A particle P of mass 0.2 kg is released from rest at a point O above horizontal ground. At time t s after its release the velocity of P is v m s⁻¹ downwards. A vertically downwards force of magnitude 0.6t N acts on P. A vertically upwards force of magnitude ke⁻ᵗ N, where k is a constant, also acts on P.

Mark scheme

Marking points from the official mark scheme (May/June 2018). The mark or mark type is shown in brackets.

1. [M1] (i) 0.2dv/dt = 0.2 g + 0.6t - ke⁻ᵗ
2. [A1] (i) dv/dt = 10 + 3t - 5 ke⁻ᵗ
3. [M1] (ii) dv/dt = 10-5k e⁰ = 0
4. [M1] (ii) Attempts to solve the equation
5. [A1] (ii) k = 2
6. [M1] (iii) dv = ∫(10 + 3t – 5ke⁻ᵗ )dt
7. [A1] (iii) v = 10t + 3 t²/2 + 5 e⁻ᵗ + c, v = 0, t = 0 so c = − 5
8. [M1] (iii) v = 10t + 3 t²/2 + 5 e⁻ᵗ - 5
9. [M1] (iii) dx = ∫(10t + 3 t²/2 + 5 e⁻ᵗ – 5)dt
10. [A1] (iii) x = 5t² + t³/2 – 5 e⁻ᵗ – 5t + c
11. [M1] (iii) x = 0, t = 0, so c = 5 and substitutes t = 2
12. [A1] (iii) Height = 18.3 m