A-LevelMathematicsMechanics9709/42 · Oct/Nov 2025 · Q711 Marks

Four coplanar forces of magnitudes 20 N, 30 N, 15 N and 25 N act at a point P in the directions shown in Fig. 7.1. The forces act in a vertical plane. The resultant of these forces has magnitude S N and acts at an angle αº below the horizontal as shown in Fig. 7.2. [Figure 7.1] [Figure 7.2]

(a)Find the value of S and the value of α.[6]
(b)A small ring of mass 0.6 kg is threaded on a rough straight horizontal wire. The four forces shown in Fig. 7.1 act on the ring and are in the same vertical plane as the wire. The ring starts from rest and takes 3 s to travel a distance of 2 m along the wire. Find the coefficient of friction between the ring and the wire.[5]

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Mark scheme

Marking points from the official mark scheme (Oct/Nov 2025). The mark or mark type is shown in brackets.

[*M1] For attempt to resolve in any direction to form an expression or equation; Correct number of terms; allow sign errors and sin/cos mix ONLY.
[A1] [X =]±(30cos30+15cos45-20cos20); OE. +17.79351142...
[A1] [Y =]±(25+15sin45 – 20 sin 20 – 30 sin 30); OE. +13.76619885...
[DM1] S= √(their17.7935...)² + (their13.76661...)²; For attempt to find force from equations with the correct number of relevant terms (only sign errors and sin/cos mix allowed).
[DM1] α = tan⁻¹ (their 13.7661... / their 17.7935...); Allow reciprocal in tan⁻¹ only. For attempt to find angle from equations with the correct number of relevant terms (only sign errors and sin/cos mix allowed).
[A1] [S=] 22.5 [α =]37.7; S=22.497050406... α=37.72782933... Condone α=37.8. Allow S = -22.5 becoming 22.5 with no explanation. Allow α=-37.7 becoming 37.7 with no explanation. A0 for α=-37.7 ONLY, A0 for S = −22.5 ONLY
[*M1] (their17.7935...)– F = 0.6a; Using N2L with correct number of dimensionally correct terms, allow sign errors, allow sin/cos mix. their 17.7935 must have come from horizontal component from 7(a) with only sign errors and sin/cos mix allowed. May be using their F and/or their a which may be incorrect.
[*M1] 2=0+½ax3²; Use of s = ut + ½at² or other complete method to get an equation in a, using s = 2,u=0 and t=3.
[*B1FT] R = their (13.7661...)+0.6g [=19.7661...]; their 13.7661 must have come from vertical component from 7(a) with only sign errors and sin/cos mix allowed.
[DM1] (their 17.7935)...-μ×19.7661=0.6× (their 4/9); Dependent on all previous marks. Use of F = μR to get an equation in μ only.
[A1] μ = 0.887; 0.886711... Condone 0.886. Condone 0.888.