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An object of mass 35 kg fell from a boat to the seabed - Edexcel - A-Level Physics - Question 18 - 2023 - Paper 1
Question 18
An object of mass 35 kg fell from a boat to the seabed. (a) The object reached terminal velocity as it fell. (i) Show that the drag force acting on the object at t... show full transcript
Worked Solution & Example Answer:An object of mass 35 kg fell from a boat to the seabed - Edexcel - A-Level Physics - Question 18 - 2023 - Paper 1
Step 1
(i) Show that the drag force acting on the object at terminal velocity was about 200 N.
Answer
To determine the drag force acting on the object, we first calculate the weight of the object and the upthrust experienced in the fluid:
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Calculate the weight (W):
W = mg = 35 ext{ kg} imes 9.81 ext{ m/s}^2 = 343 ext{ N}
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Calculate the volume (V) and upthrust (U):
The upthrust U is equal to the weight of the fluid displaced, which can be calculated using the formula:
U = ho V g
where
ho = density of seawater = 1.03 × 10^{3} kg/m³
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V = volume of object = 1.60 × 10^{-3} m³
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g = 9.81 m/s² (acceleration due to gravity)
U = 1.03 imes 10^{3} ext{ kg/m}^3 imes 1.60 imes 10^{-3} ext{ m}^3 imes 9.81 ext{ m/s}^2 = 16.6 ext{ N}
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Calculate the drag force (D):
At terminal velocity, the drag force is equal to the weight of the object minus the upthrust:
D = W - U = 343 ext{ N} - 16.6 ext{ N} = 326.4 ext{ N},
which does not match the required force of 200 N due to incorrect calculations or approximations in the values used. Adjustments in values might be needed.
Step 2
(ii) Determine the terminal velocity of the object.
Answer
The drag force D on the object can be represented using the formula:
D = kv²
We know that at terminal velocity, D equals the net force, thus:
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Substituting known values:
At terminal velocity, D = 2.2 N s^{-1} m^{-2} imes v²
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Setting D equal to the weight (since at terminal velocity, these balance):
181 ext{ N} = 2.2 ext{ N s}^{-1} ext{ m}^{-2} imes v^{2}
v^{2} = rac{181}{2.2} ext{ m}^2/{s}^2
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Calculating terminal velocity v:
v = ext{sqrt}igg( rac{181}{2.2}igg) ext{ m/s}
= 9.1 ext{ m/s}
Therefore, the terminal velocity of the object is approximately 9.1 m/s.
Step 3
(iii) Give two reasons why Stokes' law could not be used to calculate the terminal velocity of the object.
Answer
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Object might not be spherical: Stokes' law assumes that the object is spherical, which may not be the case for all falling objects.
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Flow might not be laminar: The flow of fluid around the object may not be laminar due to the object's size or speed, violating the conditions under which Stokes' law applies.
Step 4
(b) Explain why the lift bag and object reached a maximum velocity.
Answer
The lift bag and object reached a maximum velocity due to the balance between the upward thrust provided by the lift bag and the downward forces such as weight and drag:
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Drag Force Increases: As the lift bag ascends, the drag force acting against it increases with speed. This drag force continues to rise until it equals the total weight of the lift bag and object.
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Equilibrium Condition: Once the upward thrust equals the downward gravitational force plus drag, the net force becomes zero, resulting in no further acceleration. Hence, the object and lift bag stop accelerating and maintain a constant maximum velocity.