A particle is projected vertically upwards from horizontal ground. The speed of the particle 2 seconds after it is projected is 5 m s⁻¹ and it is travelling downwards.
- (a)Find the speed of projection of the particle.[2]
- (b)Find the distance travelled by the particle between the two times at which its speed is 10 m s⁻¹.[2]
Mark scheme
Marking points from the official mark scheme (Feb/Mar 2024). The mark or mark type is shown in brackets.
- [M1] For use of constant acceleration to get an equation in u only. Using v=-5, t=2_and a=±g.
- [A1] Speed = 15 m s⁻¹. Must be positive.
- [M1] For use of constant acceleration formula(e) to find distance travelled from height with speed 10 m s⁻¹ to maximum height with speed 0 ms⁻¹, a=±g. Must be a complete method, e.g. 0² = (their15)²-2gs₁ [⇒s₁ =11.25] and 10² = (their15)² -2gs₂ [⇒ s₂ = 6.25] or 10=(their15)-gt [⇒t=0.5], s₂ = 1/2 (their 15+10)×(their 0.5) and with an attempt at s₁ - s₂ [=5]. Energy method with 2 terms, dimensionally correct.
- [A1FT] Total distance = 10 m. FT their 15 m s⁻¹ if used.
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