S5 Physics Practice Paper: Projectile And Newton Laws, Circular Motion And Gravitation, Work And Energy, Momentum, Circuits, Induction And Radioactivity (100 Marks)

S5 Physics Practice Paper: Projectile and Newton Laws, Circular Motion and Gravitation, Work and Energy, Momentum, Circuits, Induction and Radioactivity (100 marks)

2026-11-13

S5 Physics – timed practice

S5 Physics practice paper, 24 parts, 100 marks in 60 minutes, covering mechanics, energy, momentum and electricity, with answers.

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Paper details　S5 Physics practice paper with 6 sections and 24 parts on projectile motion, circular motion, energy, momentum, circuits, induction, gases and radioactivity, with full answers.

Section A: Mechanics and Newton Laws (20 marks)

Answer all four questions, showing formulae and working.

#	Question	Answer and marking notes
1	A ball is thrown horizontally at 15 m/s from a height of 20 m. Taking g as 10 m/s², find (a) the time of flight and (b) the horizontal range.	Answer: (a) 2.0 s; (b) 30 m. Vertically h = ½gt² gives t = √(2h/g) = 2.0 s, and the horizontal velocity is unchanged, so range = 15 × 2.0 = 30 m. (Marking: time 2, range 2, explanation 1)
2	A car accelerates uniformly from 8 m/s to 20 m/s in 6 s. Find (a) the acceleration and (b) the displacement during this interval.	Answer: (a) 2.0 m/s²; (b) 84 m. Using a = (v − u)/t = (20 − 8)/6 = 2.0 m/s² and s = (u + v)t/2 = 84 m. (Marking: acceleration 3, displacement 2)
3	A 4 kg block rests on a horizontal table and is pulled by a 12 N horizontal force while friction is 4 N. Find (a) the acceleration and (b) the distance travelled in 3 s.	Answer: (a) 2.0 m/s²; (b) 9.0 m. Net force = 12 − 4 = 8 N, so a = 8 ÷ 4 = 2.0 m/s², and from rest s = ½at² = 9.0 m. (Marking: net force 2, acceleration 2, distance 1)
4	A stone is released from rest and falls for 3 s with no air resistance. Taking g as 10 m/s², find (a) the speed just before landing and (b) the distance fallen.	Answer: (a) 30 m/s; (b) 45 m. v = gt = 30 m/s and s = ½gt² = 45 m. (Marking: speed 2, distance 2, formula 1)

Section B: Circular Motion and Gravitation (16 marks)

Answer all four questions and give units with every answer.

#	Question	Answer and marking notes
5	A 0.5 kg object moves at a constant speed of 2.0 m/s along a circle of radius 0.8 m. Find (a) the centripetal acceleration and (b) the centripetal force.	Answer: (a) 5.0 m/s²; (b) 2.5 N. ac = v²/r = 5.0 m/s² and F = m × ac = 2.5 N. (Marking: acceleration 2, force 2)
6	An object moves on a circle of radius 0.6 m with a period of 0.5 s. Find (a) its speed and (b) its centripetal acceleration (take π as 3.14).	Answer: (a) 7.54 m/s; (b) 94.7 m/s². Using v = 2πr/T = 7.54 m/s and ac = v²/r = 94.7 m/s². (Marking: speed 2, acceleration 2)
7	Earth mass 6.0 × 10²⁴ kg, Moon mass 7.0 × 10²² kg, centre distance 4.0 × 10⁸ m, gravitational constant 6.67 × 10⁻¹¹ N m² kg⁻². Find the gravitational force on the Moon.	Answer: about 1.8 × 10²⁰ N. F = G m₁ m₂ / r² = 1.8 × 10²⁰ N. (Marking: substitution 2, order of magnitude 1, unit 1)
8	The radius of a satellite circular orbit becomes 4 times its original value. Find the factor by which its period changes and state the reasoning.	Answer: 8 times. Kepler third law gives T² ∝ r³, so T ∝ r^1.5 and multiplying the radius by 4 multiplies the period by 4^1.5 = 8. (Marking: ratio 2, calculation 1, explanation 1)

Section C: Work, Energy and Power (16 marks)

For energy conservation questions, state the assumptions made.

#	Question	Answer and marking notes
9	A 200 N horizontal force moves a crate 5.0 m along its own direction in 4.0 s. Find (a) the work done and (b) the average power.	Answer: (a) 1000 J; (b) 250 W. W = Fd = 1000 J and P = W/t = 250 W. (Marking: work 2, power 2)
10	A 2.0 kg block slides down a 3.0 m high slope and reaches the bottom at 6.0 m/s. Taking g as 10 m/s², find (a) potential energy lost, (b) kinetic energy at the bottom and (c) energy lost to friction.	Answer: (a) 60.0 J; (b) 36.0 J; (c) 24.0 J. Ep = mgh = 60.0 J, Ek = ½mv² = 36.0 J and the difference of 24.0 J is lost to friction. (Marking: 1 mark each, explanation 1)
11	A crane lifts a 300 kg load 4.0 m at constant speed in 1.5 s. Taking g as 10 m/s², find (a) the work done and (b) the average power.	Answer: (a) 1.2 × 10⁴ J; (b) 8.0 kW. W = mgh = 1.2 × 10⁴ J and P = W/t = 8.0 kW. (Marking: work 2, power 2)
12	A spring of stiffness 2000 N/m is compressed by 0.10 m. Find (a) the elastic potential energy stored and (b) the maximum speed of a 0.5 kg ball released from it, ignoring friction.	Answer: (a) 10.0 J; (b) 6.32 m/s. E = ½kx² = 10.0 J and v = √(2E/m) = 6.32 m/s. (Marking: energy 2, speed 2)

Section D: Momentum and Collisions (16 marks)

State the positive direction you choose in each answer.

#	Question	Answer and marking notes
13	A 3.0 kg object moving at 4.0 m/s strikes a stationary 1.0 kg object and they stick together. Find (a) their common velocity and (b) the total momentum after the collision.	Answer: (a) 3.0 m/s; (b) 12.0 kg m/s. Momentum is conserved, so v = 3.0 m/s and the total momentum is 12.0 kg m/s. (Marking: velocity 3, momentum 1)
14	A 0.15 kg ball hits a wall at 20 m/s and rebounds at 15 m/s in the opposite direction, the contact lasting 0.020 s. Find (a) the magnitude of the change in momentum and (b) the average force from the wall.	Answer: (a) 5.25 kg m/s; (b) 262.5 N. Taking the rebound direction as positive, Δp = 0.15 × (15 + 20) = 5.25 kg m/s and F = Δp/Δt = 262.5 N. (Marking: momentum change 2, average force 2)
15	A constant force of 250 N acts for 0.020 s. Find (a) the impulse and (b) the speed gained by a 0.50 kg object initially at rest.	Answer: (a) 5.0 N s; (b) 10.0 m/s. J = FΔt = 5.0 N s and v = J/m = 10.0 m/s. (Marking: impulse 2, speed 2)
16	A 2.0 kg object moving at 6.0 m/s hits a stationary 4.0 kg object and comes to rest. Find (a) the velocity of the second object and (b) the total kinetic energy before and after, and state whether the collision is elastic.	Answer: (a) 3.0 m/s; (b) 36.0 J before and 18.0 J after, so kinetic energy falls and the collision is inelastic. (Marking: velocity 2, energy comparison 1, conclusion 1)

Section E: Electric Fields and d.c. Circuits (16 marks)

Describe the circuit and write the formulae used.

#	Question	Answer and marking notes
17	A 6 Ω and a 3 Ω resistor are joined in parallel, and the pair is connected in series with a 5 Ω resistor across a 14 V supply of negligible internal resistance. Find (a) the total resistance and (b) the total current.	Answer: (a) 7.0 Ω; (b) 2.0 A. The parallel part is 2.0 Ω, so the total is 7.0 Ω and I = 14 ÷ 7.0 = 2.0 A. (Marking: parallel part 2, total 1, current 1)
18	A point charge of 2.0 × 10⁻⁶ C is placed in a uniform electric field of strength 5.0 × 10³ N/C. Find (a) the electric force on it and (b) its acceleration if its mass is 2.0 × 10⁻⁴ kg.	Answer: (a) 0.010 N; (b) 50 m/s². F = qE = 0.010 N and a = F/m = 50 m/s². (Marking: force 2, acceleration 2)
19	An appliance runs on 12 V and draws 3.0 A for 5.0 minutes. Find (a) its power and (b) the electrical energy it uses.	Answer: (a) 36.0 W; (b) 1.1 × 10⁴ J. P = VI = 36.0 W and with t = 300 s, E = Pt = 1.1 × 10⁴ J. (Marking: power 2, energy 2)
20	A 20 Ω resistor carries a current of 0.50 A for 10 minutes. Find (a) its power and (b) the heat generated.	Answer: (a) 5.0 W; (b) 3000 J. P = I²R = 5.0 W and E = Pt = 3000 J. (Marking: power 2, heat 2)

Section F: Induction, Gases and Radioactivity (16 marks)

Answer all four questions, giving values to sensible significant figures.

#	Question	Answer and marking notes
21	An ideal transformer has 1200 turns in its primary connected to a 240 V supply and 60 turns in its secondary. Find (a) the secondary voltage and (b) the primary current when the secondary current is 4.0 A.	Answer: (a) 12.0 V; (b) 0.20 A. Vs/Vp = Ns/Np gives 12.0 V, and equal power of 48.0 W gives Ip = 0.20 A. (Marking: voltage 2, current 2)
22	Gas sealed in a rigid container is at 300 K with a pressure of 1.0 × 10⁵ Pa. It is heated to 450 K at constant volume. Find (a) the new pressure and (b) name the gas law used.	Answer: (a) 1.5 × 10⁵ Pa; (b) the pressure law, since p/T is constant at fixed volume. p₂ = 1.0 × 10⁵ × 450 ÷ 300 = 1.5 × 10⁵ Pa. (Marking: pressure 3, law 1)
23	A radioisotope has a half-life of 8 days. After 48 days, find (a) the number of half-lives elapsed and (b) the fraction of atoms remaining.	Answer: (a) 6 half-lives; (b) 1/64 of the original, about 1.6 per cent. 48 ÷ 8 = 6 and (1/2)^6 = 1/64. (Marking: count 2, fraction 2)
24	The flux through a 200-turn coil falls from 0.20 Wb to 0.05 Wb in 0.30 s. Find (a) the magnitude of the induced e.m.f. and (b) name the law used.	Answer: (a) 100 V; (b) Faraday law of electromagnetic induction, with Lenz law giving the direction. E = N ΔΦ/Δt = 100 V. (Marking: formula 2, value 1, law 1)

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