sound energy and energy of deformation, so KE is not conserved, although momentum is. [2 marks] Marking notes: momentum is a vector; a reversed direction carries a negative sign.

Section B: Circular Motion, Gravity and Energy (Questions 5-8, 36 marks)

Each question carries 9 marks; state the direction in circular motion and gravity questions.

#	Question	Answer and marking notes
5	A car of mass 1,200 kg travels at 20 m/s on a horizontal circular track of radius 50 m. Take g = 10 m s⁻². (a) Find the centripetal acceleration. (b) Find the centripetal force provided by the road. (c) Find the minimum coefficient of static friction needed. (d) Explain how banking the track changes the friction required.	(a) a = v²/r = 400/50 = 8.0 m s⁻². [2 marks] (b) F = ma = 1,200 × 8.0 = 9,600 N. [2 marks] (c) μmg is the friction limit, so μ = v²/(gr) = 400/500 = 0.80. [3 marks] (d) On a banked track a component of the normal force acts towards the centre, so less friction is needed; a larger angle needs less friction, but at low speed a steep bank makes the car slide inwards. [2 marks] Marking notes: centripetal force is the resultant force, not an extra force; μ is independent of mass.
6	A 0.20 kg ball on a string of length 0.50 m moves in a vertical circle. Take g = 10 m s⁻². (a) Find the centripetal acceleration at the top when the speed is 3.0 m/s. (b) Find the tension in the string then. (c) Find the minimum speed needed to keep the string taut at the top. (d) Explain why the tension at the bottom is greater than at the top.	(a) a = v²/r = 9.0/0.50 = 18 m s⁻². [2 marks] (b) At the top mg + T = mv²/r, so T = 0.20 × 18 – 0.20 × 10 = 1.6 N. [3 marks] (c) With T = 0, v = √(gr) = √5.0 = 2.24 m/s. [2 marks] (d) At the bottom the tension must both support the weight and provide the centripetal force (T – mg = mv²/r), and the speed is greatest there, so the tension is greatest. [2 marks] Marking notes: at the top weight and tension act in the same direction; at the bottom they are opposite.
7	A satellite orbits the Earth in a circle of radius 7.0 × 10⁶ m. GM for the Earth is 4.0 × 10¹⁴ m³ s⁻². (a) Find the orbital speed. (b) Find the period in minutes. (c) If the radius doubles, by what factor does the speed change? (d) Explain why astronauts in orbit feel weightless.	(a) v = √(GM/r) = √(5.7 × 10⁷) ≈ 7.6 × 10³ m/s. [2 marks] (b) T = 2πr/v ≈ 5,820 s ≈ 97 minutes. [3 marks] (c) v is proportional to 1/√r, so doubling r gives about 0.71 times the speed. [2 marks] (d) Astronaut and spacecraft fall with the same acceleration, so the normal force on the astronaut is zero; gravity still acts as the centripetal force. [2 marks] Marking notes: gravity provides the centripetal force; use T = 2πr/v.
8	A 500 kg roller-coaster car starts from rest and slides 20 m down a slope. Take g = 10 m s⁻². (a) Find its speed at the bottom if friction is ignored. (b) Find its kinetic energy there. (c) If 20% of the mechanical energy is lost to friction, find the actual speed at the bottom. (d) Explain why the actual speed is lower.	(a) mgh = 0.5mv², so v = √(2gh) = √400 = 20 m/s. [2 marks] (b) KE = 0.5 × 500 × 400 = 1.0 × 10⁵ J. [2 marks] (c) Usable energy is 80%, so v = √(0.80 × 400) = √320 ≈ 17.9 m/s. [3 marks] (d) Friction converts mechanical energy into thermal and sound energy, so mechanical energy is not conserved and the speed is lower. [2 marks] Marking notes: energy conservation applies directly only without friction; the loss is a fraction of the total mechanical energy.

Section C: Moments, Springs and Impacts (Questions 9-12, 32 marks)

Each question carries 8 marks; state the pivot in moment questions and define a positive direction in impact questions.

#	Question	Answer and marking notes
9	A uniform 3.0 m plank of mass 20 kg is pivoted at its centre. A 40 kg child sits 1.0 m from the pivot. Take g = 10 m s⁻². (a) Find the moment of the child about the pivot. (b) Find where a 30 kg mass must be placed on the other side to balance the plank. (c) Does the weight of the plank itself produce a moment? Explain. (d) If the pivot moves towards the child, describe how the plank turns.	(a) Moment = 40 × 10 × 1.0 = 400 N m. [2 marks] (b) 30 × 10 × x = 400, so x ≈ 1.33 m. [3 marks] (c) The centre of gravity of a uniform plank is at the pivot and its line of action passes through the pivot, so the moment arm is zero and there is no moment. [1 mark] (d) Moving the pivot towards the child shortens the child moment arm and lengthens that of the 30 kg mass, so the plank turns towards the 30 kg side. [2 marks] Marking notes: moment = force × perpendicular distance; balance means the clockwise moment equals the anticlockwise moment.
10	A 0.50 kg mass hangs at rest from a spring of stiffness 200 N/m. Take g = 10 m s⁻². (a) Find the extension of the spring. (b) Find the period of oscillation after the mass is pulled down slightly and released. (c) Find the frequency. (d) State how the period changes if a stiffer spring is used.	(a) mg = ke, so e = 5.0/200 = 0.025 m = 2.5 cm. [2 marks] (b) T = 2π√(m/k) = 2π × 0.050 ≈ 0.31 s. [3 marks] (c) f = 1/T ≈ 3.2 Hz. [1 mark] (d) A stiffer spring has a larger k, so T = 2π√(m/k) is smaller: the oscillation is faster and the frequency higher. [2 marks] Marking notes: the period depends only on mass and stiffness, not on amplitude.
11	A coin lies on a horizontal turntable of radius 0.15 m. The coefficient of static friction is 0.40 and g = 10 m s⁻². (a) Find the maximum angular speed at which the coin can turn with the table. (b) Find the linear speed then. (c) Find the maximum number of revolutions per second. (d) Explain why the coin slides outwards when the table spins too fast.	(a) μmg = mω²r, so ω = √(μg/r) = √26.7 ≈ 5.2 rad/s. [3 marks] (b) v = ωr = 5.16 × 0.15 ≈ 0.77 m/s. [2 marks] (c) Revolutions per second = ω/(2π) ≈ 0.82. [1 mark] (d) The required centripetal force mω²r rises with the square of ω; once it exceeds the maximum static friction the coin can no longer stay in the circle and slides outwards. [2 marks] Marking notes: the mass cancels; the required force grows with the square of ω.
12	A ball of mass 0.15 kg hits a wall at 20 m/s and rebounds along the same line at 15 m/s. The contact time is 0.050 s. (a) Find the change in momentum of the ball. (b) Find the average force exerted by the wall. (c) Find the impulse. (d) Explain why part (b) gives only an average force.	(a) Taking away from the wall as positive, change = 0.15 × 15 – 0.15 × (-20) = 5.25 kg m s⁻¹ away from the wall. [2 marks] (b) F = Δp/t = 5.25/0.050 = 105 N. [3 marks] (c) Impulse = Δp = 5.25 N s away from the wall. [1 mark] (d) The force varies during contact and is greatest at maximum compression, so applying Newton’s second law over the whole contact time gives the average value. [2 marks] Marking notes: momentum and impulse are vectors; a rebound reverses the sign of the velocity.

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