DSE Physics Practice Paper: Mechanics And Motion (12 Questions, 50 Marks, 75 Minutes)

DSE Physics Practice Paper: Mechanics and Motion (12 Questions, 50 Marks, 75 Minutes)

2026-11-02

DSE Physics: Mechanics and Motion

DSE Physics paper: 12 questions, 50 marks, 75 minutes on uniformly accelerated motion, projectiles, Newton laws, momentum, circular motion, work, energy and pulley systems.

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Paper details　Level: S6 | Subject: Physics | Time: 75 minutes | Marks: 50 | Questions: 12

Section A (Short Questions)

Each question carries 2 marks. Show the equation and give units.

#	Question	Answer and marking notes
1	A car accelerates uniformly from rest to 20 m/s in 5 s. Find (a) the acceleration; (b) the displacement during this time.	a = (20 – 0) / 5 = 4 m/s2; s = (0 + 20) / 2 x 5 = 50 m. [1 mark equation, 1 mark answer with unit]
2	A ball is released from rest at a height of 45 m (take g = 10 m/s2, ignore air resistance). Find (a) the time of fall; (b) the speed on hitting the ground.	(a) s = 0.5 g t^2, so 45 = 0.5 x 10 x t^2 and t = 3 s. (b) v = g t = 10 x 3 = 30 m/s. [1 mark each]
3	A body of mass 2 kg is acted on by a net force of 12 N for 3 s. Find (a) the acceleration; (b) the change in momentum.	(a) a = F / m = 12 / 2 = 6 m/s2. (b) change in momentum = F t = 12 x 3 = 36 kg m/s. [1 mark each]
4	A car of mass 1200 kg travelling at 15 m/s stops in 30 m under braking. Find (a) its kinetic energy before braking; (b) the average braking force.	(a) KE = 0.5 m v^2 = 0.5 x 1200 x 225 = 135000 J. (b) F = KE / s = 135000 / 30 = 4500 N. [1 mark each]
5	A ball is projected horizontally at 12 m/s from a height of 20 m (take g = 10 m/s2). Find (a) the time of flight; (b) the horizontal range.	(a) Vertically 20 = 0.5 x 10 x t^2, so t = 2 s. (b) Range = u t = 12 x 2 = 24 m. [1 mark each]
6	A block of mass 5 kg on a horizontal floor is pulled by a horizontal force of 30 N against a friction of 10 N. Find (a) the acceleration; (b) the speed after 4 s and the displacement in that time.	(a) a = (30 – 10) / 5 = 4 m/s2. (b) v = a t = 16 m/s; s = 0.5 x 4 x 16 = 32 m. [1 mark each]

Section B (Multi-step Calculations)

Each question carries 4 marks. Show all working.

#	Question	Answer and marking notes
7	A body of mass 2 kg moving at 6 m/s collides with a stationary body of mass 4 kg and they stick together. Find (a) the common velocity after impact; (b) the kinetic energy lost, and state where that energy goes.	(a) Momentum: 2 x 6 = (2 + 4) v, so v = 2 m/s. (b) KE before = 36 J, after = 12 J, loss = 24 J, converted to heat, sound and deformation energy. [2 marks velocity, 2 marks energy]
8	A ball of mass 0.5 kg moves at 3 m/s in a horizontal circle of radius 1.2 m. Find (a) the centripetal acceleration; (b) the centripetal force; (c) the period.	(a) a = v^2 / r = 9 / 1.2 = 7.5 m/s2. (b) F = m a = 0.5 x 7.5 = 3.75 N towards the centre. (c) T = 2 pi r / v = 2.51 s. [(a) 1 mark, (b) 2 marks, (c) 1 mark]
9	A runner of mass 60 kg runs up 300 steps, each 0.18 m high, in 240 s. (a) Find the work done against gravity. (b) Find the average power. (c) If his useful output is 108 W, find the efficiency (1 d.p.).	(a) Height = 300 x 0.18 = 54 m; W = m g h = 60 x 10 x 54 = 32400 J. (b) P = W / t = 32400 / 240 = 135 W. (c) Efficiency = 108 / 135 x 100% = 80%. [(a) 2, (b) 1, (c) 1 marks]

Section C (Integrated and Essay Questions)

Follow the instructions and support answers with physical principles and calculations.

#	Question	Answer and marking notes
10	A car of mass 1500 kg hits a wall at 20 m/s and stops in 0.5 s, with the front crumpling by 5 m. (a) Use impulse to find the average impact force. (b) Find the average force again from energy and distance, and compare. (c) Explain why modern cars have crumple zones and seat belts instead of a rigid body.	(a) Impulse = m dv = 1500 x 20 = 30000 N s; F = 30000 / 0.5 = 60000 N (about 6.0 x 10^4 N). (b) KE = 0.5 x 1500 x 400 = 300000 J; F = 300000 / 5 = 60000 N, the same as (a) because both give the average force. (c) A crumple zone lengthens the stopping distance and time, so for the same change in momentum the average force falls sharply; the belt spreads the force over strong bones and prevents the occupant hitting the interior. A rigid body would make the stopping time very short and the force enormous. [(a) 3, (b) 3, (c) 2 marks]
11	A block of mass 4 kg slides from rest down a slope of length 5 m inclined at 30 degrees, with kinetic friction coefficient 0.25 (take g = 10 m/s2). (a) Find the component of weight along the slope and the normal force. (b) Find the friction and the net force. (c) Find the acceleration and the speed at the bottom. (d) Check (c) by energy conservation and find the energy lost to friction.	(a) Along slope = m g sin30 = 40 x 0.5 = 20 N; normal = m g cos30 = 34.64 N. (b) f = mu N = 0.25 x 34.64 = 8.66 N; net = 20 – 8.66 = 11.34 N down the slope. (c) a = 11.34 / 4 = 2.83 m/s2; v = sqrt(2 x 2.835 x 5) = 5.32 m/s. (d) Gravity does m g h = 4 x 10 x 2.5 = 100 J; friction takes f L = 43.3 J; remaining KE = 56.7 J, giving v = 5.32 m/s, the same as (c). [(a) 2, (b) 3, (c) 2, (d) 2 marks]
12	Two blocks of mass 3 kg and 2 kg are connected by a light string over a frictionless, massless fixed pulley, the heavier one hanging down. Take g = 10 m/s2. (a) Find the acceleration of the system. (b) Find the string tension and verify it with the other block. (c) Explain why the tension is not equal to the weight of either block. (d) If the pulley had mass, would the actual acceleration be larger or smaller? Explain.	(a) a = (m1 – m2) g / (m1 + m2) = (3 – 2) x 10 / 5 = 2 m/s2, with the heavier block moving down. (b) For the 2 kg block: T – m2 g = m2 a, so T = 2 x (10 + 2) = 24 N; for the 3 kg block 30 – 24 = 6 = 3 x 2, which checks. (c) Both blocks accelerate, so the tension must exceed the weight of the lighter block (20 N) to give it an upward net force, yet be less than the weight of the heavier block (30 N); if the tension equalled either weight that block could not accelerate. (d) With a pulley of mass, extra torque is needed to spin it and some energy becomes rotational kinetic energy, so the actual acceleration is smaller. [(a) 3, (b) 3, (c) 2, (d) 1 marks]

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