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Date May 2012 Marks available 2 Reference code 12M.2.hl.TZ2.8
Level HL only Paper 2 Time zone TZ2
Command term What Question number 8 Adapted from N/A

Question

Each time a ball bounces, it reaches 95 % of the height reached on the previous bounce.

Initially, it is dropped from a height of 4 metres.

What height does the ball reach after its fourth bounce?

[2]
a.

How many times does the ball bounce before it no longer reaches a height of 1 metre?

[3]
b.

What is the total distance travelled by the ball?

[3]
c.

Markscheme

height \( = 4 \times {0.95^4}\)     (A1)

= 3.26 (metres)     A1

[2 marks]

a.

\(4 \times {0.95^n} < 1\)     (M1)

\({0.95^n} < 0.25\)

\( \Rightarrow n > \frac{{\ln 0.25}}{{\ln 0.95}}\)     (A1)

\( \Rightarrow n > 27.0\) 

Note: Do not penalize improper use of inequalities.

 

\( \Rightarrow n = 28\)     A1 

Note: If candidates have used n – 1 rather than n throughout penalise in part (a) and treat as follow through in parts (b) and (c).

 

[3 marks]

b.

METHOD 1

recognition of geometric series with sum to infinity, first term of \(4 \times 0.95\) and common ratio 0.95     M1

recognition of the need to double this series and to add 4     M1

total distance travelled is \(2\left( {\frac{{4 \times 0.95}}{{1 - 0.95}}} \right) + 4 = 156{\text{ (metres)}}\)     A1

[3 marks] 

Note: If candidates have used n – 1 rather than n throughout penalise in part (a) and treat as follow through in parts (b) and (c).

 

METHOD 2

recognition of a geometric series with sum to infinity, first term of 4 and common ratio 0.95     M1

recognition of the need to double this series and to subtract 4     M1

total distance travelled is \({\text{2}}\left( {\frac{4}{{1 - 0.95}}} \right) - 4 = 156{\text{ (metres)}}\)     A1

[3 marks]

c.

Examiners report

The majority of candidates were able to start this question and gain some marks, but only better candidates gained full marks. In part (a) the common error was to assume the wrong number of bounces and in part (b) many candidates lost marks due to rounding the inequality in the wrong direction. Part (c) was found difficult with only a limited number recognising the need for the sum to infinity of a geometric sequence and many of those did not recognise how to link the sum to infinity to the total distance travelled.

a.

The majority of candidates were able to start this question and gain some marks, but only better candidates gained full marks. In part (a) the common error was to assume the wrong number of bounces and in part (b) many candidates lost marks due to rounding the inequality in the wrong direction. Part (c) was found difficult with only a limited number recognising the need for the sum to infinity of a geometric sequence and many of those did not recognise how to link the sum to infinity to the total distance travelled.

b.

The majority of candidates were able to start this question and gain some marks, but only better candidates gained full marks. In part (a) the common error was to assume the wrong number of bounces and in part (b) many candidates lost marks due to rounding the inequality in the wrong direction. Part (c) was found difficult with only a limited number recognising the need for the sum to infinity of a geometric sequence and many of those did not recognise how to link the sum to infinity to the total distance travelled.

c.

Syllabus sections

Topic 1 - Core: Algebra » 1.1 » Arithmetic sequences and series; sum of finite arithmetic series; geometric sequences and series; sum of finite and infinite geometric series.
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