A light string AB is fixed at A and has a particle of weight 80N attached at B. A horizontal force of magnitude PN is applied at B such that the string makes an angle θ° to the vertical [Figure 5.1].

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2023). The mark or mark type is shown in brackets.

1. [M1] Attempt to resolve in one direction and form equation. Must have correct number of relevant terms (forces must have components as required). Allow sin/cos mix. Allow sign errors. If only one equation shown and it involves 32, it must be 32, not P.
2. [A1] Tsino=32 and Tcos0=80 or 0=80sin0-32cose and T=80cos0+32sin0 For both horizontal and vertical, or both parallel and perpendicular.
3. [M1] Attempt to solve for T or θ Must get to T or θ; e.g. T = √32² +80² or θ = tan⁻¹(32/80). Condone, e.g. θ = tan⁻¹(80/32). Must come from equations with correct number of relevant terms.
4. [A1] T=86.2 [N 86.1626....] or 16√29 or √7424 and θ=21.8 [21.801...] For both.
5. [M1] [Alternative method using triangle of forces for 5(a)] T² = 80² +32²[-2×80×32cos90] or Tsinθ=32 or Tcosθ=80 For any of the five; allow sign errors.
6. [A1] 80 tan θ=32 or T =80cosθ+32sinθ oe For any two equations.
7. [M1] Attempt to solve for T or θ Must get to T or θ; e.g. T = √32² +80² or θ = tan⁻¹(32/80).
8. [A1] T=86.2 [N 86.1626....] or 16√29 or √7424 and θ=21.8 [21.801...] For both.
9. [M1] [Alternative Triangle of forces method using sine rule for 5(a)] T/sin90 = 32/sin(θ) = 80/sin(90-θ) For any two.
10. [A1] For all three.
11. [M1] Attempt to solve for T or θ e.g. θ = tan⁻¹(32/80)
12. [A1] T=86.2 [N 86.1626....] or 16√29 or √7424 and θ=21.8 [21.801...] For both.
13. [M1] [For 5(b)] Attempt to resolve in one direction and form equation. Must have correct number of relevant terms (forces must have components as required). Allow sin/cos mix. Allow sign errors. Must use 120, not T.
14. [A1] 120sinθ=P and 120cosθ=80 or 0=80sinθ-Pcosθ and 120=80cosθ+Psinθ For both horizontal and vertical, or both parallel and perpendicular.
15. [M1] Attempt to solve for P or θ Must get to P or θ; e.g. P=√120²-80² or θ = cos⁻¹(80/120). Must come from equations with correct number of relevant terms.
16. [A1] P=89.4 [89.4427] or 40√5 or √8000 θ=48.2 [48.1896...] For both; allow P=89.5 (from 120sin48.2).
17. [M1] [Alternative method using triangle of forces for 5(b)] 120² = P² +80² [-2×80×Pcos90] or 120sinθ=P or 120cosθ=80 For any of the five; allow sign errors.
18. [A1] 80 tan θ = P or 120=80cosθ+Psinθ oe For any two equations.
19. [M1] Attempt to solve for P or θ Must get to P or θ; e.g. P=√120²-80² or θ = cos⁻¹(80/120) , oe.
20. [A1] P=89.4 [89.4427] or 40√5 or √8000 θ=48.2 [48.1896...] For both; allow P=89.5 (from 120sin48.2).
21. [M1] [Alternative Triangle of forces method using sine rule for 5(b)] 120/sin90 = P/sin(θ) = 80/sin(90-θ) For any two.
22. [A1] For all three.
23. [M1] Attempt to solve for P or θ e.g. θ=90-sin⁻¹(80/120) or θ = cos⁻¹(80/120).
24. [A1] P=89.4 [89.4427] or 40√5 or √8000 θ=48.2 [48.1896...] For both; allow P=89.5 (from 120sin48.2).