Cambridge Physics for the IB Diploma
Answers to Coursebook questions – Chapter 1.5
1 Let a be the side of the square and r the radius of the circle. Then 4 a

2π r a
The ratio of circle to square area is
π a r
2
2

π r
2
π
( )
2 r
2
π r
4
1 , so the circle has the greater
π
2 area.
.
2 Let a be the side of the cube and r the radius of the sphere. Then,
6 a
2 
4π r
2 a r
2π
.
3
The ratio of sphere to cube volume is
4π r
3 a
3
3
4π r
3
 r 3
3
2π
( ) 3/ 2
3

6

1 ,
π so the sphere has the greater volume.
3 We have cos x 1 x
2
2
and so x
 x
2
2
.
4 a The initial voltage V is such that
0 ln V
0
4 V
0
 e
4 
55 V . b When find
V
  t

7 s
V
2
0
.
27 V , ln V
 ln 27

3.29
. From the graph, when ln V

3.29
we c Since V

V
0 e
 t RC , taking logs, ln V
 ln V
0
 t
RC
, so a graph of ln V versus time gives a straight line with slope equal to

1
RC
. The slope of the given graph is approximately
 
0.10
.
Hence,

1
RC
 
0.10
1
0.10

C

1

6
2 10
6 
.
Copyright Cambridge University Press 2011. All rights reserved. Page 1 of 6

Cambridge Physics for the IB Diploma
5 We expect v
 v
T
(1 e
 at
) . The velocity approaches the terminal value v
T

0.2 m s

1 and so ln(0.2
 v )
 ln 0.2
 at v

0.2(1 e
. Graphing
 at
) . Hence, ln(0.2
 v ) v

0.2
versus t
 
0.2e
 at
and so
gives a straight line with slope
 a .
The slope of this line is v
  
0.5
t
0.2(1 e ) .
 
0.5
; hence a

0.5 s

1
, so the equation is
6 We expect L
 kM

, and so ln shown below. The slope is

.
L
 ln k
  ln M . A graph of ln L versus ln M is
Drawing a line of best fit gives:
Measuring the slope gives
 
3.4
.
Copyright Cambridge University Press 2011. All rights reserved. Page 2 of 6

Cambridge Physics for the IB Diploma
7 We can plot either y versus x
2.5
to get a straight line with slope c or ln y versus ln x with slope 2.5 and vertical intercept ln c . Doing the latter gives:
The vertical intercept is about 0.70, and so ln c

0.70
e 0.70

2.0
8 We can plot either y versus e x
2
or ln y versus x
2 .
9 S

(120

60) (5 3)

180 8 . D

(120 60) (5 3)

60 8 .
Copyright Cambridge University Press 2011. All rights reserved. Page 3 of 6

Cambridge Physics for the IB Diploma
10 a Q

Q
0
 
Q . Q
0

20

10
2.0
.
 a

1

20
0.05
and
 b

1

10
0.10
. a
0 b
0
Hence,

Q

 a


Q a b
0 0 0 b

 
Q
0

0.05 0.10

0.15
0.15

2.0 0.15

0.30
. Hence,
. Hence, Q

2.0

0.3
. (The uncertainty is expressed in 1 significant figure.) b Q

Q
0
 
Q . Q
0
2 20 3 15 85
Q
.
2 a 3 b 2 2 3 3 13 10 . Hence, Q

85 10

80 10 . (The uncertainty is expressed in 1 significant figure. The uncertainty is large and affects the hundreds digit of the answer. Hence there is no point in quoting the answer in the units digit, hence the rounding.) c Q

Q
0
 
Q . Q
   
0
50 2 24 2 . Q a 2 b 1 3 0.5

2.5

2 .
Hence, Q 2 2 . (The answer has no decimal places, so the uncertainty must be kept to the units digit.) d Q

Q
0
 
Q . Q
0

10
2 
100 .
 a a
0

0.3
10.0

0.03
.
Hence,

Q
Q
0
2
 a
0 a
2 0.03

0.06
Hence, Q

100 6 .
. Hence,
 
Q
0

0.06 100 0.06

6 . e Q

Q
0
 
Q . Q

0
100
2
20
2

25 .
 a a
0

5
100

0.05
and
 b b
0

2

20
0.10
.
Hence,

Q
Q
0
2
 a
0 a
2
 b
0 b

0.10 0.20

0.30
.
Hence,
 
Q
0

0.30

25 0.30

7.5

8 . Hence, Q

25 8 .
11 F
0


 
8.0
F m
2

F m
0 0

68.6 N . F

F
0
 
F .
2
 v

 v r
0 0 r

0.1
2.8
2
2

0.2
14 8.0

0.346
.
Hence,
 
F
0

0.346
  
23.7

20 to one significant figure.
Hence, F

(68.6

20) N . Since the uncertainty is in the tens digit there is no point in quoting the answer beyond the tens digit. Hence, F

(70

20) N . Notice that, whereas each of mass, speed and radius are given to 2 s.f., the uncertainty in the force does not warrant keeping 2 s.f. in the answer.
Copyright Cambridge University Press 2011. All rights reserved. Page 4 of 6

Cambridge Physics for the IB Diploma
12 The volume of the cube is
V
0
  
3   
3   
3   
4 m
3 .
Hence, the density is

0
 m
0

V
0
2.2

4
  4
1.8 10 kg m

3
The uncertainty in the density satisfies



0

 m

 a

 b

 m
0 a
0 b
0 c
0 c

0.2

3

2

2
2.2
60 50 40

0.231
.
.
Hence,
 
0
   4 
0.231
  3   4
Hence,

0

(1.8
  
3 .
.
13 a The area is A
0
 π r
0
2
The uncertainty is
  2  18.1 cm 2

A
A
0
2
 r
0 r
2
0.1
2.4

.
0.083
 
A
0

.
, i.e.
The answer must have at most 2 s.f., so A

(18 2) cm
2
. b The circumference is C
0

2π r
0
  
C 2π r 2π 0.1 0.628 cm
, and
. Hence, C

(15 1) cm .
14 a A

4π r
2
We have
, and so

A
A

2 r
A
  2  6475.3 cm 2
 r
, i.e.

A
A
0.2
22.7

.
0.01762
We must express the uncertainty to 1 s.f., i.e.
114 cm
2
A
  2
100 1 10 cm
2
.
.
This means that A
   2
(65 1) 10 cm
2
or A
   3
(6.5 0.1) 10 cm
2
. b Similarly,

V
V
V

4π r
3
3

0.2
22.7

0.02643
3
3

48997 cm 3 and
1295 cm
3
.

V
V

3
 r r
so that
We must express the uncertainty to 1 s.f., i.e.
  3
1000 1 10 cm
3
. This means that V
   3
(49 1) 10 cm
3
or V
   4
(4.9 0.1) 10 cm
3
.
Copyright Cambridge University Press 2011. All rights reserved. Page 5 of 6

Cambridge Physics for the IB Diploma
15 The area is
A
  
37.4 cm
2

A

 a


A a b b

0.2

0.3
4.4
8.5

0.0807
3.02 cm 2
We must express the uncertainty to 1 s.f., i.e.
3 cm
2
This means that
A
  2 b The perimeter is P

2 a

2 b 2 4.4
2 8.5

25.8 cm .
P 2 a 2 b 2 0.2
2 0.3 1.0 cm
This means that P

26 1 cm .
.
16 By a factor of 2 .
17

T
T

1

L
2 L
and so

T
T
1
2%

1%
2
.
18 T

2π
L g
 g

 g L
L

2

T
T
4π 2 L
T
2
. This implies that

0.5 2 0.6 1.7%

2% .
Copyright Cambridge University Press 2011. All rights reserved. Page 6 of 6