IB PHYSICS HL(Higher level) – 2024 – Practice Questions- All Topics
Topic 2.2 – Forces
Topic 2 Weightage : 9 %
All Questions for Topic 2.3 – Kinetic energy ,Gravitational potential energy ,Elastic potential energy ,Work done as energy transfer ,Power as rate of energy transfer ,Principle of conservation of energy ,Efficiency
Question
A company delivers packages to customers using a small unmanned aircraft. Rotating horizontal blades exert a force on the surrounding air. The air above the aircraft is initially stationary.
The air is propelled vertically downwards with speed v. The aircraft hovers motionless above the ground. A package is suspended from the aircraft on a string. The mass of the aircraft is
0.95 kg and the combined mass of the package and string is 0.45 kg. The mass of air pushed downwards by the blades in one second is 1.7 kg.
(a) (i) State the value of the resultant force on the aircraft when hovering. [1]
(ii) Outline, by reference to Newton’s third law, how the upward lift force on the aircraft is achieved. [2]
(iii) Determine ν. State your answer to an appropriate number of significant figures. [3]
(iv) Calculate the power transferred to the air by the aircraft. [2]
(b) The package and string are now released and fall to the ground. The lift force on the aircraft remains unchanged. Calculate the initial acceleration of the aircraft. [2]
▶️Answer/Explanation
Ans:
1. a i zero
1 a ii
Blades exert a downward force on the air
air exerts an equal and opposite force on the blades «by Newton’s third law»
OR air exerts a reaction force on the blades «by Newton’s third law»
1 a iii
«lift force/change of momentum in one second»
=1.7 v 1.7 (0.95 + 0.45) × 9.81 v = 8.1 ms− = « » AND answer expressed to 2 sf only
1. a iv
ALTERNATIVE 1
power «= rate of energy transfer to the air = \(\frac{1}{2}\frac{\Delta m}{\Delta t}\)v2 » \(\frac{1}{2}\) × 1.7 ×8.12 = 56 «W»
ALTERNATIVE 2
Power« = Force x v ave» (0.95+0.45) × 9.81 × \(\frac{8.1}{2}\)
1 b
vertical force= lift force – weight OR = 0.45 ×9.81
OR = 4.4 «N» acceleration = \(\frac{0.45\times 9.81}{0.95}\)
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