A particle P of mass 0.4kg is in limiting equilibrium on a plane inclined at 30° to the horizontal.

Mark scheme

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

1. [B1] Normal reaction force R = 0.4g cos 30°.
2. [M1] Apply Newton's Second Law along the plane in limiting equilibrium: 0.4g sin 30° – μ R = 0 (or 0.4g sin 30° – μ (0.4g cos 30°) = 0). For using F = µR. Allow sin/cos mix, sign errors, or g omitted.
3. [A1] μ = ¹⁄₃√3 (AG, exact answer only).
4. [SC B1] If zero scored, for stating angle of friction = 30° so μ = tan 30° = ¹⁄₃√3.
5. [M1] Apply Newton's Second Law up the plane: Applied force - component of weight - friction = mass × acceleration. So, 7.2 - 0.4g sin 30° - μR = 0.4a. Four terms. Second term must be a component of their weight. F (friction) ≠ 0 and F ≠ μ. Allow sin/cos mix, sign errors, or g omitted. F must be a numerical expression (using μ from part (a)).
6. [A1] Calculate acceleration a = 8 ms⁻².
7. [M1] For use of a constant acceleration formula (suvat) and solving for time t. e.g., s = ut + ½at². 1 = 0 + ½ × (their positive 8) × t². Ensure a≠±10, a≠±g. Allow if acceleration from part (a) was negative and its absolute value is used here.
8. [A1] Time = 0.5 s.