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NASSection 6 Solutions to Practice Questions

NAS Chemistry Teachers Guide 2005 Nelson Thornes Ltd.

Unit 1Mechanics and RadioactivitySolutions to Practice QuestionsChapter 11 Length = (74.4 + 0.4) mm = 74.8 mm Width = (32.7 + 0.4) mm = 33.1 mm Height = (28.9 + 0.4) mm = 29.3 mm 2 Close micrometer and check for any zero error Use it to measure combined thickness of all 56 pages (not the covers) Page thickness = measurement/56 [ 5 mm/56 0.09 mm] 3 4 5 E.g. the metre rule may have shrunk or its end may be 3 mm short Percentage uncertainty = 0.05 mm 100/(1.23 mm) = 4.1 % Uncertainty = 4.7 k 2/100 = 94 100 = 0.1 k Possible values range from 4.6 k to 4.8 k

Chapter 21 2 See experiment description on page 4 Mass = density volume (a) Volume of room = 4 m 3 m 2 m = 24 m3 Mass of air = 1.3 kg m3 24 m3 = 31.2 kg = 31 kg (b) Volume of Earth = 4r3/3 = 4 (6.35 106 m)3/3 = 8.04 1020 m3 Mass of Earth = 5500 kg m3 8.04 1020 m3 = 4.42 1024 kg (c) Volume of rod = r2 l = (0.2 cm)2 24 cm = 3.02 cm3 Mass of rod = 8.0 g cm3 3.02 cm3 = 24 g 3 Volume = mass/density = 170 g/(2.7 g cm3) = 63 cm3 Length3 = 63 cm3 Length of each side = 3.98 cm = 4.0 cm 4 Volume of cube = 5 cm 5 cm 5 cm = 125 cm3Mass = density volume = 2.5 g cm3 125 cm3 = 313 g

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Estimates: diameter = 10 cm; thickness = 1 mm; mass = 20 g

Volume of DVD = r2 tSimilar to that of glass

(5 cm)2 0.1 cm = 8 cm3

Density = mass/volume = 20 g/(8 cm3) = 2.5 g cm3 = 2500 kg m3

NAS Physics Teachers Guide 2005 Nelson Thornes Ltd.

Unit 1Mechanics and RadioactivitySolutions to Practice Questions6 Volume of mercury = mass/density = 10.1 g/(13.6 g cm3) = 0.743 cm3 Volume of cylinder = r2 l = 0.743 cm3 r = (0.743 cm3/( 10.5 cm)) = 0.150 cm Internal diameter = 2r = 2 0.150 cm = 0.300 cm

Chapter 31 2 See Table 3.1 on page 7 Metre the distance an electromagnetic wave travels in a vacuum in a time of 1/(299 792 458) s Using this definition has the advantages that the metre can be reproduced anywhere in the world and it does not vary with temperature like the original standard bar A disadvantage is the difficulty imagining the distance travelled by such a fast wave in such a short time compared with observing the actual length of the original standard bar 3 A caesium atomic clock makes 9 192 631 770 oscillations every second So in 1 day Number of oscillations = 9 192 631 770 s1 24 hour 3600 s hour1 = 794 243 384 928 000 4 Examples: 75 kg 32 mm 5.4 m s1

75 kg = 75 kg 5 13 Mm/(13 m) = 13 106 m/(13 106 m) = 1 1012

Chapter 41 All quantities, other than base quantities, are called derived quantities All derived quantities can be produced by suitable combinations of base quantities 2 Speed Area Volume 3 m s1 m2 m3

Density = mass/volume Units are kg/m3 = kg m3

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Homogeneous means the same type Can only equate or add together quantities which are of the same type

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e.g. Density = 3 mass/volume Speed = distance/(4 time)

NAS Physics Teachers Guide 2005 Nelson Thornes Ltd.

Unit 1Mechanics and RadioactivitySolutions to Practice QuestionsChapter 51 Total distance = 3.5 km + 5.5 km = 9.0 km Final displacement from home = 0 m 2 A scalar is a physical quantity where the magnitude is not associated with any particular direction a scalar has only size while a vector has both size and direction Scalars: distance, energy, volume, speed, mass Vectors: acceleration, weight, displacement, velocity, force 3 Average speed = total distance travelled/total time taken (a) Average speed = 100 m/(10 s) = 10 m s1 (b) Average speed = 42 500 m/(2.25 hour 3600 s hour1) = 5.25 m s1 4 Attach a measured length of card centrally to the trolley Position the light gate so that the card blocks its beam as the trolley passes Use an electronic timer to record the time interval for which the beam is blocked Average speed = length of card/recorded time 5 Average speed = total distance travelled/total time taken Both journeys are the same length L, so total distance is 2L For each journey, time = distance/speed Time to work = (L/3) seconds time to home = (L/9) seconds Total time = (L/3) + (L/9) = (3L/9) + (L/9) = (4L/9) seconds Average speed = 2L/(4L/9) = 2 9/4 = 4.5 m s1 [OR chose a journey of, say, 90 m [Time to work = 90 m/(3 m s1) = 30 s [Time to home = 90 m/(9 m s1) = 10 s [Total time = 30 s + 10 s = 40 s [Average speed = 180 m/(40 s) = 4.5 m s1 ] ] ] ] ]

Chapter 61 Average acceleration = change in velocity/time taken a = (25.1 m s1 3.4 m s1)/(6.2 s) = 21.7 m s1/(6.2 s) = 3.5 m s2 2 Time taken = change in velocity/acceleration t = (330 m s1 75 m s1)/(5 m s2) = 255 m s1/(5 m s2) = 51 s

NAS Physics Teachers Guide 2005 Nelson Thornes Ltd.

Unit 1Mechanics and RadioactivitySolutions to Practice Questions3 Attach a double interrupter card of measured prong length x centrally to the trolley Position the light gate so that the prongs block its beam as the trolley passes Use an electronic timer to record: the time intervals for which the beam is blocked t1 t2 the time interval between the interruptions t3 Average velocity = length of prong/recorded time v1 = x/t1 v2 = x/t2 Acceleration a = (v1 v2)/t3 4 rate of means divided by time so rate of doing work means work done divided by time (which is power) 5 Average acceleration = change in velocity/time taken a = (30 m s1 0 m s1)/(8 s) = 4 m s2

Chapter 71 The gradient of a displacement-time graph is the instantaneous velocity The gradient of a velocity-time graph is the instantaneous acceleration The area of a velocity-time graph is the change in displacement The area of an acceleration-time graph is the change in velocity 2 Since body moves 20 m in 15 s and graph is a straight line (a) (10 s/15 s) 20 m = 13.3 m (b) (8 m/20 m) 15 s = 6.0 s (c) Average speed = total distance/total time = 20 m/(15 s) = 1.3 m s1 3 (a) Object is first accelerating, then constant velocity and then decelerating (b) Stage 1 Acceleration = change in velocity/time = 12 m s1/(5 s) = 2.4 m s2 Distance = area under graph up to 5 s = Stage 2 Acceleration = 0 m s2 (since constant velocity) Distance = area under graph from 5 s to 10 s = 12 m s1 5 s = 60 m Stage 3 Acceleration = 12 m s1/(12 s) = 1 m s2 Distance = area under graph from 10 s to 22 s = (c) Total distance = 30 m + 60 m + 72 m = 162 m Average speed = total distance/total time = 162 m/(22 s) = 7.4 m s11 2 1 2

12 m s1 5 s = 30 m

12 m s1 12 s = 72 m

NAS Physics Teachers Guide 2005 Nelson Thornes Ltd.

Unit 1Mechanics and RadioactivitySolutions to Practice Questions42.5 2 Acceleration/m s2 1.5 1 0.5 0 0.5 1 1.5 0 5 10 Time/s 15 20 25

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(a) 5 4 3 2 1 0 0 (b) 1 Time/s 12 2 3

Acceleration/m s2 Velocity/m s1

8

4

0 0 (c) 20 Displacement 15 10 5 0 0 1 Time/s 2 3 1 Time/s 2 3

NAS Physics Teachers Guide 2005 Nelson Thornes Ltd.

Unit 1Mechanics and RadioactivitySolutions to Practice QuestionsChapter 81 Record motion of ball in front of a vertical metre rule using a video camera Replay the video a frame at a time and record displacement from scale at 0.04 s intervals 2

Velocity

Time

Height of second bounce = either of the shaded areas Acceleration of gravity = gradient of negative sloping lines 3

Velocity/m s1

Time/s

The two graphs are the same for the times for which the two balls are in the air

NAS Physics Teachers Guide 2005 Nelson Thornes Ltd.

Unit 1Mechanics and RadioactivitySolutions to Practice Questions4 Centre of the track (since both positive and negative displacements)

Displacement

Time

Velocity

Time

Acceleration

Time

NAS Physics Teachers Guide 2005 Nelson Thornes Ltd.

Unit 1Mechanics and RadioactivitySolutions to Practice Questions5Displacement

Time

Velocity

Time

Acceleration

Time

Chapter 91 Using v = u + at t = (v u)/a = (18 m s1 0 m s1)/(4.5 m s2) = 4.0 s 2 Using v = u + at a = (v u)/t = (0 m s1 18 m s1)/(4.5 s) = 4.0 m s2 Using x = x= 31 2 1 2

(u + v)t

(18 m s1 + 0 m s1) 4.5 s = 41 m1 2

Using x = ut +

at21

x = (3.6 m s1 4.5 s) + [ 2 1.4 m s2 (4.5 s)2] = 16.2 m + 14.2 m = 30.4 m 41 Using x = ut + at2 2 1

u = (x 2 at2)/t = {60 m [ 2 35 m s2 (1.6 s)2]}/(1.6 s) = (60 m 44.8 m)/(1.6 s) = 15.2 m/(1.6 s) = 9.5 m s1 Using v = u + at v = 9.5 m s1 + (35 m s2 1.6 s) = 9.5 m s1 + 56.0 m s1 = 65.5 m s1 = 66 m s1

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NAS Physics Teachers Guide 2005 Nelson Thornes Ltd.

Unit 1Mechanics and RadioactivitySolutions to Practice Questions5 (a) Using v2 = u2 + 2ax a = [v2 u2]/2x = [(9.0 105 m s1)2 (1.0 105 m s1)2]/(2 0.20 m) = 2.0 1012 m s2 (b) Using x =1 2

(u + v)t

t = 2x/(u + v) = 2 0.20 m/(10.0 105 m s1) = 4.0 107 s

Chapter 101 2 See pages 22 and 231 Using x = ut + at2 2

from rest, x = at2 = 2 3 Using x =1 2

1

1 2

9.81 m s2 (1.9 s)2 = 17.7 m

at2 (from rest)

t = (2x/a) = [2 30 m/(9.81 m s2)] = (6.12 s2) = 2.5 s Using v2 = 2ax (from rest) v2 = 2 9.81 m s2 30 m = 588.6 m2 s2 v = (588.6 m2 s2) = 24.3 m s1 4 Using v2 = u2 + 2ax and taking upwards as positive u2 = v2 2ax = 02 (2 9.81 m s2 200 m) = 3924 m2 s2 u = (3924 m2 s2) = 62.6 m s1 Using x =1 2

at2 (from rest)

Time to rise = time to fall = t = (2x/a) = [2 200 m/(9.81 m s2)] = (40.8 s2) = 6.4 s Total distance travelled = 200 m + 200 m = 400 m Final displacement = 0 m 5 (a) Using v = u + at Speed = at = 150 m s2 6 s = 900 m s1 Distance = 2 at2 =1 1 2

150 m s2 (6 s)2 = 2700 m

(b) Rocket then decelerates at 9.81 m s2 Time to slow down = 900 m s1/(9.81 m s2) = 91.7 s Total time to reach top = 6 s + 91.7 s = 97