Pearson Edexcel International Advanced Level Mathematics: Models, vectors and constant acceleration (YMA01) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions follow Unit M1: Mechanics 1, sections 1.1, 2.1, 2.2 and 3.1 of the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and Pure Mathematics specification (Issue 3, April 2019). Under the specification, M1 is optional: it can be taken within IAS Mathematics, IAS Further Mathematics, IAL Mathematics and IAL Further Mathematics. Learn the content first from the study guide or the revision notes.

Course links: the Edexcel A Level Mathematics hub, the printable checklist and the free diagnostics. Use g = 9.8 m s⁻² where needed. Vectors use i (east) and j (north) unless stated.

Questions

1. A block rests on a rough horizontal table. A string attached to the block passes over a pulley at the table edge and carries a hanging mass.

(a) Why can the block be modelled as a particle? [1] (b) State the consequence of modelling the string as (i) light and (ii) inextensible. [2] (c) State the consequence of modelling the pulley as smooth. [1]

2. (a) The force F = (−4i + 7j) N. Find its magnitude and the angle it makes with the positive j direction, to 1 decimal place. [4] (b) A rope pulls a crate with force 40 N at 25° above the horizontal. Taking i horizontal and j upwards, write the force as a vector (3 significant figures). [2] (c) A boat travels at 7 m s⁻¹ on a bearing of 250°. Write its velocity in the form ai + bj, to 3 significant figures. [2]

3. Three forces act on a particle: F₁ = (3i + pj) N, F₂ = (qi − 5j) N and F₃ = (−2i + 3j) N. Their resultant is (8i + 4j) N. (a) Find p and q. [3] (b) Find the magnitude of the resultant and the angle it makes with i, giving the angle to 1 decimal place. [3]

4. At time t = 0 a particle P is at the point with position vector (−10i + 12j) m, moving with constant velocity (4i − 3j) m s⁻¹. (a) Find the speed of P. [2] (b) Find the position vector of P when t = 6. [2] (c) Find when P is due north of the origin O, and its distance from O then. [3]

5. A particle moves with constant acceleration. Its velocity is (2i + 7j) m s⁻¹ at one instant and (−8i + 2j) m s⁻¹ five seconds later. Find the acceleration as a vector, its magnitude, and the acute angle it makes with the negative i direction (1 decimal place). [4]

6. A cyclist increases her speed uniformly from 3 m s⁻¹ to 9 m s⁻¹ while travelling 60 m in a straight line. (a) Find her acceleration. [2] (b) Find the time taken. [2]

7. A ball is thrown vertically upwards at 21 m s⁻¹ from a point 3.5 m above horizontal ground. It moves freely under gravity. (a) Find the greatest height of the ball above the ground. [3] (b) Find the time taken to reach the ground, to 3 significant figures. [2]

8. A tram leaves a stop and accelerates uniformly at 0.8 m s⁻² for 15 s. It then travels at constant speed, and finally decelerates uniformly at 1.2 m s⁻² to rest at the next stop, 1050 m from the first. (a) Find the maximum speed and the time spent decelerating. [2] (b) Find the time spent at constant speed. [3] (c) Find the total journey time and the average speed. [2]

9. A particle moves on a straight line. Its velocity is 4 m s⁻¹ at t = 0 and increases uniformly to 12 m s⁻¹ at t = 5. It stays at 12 m s⁻¹ until t = 8, then decreases uniformly to −6 m s⁻¹ at t = 11. (a) Find the acceleration in each of the three stages. [3] (b) Find the time at which the particle is instantaneously at rest. [1] (c) Find the displacement from the start at t = 11. [3] (d) Find the total distance travelled. [1]

10. Car A passes a point O at t = 0 travelling at a constant 6 m s⁻¹ along a straight road. Car B passes O from rest at t = 6, accelerating uniformly at 1.5 m s⁻² in the same direction. Time t is in seconds. (a) Show that B’s distance from O at time t > 6 is 0.75(t − 6)². [2] (b) Show that the cars are level when t² − 20t + 36 = 0. [2] (c) Solve this equation and state, with a reason, which root is valid. [3] (d) Find the speed of B when the cars are level. [2] (e) State one modelling assumption about the cars. [1]

11. A drone moves from O with constant velocity (3i + 4j) m s⁻¹ for 20 s, then with constant velocity (−4i + 6j) m s⁻¹. Time t is in seconds from the start. (a) Find the position vector of the drone when t = 20. [2] (b) Show that for t > 20 its position vector is (140 − 4t)i + (6t − 40)j. [2] (c) Find when the drone is due north of O, and how far from O it is then. [3] (d) Find its speed during the second stage, to 3 significant figures. [2]

Answers

1. (a) Its size and rotation do not matter, so its mass is treated as at one point. [1] (b) (i) Light: negligible mass, so the tension is the same throughout the string. [1] (ii) Inextensible: the string does not stretch, so block and hanging mass have the same speed and acceleration magnitude. [1] (c) Smooth pulley: no friction acts on the string there, so the tension is unchanged as it passes over the pulley. [1] Examiner insight: Link each assumption to its consequence for the equations; naming the word alone gives little to be credited.

2. (a) Magnitude = √((−4)² + 7²) [1] = √65 = 8.1 N [1] Angle with j: tan θ = 4/7 [1], so θ = 29.7° [1] (b) Horizontal 40 cos 25° [1]; force = (36.3i + 16.9j) N [1] (c) Bearing 250° gives east 7 sin 250° [1] and north 7 cos 250°, so velocity = (−6.58i − 2.39j) m s⁻¹ [1] Examiner insight: Sketch the right-angled triangle before choosing sin or cos, and state which line an angle is measured from.

3. (a) i: 3 + q − 2 = 8 [1], so q = 7. j: p − 5 + 3 = 4 [1], so p = 6 [1] (b) Magnitude = √(8² + 4²) [1] = √80 = 8.94 N [1] Angle: tan α = 4/8, so α = 26.6° [1] Examiner insight: Equate i and j components on separate lines so each step is easy to follow.

4. (a) Speed = √(4² + (−3)²) [1] = 5 m s⁻¹ [1] (b) r = (−10i + 12j) + 6(4i − 3j) [1] = (14i − 6j) m [1] (c) Due north means the i component is zero: −10 + 4t = 0 [1], so t = 2.5 s. j component = 12 − 3(2.5) = 4.5, so distance = 4.5 m [1] from O. [1] Examiner insight: Position is r₀ + vt. Speed is the magnitude of velocity, so it has no direction. Answer both parts of (c).

5. a = (change in velocity)/time = ((−8 − 2)i + (2 − 7)j)/5 [1]; a = (−2i − j) m s⁻² [1] Magnitude = √(4 + 1) = 2.24 m s⁻² [1] tan α = 1/2, so α = 26.6° below the negative i direction. [1] Examiner insight: Subtract initial from final velocity, not the reverse; a sign error here affects everything after it.

6. (a) v² = u² + 2as: 81 = 9 + 120a [1], so a = 0.6 m s⁻² [1] (b) v = u + at: 9 = 3 + 0.6t [1], so t = 10 s [1] Examiner insight: Pick the formula that avoids the unknown you do not need; listing s, u, v, a, t first shows why.

7. (a) At the top v = 0: 0 = 21² − 2(9.8)s [1]; s = 441/19.6 = 22.5 m [1]; height above ground = 22.5 + 3.5 = 26.0 m [1] (b) Take up as positive: −3.5 = 21t − 4.9t² [1]. Then 4.9t² − 21t − 3.5 = 0 gives t = (21 + √509.6)/9.8 = 4.45 s, rejecting the negative root. [1] Examiner insight: State your sign convention; the ground is below the start, so the displacement is −3.5.

8. (a) v = 0.8 × 15 = 12 m s⁻¹ [1]; decelerating time = 12/1.2 = 10 s [1] (b) Acceleration stage: ½(12)(15) = 90 m [1]; deceleration stage: ½(12)(10) = 60 m [1]. Constant stage: 12T = 1050 − 150, so T = 75 s [1] (c) Total time = 15 + 75 + 10 = 100 s [1]; average speed = 1050/100 = 10.5 m s⁻¹ [1] Examiner insight: A speed-time sketch keeps the areas organised. Average speed is total distance over total time, not the mean of the speeds.

9. (a) Stage 1: (12 − 4)/5 = 1.6 m s⁻² [1]. Stage 2: 0 [1]. Stage 3: (−6 − 12)/3 = −6 m s⁻² [1] (b) 12 − 6s = 0 after t = 8, so s = 2 and t = 10 [1] (c) Areas under the graph: ½(4 + 12)(5) = 40 [1]; 12 × 3 = 36; 8 to 10 gives ½(12)(2) = 12; 10 to 11 gives −½(6)(1) = −3 [1]. Displacement = 85 m [1] (d) Distance = 40 + 36 + 12 + 3 = 91 m [1] Examiner insight: Treat the region below the axis separately: adding it gives distance, subtracting it gives displacement.

10. (a) B starts from rest at t = 6, so it has travelled for (t − 6) s. [1] s = ½(1.5)(t − 6)² = 0.75(t − 6)² [1] (b) A’s distance is 6t. Level when 6t = 0.75(t − 6)² [1]. Then 8t = (t − 6)², so t² − 12t + 36 − 8t = 0, giving t² − 20t + 36 = 0. [1] (c) (t − 2)(t − 18) = 0 [1], so t = 2 or 18 [1]. t = 2 is invalid because B only passes O at t = 6, so t = 18. [1] (d) B has been moving for 12 s: v = 1.5 × 12 [1] = 18 m s⁻¹ [1] (e) The cars are modelled as particles. [1] Examiner insight: After solving a quadratic, say why you reject a root, tied to the situation. Here t is measured from A’s start, not B’s.

11. (a) r = 20(3i + 4j) [1] = (60i + 80j) m [1] (b) r = (60i + 80j) + (t − 20)(−4i + 6j) [1]. i: 60 − 4t + 80 = 140 − 4t. j: 80 + 6t − 120 = 6t − 40. Hence shown. [1] (c) Due north: 140 − 4t = 0 [1], so t = 35 s. j component = 6(35) − 40 = 170 [1], so the distance is 170 m. [1] (d) Speed = √((−4)² + 6²) [1] = √52 = 7.21 m s⁻¹ [1] Examiner insight: In a show-that question, write the substitution and expanded components. The second stage starts at t = 20.

Where marks are usually lost

• Naming a modelling assumption without its effect on the motion.
• Giving an angle without saying which direction it is measured from.
• Using sin where cos is needed, because no triangle was drawn.
• Subtracting velocities in the wrong order when finding acceleration.
• Confusing speed with velocity, or distance with displacement.
• Using time from the start of the whole journey in a formula for a later stage.
• Counting an area below the axis as positive when displacement is wanted.
• Forgetting to reject a root that is physically impossible.
• Leaving the sign convention unstated in vertical motion.

Next steps

• Revise with the revision notes.
• Re-read the study guide for weak topics.
• Go to the course hub and the checklist.
• Try the free diagnostics.
• Book a free trial class on the trial page.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and Pure Mathematics, Specification, Issue 3, April 2019, Pearson Education Limited. Unit M1 sections 1.1, 2.1, 2.2 and 3.1.