The diagram shows the vertical cross-section ABC of a rough waterslide. The section AB is a straight line of length 5m inclined at an angle of θ to the horizontal, where sinθ = 0.8. The point B is 2.5m above the level of C. A man of mass 80kg, modelled as a particle, slides down the waterslide, starting from rest at A. The coefficient of friction between the man and the straight section of the waterslide is 0.1. [Figure with waterslide]

Mark scheme

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

1. [B1] R = 80g×0.6[=480] Allow 80g × cos 53 (or better for θ=53.1301...)
2. [*M1] F=0.1×80g×0.6[=48] For use of F=0.1R with R = 80g×0.6 or R=80g×0.8 or equivalent with cos53 or sin53 or better.
3. [*M1] 80g×0.8-F = 80a [⇒ a=7.4] For attempt to find an equation for a using N2L with 3 terms; allow sign errors; allow sin/cos mix for the weight component with cos53 or sin53 or better. Allow F or their F.
4. [DM1] v² = (0+)2×(theira)×5 For attempt to find v² or v using their positive a.
5. [A1] Velocity = 8.60 m s⁻¹ Allow √74 but A0 for 8.6 if 3sf or better (8.6023...) answer not seen.
6. [B1FT] Change in KE = ±(1/2×80×11² - 1/2×80×(their v²)) FT their v² from part (a).
7. [B1] Change in PE = ±80g×2.5 [=±2000] Including PE from A is B0.
8. [M1] 80g×2.5-W = 1/2×80×11² - 1/2×80×(their v²) [2000-W = 4840–40×(theirv²)] For attempt at work energy equation. 4 relevant terms; allow sign errors but must be dimensionally correct. M0 if using change in PE from A to C.
9. [A1] Work done = 120 J Allow 118(.4) from using 8.6(0) from part (a). Working must lead to a positive answer for the work done (so-120 oe is A0).