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Edexcel IAL M1 June 2023 mark scheme and question paper (WME01/01)

Mechanics 1 (M1), sat in June 2023. The buttons open Pearson's own files.

The official paper and mark scheme

Sit it under exam conditions: open the paper, press Start timer, press Finish when time is up, then mark it against the mark scheme.

Marked it? Worked solutions to 3 similar M1 questions are just below, then every M1 topic to practise.

3 M1 questions like these, with worked solutions

Written and checked by us, not taken from the Pearson paper. Try each one, then tap "Worked solution" to see every step and its marks.

  1. Kinematics · 5 marks

    A particle moving with constant acceleration passes three points A, B and C, in that order, on a straight line. The distance AB is 20 m and the distance AC is 84 m. The speed of the particle at B is 12 m s−1 and its speed at C is 20 m s−1. Find the acceleration of the particle and its speed at A.

    Worked solution
    1. M1 for BC=84−20=64 m.
    2. M1 for v2=u2+2as from B to C: 202=122+2a(64).
    3. A1 for 128a=256⇒a=2 m s−2.
    4. M1 for v2=u2+2as from A to B: 122=uA2+2(2)(20).
    5. A1 for uA2=64⇒uA=8 m s−1.
  2. Vectors (Mechanics context) · 7 marks

    At time t hours after 09:00, a boat P has position vector ((1+2t)i+(5−t)j) km relative to a fixed origin. A lighthouse L is at the point with position vector (8i+7j) km.

    (a) Find PL→ in terms of t. [2]

    (b) Show that PL2=5t2−24t+53. [2]

    (c) Find the shortest distance between the boat and the lighthouse, and the time at which it occurs. [3]

    Worked solution

    (a) PL→=(8i+7j)−((1+2t)i+(5−t)j) M1

    (7−2t)i+(2+t)j A1

    (b) PL2=(7−2t)2+(2+t)2 M1

    =49−28t+4t2+4+4t+t2=5t2−24t+53 A1 (cso)

    (c) 5t2−24t+53=5(t−2.4)2+24.2 (or any complete method for the minimum) M1

    t=2.4, i.e. 11:24 A1

    Shortest distance 24.2=4.92 km A1

  3. Momentum and Impulse · 4 marks

    A ball of mass 0.5 kg is dropped from rest from a height of 19.6 m above horizontal ground. It rebounds to a height of 4.9 m. Calculate the magnitude of the impulse exerted by the ground on the ball. (Take g=9.8 m s−2)

    Worked solution
    1. M1 for finding the velocity just before impact using v2=u2+2as: v12=0+2(9.8)(19.6)=19.62⟹v1=19.6 m s−1 downwards.
    2. M1 for finding the velocity just after impact using v2=u2+2as: 0=v22−2(9.8)(4.9)⟹v22=9.82⟹v2=9.8 m s−1 upwards.
    3. M1 for calculating the impulse: I=m(v2−v1)=0.5(9.8−(−19.6)).
    4. A1 for evaluating I=0.5(29.4)=14.7 N s.

More on these topics: Kinematics · Vectors (Mechanics context) · Momentum and Impulse

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What's on this paper

What this paper is like

WME01/01: question by question

Labels and topics are ours, read from the WME01/01 paper (checked 2026-11-01); marks as printed. Tap a number to practise that topic.

  1. Q1Two particles colliding (7 marks) · M1 Momentum and Impulse
  2. Q2Velocity and position with vectors (10 marks) · M1 Vectors (Mechanics context)
  3. Q3Book falling from a shelf (8 marks) · M1 Kinematics
  4. Q4Beam suspended by two ropes (12 marks) · M1 Moments
  5. Q5Speed-time graph of a journey (11 marks) · M1 Kinematics
  6. Q6Particle on a rough surface pulled at an angle (7 marks) · M1 Statics of a Particle, M1 Forces and Newton's Laws
  7. Q7Car towing a trailer up a slope (11 marks) · M1 Dynamics of a Particle
  8. Q8Motion across a square floor with vectors (9 marks) · M1 Vectors (Mechanics context)

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Where is the M1 June 2023 mark scheme?

The mark scheme button above opens Pearson's own copy.

What are the grade boundaries for IAL M1 June 2023?

On Pearson's grade boundaries page (opens in a new tab), under June 2023.