A-LevelMathematicsMechanics9709/51 · Oct/Nov 2017 · Q47 Marks

A particle P is projected with speed 25 m s⁻¹ at an angle of 30° above the horizontal from a point O on horizontal ground. At time t s after projection the horizontal and vertically upwards displacements of P from O are x m and y m respectively.

(i)Express x and y in terms of t and hence show that the equation of the trajectory of P is y = x/√3 - 4x²/375[4]
(ii)Find the horizontal distance between the two points at which P is 5 m above the ground.[3]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2017). The mark or mark type is shown in brackets.

[B1] x = (25cos30)t Horizontal motion.
[B1] y = (25sin30)t -gt²/2 Vertical motion.
[M1] y = (25sin30)x / (25cos30) -5[x/(25cos30)]² Attempts to eliminate t.
[A1] y = x/√3 - 4x²/375 AG
[M1] 5 = x/√3 − 4x²/375 (leads to 4x² – 216.5x + 1875 = 0) Substitutes y = 5 into the trajectory equation.
[A1] x = 43.3,10.8 Solves the quadratic equation.
[A1] Distance = 43.3 - 10.8 = 32.5 m

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Topic: Mechanics, A-Level Mathematics (syllabus 9709). Structured question, 7 marks.
Source: A-Level Mathematics, Oct/Nov 2017 Paper 5 Variant 1, question 4.

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