S3 Physics Paper: Speed And Acceleration, Forces And Newton Laws, Moments And Pressure, Work And Energy, Heat Transfer

S3 Physics Paper: Speed and Acceleration, Forces and Newton Laws, Moments and Pressure, Work and Energy, Heat Transfer

2026-11-13

S3 Physics – motion, forces and energy

S3 Physics paper: 6 questions, 24 parts, 100 marks, 60 minutes on speed, acceleration, forces, moments, pressure, work, energy and heat transfer, with full marked answers.

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Paper details　S3 Physics practice paper with 6 questions and 24 parts on speed and acceleration, Newton laws, moments, pressure, work, energy and heat transfer, with marked answers.

Section A: Speed, Acceleration and Motion Graphs (16 marks)

Answer the following questions on motion and show your working.

#	Question	Answer and marking notes
1	A car travels 2.7 km in 3 minutes. Find its average speed in m/s and in km/h.	Answer: time = 180 s, distance = 2700 m, speed = 2700 / 180 = 15 m/s; in km/h, 15 x 3.6 = 54 km/h. (Marking: m/s 2, km/h 2)
2	A car accelerates uniformly from rest and reaches 20 m/s after 8 s. Find the acceleration and the distance travelled in these 8 s.	Answer: acceleration = (20 – 0) / 8 = 2.5 m/s2; distance = average speed x time = (0 + 20) / 2 x 8 = 80 m. (Marking: acceleration 2, distance 2)
3	A ball falls freely from a height of 20 m (air resistance neglected, g = 10 m/s2). Find the time of fall and the speed on landing.	Answer: from h = 0.5 x g x t x t, t = square root of (2 x 20 / 10) = 2.0 s; speed v = g x t = 10 x 2.0 = 20 m/s. (Marking: time 2, speed 2)
4	The velocity-time graph of a car shows: 0 to 4 s uniform acceleration from 0 to 12 m/s; 4 to 10 s constant 12 m/s; 10 to 15 s uniform deceleration to rest. Find the total distance travelled in 15 s.	Answer: accelerating part = 0.5 x 4 x 12 = 24 m; constant part = 12 x 6 = 72 m; decelerating part = 0.5 x 5 x 12 = 30 m; total = 126 m. (Marking: three parts 1 each, total 1)

Section B: Forces and Newton Laws of Motion (16 marks)

Take g = 10 m/s2 and answer the following questions on forces.

#	Question	Answer and marking notes
5	A 5 kg box on a horizontal surface is pushed by a horizontal force of 20 N while friction is 5 N. Find the net force and the acceleration.	Answer: net force = 20 – 5 = 15 N; acceleration = 15 / 5 = 3 m/s2. (Marking: net force 2, acceleration 2)
6	A person of mass 60 kg stands on a weighing scale. Find the weight of the person.	Answer: weight W = m x g = 60 x 10 = 600 N, acting vertically downwards. (Marking: formula 2, value and direction 2)
7	A 10 kg object is acted on by two opposite forces of 30 N and 10 N. Find the net force and the acceleration.	Answer: net force = 30 – 10 = 20 N in the direction of the 30 N force; acceleration = 20 / 10 = 2 m/s2. (Marking: net force and direction 2, acceleration 2)
8	Use Newton third law to explain how a person walking is pushed forward, and give one other everyday example.	Answer: the foot pushes the ground backwards, so the ground pushes the foot forwards with an equal and opposite force, moving the person forward. Another example: a rocket pushes gas downwards and the gas pushes the rocket upwards. (Marking: action-reaction description 2, walking 1, example 1)

Section C: Moments, Pressure and Upthrust (18 marks)

Take g = 10 m/s2 and the density of water as 1000 kg/m3.

#	Question	Answer and marking notes
9	A crowbar is used to lift a load: a force of 200 N is applied 0.2 m from the pivot and the load is 0.8 m from the pivot. Find the weight of load that can be lifted (weight of the crowbar neglected).	Answer: taking moments about the pivot, 200 x 0.2 = W x 0.8, so W = 40 / 0.8 = 50 N. (Marking: moment equation 2, answer 2)
10	A person of weight 600 N stands on both feet with a total contact area of 0.03 m2. Find the pressure on the ground in Pa and in kPa.	Answer: pressure = 600 / 0.03 = 20000 Pa = 20 kPa. (Marking: formula and substitution 2, each unit 1)
11	The water in a tank is 3 m deep. Find the pressure at the bottom, and the force of the water on a valve of area 0.02 m2 at the bottom.	Answer: pressure = rho x g x h = 1000 x 10 x 3 = 30000 Pa = 30 kPa; force = p x A = 30000 x 0.02 = 600 N. (Marking: pressure 2, unit 1, force 2)
12	A stone of volume 0.002 m3 is completely immersed in water (its density is greater than that of water). Find the upthrust on the stone and state how much its apparent weight is reduced.	Answer: upthrust = rho x g x V = 1000 x 10 x 0.002 = 20 N; the apparent weight is reduced by 20 N. (Marking: formula 2, value 2, apparent weight 1)

Section D: Work, Power and Energy (18 marks)

Take g = 10 m/s2 and answer the following questions on work and energy.

#	Question	Answer and marking notes
13	A horizontal force of 250 N pushes a box 4 m along a horizontal floor in the direction of the force. Find the work done.	Answer: work = F x s = 250 x 4 = 1000 J. (Marking: formula 2, answer with unit 2)
14	A machine does 3000 J of work in 20 s. Find its power.	Answer: power = work / time = 3000 / 20 = 150 W. (Marking: formula 2, answer with unit 2)
15	A 2 kg object falls freely from a height of 5 m (air resistance neglected). Find the potential energy lost and the speed on landing.	Answer: potential energy = m x g x h = 2 x 10 x 5 = 100 J; all becomes kinetic energy, 0.5 x m x v x v = 100, v = square root of (2 x 100 / 2) = 10 m/s. (Marking: PE 2, energy conversion 1, speed 2)
16	A motor takes in 500 J of electrical energy and delivers 400 J of useful work. Find its efficiency and state where the rest of the energy goes.	Answer: efficiency = 400 / 500 x 100% = 80%; the remaining 100 J is lost as internal energy due to friction and resistance heating. (Marking: efficiency 3, energy loss 2)

Section E: Heat and Energy Transfer (16 marks)

The specific heat capacity of water is 4200 J kg-1 K-1. Answer the following.

#	Question	Answer and marking notes
17	2 kg of water is heated from 20 C to 30 C. Find the heat absorbed, in J and in kJ.	Answer: temperature rise = 10 K; Q = m x c x dT = 2 x 4200 x 10 = 84000 J = 84 kJ. (Marking: formula 2, each answer 1)
18	0.5 kg of water at 80 C is mixed with 0.5 kg of water at 20 C (heat loss neglected). Find the final temperature.	Answer: the masses and specific heat capacities are equal, so the final temperature is the average, (80 + 20) / 2 = 50 C. (Marking: thermal equilibrium 2, answer 2)
19	A 2 kW heater raises 2 kg of water from 20 C to 100 C (heat loss neglected). Find the energy needed and the heating time.	Answer: Q = 2 x 4200 x 80 = 672000 J; time = 672000 / 2000 = 336 s. (Marking: energy 2, time 2)
20	Describe conduction, convection and radiation, and give one everyday example of each.	Answer: conduction transfers energy by particle vibration in solids, e.g. a metal spoon in hot soup; convection transfers heat by density-driven flow in fluids, e.g. an air conditioner cooling a room; radiation transfers energy by electromagnetic waves without a medium, e.g. sunlight warming the ground. (Marking: three methods 1 each, examples 0.5 each)

Section F: Mixed Applications (16 marks)

Answer the following questions and show your working.

#	Question	Answer and marking notes
21	A car travels at 20 m/s. The driver reaction time is 0.5 s, after which the car decelerates uniformly at 5 m/s2 to rest. Find the reaction distance, braking distance and total stopping distance.	Answer: reaction distance = 20 x 0.5 = 10 m; braking distance = v x v / (2 x a) = 400 / 10 = 40 m; total = 50 m. (Marking: reaction 1, braking 2, total 1)
22	A crane lifts a load 6 m at constant speed with a force of 500 N in 10 s. Find the work done and the power.	Answer: work = 500 x 6 = 3000 J; power = 3000 / 10 = 300 W. (Marking: work 2, power 2)
23	A 0.5 kg stone falls freely from 20 m (air resistance neglected). Find its kinetic energy and speed on landing.	Answer: kinetic energy = potential energy = m x g x h = 0.5 x 10 x 20 = 100 J; v = square root of (2 x 100 / 0.5) = 20 m/s. (Marking: kinetic energy 2, speed 2)
24	A motor takes in 2400 J of electrical energy and lifts a 30 kg load 6 m at constant speed. Find the useful work and the efficiency.	Answer: useful work = m x g x h = 30 x 10 x 6 = 1800 J; efficiency = 1800 / 2400 x 100% = 75%. (Marking: useful work 2, efficiency 2)

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