A small remote-controlled model aircraft has two propellers, each of diameter 16cm. Fig. 3.1 is a side view of the aircraft when hovering. [Figure 3.1] Air is propelled vertically downwards by each propeller so that the aircraft hovers at a fixed position. The density of the air is 1.2kgm⁻³. Assume that the air from each propeller moves with a constant speed of 7.6ms⁻¹ in a uniform cylinder of diameter 16cm. Also assume that the air above each propeller is stationary.
- (a)Show that, in a time interval of 3.0s, the mass of air propelled downwards by one propeller is 0.55 kg. [3][3]
- (b)(i)Calculate: the increase in momentum of the mass of air in (a) increase in momentum = ............................................. Ns [1][1]
- (b)(ii)the downward force exerted on this mass of air by the propeller. force = .................................................... N [1][1]
- (c)(i)State: the upward force acting on one propeller force = .................................................... N [1][1]
- (c)(ii)the name of the law that explains the relationship between the force in (b)(ii) and the force in (c)(i). [1][1]
- (d)Determine the mass of the aircraft. mass = .................................................... kg [1][1]
- (e)In order for the aircraft to hover at a very high altitude (height), the propellers must propel the air downwards with a greater speed than when the aircraft hovers at a low altitude. Suggest the reason for this. [1][1]
- (f)When the aircraft is hovering at a high altitude, an electric fault causes the propellers to stop rotating. The aircraft falls vertically downwards. When the aircraft reaches a constant speed of 22 m s⁻¹, it emits sound of frequency 3.0 kHz from an alarm. The speed of the sound in the air is 340 ms⁻¹. Determine the frequency of the sound heard by a person standing vertically below the falling aircraft. frequency = .................................................... Hz [2][2]
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Mark scheme
Marking points from the official mark scheme (Oct/Nov 2019). The mark or mark type is shown in brackets.
- [C1] ρ = m/V
- [C1] V = π × (0.16 / 2)² × 7.6 × 3.0 (= 0.458 m³)
- [A1] m = π × (0.16 / 2)² × 7.6 × 3.0 × 1.2 = 0.55 kg
- [A1] Δp = 0.55 × 7.6; = 4.2 N s
- [A1] F = 4.2/3.0 or 0.55 × 7.6 / 3.0; = 1.4 N
- [A1] F = 1.4 N
- [B1] Newton's third law (of motion)
- [A1] 2 x 1.4 = m x 9.81; m = 0.29 kg
- [B1] the density of air is less at high altitude
- [C1] f₀ = fsv / (v - vs); = 3000 × 340 / (340 – 22)
- [A1] = 3200 Hz
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