S4 Physics Paper: Kinematics, Forces, Work And Energy, Momentum, Parallel Circuits And Heat (100 Marks)

S4 Physics Paper: Kinematics, Forces, Work and Energy, Momentum, Parallel Circuits and Heat (100 marks)

2026-11-13

S4 Physics – mechanics, circuits and heat

S4 Physics paper: 6 questions and 24 parts on accelerated motion, Newton laws, work and energy, momentum, circuits and heat. Values are python-checked.

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Paper details　S4 Physics practice paper with 6 questions and 24 parts on kinematics, Newton laws with friction, work and energy, momentum, parallel circuits and specific heat.

Section A: Kinematics (16 marks)

A car starts from rest and accelerates uniformly at 2.5 m/s2 for 8 s, then decelerates uniformly to rest in 5 s.

#	Question	Answer and marking notes
1	Find the velocity of the car after accelerating for 8 s.	Answer: v = u + at = 0 + 2.5 x 8 = 20 m/s. (Marking: formula 2, answer 2)
2	Find the distance travelled while accelerating.	Answer: s = at2/2 = 0.5 x 2.5 x 64 = 80 m. (Marking: formula 2, substitution and answer 2)
3	Find the magnitude of the deceleration and the distance travelled while slowing down.	Answer: a = 20 / 5 = 4 m/s2 opposite to the motion; s = 0.5 x 20 x 5 = 50 m. (Marking: acceleration 2, distance 2)
4	Find the average speed for the whole journey and explain why the mean of initial and final velocity cannot be used.	Answer: total distance 130 m over 13 s gives 130 / 13 = 10 m/s. Averaging the two end velocities is valid only for uniform acceleration throughout; here the two stages last different times. (Marking: answer 2, explanation 2)

Section B: Forces and Newton Laws (17 marks)

A 4 kg box on a horizontal table is pulled by a 30 N horizontal force. Friction equals 0.25 times the normal reaction. Take g = 10 m/s2.

#	Question	Answer and marking notes
5	Find the weight of the box and the normal reaction from the table.	Answer: W = mg = 4 x 10 = 40 N; on a horizontal surface R = W = 40 N. (Marking: weight 2, normal reaction 2)
6	Find the friction acting on the box.	Answer: f = 0.25 x 40 = 10 N, acting against the motion. (Marking: formula 2, value and direction 2)
7	Find the acceleration of the box.	Answer: net force = 30 – 10 = 20 N; a = 20 / 4 = 5 m/s2. (Marking: net force 2, acceleration 2)
8	If the pull is at 30 degrees above the horizontal, find the new acceleration and explain why the normal reaction changes.	Answer: horizontal component = 30 cos 30 = 26.0 N, vertical = 30 sin 30 = 15 N; R = 40 – 15 = 25 N because the upward component relieves the table; friction = 0.25 x 25 = 6.25 N; net force = 19.7 N, so a = 4.9 m/s2. (Marking: resolve forces 2, new normal reaction 2, answer 1)

Section C: Work, Energy and Power (17 marks)

A 5 kg box at rest is pushed 6 m across a horizontal floor by a 24 N horizontal force. Friction on the box is 10 N.

#	Question	Answer and marking notes
9	Find the work done by the pushing force.	Answer: W = Fs = 24 x 6 = 144 J. (Marking: formula 2, answer 2)
10	Find the work done against friction.	Answer: W = fs = 10 x 6 = 60 J, which becomes thermal energy. (Marking: calculation 3, thermal energy 1)
11	Find the speed of the box after 6 m.	Answer: net work = 84 J becomes kinetic energy; 0.5 x 5 x v2 = 84, so v = 5.8 m/s. (Marking: net work 2, kinetic energy equation 1, answer 1)
12	If the process takes 4 s, find the average power of the push and the energy efficiency of the process.	Answer: P = 144 / 4 = 36 W; efficiency = 84 / 144 x 100 = 58.3 percent. (Marking: power 2, efficiency formula 2, answer 1)

Section D: Momentum and Collisions (17 marks)

Trolley A of mass 2 kg moves at 3 m/s on a smooth horizontal track and hits a stationary trolley B of mass 1 kg. They stick together after the collision.

#	Question	Answer and marking notes
13	Find the total momentum of the trolleys before the collision.	Answer: p = 2 x 3 + 0 = 6 kg m/s in the direction of A. (Marking: calculation 3, direction 1)
14	Find the common velocity after the collision.	Answer: momentum is conserved, 6 = 3 x v, so v = 2 m/s. (Marking: conservation equation 2, answer 2)
15	Find the kinetic energy lost in the collision.	Answer: before 0.5 x 2 x 9 = 9 J; after 0.5 x 3 x 4 = 6 J; loss 3 J. (Marking: 1 each for the two values, 2 for the difference)
16	Explain why momentum is conserved but kinetic energy is not, and identify the type of collision.	Answer: during the collision the action and reaction between the trolleys are equal, opposite and last the same time, so the net impulse is zero and total momentum is conserved. Some kinetic energy becomes sound and thermal energy as the trolleys deform, so kinetic energy is not conserved. This is a perfectly inelastic collision. (Marking: reason for momentum 2, energy transfer 2, type 1)

Section E: Parallel Circuits and Power (17 marks)

A 12 V battery of negligible internal resistance is connected to a 6 ohm and a 3 ohm resistor in parallel.

#	Question	Answer and marking notes
17	Find the equivalent resistance of the parallel combination.	Answer: 1/R = 1/6 + 1/3 = 1/2, so R = 2 ohm. (Marking: parallel formula 2, answer 2)
18	Find the total current from the battery.	Answer: I = 12 / 2 = 6 A. (Marking: formula 2, answer 2)
19	Find the current through each resistor.	Answer: both resistors have 12 V across them, so the 6 ohm branch takes 2 A and the 3 ohm branch 4 A, giving 6 A in total. (Marking: 1.5 each for the currents, 1 for the check)
20	Find the power dissipated in the 6 ohm resistor and the total power of the circuit.	Answer: P = 144 / 6 = 24 W; total power = 12 x 6 = 72 W. Check: 24 + 144/3 = 72 W. (Marking: 2 for the 6 ohm power, 2 for the total, 1 for the check)

Section F: Heat and Heating Efficiency (16 marks)

A 700 W heater warms 0.5 kg of water from 20 °C to 80 °C. The specific heat capacity of water is 4200 J/(kg °C).

#	Question	Answer and marking notes
21	Find the energy gained by the water.	Answer: E = mc dT = 0.5 x 4200 x 60 = 126000 J, that is 1.26 x 10^5 J. (Marking: formula 2, answer 2)
22	Assuming no heat loss, find the heating time required.	Answer: t = 126000 / 700 = 180 s, which is 3 minutes. (Marking: formula 2, answer 2)
23	If the heater is only 80 percent efficient, find the actual heating time.	Answer: useful power = 700 x 0.8 = 560 W; t = 126000 / 560 = 225 s. (Marking: useful power 2, time 2)
24	Give two reasons why the real heating time is longer than the theoretical value.	Answer: (1) heat is lost from the water to the surrounding air; (2) part of the energy heats the container itself rather than the water. (Marking: 2 each)

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