Edexcel A-Level Physics 9PH0
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Materials
Terminal velocity of a sphere falling through a liquid
- 1 thing to know
- Papers 2 and 3
Overview
Recall questions on the three forces acting, the condition at terminal velocity, the expression that follows, and how terminal velocity depends on radius.
What you need to know
Answer each one before you reveal it.
Questions from this subtopic
Each one with the model answer underneath.
- 1
How does the viscous drag on a sphere change as it speeds up in a liquid?
Model answer
The drag increases in direct proportion to the speed, as long as the flow stays laminar.
- 2
For a sphere falling through a liquid, which forces balance at terminal velocity?
Model answer
Terminal velocity is reached when the upward viscous drag and the upthrust together equal the weight, so the resultant force and the acceleration are zero.
- 3
What three forces act on a small sphere falling through a liquid, and what is the condition for terminal velocity?
Model answer
Weight acts downwards while upthrust and viscous drag act upwards, and at terminal velocity weight = upthrust + viscous drag.
- 4
What expression can be derived for the terminal velocity of a sphere of density ρₛ falling through a liquid of density ρₗ?
Model answer
Setting (4/3)πr³ρₛg = (4/3)πr³ρₗg + 6πηrv and rearranging gives v = 2r²g(ρₛ − ρₗ)/(9η).
- 5
How does the terminal velocity of a small sphere in a liquid depend on its radius?
Model answer
The terminal velocity is proportional to r², because the weight grows as r³ while the drag grows only as r.
- 6
Two ball bearings of the same material are dropped into the same oil, one having twice the radius of the other. What is the ratio of their terminal velocities?
Model answer
Since terminal velocity is proportional to r², the larger ball reaches four times the terminal velocity of the smaller one.
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