DSE Physics Practice Paper: Mechanics And Electricity (12 Questions, 100 Marks, 75 Minutes)

DSE Physics Practice Paper: Mechanics and Electricity (12 questions, 100 marks, 75 minutes)

2026-11-03

DSE Physics: Mechanics and Electricity

DSE Physics practice paper on forces, momentum, energy and circuits: 12 questions, 100 marks, 75 minutes, with full worked answers and marking points.

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

Section A: Kinematics and forces (Questions 1-4, 24 marks)

Answer all questions. Write down the formula, substitute values and give units.

#	Question	Answer and marking notes
1	A car starts from rest and accelerates uniformly at 2.5 m/s² for 8 s. (a) Find its speed after 8 s. (b) Find the distance travelled in these 8 s. (6 marks)	(a) v = u + at = 0 + 2.5 x 8 = 20 m/s. (b) s = ut + at²/2 = 0 + 2.5 x 64 / 2 = 80 m. Marks: 3 for speed, 3 for distance (working, substitution and units required).
2	A 5 kg object is pushed by a horizontal force of 20 N while a friction force of 8 N acts on it. (a) Find the acceleration. (b) Find its kinetic energy 3 s after starting from rest. (6 marks)	(a) Net force = 20 – 8 = 12 N, so a = 12 / 5 = 2.4 m/s². (b) v = at = 2.4 x 3 = 7.2 m/s, kinetic energy = 5 x 7.2² / 2 = 129.6 J. Marks: 3 for acceleration, 3 for kinetic energy.
3	A stone is dropped from a height of 45 m. Take g = 10 m/s² and ignore air resistance. (a) Find the time of fall. (b) Find the speed on landing. (c) If the mass of the stone is doubled, how does the time of fall change? Explain. (6 marks)	(a) s = gt²/2, so 45 = 5t² and t = 3 s. (b) v = gt = 10 x 3 = 30 m/s. (c) The time is unchanged, because the acceleration of free fall does not depend on mass; the stone still takes 3 s. Marks: 2 time, 2 speed, 2 explanation.
4	A 12 V battery is connected to a 4 Ω resistor and a 6 Ω resistor in parallel. (a) Find the equivalent resistance. (b) Find the total current drawn from the battery. (c) Find the current through the 4 Ω resistor. (6 marks)	(a) 1/R = 1/4 + 1/6, so R = 2.4 Ω. (b) I = V / R = 12 / 2.4 = 5 A. (c) Current through 4 Ω = 12 / 4 = 3 A, the other 2 A passing through 6 Ω. Marks: 2 equivalent resistance, 2 total current, 2 branch current.

Section B: Momentum, energy and electrical power (Questions 5-8, 32 marks)

Show full working in energy and power questions; write explanations in complete sentences.

#	Question	Answer and marking notes
5	A 2 kg trolley moving at 3 m/s collides with a stationary 1 kg trolley and they stick together. (a) Find the common velocity after the collision. (b) Find the total kinetic energy before and after, and the loss. (c) Explain why momentum is conserved but kinetic energy is not. (8 marks)	(a) By conservation of momentum, 2 x 3 = (2 + 1) x v, so v = 2 m/s. (b) Before: 2 x 3² / 2 = 9 J; after: 3 x 2² / 2 = 6 J; loss = 3 J. (c) The trolleys stick together, so the collision is inelastic and some kinetic energy becomes heat and sound. Momentum is still conserved because no external force acts on the system. Marks: 2 velocity, 3 energy, 3 explanation.
6	Water of mass 50 kg is lifted through 12 m in 20 s. Take g = 10 m/s². (a) Find the work done by the pump. (b) Find its output power. (c) If the motor input power is 400 W, find the efficiency. (8 marks)	(a) W = mgh = 50 x 10 x 12 = 6000 J. (b) P = W / t = 6000 / 20 = 300 W. (c) Efficiency = 300 / 400 x 100% = 75%. Marks: 3 work, 2 power, 3 efficiency.
7	A 1500 W electric kettle is used for 0.4 h every day. Electricity costs 1.2 dollars per kWh. (a) Find the daily energy used in kWh. (b) Find the daily cost. (c) Find the total cost for 30 days. (8 marks)	(a) E = 1.5 kW x 0.4 h = 0.6 kWh. (b) Cost = 0.6 x 1.2 = 0.72 dollars. (c) For 30 days, 0.72 x 30 = 21.6 dollars. Marks: 3 energy, 2 daily cost, 3 total cost (kW and h must be used).
8	Two 6 Ω resistors are connected to a 12 V supply, first in series and then in parallel. (a) Find the equivalent resistance in each case. (b) Find the total power in each case. (c) Explain why the power is greater in parallel. (8 marks)	(a) Series: 6 + 6 = 12 Ω; parallel: 3 Ω. (b) Series power = 12² / 12 = 12 W; parallel power = 12² / 3 = 48 W. (c) The parallel combination has a smaller equivalent resistance, so it draws a larger total current at the same voltage, and P = V² / R is larger. Marks: 2 resistance, 3 power, 3 explanation.

Section C: Energy conservation, resistance and circular motion (Questions 9-12, 44 marks)

Answer the long questions in points; give numerical answers to three significant figures.

#	Question	Answer and marking notes
9	A 20 kg crate slides from rest down a 5 m long ramp inclined at 30 degrees to the horizontal. The friction force is 40 N. Take g = 10 m/s². (a) Find the height of the ramp. (b) Find the loss in gravitational potential energy. (c) Find the work done against friction. (d) Find the speed of the crate at the bottom. (11 marks)	(a) h = 5 sin 30 degrees = 2.5 m. (b) Loss in potential energy = mgh = 20 x 10 x 2.5 = 500 J. (c) Work against friction = 40 x 5 = 200 J. (d) Kinetic energy = 500 – 200 = 300 J, so v = sqrt(2 x 300 / 20) = sqrt 30 = 5.48 m/s. Marks: 2 height, 3 potential energy, 3 friction work, 3 speed.
10	In an experiment on Ohm law, the current through a resistor is 0.25 A when the voltage across it is 6.0 V. (a) Find the resistance. (b) Find the power dissipated. (c) If the voltage is doubled, find the new current and power. (d) Explain why several readings are taken and a current against voltage graph is plotted. (11 marks)	(a) R = V / I = 6.0 / 0.25 = 24 Ω. (b) P = VI = 6.0 x 0.25 = 1.5 W. (c) Doubling the voltage doubles the current to 0.5 A; the power becomes 12 x 0.5 = 6 W, four times as large. (d) Several readings reduce random error, and the graph shows whether the current is proportional to the voltage; the slope also gives a more reliable resistance. Marks: 2 resistance, 2 power, 3 doubling, 4 experimental explanation.
11	A 0.2 kg ball moves at 4 m/s in a horizontal circle of radius 0.5 m. (a) Find the centripetal acceleration. (b) Find the centripetal force. (c) Find the period of the motion. (d) If the string suddenly breaks, describe the subsequent motion of the ball. (11 marks)	(a) Centripetal acceleration = v² / r = 16 / 0.5 = 32 m/s². (b) Centripetal force = m x a = 0.2 x 32 = 6.4 N, directed towards the centre. (c) Period = 2 pi r / v = 2 pi x 0.5 / 4 = 0.785 s. (d) With no centripetal force, the ball moves in a straight line along the tangent at 4 m/s, by Newton first law. Marks: 2 acceleration, 2 force, 3 period, 4 motion.
12	A 1200 kg car travels at 20 m/s. The driver reaction time is 0.8 s and the braking force is 6000 N. (a) Find the distance travelled during the reaction time. (b) Find the deceleration while braking. (c) Find the braking distance. (d) Find the total stopping distance. (e) If the speed is doubled, by what factor does the total stopping distance increase? (11 marks)	(a) Reaction distance = 20 x 0.8 = 16 m. (b) a = F / m = 6000 / 1200 = 5 m/s², opposite to the motion. (c) Braking distance = v² / (2a) = 400 / 10 = 40 m. (d) Total stopping distance = 16 + 40 = 56 m. (e) At double speed the reaction distance is 32 m and the braking distance 160 m, giving 192 m, about 3.4 times the original. Marks: 2 reaction distance, 2 deceleration, 2 braking distance, 2 total, 3 factor.

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