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Date November Example questions Marks available 2 Reference code EXN.1.SL.TZ0.9
Level Standard Level Paper Paper 1 (without calculator) Time zone Time zone 0
Command term Show that Question number 9 Adapted from N/A

Question

The following diagram shows a ball attached to the end of a spring, which is suspended from a ceiling.

The height, h metres, of the ball above the ground at time t seconds after being released can be modelled by the function ht=0.4cosπt+1.8 where t0.

Find the height of the ball above the ground when it is released.

[2]
a.

Find the minimum height of the ball above the ground.

[2]
b.

Show that the ball takes 2 seconds to return to its initial height above the ground for the first time.

[2]
c.

For the first 2 seconds of its motion, determine the amount of time that the ball is less than 1.8+0.22 metres above the ground.

[5]
d.

Find the rate of change of the ball’s height above the ground when t=13. Give your answer in the form pπqms-1 where p and q+.

[4]
e.

Markscheme

* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.

attempts to find h0       (M1)

h0=0.4cos0+1.8=2.2

2.2 (m) (above the ground)       A1

 

[2 marks]

a.

EITHER

uses the minimum value of cosπt which is -1       M1

0.4-1+1.8 (m)

 

OR

the amplitude of motion is 0.4 (m) and the mean position is 1.8 (m)         M1

 

OR

finds h't=-0.4πsinπt, attempts to solve h't=0 for t and determines that the minimum height above the ground occurs at t=1,3,        M1

0.4-1+1.8 (m)

 

THEN

1.4 (m) (above the ground)       A1

 

[2 marks]

b.

EITHER

the ball is released from its maximum height and returns there a period later       R1

the period is 2ππ=2 s       A1

 

OR

attempts to solve ht=2.2 for t       M1

cosπt=1

t=0,2,       A1

 

THEN

so it takes 2 seconds for the ball to return to its initial position for the first time       AG

 

[2 marks]

c.

0.4cosπt+1.8=1.8+0.22       (M1)

0.4cosπt=0.22

cosπt=22       A1

πt=π4,7π4       (A1)

 

Note: Accept extra correct positive solutions for πt.

t=14,74 0t2       A1

 

Note: Do not award A1 if solutions outside 0t2 are also stated.

the ball is less than 1.8+0.22 metres above the ground for 74-14(s)

1.5(s)       A1

  

[5 marks]

d.

EITHER

attempts to find h't        (M1)

 

OR

recognizes that h't is required        (M1)

 

THEN

h't=-0.4πsinπt        A1

attempts to evaluate their h'13        (M1)

h'13=-0.4πsinπ3

=0.2π3 ms-1        A1

 

Note: Accept equivalent correct answer forms where p. For example, -15π3.

 

[4 marks]

e.

Examiners report

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b.
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d.
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e.

Syllabus sections

Topic 3— Geometry and trigonometry » SL 3.7—Circular functions: graphs, composites, transformations
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Topic 3— Geometry and trigonometry

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