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Date May Specimen paper Marks available 4 Reference code SPM.2.AHL.TZ0.7
Level Additional Higher Level Paper Paper 2 Time zone Time zone 0
Command term Find Question number 7 Adapted from N/A

Question

An object is placed into the top of a long vertical tube, filled with a thick viscous fluid, at time  t = 0  seconds.

Initially it is thought that the resistance of the fluid would be proportional to the velocity of the object. The following model was proposed, where the object’s displacement, x , from the top of the tube, measured in metres, is given by the differential equation

d 2 x d t 2 = 9.81 0.9 ( d x d t ) .

The maximum velocity approached by the object as it falls is known as the terminal velocity.

An experiment is performed in which the object is placed in the fluid on a number of occasions and its terminal velocity recorded. It is found that the terminal velocity was consistently smaller than that predicted by the model used. It was suggested that the resistance to motion is actually proportional to the velocity squared and so the following model was set up.

d 2 x d t 2 = 9.81 0.9 ( d x d t ) 2

At terminal velocity the acceleration of an object is equal to zero.

By substituting  v = d x d t into the equation, find an expression for the velocity of the particle at time t . Give your answer in the form v = f ( t ) .

[7]
a.

From your solution to part (a), or otherwise, find the terminal velocity of the object predicted by this model.

[2]
b.

Write down the differential equation as a system of first order differential equations.

[2]
c.

Use Euler’s method, with a step length of 0.2, to find the displacement and velocity of the object when  t = 0.6 .

[4]
d.

By repeated application of Euler’s method, find an approximation for the terminal velocity, to five significant figures.

[1]
e.

Use the differential equation to find the terminal velocity for the object.

[2]
f.

Use your answers to parts (d), (e) and (f) to comment on the accuracy of the Euler approximation to this model.

[2]
g.

Markscheme

d v d t = 9.81 0.9 v         M1

1 9.81 0.9 v d v = 1 d t         M1

1 0.9 ln ( 9.81 0.9 v ) = t + c        A1

9.81 0.9 v = A e 0.9 t        A1

v = 9.81 A e 0.9 t 0.9        A1

when  t = 0 v = 0 hence  A = 9.81        A1

v = 9.81 ( 1 e 0.9 t ) 0.9

v = 10.9 ( 1 e 0.9 t )        A1

[7 marks]

a.

either let t tend to infinity, or  d v d t = 0         (M1)

v = 10.9        A1

[2 marks]

b.

d x d t = y        M1

dy d t = 9.81 0.9 y 2       A1

[2 marks]

c.

x n + 1 = x n + 0.2 y n y n + 1 = y n + 0.2 ( 9.81 0.9 ( y n ) 2 )        (M1)(A1)

x = 1.04 d x d t = 3.31        (M1)A1

[4 marks]

d.

3.3015      A1

[1 mark]

e.

0 = 9.81 0.9 ( v ) 2      M1

v = 9.81 0.9 = 3.301511 ( = 3.30 )      A1

[2 marks]

f.

the model found the terminal velocity very accurately, so good approximation        R1

intermediate values had object exceeding terminal velocity so not good approximation        R1

[2 marks]

g.

Examiners report

[N/A]
a.
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b.
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c.
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d.
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e.
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f.
[N/A]
g.

Syllabus sections

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