DSE Physics Practice Paper: Mechanics And Motion

DSE Physics Practice Paper: Mechanics and Motion

2026-11-03

DSE Physics: Indices, Logs and Sequences

DSE Physics practice paper: 12 questions, 100 marks, 75 minutes, mechanics and motion, covering acceleration, projectiles, friction and momentum, with marking points.

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

Section A: Uniform acceleration and momentum (Questions 1-4, 24 marks)

Answer all questions. In calculations write the formula, substitute the values and state the units, giving 3 significant figures.

#	Question	Answer and marking notes
1	A car starts from rest and accelerates uniformly to 20 m/s in 8 s. (a) Find the acceleration of the car. (b) Find the distance travelled in these 8 s. (6 marks)	Answer: (a) a = (v – u) / t = (20 – 0) / 8 = 2.5 m/s^2 (3 marks). (b) s = (u + v) t / 2 = 20 x 8 / 2 = 80 m (3 marks). Marking: 1 mark for the formula, 1 for substitution, 1 for the answer with unit.
2	A small ball is released from rest from a platform 45 m above the ground. Air resistance is negligible and g = 10 m/s^2. (a) Find the time the ball takes to land. (b) Find its speed just before landing. (6 marks)	Answer: (a) From s = g t^2 / 2, 45 = 5 t^2, so t = 3 s (3 marks). (b) v = g t = 10 x 3 = 30 m/s (3 marks). Marking: 1 mark for the formula, 1 for substitution, 1 for the answer with unit; deduct 1 mark for a missing unit.
3	A lorry of mass 1 200 kg travels at 25 m/s. After braking it decelerates uniformly to rest in 50 m. (a) Find the deceleration. (b) Find the size of the braking force. (6 marks)	Answer: (a) From v^2 = u^2 + 2 a s, 0 = 625 + 100 a, so a = -6.25 m/s^2 and the deceleration is 6.25 m/s^2 (3 marks). (b) F = m a = 1 200 x 6.25 = 7 500 N (3 marks). Marking: 1 mark for correct sign handling, 1 for stating that the braking force opposes the motion.
4	A ball of mass 0.5 kg hits a wall at 8 m/s and rebounds at 6 m/s in the opposite direction. The contact time is 0.05 s and the original direction is taken as positive. (a) Find the impulse on the ball. (b) Find the average force exerted by the wall on the ball. (6 marks)	Answer: (a) Impulse = m (v – u) = 0.5 x (-6 – 8) = -7 kg m/s, so the magnitude is 7 kg m/s opposite to the original motion (3 marks). (b) F = impulse / t = -7 / 0.05 = -140 N, so the average force is 140 N opposite to the original motion (3 marks).

Section B: Projectiles, slopes and circular motion (Questions 5-8, 32 marks)

In vector questions state the direction and use signs to show opposite directions.

#	Question	Answer and marking notes
5	A small ball leaves the edge of a horizontal table 20 m above the floor with a horizontal velocity of 15 m/s. Take g = 10 m/s^2 and ignore air resistance. (a) Find the time of flight. (b) Find the horizontal distance travelled. (c) Find the speed of the ball on landing. (8 marks)	Answer: (a) Vertically 20 = 5 t^2, so t = 2 s (3 marks). (b) Range = 15 x 2 = 30 m (2 marks). (c) Vertical speed on landing = 20 m/s, so speed = sqrt(225 + 400) = 25 m/s (3 marks). Marking: horizontal and vertical motions must be treated separately.
6	A block of mass 4 kg slides down a slope of 30 degrees. The coefficient of kinetic friction is 0.25 and g = 10 m/s^2. (a) Find the normal force on the block. (b) Find the friction force. (c) Find the acceleration of the block down the slope. (8 marks)	Answer: (a) N = m g cos 30 = 4 x 10 x 0.866 = 34.6 N (3 marks). (b) f = 0.25 x 34.6 = 8.7 N (2 marks). (c) Net force along the slope = m g sin 30 – f = 20 – 8.7 = 11.3 N, so a = 11.3 / 4 = 2.8 m/s^2 (3 marks).
7	A body of mass 2 kg moves at 6 m/s on a smooth horizontal surface and collides with a stationary body of mass 4 kg. The two bodies stick together. (a) Find their common velocity. (b) Find the change in kinetic energy and state the nature of the collision. (8 marks)	Answer: (a) By conservation of momentum, 2 x 6 = 6 v, so v = 2 m/s (3 marks). (b) Kinetic energy before = 36 J and after = 12 J, so 24 J is lost and the collision is perfectly inelastic (5 marks: 3 for the energy calculation, 2 for the classification).
8	A car of mass 1 000 kg travels at 15 m/s around a horizontal bend of radius 50 m. (a) Find the centripetal force required. (b) If the coefficient of static friction between the tyres and the road is 0.6, find the maximum safe speed. (8 marks)	Answer: (a) F = m v^2 / r = 1 000 x 225 / 50 = 4 500 N (4 marks). (b) Maximum friction = 0.6 x 1 000 x 10 = 6 000 N, so v = sqrt(0.6 x 10 x 50) = 17.3 m/s (4 marks). Marking: 1 mark for stating that friction provides the centripetal force.

Section C: Work, energy, power and experimental analysis (Questions 9-12, 44 marks)

In experimental questions state the sources of error and how each can be reduced.

#	Question	Answer and marking notes
9	A student of mass 60 kg climbs 5 m vertically up a staircase in 4 s. Take g = 10 m/s^2. (a) Find the work done against gravity. (b) Find the average power of the student. (c) A second student of mass 75 kg works at the same power for 5 s. Find the vertical height he rises. (10 marks)	Answer: (a) W = m g h = 60 x 10 x 5 = 3 000 J (3 marks). (b) P = W / t = 3 000 / 4 = 750 W (3 marks). (c) Work done = 750 x 5 = 3 750 J = 75 x 10 x h, so h = 5 m (4 marks). Marking: the height must come from a calculation, not be quoted.
10	A crate of mass 20 kg is pulled 8 m along horizontal ground by a force of 150 N acting at 30 degrees above the horizontal. The coefficient of kinetic friction is 0.2 and g = 10 m/s^2. (a) Find the work done by the pulling force. (b) Find the work done against friction. (c) Find the speed of the crate after 8 m, starting from rest. (10 marks)	Answer: (a) W = F s cos 30 = 150 x 8 x 0.866 = 1 039 J (3 marks). (b) N = 200 – 75 = 125 N, friction = 25 N, so work against friction = 200 J (4 marks). (c) Net work = 1 039 – 200 = 839 J = 20 v^2 / 2, so v = 9.2 m/s (3 marks).
11	A train starts from rest and accelerates uniformly at 2 m/s^2 for 10 s. It then travels at constant speed for 30 s and finally decelerates uniformly to rest in 15 s. (a) Find the maximum speed. (b) Find the total distance travelled. (c) Find the average speed for the whole journey. (d) State the physical meaning of the area under a speed-time graph. (12 marks)	Answer: (a) v = a t = 2 x 10 = 20 m/s (2 marks). (b) Distance = 100 + 600 + 150 = 850 m (6 marks, 2 per stage). (c) Average speed = 850 / 55 = 15.5 m/s (2 marks). (d) The area represents the displacement (2 marks).
12	In a laboratory the acceleration due to gravity is measured with an electronic timer. A steel ball is released from rest and falls 1.8 m in 0.60 s. (a) Calculate g from these data. (b) State two sources of error and how each can be reduced. (c) Predict the time of fall if the ball is released from 3.2 m. (12 marks)	Answer: (a) g = 2 s / t^2 = 3.6 / 0.36 = 10.0 m/s^2 (4 marks). (b) Reaction time when starting the timer and air resistance: use a light gate or average several runs to cut random error, and use a heavier ball or a shorter drop to cut the effect of air resistance (4 marks, 2 per source). (c) t = sqrt(2 x 3.2 / 10) = 0.80 s (4 marks).

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