IB Physics Tsokos solution version 6 edition

Answers to test yourself questions Topic 1 1.1 Measurement in physics 1 Taking the diameter of a proton to be order 10 −15 m we find

10 −15 −23 = 3 × 10 −24 ≈ 10 −24 s . 8 = 0.3 × 10 3 × 10

2 The mass of the Earth is about 6 × 10 24 kg and the mass of a hydrogen atom about 2 × 10 −27 kg so we need 6 × 10 24 = 3 × 1051 ≈ 1051. 2 × 10 −27 3

1017 = 1060 10 −43

4 A heartbeat lasts or 1 s so

5

1041 = 1011 10 30

6

10 21 ≈ 1010 1.5 × 1011

75 × 365 × 24 × 3600 ≈ 8 × 4 × 2 × 4 × 107 ≈ 2.6 × 109 ≈ 109. 1

7 There are 300 g of water in the glass and hence is 15 × 6 × 10 23 = 90 × 10 23 ≈ 10 25.

300 300 ≈ = 15 moles of water. Hence the number of molecules 18 20

8 There are 6 × 104 g of water in the body and hence

6 × 104 ≈ 0.3 × 104 = 3 × 10 3 moles of water. Hence the 18

number of molecules is 3 × 10 3 × 6 × 10 23 = 18 × 10 26 ≈ 10 27 . 9 The mass is about 1.7 × 10 −27 kg and the radius about 10 −15 m so the density is 1.7 × 10 −27 1.7 × 10 −27 18 17 −3 ≈ −45 = 0.5 × 10 = 5 × 10 kg m . 4π −15 3 4 × 10 × (10 ) 3 10

10 21 ≈ 0.3 × 1013 = 3 × 1012 s ≈ 105 yr 3 × 108

−19 −19 J 11 a E = 2.5 × 1.6 × 10 = 4.0 × 10

b E =

8.6 × 10 −18 = 54 eV 1.6 × 10 −19

12 V = (2.8 × 10 −2 )3 = 2.2 × 10 −5 m 3 13 a = (588 × 10 −9 )1/3 = 8.38 × 10 −3 m

physics for the IB Diploma © Cambridge University Press 2015

ANSWERS TO TEST YOURSELF QUESTIONS 1

1

14 a 200 g b 1 kg c 400 g 15 The mass is about10 30 kg and the radius is 6.4 × 106 m so the density is of about 10 30

4π (6.4 × 106 )3 3

≈ 9 × 108 ≈ 109 kg m −3 .

105 10 3 2 3600 = 4 × 10 3 = 10 ≈ 6.25 m s −2 ≈ 0.7g . 16 In SI units the acceleration is 4 16 4 17 Assuming a mass of 70 kg made out of water we have 7 × 104 g of water in the body and 100 ×

7 × 104 ≈ 0.5 × 104 = 5 × 10 3 moles of water. Hence the number of molecules is 18 5 × 10 3 × 6 × 10 23 = 30 × 10 26 ≈ 3 × 10 27 . Each molecule contains 2 electrons from hydrogen and 8 from oxygen for a total of 10 × 3 × 10 27 ≈ 10 28 electrons. hence

18 The ratio is

Fe ke 2 9 × 109 × (1.6 × 10 −19 )2 9 × 109 × 3 × 10 −38 3 × 1044 1044 = = ≈ ≈ ≈ ≈ 5 × 1042. −11 −31 2 −11 −62 2 Fg Gm 63 20 6.7 × 10 × (9.1 × 10 ) 7 × 10 × 81 × 10

19 f = cm x k y . The units of m is kg i.e. M and those of k are

T = M x (M T −2 )y = M x + y T 2 y .

From this we deduce that x+y =0 1 1 2y = 1 ⇒ y = ⇒ x = − 2 2

Thus, f = c

−2 N kg m s = = kg s −2 = M T −2 . Hence m m

k . m

1.2 × 9.81 × 5.55 = 2.6667 × 101 W . The answer must be given to 2 s.f. and so 2.450 1.2 × 9.81 × 5.55 P= = 2.7 × 101 W . 2.450

20 P =

21 E K =

1 2 × 5.00 × 12.52 = 3.9063 × 10 2 J . The answer must be given to 3 s.f. and so E K = 3.91 × 10 J . 2

243 250 ≈ =5 43 50 b 2.80 × 1.90 ≈ 3 × 2 = 6 312 × 480 300 × 500 c ≈ = 1000 160 150

22 a

2

d

8.99 × 109 × 7 × 10 −16 × 7 × 10 −6 1010 × 50 × 10 −22 ≈ ≈ 10 −16 (8 × 10 2 )2 60 × 104

e

6.6 × 10 −11 × 6 × 10 24 50 × 1013 ≈ ≈ 10 (6.4 × 106 )2 40 × 1012



1.2 Uncertainties and errors 23 sum = (180 ± 8) N = (1.8 ± 0.8) × 10 2 N

dif = (60 ± 8) N = (6.0 ± 0.8) × 101 N

24 a Q0 =

a 20 ∆Q ∆a ∆b 1 1 = = 2; = + = + = 0.15 ⇒ ∆Q = 2.0 × 0.15 = 0.30 . Hence Q = 2.0 ± 0.3 . b 10 Q0 a b 20 10

b Q0 = 2 × 20 + 3 × 15 = 85; ∆Q = 2 × 2 + 3 × 3 = 13. Hence Q = 85 ± 13 ≈ (8.5 ± 0.1) × 101

c Q0 = 50 − 2 × 24 = 2; ∆Q = 1 + 2 × 1 = 3. Hence Q = 2 ± 3

d Q0 = 1.00 × 10 2 ;

∆Q ∆a 0.3 = 2× = 2× = 6.00 × 10 −2 ⇒ ∆Q = 100 × 6.00 × 10 −2 = 0.06 × 10 −2. Q0 a 10.0

Hence Q = 1.00 × 10 2 ± 0.06 × 10 2 = (1.00 ± 0.06) × 10 2

100 2 ∆Q ∆a ∆b 5 2 = 2× +2× = 2× +2× = 3.0 × 10 −1 ⇒ ∆Q = 25 × 3.0 × 10 −1 = 7.5 ≈ 8 2 = 25; 20 Q0 a b 100 20 Hence Q = 25 ± 8

e Q0 =

2.8 × 14 2 = 68.6 N 8.0 ∆F ∆m ∆v ∆r 0.1 2 0.2 = +2× + = +2× + = 0.3464 ⇒ ∆F = 68.6 × 0.3464 = 23.7 ≈ 20 N. F0 m v r 2.8 14 8.0

25 F0 =

1 Hence F = (68.6 ± 20) N ≈ (7 ± 2) × 10 N

26 a A0 = π R 2 = 18.096 cm 2.

∆A ∆R 0.1 =2× = 2× = 0.0833 ⇒ ∆ A = 18.096 × 0.0833 = 1.51 ≈ 2 cm 2 . A0 R 2.4

2 2 Hence A = (18.096 ± 2) cm ≈ (18 ± 2) cm . ∆S ∆R 0.1 b S0 = 2π R = 15.08 cm. = = = 0.04167 ⇒ ∆S = 15.08 × 0.04167 = 0.628 cm 2. S0 R 2.4 Hence S0 = (15.08 ± 0.628) cm ≈ (15 ± 1) cm .

27 A0 = ab = 37.4 cm 2.

∆ A ∆a ∆b 0.2 0.3 = + = + = 0.080749 ⇒ ∆ A = 37.4 × 0.080749 = 3.02 ≈ 3 cm 2. A0 a0 b0 4.4 8.5

2 2 Hence A = (37.4 ± 3) cm ≈ (37 ± 3) cm .

P0 = 2(a + b ) = 25.8 cm . ∆P = 2 × ∆a + 2 × ∆b = 2 × 0.2 + 2 × 0.3 = 1.0 cm . Hence P = (25.8 ± 1) cm ≈ (26 ± 1) cm.

28

∆T 1 ∆L ∆T 1 (assuming g is accurately known). Hence = = × 2% = 1% . T0 2 L0 T0 2

29

∆V ∆R ∆h = 2× + = 2 × 4% + 4% = 12% V0 R0 h0



3

30 The line of best-fit does not go through the origin. There is a vertical intercept of about 4 mA. Lines of maximum and minimum slope give intercepts of about 0 and 9 mA implying an error in the intercept of about 4 mA. The intercept is thus (4 ± 4) mA . This just barely includes the origin so the conclusion has to be that they can be proportional. 100 linear fit for: data set: current / mA y = mx + b m(slope): 209.0 b(Y-intercept): 5.000 correlation: 0.9976

Current / mA

80

60

40

20

0

0.0

0.1

0.2

0.3

0.4

0.5 x

31 The vertical intercept is about 10 mA. No straight line can be made to pass through the origin and the error bars unless a systematic error of about 10 mA in the current is invoked. 100 linear fit for: data set: current / mA y = mV + b m(slope): 169.0 b(Y-intercept): 10.00 correlation: 0.9998

Current / mA

80

60

40

20

0

0.0

0.1

0.2 0.3 Voltage / mV

0.4

0.5

However, a line of best fit that is a curve can also be fitted through the data and that does go through the origin. (However, it may be objected that this particular functional form is chosen – at low voltages we might expect a straight line (Ohm’s law). So a different functional form may have to be tried.) 100

Current / mA

auto fit for: data set: current / mA y = AV^B A: 158.4 + / – 7.459 B: 0.7878 + / – 0.03784 RMSE: 1.523

50

0

4

0.0

0.1

0.2 0.3 Voltage / mV


0.4

0.5


P and the side of the 2π 2 P2 P 2 P2 P  P  square 4a = P ⇒ a = . The circle area is then Ac = π   = . The square area is As =   = and is  4 2π 4π 4 16 smaller.

32 Let P the common perimeter. Then the radius of the circle satisfies 2π R = P ⇒ R =

33 a The initial voltage V0 is such that lnV0 = 4 ⇒ V0 = e 4 = 55 V .

V0 ≈ 27 V , lnV = ln 27 ≈ 3.29 . From the graph when lnV ≈ 3.29 we find t ≈ 7 s . 2 t so a graph of lnV versus time gives a straight c Since V = V0e − t /RC , taking logs, lnV = lnV0 − RC 1 4−2 line with slope equal to − . The slope of the given graph is approximately = −0.10. Hence RC 0 − 20 1 1 1 − = −0.10 ⇒ R = = = 2 × 106 Ω . RC 0.10 × C 0.10 × 5 × 10 −6 b When V =

34 We expect L = kM α and so ln L = ln k + α ln M . A graph of ln L versus ln M is shown below. The slope is a. lnL 10 9 8 7 6 5 0

0.5

1

1.5

2

–5

Drawing a best-fit line gives:

2.5

3

lnM

lnL 10 8 6 4 2

0.5

1

1.5

2

2.5

3 lnM

Measuring the slope gives α = 3.4 .



5

1.3 Vectors and scalars 35

  36 a A + B:

length 9 cm

⇒ F  18 N

q  49°

A+B

A–B

B A

  b A − B:

length 4.5 cm

⇒ F  9 N

 Θ  14° below horizontal.  c A − 2B :

length 6.1 cm

⇒ F  12.2 N

Θ  50° below horizontal.

scale: 1 cm ↔ 2 N

A – 2B

2B

A

6



37 The components are: Ax = 12 × cos 30° = 10.39

Ay = 12 × sin 30° = 6.00

Bx = 8.00 × cos 80° = 1.389 Ay = 8.00 × sin 80° = 7.878

Hence a ( A + B )x = 10.39 + 1.389 = 11.799 ( A + B )y = 6.00 + 7.878 = 13.878

  The vector A + B has magnitude

θ = arctan

11.799 2 + 13.878 2 = 18.2 and is directed at an angle

13.878 = 49.6° to the horizontal. 11.799

b ( A − B )x = 10.39 − 1.389 = 9.001 ( A − B )y = 6.00 − 7.878 = −1.878   The vector A − B has magnitude 9.0012 + 1.878 2 = 9.19 and is directed at an angle 1.878 θ = arctan − = −11.8° (below) the horizontal. 9.001 c ( A − 2B )x = 10.39 − 2 × 1.389 = 7.612

( A − 2B )y = 6.00 − 2 × 7.878 = −9.756   The vector A − 2B has magnitude 7.612 2 + 9.756 2 = 12.4 and is directed at an angle 9.756 θ = arctan − = −52.0° (below) the horizontal. 7.612 38 a

4.0 2 + 4.0 2 = 5.66 cm in a direction θ = 180° + arctan

2 2 b 124 + 158 = 201 km in a direction θ = arctan −

c

0 2 + 5.0 2 = 5.0 m at θ = 270° or θ = −90°.

d

8.0 2 + 0 2 = 8.0 N at θ = 0°.

39 a

4.0 = 225°. 4.0

158 = −52°. 124

3.00 = 56.3° 2.00 5.00 = 112° 2.00 2 + 5.00 2 = 5.39 at θ = 180° − arctan 2.00 2.00 2 + 3.00 2 = 3.61 at θ = arctan

b

c

0 2 + 8.00 2 = 8.00 at θ = 90°

d

4.00 2 + 2.00 2 = 4.47 at θ = arctan −

e

2.00 = −26.6° 4.00 1.00 = 9.46° 6.00 2 + 1.00 2 = 6.08 at θ = arctan 6.00

40 The displacement has components ∆rx = 4 − 2 = 2 and ∆ry = 8 − 2 = 6. 41 A diagram is:

change

initial

final



7

The magnitude of the change n the velocity vector is with the horizontal as shown in the diagram.

10 2 + 10 2 = 14.1 m s −1. The vector makes an angle of 45°

42 A diagram is: change

initial

final 30˚

The other two angles of the triangle are each ∆p

43

sin 30°

=

1 (180° − 30°) = 75°. Using the sine rule we find 2

p sin 30° ⇒ ∆p = p × = 0.518 p ≈ 0.52 p. sin 75° sin 75°

The components of the velocity vector at the various points are: A: v Ax = −4.0 m s −1 and v Ay = 0 B: vBx = +4.0 m s −1 and vBy = 0 C: vCx = 0 and vCy = 4.0 m s −1 Hence a From A to B the change in the velocity vector has components vBx − v Ax = +4.0 − ( −4.0) = 8.0 m s −1 and vBy − v Ay = 0 − 0 = 0 . b From B to C the change in the velocity vector has components vCx − vBx = 0 − 4.0 = −4.0 m s −1 and vCy − vBy = 4.0 − 0 = 4.0 m s −1. c From A to C the change in the velocity vector has components vCx − v Ax = 0 − ( −4.0) = +4.0 m s −1 and vCy − v Ay = 4.0 − 0 = 4.0 m s −1. The change in the vector from A to C is the sum of the change from A to B plus the change from B to C.

= −10.0 cos 40° = −7.66 and Ay = −10.0 sin 40° = +6.43 = −10.0 cos 35° = −8.19 and Ay = −10.0 sin 35° = −5.74 = +10.0 cos 68° = +3.75 and Ay = −10.0 sin 68° = −9.27 = +10.0 cos(90° − 48°) = +7.43 and Ay = −10.0 sin(90° − 48°) = −6.69 = −10.0 cos(90° − 30°) = −5.00 and Ay = −10.0 sin(90° − 30°) = −8.66      45 The vector we want is C = −( A + B ) . The components of A and B are: Ax = 6.0 cos 60° = +3.0 and Ay = 6.0 sin 60° = +5.20; Bx = 6.0 cos 120° = −3.0 and Ay = 6.0 sin 120° = +5.20. Hence  and C y = −( +5.20 + 5.20) = −10.4 . The magnitude of the vector C therefore is 10.4 units C x = −( +3.0 − 3.0) = 0 and is directed along the negative y – axis. 44

A Ax B Ax C Ax D Ax E Ax

B

A

B

C C

A

8



46 a Ax = 12.0 cos 20° = +11.28 and Ay = 12.0 sin 20° = +4.10; B = 14.0 cos 50° = +9.00 and Ay = 14.0 sin 50° = +10.72. Hence the sum has components: x Sx = +11.28 + 9.00 = 20.28 and Sy = +4.10 + 10.72 = 14.82 . The magnitude of the sum is thus 14.82 20.28 2 + 14.82 2 = 25.1. Its direction is θ = arctan = 36.2°. 20.28 b Ax = 15.0 cos 15° = +14.49 and Ay = 15.0 sin 15° = +3.88; B = 18.0 cos 105° = −4.66 and By = 18.0 sin 105° = +17.39. Hence the sum has components: x Sx = 14.49 − 4.66 = 9.83 and Sy = +3.88 + 17.39 = 21.27. The magnitude of the sum is thus 21.27 = 65.2°. 9.832 + 21.27 2 = 23.4 . Its direction is θ = arctan 9.83

c Ax = 20.0 cos 40° = +15.32 and Ay = 20.0 sin 40° = +12.86; B = 15.0 cos 310° = +9.64 and By = 15.0 sin 310° = −11.49. Hence the sum has components: x Sx = 15.32 + 9.64 = +24.96 and Sy = +12.86 − 11.49 = +1.37. The magnitude of the sum is thus 1.37 24.96 2 + 1.37 2 = 25.0. Its direction is θ = arctan = 3.14°. 24.96



9

IB Physics Tsokos solution version 6 edition

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