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CIE A Level Mathematics 9709 Paper 4 (Mechanics) Practice Paper | 50 marks, 1 hour 15 minutes
CIE A Level Mathematics 9709 Paper 4 (Mechanics)
CIE A Level Maths 9709 Paper 4 Mechanics practice paper: 9 questions, 50 marks, 1 hour 15 minutes with answer key and worked steps; PDFs download instantly. All original.
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Paper details Level: CIE A Level ｜ Topic: mechanics ｜ Time: 1 hour 15 minutes ｜ Marks: 50 ｜ Questions: 9 ｜ Calculator allowed
Section A — Kinematics, momentum and force (25 marks)
Answer all questions. Show all working and state the units.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | A particle P moves along a straight line. Its velocity is v = 3t² − 12t + 9 m/s at time t seconds. (a) Find the two values of t at which P is instantaneously at rest. (b) Find the distance travelled by P in the first 4 seconds. | (a) 3t² − 12t + 9 = 0 gives t² − 4t + 3 = 0, i.e. (t − 1)(t − 3) = 0, so t = 1 s and t = 3 s. (b) Integrating: s = t³ − 6t² + 9t. Displacement in 4 s = 4 m. Since v changes sign at t = 1 s and t = 3 s, distance = 4 + 4 + 4 = 12 m (displacement alone underestimates). |
| 2 | A train travelling at 30 m/s decelerates uniformly and comes to rest in 200 m. Find (a) the deceleration, (b) the time taken to stop. | (a) v² = u² + 2as: 0 = 30² + 2a(200), so a = -2.25 m/s² (deceleration 2.25 m/s²). (b) v = u + at: 0 = 30 − 2.25t, so t = 13.33 s. Check: average speed 15 m/s × 13.33 s ≈ 200 m. |
| 3 | Particle A of mass 3 kg moves at 5 m/s towards particle B of mass 2 kg, which is moving at 2 m/s in the opposite direction. The particles coalesce. Find (a) their common velocity, (b) the kinetic energy lost. | (a) Taking A direction as positive: 3(5) + 2(−2) = 11 = 5v, so v = 2.2 m/s, i.e. 2.2 m/s in the original direction of A. (b) KE before = 41.5 J; KE after = 12.1 J; loss = 29.4 J. |
| 4 | A ball is projected vertically upwards from ground level with speed 14 m/s. Taking g = 9.8 m/s² and ignoring air resistance, find (a) the greatest height reached, (b) the total time before the ball returns to the ground. | (a) 0 = 14² − 2(9.8)h, so h = 10 m. (b) Time to the top = 14/9.8 = 1.429 s, so total time = 2.86 s. Check: 14(2.857) − 0.5(9.8)(2.857)² ≈ 0. |
| 5 | A lift of mass 800 kg is pulled upwards by a vertical cable with tension 9000 N. Find (a) its acceleration, (b) the tension needed for the lift to move upwards at constant speed. | (a) T − mg = ma: 9000 − 800(9.8) = 800a, so a = 1.45 m/s². (b) At constant speed a = 0, so T = mg = 7840 N. |
Section B — Applied mechanics (25 marks)
Answer all questions. Take g = 9.8 m/s².
| # | Question | Answer and marking notes |
|---|---|---|
| 6 | A block of mass 5 kg rests on rough horizontal ground. The coefficient of friction is 0.3. A horizontal force of 25 N is applied to the block. (a) Show that the block moves. (b) Find the friction force on the block. (c) Find its acceleration. | (a) Greatest possible friction = μR = 0.3(5)(9.8) = 14.7 N, which is smaller than 25 N, so the block does move. (b) Friction = 14.7 N acting backwards. (c) F − friction = ma: 25 − 14.7 = 5a, so a = 2.06 m/s². |
| 7 | A uniform rod AB has length 6 m and mass 20 kg. It rests horizontally on two supports, one at A and one at C, where AC = 4 m. Find the normal reaction at each support. | The weight 196 N acts at the midpoint of the rod, 3 m from A. Taking moments about A: R_C(4) = 196(3), so R_C = 147 N. Resolving vertically: R_A + R_C = 196, so R_A = 49 N. |
| 8 | A car of mass 1200 kg travels at a constant 25 m/s on a level road with the engine working at 30 kW. (a) Find the total resistance. (b) The car then climbs a straight slope with sin θ = 0.05, the power and the resistance being unchanged. Find the steady speed up the slope. | (a) P = Fv: 30000 = F(25), so F = 1200 N. (b) Component of weight up the slope = 1200(9.8)(0.05) = 588 N; the total force needed is 1788 N, so v = 30000/1788 = 16.78 m/s. |
| 9 | Particles of mass 4 kg and 2 kg are connected by a light inextensible string which passes over a smooth fixed pulley. Both particles hang freely. Find (a) the acceleration of each particle, (b) the tension in the string. | (a) (4 − 2)g = (4 + 2)a: 2(9.8) = 6a, so a = 3.27 m/s². (b) For the lighter particle: T − 2(9.8) = 2(3.27), so T = 26.13 N. (T is the same on both sides, as the string is light.) |
A TutorZone practice paper; original questions and passages, not publisher material.
