Mechanics 1

Question bank

Real questions, real marking

2 real questions this paper has actually set, drawn only from genuine Pearson question papers. Each one links to its real mark scheme, its real examiner report where one has been published, and our own level-by-level breakdown of how marks are actually earned — write your own answer first, then compare.

Jan 2025 · Q1(b)

4 marks

Combining a fourth unknown force with a known resultant DIRECTION, via a ratio equation in one unknown

A particle of mass 2.5 kg moves on a smooth horizontal plane under the action of three horizontal forces, F1, F2 and F3, where F1 = (6i + 8j) N, F2 = (–16i + 2j) N, F3 = (–2i + 8j) N. A fourth force, F4 = (pi + pj) N, where p is a constant, is added. The resultant of the four forces acts in a direction that is parallel to the vector (7i + 2j). Find the value of p.

Our analysis

A particle of mass 2.5 kg moves under three forces F1 = (6i+8j) N, F2 = (–16i+2j) N, F3 = (–2i+8j) N. A fourth force F4 = (pi+pj) N is added. The resultant of the four forces acts in a direction parallel to (7i+2j). Find the value of p.

4 marks available

  1. 01

    (−12i + 18j) + (pi + pj) = (−12+p)i + (18+p)j

    Finds the resultant force of all four forces (the three-force resultant from part (a), −12i+18j, plus the new pi+pj) and collects the i and j components into one vector in p only.

    M1
  2. 02

    (−12+p) / (18+p) = 7 / 2 [equivalently: (−12+p)i+(18+p)j = λ(7i+2j), equate components, eliminate λ]

    Uses the given DIRECTION 7i+2j — not the resultant's own literal value — together with the i and j components of the resultant to form a ratio equation in p only. This is the step the real examiner report identifies as the one a minority of candidates skip: 'parallel to' constrains the ratio of components, not their exact size.

    M1
  3. 03

    2(−12+p) = 7(18+p) ⟹ −24 + 2p = 126 + 7p

    Correct unsimplified equation in p only, from cross-multiplying the ratio.

    A1
  4. 04

    −5p = 150 ⟹ p = −30

    Correct final answer.

    A1

Jan 2025 · Q7(c)

6 marks

Two-stage kinematics after a pulley system's string goes slack — the SAME particle under two different accelerations in sequence

One end of a light inextensible string is attached to a particle A of mass 2m. The other end is attached to a particle B of mass 3m. The string passes over a small smooth fixed pulley. The string is taut and both straight parts of the string are vertical. Both particles are held at a distance 0.5 m above a horizontal surface. The system is released from rest and B moves downwards. In the subsequent motion, A does not reach the pulley and B does not rebound after it strikes the surface. Find the time from the instant when the system is released from rest to the instant when A first reaches a height of 1.06 m above the surface.

Our analysis

Particles A (2m) and B (3m) are connected over a pulley and held 0.5 m above the ground, then released from rest; B moves down and lands, the string then goes slack, and A does not rebound off the pulley. Using the results already established earlier in this question — the system's acceleration while the string is taut is a = g/5 (part (a)), and B (and therefore A) has speed 1.4 m/s the instant B lands (part (b)) — find the time from release to the instant A first reaches a height of 1.06 m above the surface.

6 marks available

  1. 01

    Stage 1 (string taut, both particles move together at a = g/5): 0.5 = ((0 + 1.4)/2) × t1

    Forms a complete equation for the time to travel the first 0.5 m, using the candidate's OWN acceleration for the pulley system (g/5) — not g, and not an assumed value.

    M1
  2. 02

    t1 = 5/7 = 0.714285... s

    Correct equation in t1 only. The real mark scheme flags that solving it gives exactly 5/7 — the examiner report notes almost no candidate left it as this exact fraction.

    A1
  3. 03

    Stage 2 (B has landed, string now slack; A continues upward under gravity ALONE — a = −g, not g/5 — from its stage-1 height of 1.0 m, needing a further 0.06 m to reach 1.06 m): 0.06 = 1.4 t2 − 4.9 t2²

    Forms a complete equation for the time to cover the FURTHER 0.06 m (not the whole 0.56 m, and not treating u or v as 0), using g — not g/5 — for A's deceleration once the string is slack. This is the exact modelling step the real examiner report says a large number of candidates never reached.

    M1
  4. 04

    0.06 = 1.4 t2 − 4.9 t2² [ft on the candidate's own value of 1.4 from part (b)]

    Correct equation in t2 only, follow-through on the candidate's own answer to part (b) rather than requiring the exact printed value.

    A1ft
  5. 05

    t2 = 0.0525069... s = (5 − √10) / 35 [the other root, 0.233 s, is rejected — it would put A back below the height reached during stage 1]

    Correct positive root of the quadratic for t2. The real mark scheme explicitly flags the other root (0.233 s) as one that must be rejected, and notes this mark can also be earned implicitly by a correct final total time.

    A1
  6. 06

    Total time = t1 + t2 = 0.714285... + 0.052507... = 0.77 s (0.767 s to 3 s.f.)

    Correct total time, combining both stages, given to 2 or 3 significant figures as the front of the question paper requires.

    A1