A slide in a playground descends at a constant angle of 30° for 2.5 m. It then has a horizontal section in the same vertical plane as the sloping section. A child of mass 35 kg, modelled as a particle P, starts from rest at the top of the slide and slides straight down the sloping section. She then continues along the horizontal section until she comes to rest (see diagram). There is no instantaneous change in speed when the child goes from the sloping section to the horizontal section. The child experiences a resistance force on the horizontal section of the slide, and the work done against the resistance force on the horizontal section of the slide is 250 J per metre.

Mark scheme

Marking points from the official mark scheme (May/June 2021). The mark or mark type is shown in brackets.

1. [M1] PE =35g×2.5 sin 30
2. [M1] ½ ×35v² = 35g×2.5sin30 Use of conservation of energy, 2 terms, correct dimensions
3. [A1] v=5 ms⁻¹
4. Alternative method for Question 7(a)(i)
5. [M1] mg sin 30 = ma leading to a=5 For applying Newton's 2nd law down the plane, 2 terms, correct dimensions
6. [M1] v² = 0+2×5×2.5 For using v² = u² + 2as, using their a ≠ ±g
7. [A1] v=5 ms⁻¹
8. [M1] ½ ×35×5² = 250d Use of work-energy from the bottom of the slide until motion stops, 2 terms, correct dimensions, using their v
9. [A1] d=1.75 m
10. Alternative method for Question 7(a)(ii)
11. [M1] 35g×2.5 sin 30 = 250d Use of work-energy from the start until motion stops, 2 terms, correct dimensions.
12. [A1] d =1.75 m
13. Alternative method for Question 7(a)(ii)
14. [M1] -250=35a leading to a = -50/7=-7.14 Newton's 2nd law on the horizontal section with resistance = 250 N to find a and use v² =u² + 2as with v=0, u=5 and s=d.
15. [A1] d=1.75m
16. [B1] ½ ×35v² = 250×1.05 [v² =15] Either use the correct work energy equation for motion on the horizontal section or use the fact that the frictional force on the horizontal section is 250 N in order to set up an equation that would lead to finding the speed at the bottom of the slide.
17. [B1] R = 35g cos30[=303.11]
18. [M1] v² =0+2×a×2.5=15 leading to a = 3 For using v² = u² + 2as, with their v² to set up an equation that would lead to finding a.
19. [M1] 35g sin 30-F = 35a or [175-F=35a] For using Newton's 2nd law down the slope with correct dimensions. or For using energy equation, 3 relevant terms with correct dimensions. 35g×2.5sin30=Fx2.5+½×35×15 [437.5=Fx2.5+262.5]
20. [M1] F = μ×R For using F = µR, where R is a component of 35g.
21. [A1] μ = 0.231 Allow μ = 2√3/15 OE