A particle of mass 18 kg is on a plane inclined at an angle of 30° to the horizontal. The particle is projected up a line of greatest slope of the plane with a speed of 20 m s⁻¹.

Mark scheme

Marking points from the official mark scheme (May/June 2019). The mark or mark type is shown in brackets.

1. [B1] (PE gain =) 18gdsin30° or (KE loss =) ½ × 18 × 20²
2. [B1] (PE gain =) 18gdsin30° and (KE loss =) ½ × 18 × 20²
3. [M1] Energy equation (PE gain = KE loss): [18gdsin30° = ½ × 18 × 20²] or [18gh = ½ × 18 × 20²]
4. [A1] Distance up plane = 40 m
5. [B1] R = 18gcos30° (90√3 or 155.884...)
6. [M1] Use of F = μR: [F = 0.25(18gcos30°)] (45√3/2 or 38.971...)
7. [M1] Newton's Second Law (3 term equation, going up): [18gsin30° + 0.25(18gcos30°) = −18a → a = ...] (a = -7.165..)
8. [M1] Use of suvat to find s (going up): [0² = 20² + 2 x −7.165.. × s → s = ...]
9. [A1] s = 27.913...
10. [M1] Alternative Method 1 for 5(ii): Work done against friction (down plane): [WD = 45cos30° × 27.91...]
11. [M1] Alternative Method 1 for 5(ii): Work/Energy equation (down plane): [½ × 18v² = (18gsin30°) × 27.91.. – 45cos30° × 27.91...]
12. [M1] Alternative Method 2 for 5(ii): WD against friction (up and down): [WD = 2 x 45cos30° × 27.91...]
13. [M1] Alternative Method 2 for 5(ii): Uses KE loss = total WD against friction: [½ × 18 × 20² - ½ × 18v² = 2 × 45cos30° × 27.91...]
14. [A1] v = 12.6 ms⁻¹ (12.580...)