A particle of mass 0.3 kg is held at rest by two light inextensible strings. One string is attached at an angle of 60° to a horizontal ceiling. The other string is attached at an angle a° to a vertical wall [Figure X.X]. The tension in the string attached to the ceiling is 4N.

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2022). The mark or mark type is shown in brackets.

1. [M1] Attempt to resolve forces in horizontal and/or vertical directions. Correct number of terms (weight, 4N tension, T tension). Allow sin/cos mix, sign errors, or g missing.
2. [A1] Formulate vertical equilibrium equation: T cos a° + 0.3g - 4 sin 60° = 0 (or equivalent, e.g. T cos a° = 4 sin 60° - 0.3g, using g=10, T cos a° = 2√3 - 3 ≈ 0.464).
3. [A1] Formulate horizontal equilibrium equation: T sin a° - 4 cos 60° = 0 (or equivalent, e.g. T sin a° = 4 cos 60° = 2).
4. [M1] Attempt to solve the two equations for a (e.g. by dividing one by the other to get tan a). No missing/extra terms. Allow g missing. Must get to 'a ='.
5. [M1] Attempt to solve for T using the determined value of a. No missing/extra terms. Allow g missing. Must get to ‘T='.
6. [A1] Tension = 2.05 N (AWRT 2.05) and α = 76.9° (AWRT 76.9).
7. [M1] Alternative method (Triangle of Forces): Must use lengths 4 and 0.3g with a suitable angle. Allow g missing. T² = 4² + (0.3g)² - 2 × 4 × (0.3g) × cos 30°
8. [A1] T = 2.05 N (AWRT 2.05).
9. [M1] Alternative method (Sin Rule): Must have angle 30° and another angle in terms of a with correct numerators, but allow g missing. Their T / sin 30° = 4 / sin (180° - a) = 0.3g / sin (a - 30°).
10. [A1] Correct equation: T / sin 30° = 4 / sin a.
11. [A1] α = 76.9° (AWRT 76.9).