A-LevelPhysicsKinematics9702/22 · Feb/Mar 2019 · Q211 Marks

Define: displacement

(a)(i)Define: displacement[1]
(a)(ii)Define: acceleration.[1]
(b)A man wearing a wingsuit glides through the air with a constant velocity of 47ms⁻¹ at an angle of 24° to the horizontal. The path of the man is shown in Fig. 2.1. The total mass of the man and the wingsuit is 85 kg. The man takes a time of 2.8 minutes to glide from point A to point B.
(b)(i)With reference to the motion of the man, state and explain whether he is in equilibrium.[2]
(b)(ii)Show that the difference in height h between points A and B is 3200 m.[1]
(b)(iii).1For the movement of the man from A to B, determine: the decrease in gravitational potential energy[2]
(b)(iii).2For the movement of the man from A to B, determine: the magnitude of the force on the man due to air resistance.[2]
(b)(iv)The pressure of the still air at A is 63 kPa and at B is 92 kPa. Assume the density of the air is constant between A and B. Determine the density of the air between A and B.[2]

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Mark scheme

Marking points from the official mark scheme (Feb/Mar 2019). The mark or mark type is shown in brackets.

[1] (a)(i) distance in a specified direction (from a point)
[1] (a)(ii) change in velocity / time (taken)
[1] (b)(i) constant velocity so no resultant force
[1] (b)(i) no resultant force so in equilibrium
[1] (b)(ii) (difference in height =) 47 × 2.8 × 60 × sin24° = 3200 m
[1] (b)(iii) 1 (△)Ε = mg(△)h
[1] (b)(iii) 1 = 85 x 9.81 × 3200 = 2.7 × 10⁶ J
[1] (b)(iii) 2 In terms of energy: work done = 2.7 × 10⁶ J
[1] (b)(iii) 2 force = 2.7 x 10⁶ / (47 × 2.8 × 60) = 340 N
[1] (b)(iii) 2 In terms of forces: component of weight along path = force due to air resistance
[1] (b)(iii) 2 force = 85 × 9.81 x sin24° = 340 N
[1] (b)(iv) (Δ)p = ρg(Δ)h (92 - 63) × 10³ = ρ × 9.81 × 3200
[1] (b)(iv) ρ = 0.92 kg m⁻³

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