Phy Notes Manhatten 2

Chapter 8 Motion Along a Straight Line 8.1 Position, Distance, Time and Speed 8.2 Recording Motion 8.3 Displacement, Velocity and Acceleration 8.4

Uniformly Accelerated Motion Manhattan Press (H.K.) Ltd. © 2001

Section 8.1 Position, Distance, Time and Speed • Position and distance • Time • Speed Manhattan Press (H.K.) Ltd. © 2001

Position and distance

8.1 Position, distance, time and speed (SB p.2)

Position and distance Tracy

N

X 1 km

position of Tracy Chris

1 km

position of Chris Y 1 km

3

1 km

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Position and distance

8.1 Position, distance, time and speed (SB p.3)

Position ： direction and distance from the target place N

1 km east Tracy 1 km

X

1 km north

Shopping centre X from Tracy: direction: east distance: 1 km

Chris

1 km

Y 1 km 4

1 km

Shopping centre X from Chris: direction: north distance: 1 km

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Position and distance

8.1 Position, distance, time and speed (SB p.3)

Position and distance X

The position of church Y from Chris:

Chris

direction = south-east distance = 2 km 　　 = 1.41 km

N

Tracy 1 km

1 km

Y 1 km 5

1 km

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Position and distance

8.1 Position, distance, time and speed (SB p.3)

Position and distance N

Tracy X

The position of church Y from Tracy:

1 km

direction = south-east distance = 8 km 　　 = 2.83 km

Chris

1 km

Y 1 km 6

1 km

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Time

8.1 Position, distance, time and speed (SB p.3)

Time Measure the duration of an event Unit: second (s), minute (min), hour (h) a sundial

a watch

a quartz clock

7

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an atomic clock

Speed

8.1 Position, distance, time and speed (SB p.4)

Speed Distance travelled Average speed = Time taken

Unit: m s–1 or km h–1 The world record for the men’s 100 m race l00 m

100 Average speed = = 10.2 m s–1 9.79 8

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9.79 s

8.1 Position, distance, time and speed (SB p.4)

Speed Distance travelled Average speed = Time taken

If the time taken is very short

Instantaneous speed =

Distance travelled Time taken

e.g. speedometer of a car measures its instantaneous speed 9

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Speed

Speed

8.1 Position, distance, time and speed (SB p.5)

Stations of the KCR Mongkok

Hunghom

Tai Wai

Kowloon Tong

Tai Po Market

Fo Tan

Shatin

University

Fan Ling Tai Wo

Lo Wu

Sheung Shui

1→ 2→ 3→ 4→ 5→ 6→ 7→ 8→ 9→ 10→ 11→ 2 3 4 5 6 7 8 9 10 11 12

Station Distance between successive stations / km Time taken / min

2.42 3

1.80 4.50 1.10 2.00 1.80 6.70 1.15 6.25 1.50 3.50 2

4

2

2

3

6

2

5

From (4) Tai Wai to (5) Shatin Distance travelled Average speed = Time taken = 10

1.1 km 2 min

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= 9.2 m s–1

2

4

Speed

8.1 Position, distance, time and speed (SB p.6)

Class Practice 1 ： Referring to Table 8.1, find the average speed of a KCR train for the whole journey from (1) Hunghom to (12) Lo Wu. Express your answer in m s–1.

Station Distance between successive stations / km Time taken / min

1→ 2→ 3→ 4→ 5→ 6→ 7→ 8→ 9→ 10→ 11→ 2 3 4 5 6 7 8 9 10 11 12 2.42 3

1.80 4.50 1.10 2.00 1.80 6.70 1.15 6.25 1.50 3.50 2

4

2

2

3

6

2

5

2

Distance travelled Distance travelled AverageAverage speed = speed = Time taken Time taken

4

( 2.42 + 1( 　　　　　　　　　　 .8 + 4.5 + 1.1 + 2 + 1.8 + 6.7 + 1　　　　　　　　 .15 + 6.25 + 1.5 + 3.5))×103 = = ( 3 + 2 + 4 + 2 + 2 + 3 + 6 + 2 　　　　　　　　 + 5 + 2 + 4) × 60 ( 　　　　　　　　　　 ) 327 20 (m　　　　　　) = =s 2 100 ( 　　　　　　 ) 1 = 15.58 m s Manhattan Press (H.K.) Ltd. © 2001 11=

Ans wer

Section 8.2 Recording Motion • Ticker-tape timer • Tape chart and speed-time graph • Area under speed-time graph

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Ticker-tape timer

8.2 Recording motion (SB p.6)

Ticker-tape timer ─ Record a moving body 1. distance travelled 2. time taken ticker-tape timers

13

ticker-tape

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Ticker-tape timer

8.2 Recording motion (SB p.7)

Ticker-tape timer timer

ticker-tape

14

1. Pass a long tickertape through the timer 2. 50 black dots can be marked in 1 s (frequency = 50 Hz)

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8.2 Recording motion (SB p.7)

Ticker-tape timer

Experiment 8A: Motion analysis by ticker-tape timer Intro. VCD

Expt. VCD

15

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8.2 Recording motion (SB p.7)

Ticker-tape timer

Ticker-tape timer 1 = 0.02 s Time interval for 1-tick length = 50

Time interval for 5-tick length = 0.02 x 5 = 0.1 s 1-tick length

16

5-tick length takes 0.1 s

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Tape chart and speed-time graph

8.2 Recording motion (SB p.8)

Tape chart Strip length (cm)

Time (Starting point)

1. Cut the tape into strips of 5-tick length 2. Stick the strips in order side by side 3. y-axis ─ strip length x-axis ─ time Strip length / cm

strip length

time Time / s

17

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Tape chart and speed-time graph

8.2 Recording motion (SB p.9)

Speed-time graph

1. y-axis ─ speed Strip length 2. Join the mid-points of Average speed = the tops of the strips 0.1 s Tape chart Strip length / cm

speed

Speed-time graph Speed / cm s-1

region I - increasing speed region II - constant speed region III - decreasing speed

mid-point

Time / s Time / s

18

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Tape chart and speed-time graph

8.2 Recording motion (SB p.9)

Speed-time graph Speed / cm s-1

region I - increasing speed region II - constant speed

speed is increasing (changing-speed motion)

region III - decreasing speed

speed is unchanged (constant-speed motion) speed is decreasing (changing-speed motion)

Time / s

19

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8.2 Recording motion (SB p.10)

Tape chart and speed-time graph

Class Practice 2 ： The tape results below record the motions of two bodies X and Y.

Are bodies X and Y moving at constant speed or changing speed? constant X moves at a ____________ speed, and Y moves at a constant ____________ speed. double Speed of X is ____________ (half / double) that of Y. Ans wer 20

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Area under speed-time graph

8.2 Recording motion (SB p.11)

Area under speed-time graph Speed 速率 time

speed

Constant-speed motion

Distance ＝ Time × Speed time → width of rectangle speed → height of rectangle Distance ＝ Width × Height Distance ＝ Area of rectangle 　

時間 Time

21

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Area under speed-time graph

8.2 Recording motion (SB p.11)

Area under speed-time graph Speed

Changing-speed motion

Using the same principle: 　 Total distance travelled by a body 　 ＝ Area under the graph

Time

22

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8.2 Recording motion (SB p.12)

Area under speed-time graph

Class Practice 3 ： A van and a lorry move at constant speeds of 12 m s-1 and 18 m s-1 respectively in a time interval of 15 s. Complete the speed-time graphs for the van and the lorry in the given figure. Also find their distance travelled. Distance travelled by the van

Speed Speed//ms m -1s-1

12 × 15 = 180 m

＝ ˍˍˍˍˍˍˍˍˍ Distance travelled by the lorry 18 × 15 = 270 m ＝ ˍˍˍˍˍˍˍˍˍ

speed of the lorry speed of the van Time Time // ss

Ans wer 23

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Section 8.3 Displacement, Velocity and Acceleration • Displacement and distance • Velocity and speed • Acceleration • Motion graphs • Scalar and vector quantities Manhattan Press (H.K.) Ltd. © 2001

8.3 Displacement, velocity and acceleration (SB p.13)

Displacement and distance

Displacement and distance

Displacement ─ change in position of a body ─ vector quantity, has both magnitude and direction

direction negative (–)

25

direction positive (+)

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8.3 Displacement, velocity and acceleration (SB p.13)

Displacement and distance

Displacement and distance

distance displacement distance

displacement

length length depend on the travelled path 26

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independent of the travelled path

8.3 Displacement, velocity and acceleration (SB p.14)

Displacement and distance

Class Practice 4 ： Jessie starts from A and walks around a square loop as shown below. She returns to A finally. B

10 cm

10 cm

10 cm

A

C

10 cm

D

40 m 4 × 10 Total distance travelled = ˍˍˍˍˍ = ˍˍˍˍ 0m Total displacement = ˍˍˍˍˍˍ

27

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Ans wer

8.3 Displacement, velocity and acceleration (SB p.14)

Velocity and speed

Velocity and speed Displaceme nt Velocity = Time taken s v= t directional, the same direction as position Unit: m s-1 28

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8.3 Displacement, velocity and acceleration (SB p.15)

Velocity and speed

Velocity 70 km h-1

speed

70 km h-1

70 km h-1

70 km h-1 (to left) velocity = -70 km h-1

70 km h-1 70 km h-1 (to right) = 70 km h-1

Take the direction to right as positive. 29

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8.3 Displacement, velocity and acceleration (SB p.16)

Velocity and speed

Class Practice 5 ： A man is running on a road. His positions at different instants are shown in the figure below. Complete the table, and take the direction to the right as positive.

Ans wer Time interval / s

0 - 5 5 - 10 10-20 20-30 30-40

Displacement / m

15

20

30

-40

-30

Average velocity / m s-1

15 =3 5

20 =4 5

30 =3 10

− 40 = −4 10

− 30 = −3 10

30

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8.3 Displacement, velocity and acceleration (SB p.16)

Velocity and speed

Class Practice 6 ： An ant takes 10 minutes to walk from A to B along the path as shown in the figure. (a) What are the distance travelled and the average speed of the ant? 22 cm _ Distance travelled = __________ 0.22 ( ) 　　　　　　 Average speed = ( 　　　　　　 ) 10 x 60 3.7 x 10-4 m 　　　　　 = __________ __s–1 (b) What are the displacement and the average velocity of the ant? cm (due north Displacement = 2__________ _ ) 0.02 ( ) 　　　　　　 Average velocity = (　　　　　　 ) 10 x 60 Ans –1 x 10-5 m s__ (due north) 　　　　　 =3.3 __________ wer 31

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8.3 Displacement, velocity and acceleration (SB p.17)

Acceleration

Acceleration Change of velocity Acceleration = Time taken v −u a= t

Unit: m s-2 32

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8.3 Displacement, velocity and acceleration (SB p.17)

Acceleration

Acceleration

velocit y

(i)　From 0 s to10 s,

25 − 10 　　Average acceleration of the car (a) = = 2.5 m s−2 10 33

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8.3 Displacement, velocity and acceleration (SB p.17)

Acceleration

Acceleration

(ii)　From 10 − 20 s,

　Average accelerati on of the car (a) = 25 − 25 = 0 m s−2 10 34

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8.3 Displacement, velocity and acceleration (SB p.17)

Acceleration

Acceleration

(iii)　20 − 40 s,

0 − 25 　Average acceleration of the car (a) = = −1.25 m s−2 20 Note: Deceleration = 1.25 m s-2 35

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8.3 Displacement, velocity and acceleration (SB p.18)

Motion graphs

Motion graphs Displacement-time graph (s-t graph) Displacement (s)

Slope

slope

Time (t)

36

Change in displaceme nt = Change in time ∆s = ∆t = Velocity (v )

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8.3 Displacement, velocity and acceleration (SB p.19)

Motion graphs

Displacement-time graph s

Slope = 0 ∴Velocity = 0 m s −1 t

37

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8.3 Displacement, velocity and acceleration (SB p.19)

Motion graphs

Displacement-time graph s A

Slope 　 A＞B Velocity vA ＞ vB

B

t

38

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8.3 Displacement, velocity and acceleration (SB p.19)

Motion graphs

Displacement-time graph

Slope increases Velocity of body increases (accelerated motion)

39

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8.3 Displacement, velocity and acceleration (SB p.19)

Motion graphs

Displacement-time graph

Slope decreases Velocity of body decreases (decelerated motion)

40

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8.3 Displacement, velocity and acceleration (SB p.20)

Motion graphs

Class Practice 7 ： A car, which is at rest initially, accelerates from t = 0 s to t = 4 s, and then decelerates from t = 4 s to t = 8 s before it stops at t = 8 s. Sketch the displacement-time graph of the car on the graph below.

acceleration

deceleration stop

41

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Ans wer

8.3 Displacement, velocity and acceleration (SB p.21)

Motion graphs

Velocity-time graph (v-t graph) Velocity (v)

Slope =

slope

Change in velocity Change in time

∆v = ∆t = Acceleration (a) Time (t)

42

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8.3 Displacement, velocity and acceleration (SB p.21)

Motion graphs

Velocity-time graph Velocity (v)

Area under the graph slope

Displacement of the body

Time (t)

43

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8.3 Displacement, velocity and acceleration (SB p.21)

Motion graphs

Velocity-time graph

Slope = 0 Acceleration = 0 The body is moving at constant velocity

44

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8.3 Displacement, velocity and acceleration (SB p.21)

Motion graphs

Velocity-time graph

Slope A ＞ B Acceleration aA ＞ aB

45

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8.3 Displacement, velocity and acceleration (SB p.22)

Motion graphs

Velocity-time graph Slope increases Acceleration of body increases

46

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8.3 Displacement, velocity and acceleration (SB p.22)

Motion graphs

Velocity-time graph Slope decreases the body is decelerating

47

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8.3 Displacement, velocity and acceleration (SB p.22)

Motion graphs

Velocity-time graph • O - A ─ accelerates (to the right) • A - B ─ decelerates (to the right) and stops at B • B - C ─ accelerates (to the left)

48

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8.3 Displacement, velocity and acceleration (SB p.22)

Velocity-time graph

Area • • • •

I ─ distance travelled to the right II ─ distance travelled to the left (I + II) ─ total distance travelled (I - II) ─ total displacement 49

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Motion graphs

8.3 Displacement, velocity and acceleration (SB p.23)

Motion graphs

Class Practice 8 ： The v-t graph of a train is shown in the figure.

( (

) )

(a) Find the accelerations of the train in different time intervals. − 0 ＝ 0.8 m s −2 From 0 s to 25 s, a = 20 ( ) ＝　　　　　　 　　　　　 From 0 s to 25 s, a =　 m s −2 25 − 0 ( 　　　　　 ) 20 − 20　 −2 ( ) 　　　　　 FromFrom 25 s 25 tos75 s, a = ＝ 0 m s to 75 s, a = 　 ＝　　　　　　 m s −2 75 − 25 ( 　　　　　 ) 　 0 − 20 )＝ − 0.8 m s −2 ( 　　　　　 FromFrom75 75 s tos100 s, a = −2 to 100 s, a = 100 ＝ 　　　　　　 m s − 75 ) (　 　　　　　

( (

(

50

(

) )

)

)

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Ans wer

8.3 Displacement, velocity and acceleration (SB p.24)

Motion graphs

Class Practice 8 (Cont’d)

) + 100] ×___ [( 20==________ (b) Total distance travelled = __________ __________ (b) Total distance travelled = 75 − 25 1 500 m Total displacement = __________________ =2_____________ Total displacement = Total distance travelled = 1 500 m Ans wer

51

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8.3 Displacement, velocity and acceleration (SB p.24)

Motion graphs

Class Practice 9 ： The figure shows the velocity-

time graphs of two cars A and B. When a traffic light turns red, both the drivers apply their brakes and stop their cars within the same time interval. Compare the deceleration and stopping distance of the cars. uniform Car A is braking with __________ (uniform / increasing / decreasing) deceleration, while car B is braking with increasing __________ deceleration. The stopping distance of car A is shorter ____________ (longer / shorter) than that of car B because the area covered by the v-t graph for car A ________________________________. is smaller than that for car B. 52

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Ans wer

8.3 Displacement, velocity and acceleration (SB p.24)

Motion graphs

Acceleration-time graph (a-t graph)

constant velocity acceleration = 0

53

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8.3 Displacement, velocity and acceleration (SB p.25)

Acceleration-time graph

uniform acceleration

54

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Motion graphs

8.3 Displacement, velocity and acceleration (SB p.25)

Acceleration-time graph

uniform deceleration

55

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Motion graphs

8.3 Displacement, velocity and acceleration (SB p.26)

Motion graphs

Class Practice 10 ： The velocity-time graph of a car is shown in Fig. a. Complete its acceleration-time graph in Fig. b.

Fig. a 56

Fig. b Manhattan Press (H.K.) Ltd. © 2001

Ans wer

8.3 Displacement, velocity and acceleration (SB p.26)

Scalar and vector quantities

Scalar ─ has magnitude, but no direction Vector ─ has magnitude and direction (scalar) e.g. time, distance, speed (vector) e.g. displacement, velocity and acceleration

57

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Section 8.4 Uniformly Accelerated Motion • Equations of motion • Acceleration down an inclined plane • Acceleration due to gravity (g)

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8.4 Uniformly accelerated motion (SB p.26)

Equations of motion

Equations of uniformly accelerated motion Time

0

t

Velocity u

v

Uniform acceleration

59

a

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8.4 Uniformly accelerated motion (SB p.27)

Equations of motion

Equations of uniformly accelerated motion Velocity

Area of region I = area of region II ∴displaceme nt = area under the curve AB = area under the dotted line CD u + v = ×t 2 s = u + v × t ... ... (1) 2 Time

60

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8.4 Uniformly accelerated motion (SB p.27)

Equations of motion

Equations of uniformly accelerated motion Velocity

Acceleration = slope of AB v − u a= t ∴v = u + at ... ... (2)

Time

61

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8.4 Uniformly accelerated motion (SB p.27)

Equations of motion

Equations of uniformly accelerated motion Velocity

From (1), we have u + v s= ×t 2 2s = (u + v ) t 2 s t= ... ... (a) u +v Time

62

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8.4 Uniformly accelerated motion (SB p.27)

Equations of motion

Equations of uniformly accelerated motion Substitute (a ) into (2),

Velocity

v

Time

63

 2s   = u + a   u +v 

v = u + 2as u +v 2 as v −u = u +v v 2 − u 2 = 2as v 2 = u 2 + 2as ... ... (3)

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8.4 Uniformly accelerated motion (SB p.27)

Equations of motion

Equations of uniformly accelerated motion Velocity

From (1),

u + v s= t 2

2 s× =t u +v

t = 2s ... ... (b) u +v

Time

64

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8.4 Uniformly accelerated motion (SB p.27)

Equations of motion

Equations of uniformly accelerated motion Velocity

From (b ), we have  2s   t 

 − u  = u + at 

Time

65

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2s = 2u + at t 1 s = ut + at 2 ... ... (4) 2

8.4 Uniformly accelerated motion (SB p.27)

Equations of uniformly accelerated motion

v = u + at v2 = u2 + 2as s = ut + ½ at2 u — initial velocity v — final velocity s — displacement a — acceleration t — time 66

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Equations of motion

8.4 Uniformly accelerated motion (SB p.27)

Equations of motion

The sign of the quantities a v s

negative

67

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positive

8.4 Uniformly accelerated motion (SB p.28)

Equations of motion

Class Practice 11 ： A car is moving at a velocity of 70 km h-1 . The driver then sees a 50 km h-1 speed limit sign at a distance of 30 m ahead. In order not to exceed the speed limit, find the minimum deceleration of the car.

u

km h = 70 ______

–1

70 ×1000 3600

= __________ = __ ___ ___m s–1 50×1000 __________ 3600

km h = v = 50 ______ 30 m s = ______ –1

By 13.89

19.44

13.89 = _____ ___m s–1

v2 = u2 + 2as 19.44

30

( _______ )2 = (_______ )2 + 2a ( _______ ) -3

∴ a = _________ m s–2

3 m s–2 The minimum deceleration of the car is __________. 68

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Ans wer

8.4 Uniformly accelerated motion (SB p.28)

Acceleration down an inclined plane

Experiment 8B: Acceleration down an inclined runway Expt. VCD

69

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8.4 Uniformly accelerated motion (SB p.29)

Acceleration down an inclined plane

Acceleration down an inclined plane Velocity / cm s-1

v

u

Time / s

70

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8.4 Uniformly accelerated motion (SB p.29)

Acceleration down an inclined plane

Acceleration down an inclined plane

5 cm

71

u

Initial velocity of the trolley (u) = Average velocity in 1st strip = 5 cm s–1 = 0.05 m s–1

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8.4 Uniformly accelerated motion (SB p.29)

Acceleration down an inclined plane

Acceleration down an inclined plane

70 cm

72

Final velocity of the trolley (v) = Average velocity in 13th strip = 70 cm s–1 = 0.7 m s–1 v

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Acceleration down an inclined plane

8.4 Uniformly accelerated motion (SB p.29)

Acceleration down an inclined plane length

Time interval (t) = (13 - 1) × 0.1 = 1.2 s

1.2 s

t 73

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8.4 Uniformly accelerated motion (SB p.29)

Acceleration down an inclined plane

Acceleration down an inclined plane The acceleration of the trolley v −u a= t 0.7 − 0.05 = 1.2 ∴ a = 0.54 m s −2

74

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8.4 Uniformly accelerated motion (SB p.30)

Acceleration down an inclined plane

If the slope of the inclined plane increases, acceleration will increase. Velocity / cm s-1

(steep slope) (gentle slope)

Time / s

75

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8.4 Uniformly accelerated motion (SB p.30)

Acceleration down an inclined plane

Class Practice 12 ： The tape chart given records the

motion of a trolley down an inclined plane. Find the acceleration of the trolley. ( 　　　　　　 ) 0.5 cm Initial velocity (u ) = Length / cm 0.1 s ( 　　　　　　 ) 0.05 　 = __________ m s-1 4 cm ( ) 　　　　　　 Final velocity (v ) = 0.1 s ( 　　　　　　 ) -1 0.4 m s = __________ Time interval from the 1st strip to Time / s the 11th strip (t ) (11 - 1)__________ × 0.1 = 1 s _ t = __________ v −u ∴a = Ans t -2 0.4 − 0.05 0.35 m s wer ( ) 　　　　　　　　 = = __________ ______ 1 ( 　　　　　　　　) 76

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8.4 Uniformly accelerated motion (SB p.31)

Acceleration due to gravity (g)

Acceleration due to gravity (g) attraction of the earth’s gravity acceleration towards the earth acceleration due to gravity (g) 77

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8.4 Uniformly accelerated motion (SB p.31)

Acceleration due to gravity (g)

Experiment 8C ： Motion of a free falling object Expt. VCD

78

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8.4 Uniformly accelerated motion (SB p.32)

Acceleration due to gravity (g)

Acceleration due to gravity (g) The time taken for each 2 - tick length = 0.02 × 2 = 0.04 s

Length / cm Length of 12th strip = 17.5 cm

Initial velocity (u ) =

0.7 0.04

= 17.5 cm s−1 Length of 1st strip = 0.7 cm

Time / s

79

17.5 Final velocity (v ) = 0.04 = 437.5 cm s−1

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8.4 Uniformly accelerated motion (SB p.32)

Acceleration due to gravity (g)

Acceleration due to gravity (g) Time interval (t ) = (12 - 1) x 0.04 = 0.44 s v −u Acceleration due to gravity (g ) = t

437.5 − 17.5 = 0.44 = 955 cm s−2 = 9.55 m s−2 80

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8.4 Uniformly accelerated motion (SB p.35)

Acceleration due to gravity (g)

Class Practice 13 ： A ball is thrown vertically upwards at an initial velocity of 100 m s-1 . (a) Complete the following table (take the downward direction as positive). Note that t is the time elapsed, s is the displacement of the ball, and v is the velocity of the ball.

t/st/s

2

s / ms / m

-180

2

8 8 -480

-1m s -1 v / -80 -20 v/ms Direction Direction upwards upwards of motion of motion

10 10

15 15

20 20

-500

375

0

0

50

100

at rest

downwards downwards

Ans wer 81

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8.4 Uniformly accelerated motion (SB p.35)

Acceleration due to gravity (g)

Class Practice 3 (Cont’d) (b) Sketch the positions of the ball at t = 2 s, 8 s, 10 s, 15 s and 20 s in the following figure. Use a scale of 1 cm to represent 100 m in height.

Ans wer 82

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8.4 Uniformly accelerated motion (SB p.36)

Acceleration due to gravity (g)

Experiment 8D ： The “coin and feather” experiment Expt. VCD

83

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8.4 Uniformly accelerated motion (SB p.37)

Acceleration due to gravity (g)

coin feather

fall in air

84

fall in vacuum

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8.4 Uniformly accelerated motion (SB p.37)

Acceleration due to gravity (g)

gravitational force

air resistance

fall in air 85

fall in vacuum

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Chapter 9 Inertia, Force and Motion 9.1 Forces, Mass and Weight 9.2 Types of Forces 9.3 Vector Addition and Resolution of Forces 9.4 Newton’s First Law of Motion 9.5 Newton’s Second Law of Motion 9.6 Force Diagrams 9.7 Pressure Manhattan Press (H.K.) Ltd. © 2001

Section 9.1 Forces, Mass and Weight

• Mass and weight

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Mass and weight

9.1 Forces, mass and weight (SB p.50)

Force ─ cause an object start moving, stop moving, change its direction of motion Unit: newton (N) stop motion

start motion or

88

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9.1 Forces, mass and weight (SB p.51)

Mass (m) ─

a measure of the quantity of matters inside a body

Weight (W) ─ • Unit: kilogram (kg) • Scalar quantity • Measured by beam balances

89

Mass and weight

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9.1 Forces, mass and weight (SB p.51)

Mass and weight

Mass (m) ─ a measure of the quantity of matters inside a body Weight (W) ─ a measure of the attraction on a body towards the earth • Unit: newton (N) • Vector quantity • Measured by spring balances On the earth 1 kg 1N 90

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Mass and weight

9.1 Forces, mass and weight (SB p.52)

Differences between mass and weight Mass kg

Unit Physical quantity

scalar

Measuring tool

beam balance

91

Weight N

vector spring balance

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Section 9.2 Types of Forces

• Tension • Friction

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Tension

9.2 Types of forces (SB p.52)

Tension (T)

tension in a stretched string

93

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Friction

9.2 Types of forces (SB p.53)

Friction ( f ) ─ arises whenever an object slides or tends to slide over another object Reason: rough surfaces book

The direction is opposite to the motion direction of motion friction

94

table

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Friction

9.2 Types of forces (SB p.54)

Application of friction

tread patterns on tyres

shoes with studs rough road surface 95

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9.2 Types of forces (SB p.54)

Friction

Disadvantage of friction

• • • •

Waste energy in movable parts of machines Waste as heat Waste as sound Cause wear in gears

96

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9.2 Types of forces (SB p.55)

Experiment 9A ： Frictionless motion Intro. VCD

Expt. VCD

97

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Friction

Friction

9.2 Types of forces (SB p.56)

Frictionless motion turn on air blower rider floats on the layer of air

air comes out from tiny holes

rider moves to and fro several times

98

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linear air track

rider

Friction

9.2 Types of forces (SB p.56)

Frictionless motion plastic beads reduce friction between the ring puck and the tray ring puck moves on the plastic beads continuously 99

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ring puck

a layer of beads

Friction

9.2 Types of forces (SB p.56)

Ways to reduce friction 1. Bearings ball bearings

wheel wheel axle

roller bearings

100

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axle

9.2 Types of forces (SB p.57)

Ways to reduce friction 2. Lubricating oil

101

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Friction

Friction

9.2 Types of forces (SB p.58)

Ways to reduce friction 3. Air cushion

102

4. Streamlining

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Section 9.3 Vector Addition and Resolution of Forces

• Vector addition of forces • Resolution of forces

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9.3 Vector addition and resolution of forces (SB p.58)

Vector addition of forces

Vector addition of forces

─ adding several forces ─ sum of forces is called resultant force (FR) Forces on the same line F1

F2

F2

FR = F1 - F2

FR = F1 + F2

104

F1

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9.3 Vector addition and resolution of forces (SB p.59)

Forces at angle θ

Vector addition of forces

Experiment 9B ： Vector addition of forces Expt. VCD

105

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9.3 Vector addition and resolution of forces (SB p.60)

Vector addition of forces

1. Tip-to-tail method

FR

F1 X

F2

106

F1 X

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F2

9.3 Vector addition and resolution of forces (SB p.60)

Vector addition of forces

2. Parallelogram method

F1 X

F1 F2

107

FR

X

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F2

9.3 Vector addition and resolution of forces (SB p.60)

Vector addition of forces

If F1 and F2 are perpendicular to each other

F1

or

FR

θ

F1

θ F2

F2

FR = √ F12 + F22 tan θ 108

FR

=

F1 F2

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9.3 Vector addition and resolution of forces (SB p.62)

Vector addition of forces

Class Practice 1 ： Find the resultant force (FR) as shown in the figure on the right. √(5 2)2 + 42 Magnitude of FR ＝ ˍˍˍˍˍˍ 5 N tan-1 　　　　　 ＝ ˍˍˍˍˍˍ Direction of FR(θ ) ＝ ˍˍˍˍˍ 5 (4 / 3) 3 　　　　　　 ＝ ˍˍˍˍˍ 5 oN (N ∴ Resultant force (FR) = 530 E) ________________ Ans wer 109

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9.3 Vector addition and resolution of forces (SB p.62)

Resolution of forces

Resolution of forces

─ a force is resolved into two components

Fx = F cos θ Fy = F sin θ tan θ 110

= Fy Fx

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Section 9.4 Newton’s First Law of Motion

• Aristotle’s ideas concerning force and motion • Galileo’s thought experiment • Inertia • Newton’s first law of motion • Inertia and mass

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9.4 Newton’s first law of motion (SB p.63)

Aristotle’s ideas concerning force and motion

Aristotle’s ideas concerning force and motion

The Greek philosopher Aristotle proposed that for a body to move, a force must be applied to it.

112

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9.4 Newton’s first law of motion (SB p.64)

Galileo’s thought experiment

Aristotle’s deduction was turned down by Galileo Galilei

Galileo’s thought experiment (“pin-and-pendulum” experiment)

clamp

end

star t pendulum bob

113

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9.4 Newton’s first law of motion (SB p.64)

Galileo’s thought experiment

The “pin-and-pendulum” experiment ─ the bob reaches the height as before

pin

114

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9.4 Newton’s first law of motion (SB p.65)

Galileo’s thought experiment

Experiment 9C ： Galileo’s thought experiment C

Expt. VCD

A

115

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9.4 Newton’s first law of motion (SB p.66)

Inertia

Inertia ─ a measure of the tendency for a body to remain at rest or to move at a uniform velocity

at rest

uniform velocity motion or

116

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9.4 Newton’s first law of motion (SB p.66)

Experiment 9D ： Tricks with inertia Expt. VCD

117

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Inertia

9.4 Newton’s first law of motion (SB p.68)

Newton’s first law of motion

Newton’s first law of motion

If there is no net force acting on a body a body will remain in its state of motion (either at rest, or moving at a uniform velocity) uniform velocity motion

at rest or

118

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9.4 Newton’s first law of motion (SB p.69)

Newton’s first law of motion

Seat belt ─ reduce the force of throwing forwards two points seat belt

three points seat belt

Two types of seat belt 119

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9.4 Newton’s first law of motion (SB p.70)

Newton’s first law of motion

Head rest ─ prevent the neck from bending backwards

120

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9.4 Newton’s first law of motion (SB p.70)

Inertia and mass Greater mass of the ball Greater the inertia

121

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Inertia and mass

Section 9.5 Newton’s Second Law of Motion

• Friction-compensated runway • Acceleration and force • Acceleration and mass • Newton’s second law of motion • Gravitational pull Manhattan Press (H.K.) Ltd. © 2001

9.5 Newton’s second law of motion (SB p.71) Friction-compensated runway

Friction-compensated runway (a sloping runway)

If friction = mg sin θ (friction-compensated) Trolley moves down at a uniform speed (indicated by the evenly distribution of the dots on the tape) friction mg sin θ

θ The dots on the tape are evenly distributed 123

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9.5 Newton’s second law of motion (SB p.72)

Experiment 9E ： Acceleration and force Expt. VCD

124

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Acceleration and force

9.5 Newton’s second law of motion (SB p.72)

Acceleration and force

Acceleration and force Force (F) / number of elastic cords Acceleration of trolley (a) / m s-2 Acceleration / m s-2

1

2

3

0.3

0.6

0.9

Acceleration ∝ No. of elastic cords Acceleration ∝ Force applied on the trolley

a∝F Force / number of elastic cords

125

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9.5 Newton’s second law of motion (SB p.73)

Acceleration and force

Class Practice 2 ： Complete the following table for a trolley being pulled by different numbers of elastic cords down a friction-compensated runway. Force (F) / number of elastic cords Acceleration of trolley (a) / m s-2

126

1 0.2

2

0 . 4

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3

4

0 . 8

0 . 6 Ans wer

9.5 Newton’s second law of motion (SB p.73)

Experiment 9F ： Acceleration and mass

Expt. VCD

127

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Acceleration and mass

9.5 Newton’s second law of motion (SB p.74)

Acceleration and mass

Acceleration and mass Mass of trolley (m) / kg

1

2

3

1 1 ( )/kg−1 mass of trolleys m

1

0.5

0.33

0.6

0.3

0.2

Acceleration (a) / m s-2 Acceleration / ms-2

acceleration ∝

a∝ 1 mass of trolleys

128

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1 mass of trolleys

1 m /kg

−1

9.5 Newton’s second law of motion (SB p.74)

Acceleration and mass

Class Practice 3: Complete the following table for trolleys of different masses being pulled by a constant force down a friction-compensated runway.

1/ For a constant force, a ∝ _____________. m Mass of trolleys (m) / kg Acceleration of trolley (a) / m s-2

129

1

2

3

0.3

0. 1 5

0 . 1

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4

0.0 75 Ans wer

9.5 Newton’s second law of motion (SB p.75) Newton’s second law of motion

Newton’s second law of motion

at rest force

130

unbalanced force (net force) acceleration or

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9.5 Newton’s second law of motion (SB p.76) Newton’s second law of motion

Newton’s second law of motion Acceleration (a) ∝ Net force (F) Acceleration (a) ∝ 1 m F∝ma

F = ma Unit: newton (N) Note: Direction of a = Direction of F 131

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9.5 Newton’s second law of motion (SB p.76)

Gravitational pull

Gravitational pull

Weight (W) ─ gravitational pull of the earth on a body F = ma = mg (g is the acceleration due to gravity) W = mg

m = 0.8 kg 38 kg W= 8N 132

380 N

5 000 kg 50 000 N

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9.5 Newton’s second law of motion (SB p.77)

Gravitational pull

Gravitational pull g differs with positions from the earth

the gravitational force decreases as the distance from the earth increases 133

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9.5 Newton’s second law of motion (SB p.77)

Gravitational pull

Different gravitational acceleration on different planets Place

Mass (m) / kg

Earth 10

Moon Venus Jupiter

134

Gravitational acceleration (g) / m s

-2

Weight (W = mg) / N

10

100

1 of earth = 1.67 6

16.7

9 of earth = 9 10

90

2.6 of earth = 26

260

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Section 9.6 Force Diagrams

• Motion on an inclined plane • Bodies connected by string • Weightlessness

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9.6 Force diagrams (SB p.78)

Motion on an inclined plane

Motion on an inclined plane 1. Forces parallel to the plane (Wx and f ) tio c a re of n tio c e dir

136

tion o m

Wx = f

) R ( n

The block is at rest (f) n tio fric

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9.6 Force diagrams (SB p.78)

Motion on an inclined plane

Motion on an inclined plane 1. Forces parallel to the plane (Wx and f ) ion t c rea

c e r i d

of n tio

tion o m

Wx > f

( R) (f) n tio fric

The block slides down the plane Force acting on the body (F ) = ma mg sin θ − f = ma mg sin θ − f a= m

137

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9.6 Force diagrams (SB p.78)

Motion on an inclined plane

Motion on an inclined plane 2. Forces perpendicular to the plane (Wy and f ) ion t c rea

c e r i d

of n tio

tion o m

R = Wy

( R) (f) n tio fric

= mg cos θ No motion Note: R is called the reaction

138

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9.6 Force diagrams (SB p.79)

Motion on an inclined plane

Class Practice 4 ： Find the unknown forces (W and Wy) in the figure.

Wx = W sin 25º

15 / sin 25o W = ____________ 35. = ____________ 5 N W cos Wy = ____________ 25o = ____________ 32.2 N

139

Ans wer

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9.6 Force diagrams (SB p.80)

Motion on an inclined plane

Class Practice 5: A trolley of mass 1 kg runs down a runway with an acceleration of 0.2 m s-2 as shown in the figure. (a) Find the friction acting on the trolley.

mg sin θ − f = ma 1 x 10 x sin 25o

1x ___________ − f = __________ _ 0.2 4.0 3N

f = ___________ N

140

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Ans wer

9.6 Force diagrams (SB p.81)

Motion on an inclined plane

Class Practice 5 (Cont’d) (b) Find the perpendicular force exerted on the trolley by the runway. mg cos R = ________________ θ

=

o 1 x 10 x cos 25 ________________

9.06 N = ________________ Ans wer

141

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9.6 Force diagrams (SB p.81)

Bodies connected by string

Bodies connected by string

1. Forces perpendicular to the plane a

142

Consider m1

R1 = m1g

Consider m2

R2 = m2g

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9.6 Force diagrams (SB p.81)

Bodies connected by string

2. Forces parallel to the plane

143

R1

R2

m1 g

m2 g

Consider m1

T = m1a

Consider m2

F - T = m2a

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a

9.6 Force diagrams (SB p.82)

Bodies connected by string

Two or more connected bodies: consider as a whole system

a

m1 + m2 + m3

F = (m1 + m2 + m3) a

144

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9.6 Force diagrams (SB p.83)

Weightlessness

Weightlessness

R 620N

R = supporting force on the boy mg = weight of the boy = 620 N

mg 145

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9.6 Force diagrams (SB p.83)

Weightlessness

1. The lift is accelerating upwards with a = 1.4 m s-2 R a = 1.4 m s-2 707N

F

mg 146

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F = ma R − mg = ma R = m ( g + a)

R = 62 × (10 + 1.4) R = 707 N

9.6 Force diagrams (SB p.84)

Weightlessness

2. The lift is at rest, moving upwards or downwards at constant velocity R a=0

R = 62 × 10 R = 620 N

620N

F

mg 147

F = ma R − mg = 0 R = mg

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9.6 Force diagrams (SB p.84)

Weightlessness

3. The lift is accelerating downwards with a = 1.4 m s-2 F = ma mg − R = ma

R a = 1.4 m s-2 533N

F

mg 148

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R = mg − ma

R = 62 × (10 − 14) R = 533 N

9.6 Force diagrams (SB p.84)

Weightlessness

4. The lift falls freely with a = 10 m s-2 F = ma R

mg − R = ma R = mg − ma R = mg − mg

a = 10 m s-2 0N

R=0

weightlessness F

mg 149

Note: the actual weight of the boy does not change, only his feeling of weight changes

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9.6 Force diagrams (SB p.85)

Weightlessness

Class Practice 6 ： When a lift is moving upwards with increasing acceleration, the supporting force on the passenger ____________ increases (increases / does not change / decreases). When the lift is moving downwards with increasing acceleration, the supporting force on the passenger _____________ (increases / does not change / decreases). decreases When the lift falls freely under gravity, the supporting force would become __________. zero Ans wer 150

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Section 9.7 Pressure

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9.7 Pressure (SB p.86)

Pressure Force perpendicular to an area Pressure = Area F P= A

Unit: pascal (Pa)

force perpendicular to an area

F

area

A

152

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9.7 Pressure (SB p.86)

Applications of pressure

P= A↓

A↓

153

A↓ Manhattan Press (H.K.) Ltd. © 2001

F A P↑

A↓

9.7 Pressure (SB p.87)

Without skis Pressure 500 = −4 100 × 10 = 50 kPa 500 N

100 cm2

154

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9.7 Pressure (SB p.87)

With skis Pressure 500 = −4 500 × 10 = 10 kPa 500 N

500 cm2

155

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9.7 Pressure (SB p.87)

Vehicles with greater surface area reduce pressure

A↑ P= A↓

F A

A↑

P↑ 156

A↑

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9.7 Pressure (SB p.88)

Class Practice 7 ： (a) A large box of mass 10 kg is resting on a floor as shown below. Find the pressure exerted on the floor in each case. 10 x 10 = 100 N Weight of the box = _________N 0.25 x 1

0.25 m2

2 In Fig. a, area of the base = ___________ = __________ m 100 400 Pa 0.25 P = ____________ = ___________ Pa

Ans wer

157

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9.7 Pressure (SB p.88)

Class Practice 7 (Cont’d) ：

(a)

0.5 m2

0.5 x 1

In Fig. b, area of the base = ________ = ______m2 P=

158

100 0.5

=

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Pa 200 Pa

9.7 Pressure (SB p.88)

Class Practice 7 (Cont’d) ： (b) Account for the difference in pressure found in (a).

If the base area is larger, the pressure exerted on the floor is smaller ____________ (larger / smaller).

Ans wer

159

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Chapter 10 Momentum 10.1 10.2 10.3 10.4

Momentum Momentum Change, Impulsive Force and Impulse Conservation of Momentum Newton’s Third Law of Motion Manhattan Press (H.K.) Ltd. © 2001

Section 10.1 Momentum

Manhattan Press (H.K.) Ltd. © 2001

10.1 Momentum (SB p.106)

Momentum Momentum = Mass × Velocity p =mv Unit: kg m s-1 or N s vector quantity

mass (m)

162

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velocity (v)

10.1 Momentum (SB p.106)

Momentum of trolley A = mAvA =2× 3 = 6 kg m s-1 (to the right)

163

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10.1 Momentum (SB p.106)

Momentum of trolley B = mBvB = 2 × (-4) = -8 kg m s-1 (to the left)

164

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10.1 Momentum (SB p.106)

Class Practice 1 ： A jet plane of mass 50 000 kg travels at a velocity of 250 m s-1 towards east. (a) When the jet plane is moving at the velocity of 250 m s-1, -1 250 m s (due east) Velocity ＝ ˍˍˍˍˍˍˍˍ Momentum ＝ m × v 　　

50 000 x 250 ＝ ˍˍˍˍˍˍˍˍ

　　

1.25 x 107 kg m s-1 (due east) ＝ ˍˍˍˍˍˍˍˍ

(b) After the jet plane has landed on an airport, -1 0 m s Velocity ＝ ˍˍˍˍˍˍˍˍ 0 kg m s-1 Momentum ＝ ˍˍˍˍˍˍˍˍ 165

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Ans wer

Momentum Change, Impulsive Force and 10.2 Impulse Section • Momentum change • Impulsive force • Impulse Manhattan Press (H.K.) Ltd. © 2001

10.2 Momentum change, impulsive force and impulse (SB p.107) Momentum change

Momentum change

Before collision ：

momentum (pi) = mu

After collision ：

momentum (pf) = mv

Momentum change = mv - mu 167

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10.2 Momentum change, impulsive force and impulse (SB p.107)

Impulsive force (F) ─

unbalanced force acting on the ball during the collision with the wall

u F

Unit: newton (N) 168

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Impulsive force

10.2 Momentum change, impulsive force and impulse (SB p.107)

Impulsive force

Impulsive force

(Newton' s second law of motion ) F = ma ... ... (1)　　 ( ) v − u Substitute a = into (1), t ( ) F = m v − u = mv − mu t t

u F

169

Impulsive force = Rate of change of momentum mv − mu F= t

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10.2 Momentum change, impulsive force and impulse (SB p.108)

Impulsive force

Impulsive force can deform a body’s shape

F

170

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10.2 Momentum change, impulsive force and impulse (SB p.108)

A driver without wearing a seat belt (time of impact = 0.05 s)

u = 30 m s-1 v = 0 m s-1

Impulsive force on the driver (F ) ( ) m × v − u = t

( ) 60 × 0 − 30 = 0.05 = −36 000 N 171

Impulsive force

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10.2 Momentum change, impulsive force and impulse (SB p.108)

Impulsive force

A driver wearing a seat belt (time of impact =1 s)

u = 30 m s-1 v = 0 m s-1

Impulsive force on the driver (F ) ( ) m × v − u = t

( ) 60 × 0 − 30 = 1 = −1 800 N 172

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10.2 Momentum change, impulsive force and impulse (SB p.108)

A driver wearing a seat belt time of impact increases ( t ↑ )

F = mv − mu t

impulsive force reduces ( F ↓) chance of suffering serious injuries decreases 173

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Impulsive force

air bag

10.2 Momentum change, impulsive force and impulse (SB p.109)

Safety designs of cars

bumper

174

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Impulsive force

10.2 Momentum change, impulsive force and impulse (SB p.110)

Impulse

Impulse Impulse = Impulsive force (F ) × Time of impact (t ) mv − mv = ×t t

Ft = mv − mu = change in momentum Impulse is a vector quantity Unit: N s or kg m s-1 175

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10.2 Momentum change, impulsive force and impulse (SB p.111)

Impulse

Class Practice 2 ： Find the impulse of the tennis ball in Example 1.

Impulse = Change in momentum -200 × 0.0145 　　　　　 = ˍˍˍˍˍˍˍˍ -2.9 　　　　　 = ˍˍˍˍˍˍˍˍ N s Ans wer

176

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Section 10.3 Conservation of Momentum • Elastic collision • Inelastic collision • Explosion of trolleys • Recoil speed • Apparent non-conservation of momentum Manhattan Press (H.K.) Ltd. © 2001

10.3 Conservation of momentum (SB p.111)

Elastic collision

Elastic collision ─ two bodies separate after collision

178

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10.3 Conservation of momentum (SB p.112)

Experiment 10A ： Elastic collision of trolleys

Intro. VCD

Expt. VCD

179

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Elastic collision

10.3 Conservation of momentum (SB p.113)

Elastic collision

Elastic collision One trolley colliding with one trolley uA

mA = 1 kg

180

mB = 1 kg

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10.3 Conservation of momentum (SB p.113)

Elastic collision

Elastic collision Two trolleys colliding with one trolley uA

mA = 2 kg

181

mB = 1 kg

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10.3 Conservation of momentum (SB p.113)

Elastic collision

Elastic collision Three trolleys colliding with one trolley uA

mA = 3 kg

182

mB = 1 kg

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10.3 Conservation of momentum (SB p.114)

Elastic collision

Elastic collision

Mass / kg

Before collision After collision Initial Total Total Final velocity velocity momentum momentum -1 / m s / m s-1 / kg m s-1 / kg m s-1

mA

mB

uA

uB

mAuA+mBuB

vA

vB

mAvA+mBvB

1

1

0.2

0

0.2

0

0.19

0.19

2

1

0.25

0

0.5

0.085

0.33

0.5

3

1

0.3

0

0.9

0.15

0.44

0.89

Total momentum before collision = Total momentum after collision In an elastic collision, momentum is conserved 183

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10.3 Conservation of momentum (SB p.114)

Inelastic collision

Experiment 10B ： Inelastic collision of trolleys Expt. VCD

184

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Inelastic collision

10.3 Conservation of momentum (SB p.116)

Inelastic collision

Inelastic collision ─ two bodies stick together after collision Mass / kg

Before collision After collision Initial Total Total Final velocity velocity momentum momentum -1 /ms -1 -1 /ms / kg m s / kg m s-1

mA

mB

uA

uB

mAuA+mBuB

vA

vB

mAvA+mBvB

1

1

0.23

0

0.23

0.12

0.12

0.24

2

1

0.35

0

0.7

0.23

0.23

0.69

1

2

0.28

0

0.28

0.095 0.095

0.285

Total momentum before collision = Total momentum after collision In an inelastic collision, momentum is conserved Manhattan Press (H.K.) Ltd. © 2001 185

10.3 Conservation of momentum (SB p.116)

Inelastic collision

Law of conservation of momentum When there are no external forces, Total momentum before collision = Total momentum after collision mAuA + mBuB = mAvA + mBvB

186

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10.3 Conservation of momentum (SB p.118)

Inelastic collision

Class Practice 3 ： Referring to part (a) of Example 2, 2 × 3 + 1 × (-2) Total momentum before collision = ˍˍˍˍˍˍ 　　　　　　　　　　

-1 4 kg m s = ˍˍˍˍˍˍ

2× 1+1× 2 Total momentum after collision = ˍˍˍˍˍˍ 　　　　　　　　　　 = ˍˍˍˍˍˍ 4 kg m s-1 Ans 　　　　 wer

187

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10.3 Conservation of momentum (SB p.118)

Inelastic collision

Class Practice 3 (Cont’d) ： Change in momentum of trolley A mAvA - mAuA = ˍˍˍˍˍˍˍˍ -1 2 x 1 2 x 3 = -4 kg m s = ˍˍˍˍˍˍˍˍ

Change in momentum of trolley B mBvB - mBuB = ˍˍˍˍˍˍˍˍ 1 x 2 - 1 x (-2) = 4 kg m s-1 = ˍˍˍˍˍˍˍˍ

188

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Ans wer

10.3 Conservation of momentum (SB p.118)

Explosion of trolleys

Explosion of trolleys ─ separation of objects into two parts or more

cork bottle

189

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10.3 Conservation of momentum (SB p.119)

Explosion of trolleys plunger adhesive tape small stick to fix the tape on the trolley

button

Before explosion Total momentum = mAuA + mBuB 　　　　 190

= 0 kg m s-1 Manhattan Press (H.K.) Ltd. © 2001

Explosion of trolleys

10.3 Conservation of momentum (SB p.119)

Explosion of trolleys

Explosion of trolleys

It obeys law of conservation of momentum After explosion Total momentum = mAvA + mBvB 　　　　 191

= 1 × (-0.5) + 1 × 0.5 = 0 kg m s-1 Manhattan Press (H.K.) Ltd. © 2001

10.3 Conservation of momentum (SB p.119)

Recoil speed

Recoil speed ─ cannons or rifles move backwards when they are fired

192

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10.3 Conservation of momentum (SB p.120)

Recoil speed

Firing a cannon ball at rest

193

after firing

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10.3 Conservation of momentum (SB p.120)

Recoil speed

Law of conservation of momentum Total momentum after firing = Total momentum before firing Momentum of the ball + Momentum of the cannon = 0 mv + MV = 0 V = − mv M Mass of cannon (M) >> Mass of ball (m)

Recoil speed of cannon (V) << Recoil speed of ball (v) 194

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10.3 Conservation of momentum (SB p.121)

Apparent non-conservation of momentum

Apparent non-conservation of momentum m2

m1

When a boy runs forwards, does the earth remains at rest? Mass of the boy << Mass of the earth m2 << m1 The movement of the earth is unnoticeable

195

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Section 10.4 Newton’s Third Law of Motion • Action and reaction pair • Tug-of-war • Example of Newton’s third law • False examples of Newton’s third law Manhattan Press (H.K.) Ltd. © 2001

10.4 Newton’s third law of motion (SB p.122)

Action and reaction pair

Action and reaction pair Experiment 10C ： Action and reaction Expt. VCD

197

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10.4 Newton’s third law of motion (SB p.122)

Action and reaction pair

Action ─ the force (f) acting on the cardboard by the wheels of the car ─ pushes the cardboard moving to the right Reaction ─ the force (f ’) acting on the wheels by the cardboard ─ pushes the car moving to the left motion of toy car motion of cardboard cardboard 198

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10.4 Newton’s third law of motion (SB p.123)

Action and reaction pair

Newton’s third law of motion Action and reaction are equal in magnitude but opposite in direction

FA = FB

199

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10.4 Newton’s third law of motion (SB p.124)

Tug-of-war

Tug-of-war T = T’ , who will win? internal force

T

=

T’

f

f’

Determined by f and f ’ (external force) 200

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10.4 Newton’s third law of motion (SB p.124) Example of Newton’s third law

Examples of Newton’s third law 1. Sprinter and starting block F ─ action F’ ─ reaction

F

201

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F’

10.4 Newton’s third law of motion (SB p.125) Example of Newton’s third law

Examples of Newton’s third law 2. Stepping off a boat

3. Rocket and jet propulsion

F

F’

F’ F

F ─ action F’ ─ reaction 202

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10.4 Newton’s third law of motion (SB p.127)

False examples of Newton’s third law

False examples of Newton’s third law

balanced forces

203

Are the weight of Jessie and the supporting force by the chair an action and reaction pair? weight



Action and reaction – equal in magnitude （ ） supporting – opposite in direction （） force – act on two objects （ ）

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10.4 Newton’s third law of motion (SB p.127)

False examples of Newton’s third law

Are the force acting on the chair by Jessie and force acting on Jessie by chair an action and reaction pair?

action and reaction

204

force acting on Jessie by chair

Action and reaction – equal in magnitude （ ） – opposite in direction （） – act on two objects （） force acting on chair by Jessie



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10.4 Newton’s third law of motion (SB p.127)

action and reaction

force acting on Jessie by the earth



force acting on the earth by Jessie

205

False examples of Newton’s third law

Are the force acting on the earth by Jessie and force acting on Jessie by the earth an action and reaction pair?

Action and reaction – equal in magnitude （ ） – opposite in direction （） – act on two objects （）

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Chapter 11 Work, Energy and Power 11.1 Work 11.2 Different Forms of Energy 11.3 Conversion of Potential Energy and Kinetic Energy 11.4 Non-conservation of Mechanical Energy 11.5 Power Manhattan Press (H.K.) Ltd. © 2001

Section 11.1 Work • Work done

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11.1 Work (SB p.137)

Work is done when a force is applied on an object

force displacement

208

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11.1 Work (SB p.138)

Work done

Work done Work done ＝ Applied force × Displacement W ＝force F ×issparallel to the displacement The applied F

F s

Unit: J (joule) or N m 209

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11.1 Work (SB p.138)

Work done

Work done Work done (W) = Fs cos θ

F

F s F

θ

F

θ

F cosθ s

210

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F cosθ

11.1 Work (SB p.140)

Work done

Class Practice 1 ： Calculate the work done on the suitcase in each of the following cases. A force of 60 N is applied and the suitcase moves through a distance of 10 m in each case. (a) The force is applied in the same direction as the motion of the suitcase.

Fxs W = ˍˍˍˍˍˍ 60 x 10 　 = ˍˍˍˍˍˍ 600 J = ˍˍˍˍˍˍ Ans wer 211

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11.1 Work (SB p.140)

Work done

Class Practice 1 (Cont’d) ： (b) The force is applied at an angle of 30o to the direction of motion of the suitcase.

F s cos θ W= ˍˍˍˍˍˍ 60 x 10 x cos 30o 　 = ˍˍˍˍˍˍ

= ˍˍˍˍˍˍ 520 J Ans wer 212

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11.1 Work (SB p.140)

Work done

No work is done when: 1. moving with inertia 2. the body is stationary 3. the direction of motion of the body is perpendicular to that of the applied force 1.2. 3. uniform velocity

mg mg

213

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Section 11.2 Different Forms of Energy • Potential energy • Kinetic energy • Other forms of energy

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11.2 Different forms of energy (SB p.140)

Energy ─ capacity to do work Unit: joule (J) 1 kilojoule （ 1 kJ ）＝ 1 000 J ＝ 103 J 1 megajoule （ 1 MJ ）＝ 1 000 000 J ＝ 106 J Mechanical energy ─ potential energy and kinetic energy

215

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11.2 Different forms of energy (SB p.141)

Potential energy

Potential energy ─ gravitational potential energy and elastic potential energy

Gravitational potential energy ─ work done on the object by gravitational pull Work done on the object =Fs = mgh = change in potential h energy of the object 　 Potential energy (P.E.) = mgh 216

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11.2 Different forms of energy (SB p.141)

Potential energy

Potential energy (P.E.) P.E. of the object at hf = mgh P.E. of the object at hi = 0

hf

Gain in P.E. = mgh 　　　　 h

hi

217

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11.2 Different forms of energy (SB p.142)

Potential energy

Potential energy is independent of the path taken

h h

218

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mgh

mgh

11.2 Different forms of energy (SB p.142)

Potential energy

Potential energy depends on the reference level Reference level ： 1. The ground Book hA

A hB hC

hD hE

219

the ground

Potential Potential energy energy 18.9 J mAghA

B

mBghB

1.4 J

C

mCghC

9.1 J

D

mDghD

1.75 J

E

mEghE

0J

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11.2 Different forms of energy (SB p.142)

Potential energy

Reference level ： 2. Book D Book hA

A

hB hC

hD hE

the ground

Potential Potential energy energy 12.6 J mAghA

B

mBghB

0.7 J

C

mCghC

4.55 J

D

mDghD

0J

E

mEghE

-6.3 J

Note: The difference in P.E. of two books is the same for different reference levels. 220

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11.2 Different forms of energy (SB p.143)

Potential energy

Elastic potential energy

bow

trampoline

221

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11.2 Different forms of energy (SB p.143)

Kinetic energy

Kinetic energy ─ a body possesses K.E. when moving

By　v2 − u 2 = 2as

Potential energy (K.E.) = F × s

v2 a= 2s 　F= ma

mv 2 = 2

K.E. = 1 mv2 2

mv 2 ∴F = 2s u = 0 m s-1

v F

F s

222

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11.2 Different forms of energy (SB p.145)

Kinetic energy

Class Practice 2 ： A stationary golf ball of mass 0.25 kg is struck with a force of 200 N. If the club is in contact with the ball for a distance of 1 cm, find the speed of the ball when it leaves the club. Work = Change in kinetic energy 1 2 mv Fs= 2 √(2 x F x s) / m v= ˍˍˍˍˍˍˍˍ

√(2 x 200 x 0.01) / 0.25 = ˍˍˍˍˍˍˍˍ 4 m s-1 ∴ v= ˍˍˍˍˍˍ 4 m. s-1 The speed of the golf ball is ˍˍˍ 223

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Ans wer

11.2 Different forms of energy (SB p.145)

Other forms of energy

Other forms of energy wind energy

solar energy

224

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11.2 Different forms of energy (SB p.146)

Other forms of energy

Other forms of energy

nuclear energy

tidal energy

225

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Section 11.3 Conversion of Potential Energy and Kinetic Energy • • • •

Conservation of energy Energy conversion in free falling motion Energy conversion in pendulum motion Energy conversion in elastic collision Manhattan Press (H.K.) Ltd. © 2001

11.3 Conversion of potential energy and kinetic energy (SB p.146) Conservation of energy

Conservation of energy

─ energy cannot be created or destroyed, but can transform from one form to another light, heat and other forms of energy

 potential energy of water kinetic energy of 227turbines

electrical energy Manhattan Press (H.K.) Ltd. © 2001

11.3 Conversion of potential energy and kinetic energy (SB p.147) Energy conversion in free falling motion

Energy conversion in free falling motion u=0

v

Loss in P.E. = Weight × Distance = mgs v2 = u2 + 2as v2 = 2gs (½m) v2 = (½m) 2gs ½mv2 = mgs Gain in K.E. = Loss in P.E.

Note: K.E. + P.E. = constant 228

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11.3 Conversion of potential energy and kinetic energy (SB p.148) Energy conversion in free falling motion

Class Practice 3 ：

potential When a body is falling freely, its ____________ (kinetic / potential / mechanical) energy decreases and its kinetic mechanical ____________ energy increases. However, its ____________ energy remains unchanged. Ans wer

229

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11.3 Conversion of potential energy and kinetic energy (SB p.148) Energy conversion in pendulum motion

Example of pendulum motion

230

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11.3 Conversion of potential energy and kinetic energy (SB p.149) Energy conversion in pendulum motion

Experiment 11A ： Energy conversion in a simple pendulum Intro. VCD

Expt. VCD

231

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11.3 Conversion of potential energy and kinetic energy (SB p.150) Energy conversion in pendulum motion

Energy conversion in a simple pendulum

A

C B

speed increases 232

speed decreases

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11.3 Conversion of potential energy and kinetic energy (SB p.150) Energy conversion in pendulum motion

Energy conversion in a simple pendulum

B

point B (the lowest position)

Speed of the weight at B (v ) Maximum separation between successive dots = Time taken 0.0266 = = 1.33 m s−1 Manhattan Press (H.K.) Ltd. © 2001 233 0.02

11.3 Conversion of potential energy and kinetic energy (SB p.150) Energy conversion in pendulum motion

Energy conversion in a simple pendulum

Loss in P.E. = mg (hA − hB )

= 1× 10 × ( 0.2 − 0.1) =1 J 1 Gain in K.E. = mv 2 2 1 = × 1× (1.33) 2 2 = 0.88 J

max P.E.

Gain in K.E. ≈ Loss in P.E. 234

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max P.E.

max K.E.

11.3 Conversion of potential energy and kinetic energy (SB p.150) Energy conversion in pendulum motion

Energy conversion in a simple pendulum gain in kinetic energy

potential energy energy loss due to friction

235

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11.3 Conversion of potential energy and kinetic energy (SB p.152) Energy conversion in elastic collision

Energy conversion in elastic collision Before collision plunger

After collision

236

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11.3 Conversion of potential energy and kinetic energy (SB p.152) Energy conversion in elastic collision

Elastic collision

Total kinetic energy

Before collision K.E. = 1 mAuA2 + 1 mBuB2 plunger

After collision

237

2 2 1 2 = × 2 × ( 0.25) + 0 2 = 0.062 5 J

1 2 1 K.E. = mAv A + mBvB2 2 2 1 2 1 2 = × 2 × ( 0.085) + × 1× ( 0.33) 2 2 = 0.0617 J Manhattan Press (H.K.) Ltd. © 2001

11.3 Conversion of potential energy and kinetic energy (SB p.153) Energy conversion in elastic collision

Elastic collision

The total kinetic energy before and after the collision are nearly the same, the mechanical energy is conserved. The slight difference is due to ： • experimental error • presence of friction

238

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plunger

Section 11.4 Non-conservation of Mechanical Energy • Inelastic collision • Motion along rough surface

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11.4 Non-conservation of mechanical energy (SB p.153)

Inelastic collision

Inelastic collision Before collision

After collision

240

move together

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11.4 Non-conservation of mechanical energy (SB p.154)

Inelastic collision Total kinetic energy

Before collision K.E. = 1 m u 2 + 1 m u 2 A A B B 2 2 1 = × 1× ( 0.23) 2 + 0 2 = 0.026 5 J

1 1 K.E. = mAv 2 + mBv 2 2 2 1 move together = ( mA + mB ) v 2 2 1 = × (1+ 1) × ( 0.12) 2 2 = 0.014 4 J

After collision

241

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Inelastic collision

11.4 Non-conservation of mechanical energy (SB p.154)

Inelastic collision

Inelastic collision The total kinetic energy after the inelastic collision decreases, mechanical energy is not conserved.

Loss of mechanical energy → transform into internal energy and sound

242

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move together

11.4 Non-conservation of mechanical energy (SB p.154)

Motion along rough surface

243

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Motion along rough surface

11.4 Non-conservation of mechanical energy (SB p.155)

Motion along rough surface

Motion along rough surface

P.E .at top of the slide = K.E. at water surface 1 2 mgh = mv m 2 v = 2gh h = 10 m

= 2 × 10 × 10 = 14 m s−1

m v

244

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11.4 Non-conservation of mechanical energy (SB p.155)

Motion along rough surface

m s

friction ( f )

m

245

Motion along rough surface

Presence of friction • lower the speed Work done against friction W=fs

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Section 11.5 Power

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11.5 Power (SB p.156)

Power

Work done Power = Time taken

W P= t

force W ＝ mgh

247

Unit: watt (W) 1 W = 1 J s-1

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11.5 Power (SB p.156)

Power

E = 3 000 W E = 500 W E = 60 W E electrical energy 248

E kinetic energy

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11.5 Power (SB p.157)

Power 50 s

5m

40 kg

Potential energy gained Power = Time taken

mgh 　P = t 40 × 10 × 5 P= 50 = 40 W

0s

249

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11.5 Power (SB p.157)

Power W P= t F ×s = t = F ×v

v F

Power = Force × Velocity P = Fv

250

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11.5 Power (SB p.158)

Power air resistance (f’) friction by the ground (f)

driving force (F)

F =f +f' Output power of the car (P ) = F × v = (f + f ' ) × v 251

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11.5 Power (SB p.158)

Class Practice 4 ： A crane lifts a pack of steel rods at a constant speed of 0.5 m s-1 . Given that the output power of the crane is 2 500 W. Find the mass of the steel rods.

Fxv P ＝ ˍˍˍˍˍ P mxgxv ˍˍˍˍˍ ＝ ˍˍˍˍˍ m x 10 x 0.5 ˍˍˍˍˍ ＝ ˍˍˍˍˍ 2 500 ∴ m＝ ˍˍˍˍˍ 500kg Ans wer 252

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Chapter 12 Moment of a Force 12.1 Turning Effect of a Force 12.2 Principle of Moments 12.3 Parallel Forces

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Turning Effect of a Force

Section 12.1 • Moment

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12.1 Turning effect of a force (SB p.170)

Turning effect of a force ─ rotate about axes Pivot (or fulcrum) ─ position of axes axis pivot

pivot

axis

pivot pivot

axis

255

axis Manhattan Press (H.K.) Ltd. © 2001

12.1 Turning effect of a force (SB p.170)

Moment

Moment ─ the turning effect of a force Moment arm ─ perpendicular distance between the force and the pivot moment arm door hinge (pivot)

door hinge

256

Moment ＝ Force × Moment arm ＝F× d

Unit of moment: N m

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12.1 Turning effect of a force (SB p.171)

Moment

Moment Moment of F1 = F1 × d1 Moment of F2 = F2 × d2 door hinge

Same turning effect F1 × d1 = F2 × d2 As d1 > d2, 　　 F1 < F2

Note: the longer d, the smaller will be the force required. 257

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12.1 Turning effect of a force (SB p.172)

Moment

Which requires a smaller force ? (d2 > d1)

force

force pivot

case 1

258

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case 2



12.1 Turning effect of a force (SB p.172)

Moment

Moment ─ clockwise or anticlockwise clockwis e

pivot

pivot anticlockwise an anticlockwise moment

259

a clockwise moment

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Principle of Moments

Section 12.2

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12.2 Principle of moments (SB p.174)

Principle of moments Moment of F1

pivot

= 10 × 0.4 = 4 N m (anticlockwise) Moment of F2 = 5 × 0.8 = 4 N m (clockwise)

Two moments ： same in magnitude, but in opposite direction cannot turn 261

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12.2 Principle of moments (SB p.174)

Experiment 12A ： Principle of moments

Intro. VCD d2

d1

Expt. VCD lever

262

F2

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pivot

F1

12.2 Principle of moments (SB p.175)

Take the mid-point of the ruler as the pivot Total clockwise moment = 1.6 × 10 × 0.1 + 1 × 10 × 0.4 = 5.6 N m

pivot

263

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12.2 Principle of moments (SB p.175)

Take the mid-point of the ruler as the pivot Total anticlockwise moment = 0.4 × 10 × 0.2 + 1.2 × 10 × 0.4 = 5.6 N m

pivot

264

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12.2 Principle of moments (SB p.175)

Principle of moments anticlockwise moment

clockwise moment

When a body is in balance, Total clockwise moment ＝ Total anticlockwise moment 265

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12.2 Principle of moments (SB p.176)

Class Practice 1 ： Edmond, Jessie and Tracy are sitting on a seesaw at the positions shown in the figures. Given that their masses are 65 kg, 40 kg and 50 kg respectively, and the mass of the seesaw is negligible. Find the distance of Jessie from the pivot (d) when the seesaw is balanced. 1.7 m

d 1m

pivot 400 N 500 N

650 N

Take moment about the pivot, 650 x 1.7 Total clockwise moment ＝ ˍˍˍˍˍˍ 105 N m 　　　　　　　　 ＝1ˍˍˍˍˍˍ 266

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Ans wer

12.2 Principle of moments (SB p.176)

Class Practice 1 (Cont’d) ： 500 x 1 + 400 x d Total anticlockwise moment ＝ ˍˍˍˍˍˍ (500 + 400 x d) N m ＝ ˍˍˍˍˍˍ When the seesaw is balanced, Total clockwise moment ＝ Total anticlockwise moment 500 + 400 x d ˍˍˍˍˍˍ＝ ˍˍˍˍˍˍ 1 105 d＝ ˍˍˍˍˍˍ 1.51 m

Ans wer

267

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Parallel Forces

Section 12.3

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12.3 Parallel forces (SB p.178)

Parallel forces Reaction force (R) = Weight (W)

reaction force (R) by finger

no vertical movement weight (W) of ruler

269

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pivot

12.3 Parallel forces (SB p.179)

Parallel forces 1. Take the mid-point of ruler as the pivot Total clockwise moment = 16 x 0.1 + 10 x 0.4 = 5.6 N m Total anticlockwise moment = 12 x 0.4 + 4 x 0.2 = 5.6 N m 270

mid-point

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12.3 Parallel forces (SB p.179)

2. Take point A as the pivot

point A

Total clockwise moment = 12 × 0 + 4 × 0.2 + 1 × 0.4 + 16 × 0.5 + 10 × 0.8 = 17.2 N m Total anticlockwise moment = 43 x 0.4 = 17.2 N m 271

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12.3 Parallel forces (SB p.180)

When a body is in balance, Take any point as the pivot, Total anticlockwise moment = Total clockwise moment

Net moment = 0

spring balance

W= 0.1 kg

272

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12.3 Parallel forces (SB p.180)

Parallel forces Total upward force = 43 N Total downward force = 12 + 4 + 1 + 16 + 10 = 43 N Net Net force force == 00

273

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12.3 Parallel forces (SB p.180)

Conditions for a body to be in equilibrium

(i) (i) Total Total clockwise clockwise moment moment == Total Total anticlockwise anticlockwise moment moment net moment = 0

274

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12.3 Parallel forces (SB p.180)

Conditions for a body to be in equilibrium

(ii) (ii) Total Total upward upward force force == Total Total downward downward force force net force = 0

275

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12.3 Parallel forces (SB p.180)

Class Practice 2 ： Edmond and Tracy are sitting on a seesaw at the positions shown. Neglect the mass of the seesaw. Find the normal reaction (R) at the pivot in the following two ways. R Edmond 500 N

Tracy pivot 2.5 m

1.5 m (a) Take moment about Tracy. 300 N Total clockwise moment = Total anticlockwise moment Rˍˍˍˍˍˍˍˍ x 2.5 500 x (1.5 + 2.5) = ˍˍˍˍˍˍˍˍ R = ˍˍˍˍˍˍˍˍ 800 N (b) Net force = 0. Total upward force = Total downward force 500 + 300 R ˍˍˍˍˍˍˍ = ˍˍˍˍˍˍˍ Ans 800 N wer R = ˍˍˍˍˍˍˍ 276

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Chapter 13 Machines 13.1 13.2 13.3 13.4 13.5

What is a Machine? Efficiency Lever Screw Jack Inclined Plane Manhattan Press (H.K.) Ltd. © 2001

Section 13.1 What is a Machine?

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13.1 What is a machine? (SB p.185)

Machines ─ make the task become easy ─ change the magnitude or the direction of an applied force Apply a small force nutcracker (effort) lift up a heavy object (load) bottle opener

279

hammer

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Section 13.2 Efficiency

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13.2 Efficiency (SB p.186)

Efficiency of a machine 1. Ideal machine Energy input = Energy output

energy input

281

ideal machine

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useful energy output

13.2 Efficiency (SB p.186)

Efficiency of a machine 2. Real machine Energy input > Energy output energy loss

energy input

282

real machine

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useful energy output

13.2 Efficiency (SB p.186)

Principle of conservation of energy Energy input = Useful energy output + Energy loss energy loss

energy input

283

real machine

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useful energy output

13.2 Efficiency (SB p.186)

Efficiency (e) of a machine Efficiency (e) =

284

Useful energy output Energy input

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× 100%

13.2 Efficiency (SB p.186)

1. Ideal machine

2. Real machine

Energy input = Energy output Energy input > Energy output Efficiency (e) = 100%

energy input

ideal machine

285

Efficiency (e) < 100% useful energy input

energy loss

energy input

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real machine

useful energy input

13.2 Efficiency (SB p.187)

Efficiency (e) of a machine Useful energy output Efficiency (e) = × 100% Energy input Useful power output = × 100% Power input

286

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Lever

Section 13.3

• Efficiency of a lever • Types of levers

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13.3 Lever (SB p.189)

Lever ─ a device which can turn about a pivot effort load

pivot

pivot

288

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bar

13.3 Lever (SB p.189)

Lever pivot

Clockwise moment = E × E Anticlockwise moment = L × L When the lever is in equilibrium, E × E = L × L require a much smaller effort 289

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bar

E × E = L × L if 　 E >> L 　　 E << L

13.3 Lever (SB p.190)

Efficiency of a lever

Efficiency of a lever dE

dL

290

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13.3 Lever (SB p.190)

Efficiency of a lever

Efficiency of a lever

By similar triangles, ∆ ABO ≅ ∆ CDO d L lL = d E lE pivot

Efficiency (e) Work done on the load = × 100% Work done by the effort

L × dL L × lL = × 100% = × 100% E × dE E × lE 291

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13.3 Lever (SB p.191)

Efficiency of a lever

Experiment 13A ： Use of a lever and a screw jack Intro. VCD

Expt. VCD

292

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13.3 Lever (SB p.191)

Efficiency of a lever

Class Practice 1 ： Edmond uses a lever to lift up a load as

shown in the figure below. He finds that a minimum effort of 400 N is required to lift the load of 1 500 N at a uniform speed. Calculate the efficiency of the lever. load

load

pivot

) Work done on the (　　　　　 ) Efficiency = × ( 　　　　　 ) Work done by the (　　　　　 100% effort

L x L e=

( 　　　　　　　　 ) × ( 　　　　　　 ) ( 　　　　　　　　 )

100%

E x E = ____________________ (1 500 _ x 0.3) x 100% ( 400 x 1.2) = ________________ % 94% 293

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Ans wer

13.3 Lever (SB p.192)

Types of levers

Types of levers 1. Load ─ pivot ─ effort effort load

pivot

effort

pivot

294

effort

effort load pivot a crowbar

load load a pair of scissors

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pivot

a hammer

13.3 Lever (SB p.192)

Types of levers

Types of levers 2. Pivot ─ load ─ effort effort load

effort pivot a wheelbarrow

295

pivot effort

load

pivot

effort

load

pivot a nutcracker

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load a bottle opener

13.3 Lever (SB p.192)

Types of levers

Types of levers 3. Load ─ effort ─ pivot effort load

effort

load effort

effort

pivot

296

pivot

pivot

load a forearm

an ice tongs

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pivot load a fishing rod

Section 13.4 Screw Jack • Efficiency of a screw jack

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13.4 Screw jack (SB p.193)

Screw jack ─ lift a very heavy load

298

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13.4 Screw jack (SB p.193)

Experiment 13A ： Use of a lever and a screw jack Expt. VCD

299

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13.4 Screw jack (SB p.193)

When the handle is turned through a complete revolution, the load will be raised by a height of one pitch (p) raised by one p

platform

load (L)

pitch (p)

300

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one complete revolution

handle

13.4 Screw jack (SB p.194)

Efficiency of a screw jack

Work Work done done by by the the effort effort (W) (W) == EE ×× 2π 2π rr == 2π 2π rE rE Work Work done done on on the the load load (W (W ’)’) == LL ×× pp == Lp Lp

Efficiency of a screw jack the handle is turned through one complete revolution platform pitch (p)

load (L)

Efficiency (e ) handle

= =

301

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W' W

× 100%

Lp 2πrE

× 100%

Inclined Plane

Section 13.5

• Efficiency of an inclined plane

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13.5 Inclined plane (SB p.196)

Efficiency of an inclined plane

Inclined plane c e r i d

of n tio

tion o m

d loa

303

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13.5 Inclined plane (SB p.196)

Inclined plane o n o i ct e r i d d loa

f

tio o m

n

Efficiency of an inclined plane

Friction Friction is is negligible negligible EE == LL sin sin θθ == mg mg sin sin θθ << LL (as (as sin sin θθ << 1) 1) E < L Consider Consider friction friction (f) (f) EE == LL sin sin θθ ++ ff == mg mg sin sin θθ ++ ff

304

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13.5 Inclined plane (SB p.197)

Efficiency of an inclined plane ion t c dire d loa

n o i t o m f o

P.E. gained by the load 305

Efficiency of an inclined plane

Work Work done done on on the the load load == L L sin sin θθ Work Work done done by by the the effort effort == EE xx  == (L (L sin sin θθ ++ ff ))  == LL  sin sin θθ ++ ff  == mgh mgh ++ f f

Work done against f

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13.5 Inclined plane (SB p.197)

Efficiency of an inclined plane

Efficiency of an inclined plane

dire d loa

Efficiency (e) Work done on the load = ×100% Work done by the effort

Lsinθ = ×100% Manhattan Press (H.K.) Ltd. © 2001 Lsinθ +f 306

o n o i ct

f

n o i t mo

13.5 Inclined plane (SB p.197)

Efficiency of an inclined plane

Inclined at a smaller angle smaller effort is required (E = L  sinθ ) steeper slope

307

gentle slope

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Chapter 14 Wave Motion 14.1 14.2 14.3 Wave

What Is a Wave? Transverse and Longitudinal Waves Description of a

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Section 14.1

• What Is a Wave?

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14.1 What is a wave? (SB p.208)

Properties of a wave motion • A periodic motion • Transmit energy and information, e.g. light wave carries solar energy from the Sun to the Earth • E.g. water waves, sound waves, light waves, radio waves, microwaves

310

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14.1 What is a wave? (SB p.208)

Properties of a water wave • Easy to observe • Easy to produce (dropping a small stone into water forms a circular wave) • Water wave spreads radially outwards, each circular wave is called a circular pulse 311

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14.1 What is a wave? (SB p.208)

Circular waves

Circular wave — formed by circular pulses 312

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14.1 What is a wave? (SB p.209)

Propagation of a circular wave

Wavefront — a line joining a row of the peaks of pulses

wavefront 313

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14.1 What is a wave? (SB p.209)

Ray and wavefront Ray — an arrow representing the direction of propagation of a wave

The ray is perpendicular to the wavefront 314

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wavefront

ray

14.1 What is a wave? (SB p.209)

Plane waves Formed by vibration of straight pulses

315

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14.1 What is a wave? (SB p.209)

Propagation of a plane wave The ray and the wavefront are perpendicular to each other

316

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14.1 What is a wave? (SB p.209)

Ray and wavefront of a plane wave ray

wavefront

317

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14.1 What is a wave? (SB p.209)

Medium of a wave ─ for propagation of a wave Medium of a sound wave ：

Medium of a water wave ： water

318

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air

14.1 What is a wave? (SB p.209)

Mechanical wave Mechanical wave • Waves that require media to propagate • Examples: water wave, sound wave

319

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14.1 What is a wave? (SB p.209)

Speed of a wave • Speed of a wave depends only on the medium in which the wave travels • Speed of a water wave depends on the depth of water, but not the speed of throwing the stone into the water

320

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Section 14.2 • Transverse and Longitudinal Waves

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14.2 Transverse and longitudinal waves (SB p.210)

Propagation of a water wave Water waves propagate in all directions. It only transmits energy, but not the water particles

322

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14.2 Transverse and longitudinal waves (SB p.210)

Movement of cork direction of propagation of water waves

direction of oscillation of cork

The cork moves up and down about a fixed position only, but never moves along with the wave 323

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14.2 Transverse and longitudinal waves (SB p.211)

Experiment 14A Wave motion transverse wave

324

Intro. VCD

Expt. VCD

longitudinal wave

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14.2 Transverse and longitudinal waves (SB p.212)

Transverse wave

The direction of oscillation is perpendicular to the direction of propagation of wave

direction of oscillation direction of propagation

325

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14.2 Transverse and longitudinal waves (SB p.212)

Crests and troughs

crests - peaks of the wave

troughs - lowest points of the wave

326

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14.2 Transverse and longitudinal waves (SB p.212)

If the spring is flicked at a larger magnitude…... • Magnitude of pulses is larger • Speeds of pulses remain unchanged

327

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14.2 Transverse and longitudinal waves (SB p.212)

If the spring is flicked at a faster rate …... • More pulses are generated • Speed of pulses remains unchanged

328

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14.2 Transverse and longitudinal waves (SB p.212)

If the spring is extended and flicked…... • Tension of the spring increases • Speeds of pulses increase

329

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14.2 Transverse and longitudinal waves (SB p.212)

Longitudinal wave The direction of oscillation is parallel to the direction of propagation of wave

direction of oscillation

330

direction of propagation

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14.2 Transverse and longitudinal waves (SB p.212)

Rarefactions and compressions of a longitudinal wave

compressions

rarefactions

331

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14.2 Transverse and longitudinal waves (SB p.212)

If the spring is pushed at a larger magnitude …...

• Magnitude of pulses is larger • Speed of pulses remains unchanged 332

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14.2 Transverse and longitudinal waves (SB p.213)

If the spring is pushed at a faster rate …... • More pulses are generated • Speed of pulses remains unchanged

If the spring is extended and pushed…... • Tension of the spring increases • Speed of pulses increases

333

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Section 14.3 Description of a Wave • Particle motion in a transverse travelling wave • Particle motion in a longitudinal travelling wave

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14.3 Description of a wave (SB p.213)

Amplitude ( A), unit: metre (m)

A

A A

line of equilibrium positions

• Maximum displacement of a particle from its equilibrium position • The larger the amplitude, the higher is the wave energy 335

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14.3 Description of a wave (SB p.213)

Wavelength (λ ), unit: metre (m)

• In a transverse wave, the distance between two adjacent wave crests (or troughs) • In a longitudinal wave, the distance between two adjacent compressions (or rarefactions)

λ 336

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14.3 Description of a wave (SB p.213)

Period (T ), unit: second (s) • Time required to generate one complete pulse • Time for a crest (or a trough) to travel one wavelength distance

time required T 337

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14.3 Description of a wave (SB p.213)

Frequency (f ), unit: Hertz (Hz) • Number of complete pulses generated in one second

Period = T 1 Frequency(f ) = (Hz) Period 1 = (Hz) T

338

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14.3 Description of a wave (SB p.214)

Particle motion in a transverse travelling wave

Displacement-position graph of a wave

t=0

339

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14.3 Description of a wave (SB p.214)

Particle motion in a transverse travelling wave

Displacement-position graph of a wave λ /4

t = T/4

340

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14.3 Description of a wave (SB p.214)

Particle motion in a transverse travelling wave

Displacement-position graph of a wave λ /2

t = T/2

341

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14.3 Description of a wave (SB p.214)

Particle motion in a transverse travelling wave

Displacement-position graph of a wave 3λ /4

t = 3T/4

342

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14.3 Description of a wave (SB p.214)

Particle motion in a transverse travelling wave

Displacement-position graph of a wave λ

t=T

343

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14.3 Description of a wave (SB p.215)

Wave equation:

v = fλ

λ

Particle motion in a transverse travelling wave

Time for one complete oscillation = T, Distance travelled = λ Velocity =

Distance Time

=

i.e.　　　 v = fλ

344

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λ

T

=

λ

1   f 

= fλ

Class Practice 1 ： A transverse wave is travelling to the right at a speed of 0.05 m s–1 . State the time elapse for the wave.

0.3 s

1.2 s Ans wer 345

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14.3 Description of a wave (SB p.217)

Particle motion in a transverse travelling wave

Two particles having the same displacement and vibrating at the same velocity are said to be in P and R are in phase phase

346

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14.3 Description of a wave (SB p.217)

Particle motion in a transverse travelling wave

Two particles are in antiphase when their displacements and velocities are both equal in magnitude but P and Q are in antiphase opposite in direction

347

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14.3 Description of a wave (SB p.218)

Particle motion in a transverse travelling wave

Draw the displacementtime graph of particle S

displacement velocity-time graph time

348

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Class Practice 2: Fig. a is the displacementposition graph of a travelling wave at a certain instant. Fig. b is the displacement-time graph of a particular particle on the wave. Find the amplitude, wavelength, period and speed of the wave from the displacement / displacement figures./ m m position / m

time / s Fig. b

Fig. a

Amplitude = Period = 349

0.8 m

0.05 m Wavelength =

Speed 10 = s

0.8 = 0.08 m s-1 10 Manhattan Press (H.K.) Ltd. © 2001

Ans wer

14.3 Description of a wave (SB p.219)

Particle motion in a transverse travelling wave

Predict motions of particles

•Draw a waveform (in solid line), then draw another waveform (in dotted line) that is next to the original one • Motions of particles can be predicted by adding vertical arrows from the solid curve to the dotted curve particle moving upwards particle moving downwards

350

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Class Practice 3: If a wave as shown below is moving to the left, state the directions of motion of the particles A, B and C at the instant shown. Ａ：

upward B

Ｂ：

Ｃ： 351

momentarily at rest

downward Manhattan Press (H.K.) Ltd. © 2001

A

C

Ans wer

14.3 Description of a wave (SB p.220)

Particle motion in a longitudinal travelling wave

Particle motion in a longitudinal travelling wave

352

compression (C)

rarefaction (R)

wavelength (λ )

wavelength (λ )

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14.3 Description of a wave (SB p.221)

Particle motion in a longitudinal travelling wave

Particle motion in a longitudinal travelling wave

353

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14.3 Description of a wave (SB p.221)

Particle motion in a longitudinal travelling wave

Displacement-position graph of each particle at t =1/T

equilibrium positions

displacement / cm

position / cm

354

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Particle motion in a longitudinal travelling wave

14.3 Description of a wave (SB p.223)

Displacement-time graph of longitudinal wave

•

Slope represents velocity

displacement / cm

particle is stationary

negative slope time

positive slope

355

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Class Practice 4 ： Fig. a shows the equilibrium positions of some particles along a spring. A longitudinal wave is generated and travelling from right to left. Fig. b shows the positions of these particles after a short time. a)

b) direction of wave

right 9 From Fig. a to Fig. b, particle 2 has moved to ____________ the and particle has moved to ___________.

356

the left

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Ans wer

Section 14.4 Stationary Wave • Transverse stationary wave • Particle motion in a transverse stationary wave Manhattan Press (H.K.) Ltd. © 2001

14.4 Stationary wave (SB p.224)

Transverse stationary wave

Experiment 14B Transverse stationary wave

Expt. VCD

358

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14.4 Stationary wave (SB p.225)

Transverse stationary wave

Experiment 14B Increase the frequency of the vibrator gradually

One loop 359

Two loops Manhattan Press (H.K.) Ltd. © 2001

Three loops

14.4 Stationary wave (SB p.226)

Particle motion in a transverse stationary wave

Particle motion in a transverse stationary wave

360

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14.4 Stationary wave (SB p.226)

Particle motion in a transverse stationary wave

Nodes (N) and antinodes (A)

Nodes (N) - positions where particles do not vibrate at all Antinodes (A) - positions where the amplitude of particles is largest A N

A N

N

λ /2 A

361

λ /2

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N

14.4 Stationary wave (SB p.226)

Particle motion in a transverse stationary wave

Particles that are moving in phase in phase

362

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14.4 Stationary wave (SB p.226)

Particle motion in a transverse stationary wave

Particles that are moving in antiphase antiphase

363

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Class Practice 5 When a guitar string is plucked, a wave is generated on the string. The wave on the string is a _______________ (transverse / longitudinal) _______________ (travelling /

transverse

stationary) wave.

stationary

Ans wer 364

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Chapter 15 Water Waves 15.1 Ripple Tank 15.2 Stroboscope 15.3 Wave Phenomena Manhattan Press (H.K.) Ltd. © 2001

Section 15.1

• Ripple Tank

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15.1 Ripple tank (SB p.236)

Objectives • To investigate the properties of water waves

• To learn different wave phenomena (e.g. reflection, refraction, diffraction and interference) with the help of ripple tank

367

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15.1 Ripple tank (SB p.237)

Experiment 15A Circular and straight pulses

368

Intro. VCD

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Expt. VCD

15.1 Ripple tank (SB p.238)

Experiment 15A Results

369

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15.1 Ripple tank (SB p.238)

Experiment 15A

Reason

bright dark bright dark bright dark bright

370

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15.1 Ripple tank (SB p.239)

Circular wave

371

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15.1 Ripple tank (SB p.239)

Plane wave

372

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Section 15.2

• Stroboscope

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15.2 Stroboscope (SB p.241)

Experiment 15B “Freezing” the wave pattern stroboscope

374

Manhattan Press (H.K.) Ltd. © 2001

Expt. VCD

15.2 Stroboscope (SB p.241)

Motion of a water wave λ λ /4 λ λ3λ/2/4

375

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13 t == 0T TT 42

15.2 Stroboscope (SB p.242)

Rotation of a stroboscope 13 t == 0T T 24

180º 270º 360º 90º

376

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15.2 Stroboscope (SB p.242)

Viewing of a plane wave

13 tt == 0TTT 424

180º 90º 270º 360º

377

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15.2 Stroboscope (SB p.244)

Class Practice 1: A circular disc printed with an arrow, as shown, is made to rotate at a speed of 30 revolutions per second. A student holds a hand stroboscope of one a slit to view the motion of the rotating disc. Suppose the student can view the arrow at t = 0 s, sketch the pattern observed by the student if the stroboscope is rotated at a speed of: 10 rev per second 378

30 rev 60 rev 90 rev per second per second per second Ans double viewing wer Manhattan Press (H.K.) Ltd. © 2001

Section 15.3

Wave Phenomena • Reflection • Refraction • Diffraction • Interference Manhattan Press (H.K.) Ltd. © 2001

15.3 Wave phenomena (SB p.245)

Experiment 15C Reflection of water waves

380

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Reflection Expt. VCD

15.3 Wave phenomena (SB p.246)

Reflection

Reflection of water waves obeys laws of reflection

381

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Reflection

15.3 Wave phenomena (SB p.246)

Laws of reflection

r=i i

382

r

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15.3 Wave phenomena (SB p.246)

Reflection

Reflection of a circular wave

383

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15.3 Wave phenomena (SB p.247)

Reflection

Properties of reflection

After reflection of a wave • Speed, frequency and wavelength remain unchanged • Direction of propagation changes

384

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Reflection

15.3 Wave phenomena (SB p.248)

Class Practice 2 : In each of the following cases, state the angle of incidence and draw the reflected wave.

Angle of incidence = ________

385

Ans wer

0°

Angle of incidence = ________

Manhattan Press (H.K.) Ltd. © 2001

30°

15.3 Wave phenomena (SB p.248)

Class Practice 2 : (b) Draw the reflected wave for the incident wave as shown below.

Ans wer 386

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Reflection

15.3 Wave phenomena (SB p.249)

Refraction

Refraction

A wave travels from one medium to another. If it travels at different speeds in these two media, its direction of propagation changes.

387

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15.3 Wave phenomena (SB p.249)

Experiment 15D Refraction of water waves

388

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Refraction Expt. VCD

15.3 Wave phenomena (SB p.250)

Refraction of a plane wave Showing how plane waves travel from deep regions to shallow regions

389

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Refraction

Refraction

15.3 Wave phenomena (SB p.250)

Refraction of a plane wave incident ray incident wave

refracted wave shallow region

deep region refracted ray

After the refraction, the ray bends towards the normal, both the wavelength and the speed decrease 390

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Refraction

15.3 Wave phenomena (SB p.251)

Class Practice 3:

When a water wave travels from a shallow region to a deep region, it bends ________________ normal. away This fromresults from the ____________ in the wave speed.

increase

Ans wer 391

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15.3 Wave phenomena (SB p.251)

Diffraction

Diffraction Waves bend around corners and spread at slits. This phenomenon is called diffraction

water waves diffract at the gateway

392

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15.3 Wave phenomena (SB p.251)

Experiment 15E Diffraction of water waves

393

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Diffraction Expt. VCD

Diffraction

15.3 Wave phenomena (SB p.252)

Diffraction of a plane wave

a larger gap width

394

a smaller gap width

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15.3 Wave phenomena (SB p.252)

Diffraction of a plane wave

•

When d and λ are about the same, diffraction is the most prominent

• The larger the d, the less prominent is the diffraction 395

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Diffraction

Diffraction

15.3 Wave phenomena (SB p.253)

Class Practice 4 : Draw the diffracted waves in the following cases. deep region

Ans wer 396

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shallow region

15.3 Wave phenomena (SB p.253)

Diffraction － a small obstacle

397

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Diffraction

15.3 Wave phenomena (SB p.253)

Diffraction

Diffraction － a large obstacle

398

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15.3 Wave phenomena (SB p.254)

Diffraction

Class Practice 5 ： Complete the following table which compares reflection, refraction and diffraction of water waves. In the case of refraction, the wave travels from a deep region toAfter aAfter shallow region. After After refractionAfter After

reflection refraction diffraction reflection diffraction Direction Direction bends towards spread out r=i Speed normal Frequencyunchanged Speed decreases unchanged Wavelength Frequency unchanged unchanged unchanged Wavelength unchanged decreases unchanged Ans wer 399

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Interference

15.3 Wave phenomena (SB p.255)

Experiment 15F Interference of water waves by a pair of dippers

400

Expt. VCD

by two narrow gaps

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15.3 Wave phenomena (SB p.256)

Experiment 15F Results

401

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Interference

15.3 Wave phenomena (SB p.256)

Interference

Condition for generating a stable interference pattern Coherent waves are two circular waves of same frequency, wavelength, amplitude and phase

402

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15.3 Wave phenomena (SB p.256)

Interference

Coherent sources Coherent sources are the wave sources that produce coherent waves

403

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15.3 Wave phenomena (SB p.257)

Constructive interference I When a crest meets a crest, the two waves reinforce each other

404

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Interference

15.3 Wave phenomena (SB p.257)

Constructive interference II When a trough meets a trough, the two waves reinforce each other

405

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Interference

15.3 Wave phenomena (SB p.257)

Destructive interference When a crest meets a trough, the two waves cancel each other

406

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Interference

Interference

15.3 Wave phenomena (SB p.258)

Path difference Path difference at X ＝ |S1X - S2X| X

407

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Interference

15.3 Wave phenomena (SB p.259)

Constructive interference

Point P Q R V 408

Path difference 0

Path difference = nλ , n = 0, 1, 2, …

λ 2λ λ Manhattan Press (H.K.) Ltd. © 2001

Interference

15.3 Wave phenomena (SB p.259)

Destructive interference

Point U W 409

Path difference λ /2 3λ /2

1  Path difference =  n + λ , n = 0,1,2,....... 2 

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Class Practice 6 ： The figure below shows an interference pattern in a ripple tank. Identify the kind of interference at the labeled points.

Ans wer destructive interference

constructive interference

P:ˍˍˍˍˍˍˍ 　　 Q:ˍˍˍˍˍˍˍ destructive interference constructive interference R:ˍˍˍˍˍˍˍ 　　 S:ˍˍˍˍˍˍˍ 410

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Interference

15.3 Wave phenomena (SB p.261)

Antinodal lines (A)

Antinodal lines : lines that link up the points of constructive interference of the same path difference

411

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Interference

15.3 Wave phenomena (SB p.261)

Nodal lines (N)

Nodal lines : lines that link up the points of destructive interference of the same path difference

412

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Class Practice 7: The following figure is an interference pattern marked with some nodal line (N) and antinodal lines (A). If S1P is equal to 30 cm and S2P is equal to 24 cm, the wavelength of the wave is 6 __________ cm. The path difference at Q is ___________________ 6 12 λ − 5λ = 1 12 λ = 1.5 × 6 = 9 cm

Ans wer

413

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15.3 Wave phenomena (SB p.263)

Interference

Increase the density of nodal and antinodal lines increase the separation of two sources

414

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15.3 Wave phenomena (SB p.263)

Increase the density of nodal and antinodal lines decrease the wavelength

415

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Interference

15.3 Wave phenomena (SB p.263)

Interference and energy redistribution

Interference

• According to the law of conservation of energy, energy cannot be created nor destroyed • Interference is just a process of energy redistribution energy at the points of destructive interference 416

energy at the points of constructive interference

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Chapter 16 Wave Nature of Light and Electromagnetic Spectrum 16.1 Wave Nature of Light 16.2 Electromagnetic Spectrum Manhattan Press (H.K.) Ltd. © 2001

Section 16.1 Wave Nature of Light • A brief history of light • Young’s double-slit experiment Manhattan Press (H.K.) Ltd. © 2001

16.1 Wave nature of light (SB p.277)

A brief history of light

A brief history of light Newton (1642 - 1727) • Light consisted of tiny particles Huygens (1629 - 1695) • Light is a wave motion Young (1773 - 1829) • Interference of light provides evidence for the wave nature of light Fresnel (1788 - 1827) • Gave a detailed explanation of Young’s findings 419

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Experiment 16A Young’s double-slit experiment

16.1 Wave nature of light (SB p.278)

Young’s double-slit experiment Intro. VCD Expt. VCD

Young’s double-slit

420

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16.1 Wave nature of light (SB p.279)

Young’s double-slit experiment

Experiment 16A Young’s interference fringes alternate bright and dark fringes

421

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16.1 Wave nature of light (SB p.279)

Young’s double-slit experiment

Double-slit makes the sources S1 and S2 coherent diffracted beam from S1

alternate bright and dark fringes

single slit 422

double slit

diffracted beam from S2 coherent sources

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screen

16.1 Wave nature of light (SB p.280)

Young’s double-slit experiment

Interference patterns formed after inserting different colour filters As green light has a shorter wavelength than red light, the density of interference fringes is higher

423

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16.1 Wave nature of light (SB p.280)

Young’s double-slit experiment

Class Practice 1:

In Young’s double-slit experiment, if the separation more

of the slits is increased, _______________ (more / fewer) fringes will be seen. Ans wer 424

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16.1 Wave nature of light (SB p.280)

Young’s double-slit experiment

Interference in daily lives In daily lives, the diffraction or interference of light are not often seen because ... • Wavelength of light is much shorter than the common obstacles, so diffraction is not prominent • Common light sources are not coherent

425

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Section 16.2 Electromagnetic Spectrum • Light as electromagnetic wave • Electromagnetic spectrum • Radio wave • Microwave

• Visible spectrum • Ultraviolet radiation • X-ray • Gamma ray

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16.2 Electromagnetic spectrum (SB p.281)

Light as electromagnetic wave

Scotch physicist James Clerk Maxwell He suggested: • Light was a kind of electromagnetic wave • Speed of light was 3× 108 m s–1

427

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16.2 Electromagnetic spectrum (SB p.282)

Light as electromagnetic wave

Electromagnetic wave magnetic field

direction of magnetic field

electric field

direction of propagatio n

direction of electric field

direction of electromagnetic wave

• When a charged particle oscillates about an equilibrium position, a varying electric field coupled with a varying magnetic field are generated

428

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16.2 Electromagnetic spectrum (SB p.282)

Electromagnetic spectrum

Propagation of an electromagnetic wave • Electromagnetic wave propagates at a direction perpendicular to both electric and magnetic fields. So it is a transverse wave • Propagate at a speed of 3 × 108 m s–1 in vacuum • Obey the wave equation v = fλ 429

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Electromagnetic spectrum

16.2 Electromagnetic spectrum (SB p.282)

Electromagnetic spectrum

radio wave

micro wave

visible light

infrared

ultraviolet X-ray

wavelength / m

red violet

frequency / Hz

430

gamma ray

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16.2 Electromagnetic spectrum (SB p.283)

Electromagnetic spectrum

Applications of electromagnetic wave

gam

y ma ra

X-ray

ultraviolet visible infrared radiation spectrumradiation microwave

ion rial Caut ma t e e v i t oac Radi

electromagnetic spectrum 431

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radio wave

16.2 Electromagnetic spectrum (SB p.283)

Classification of radio waves

Radio wave

• Extra low, very low and low frequency waves (10 Hz to 300 kHz) • Medium waves and short waves (300 kHz to 30 MHz) • Very high and ultrahigh frequency waves (30 MHz to 800 MHz) 432

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16.2 Electromagnetic spectrum (SB p.283)

Radio wave

ELF, VLF and LF Radio waves Extra Low Frequency (ELF)

Frequency range

used in deep ocean communications as they can penetrate well in sea water Very Low Frequency 1 kHz to 10 kHz used in national (VLF) security Low Frequency (LF) 10 kHz to 300 kHz communications

433

10 Hz to 1 kHz

Applications

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16.2 Electromagnetic spectrum (SB p.284)

Radio wave

MW and SW Radio waves Medium Waves (MW) Short Waves (SW)

434

Frequency range 300 kHz to1600 kHz

6 MHz to 30 MHz

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Applications used in radio broadcasting by amplitude modulation (AM) used in radio broadcasting

Radio wave

16.2 Electromagnetic spectrum (SB p.284)

VHF and UHF Radio waves

Frequency range

Very High 30 MHz Frequency to 200 MHz (VHF) Ultrahigh several Frequency hundred MHz (UHF) 435

Applications used in radio broadcasting by frequency modulation (FM) used in television broadcasting, pagers and mobile phones communications

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Radio wave

16.2 Electromagnetic spectrum (SB p.286)

Class Practice 2: Hit Radio broadcasts with a frequency of 99.7 MHz. What is the wavelength of this radio wave? By 　　　

λ

=

λ= =

436

v ( ) 　　　　　　　 (　　　　　　　 ) f (　　　　　　　 3 × 108 ) (　　　　　　　 ) 6

99.7 × 10 3.0 m ___ ______________

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Ans wer

Radio wave

16.2 Electromagnetic spectrum (SB p.286)

Diffraction of radio waves Radio waves of long wavelengths diffract more around an obstacle than those of short wavelengths

hill

hill

hill poor reception

437

better reception

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best reception

16.2 Electromagnetic spectrum (SB p.287)

Propagation of radio waves

Radio wave

Because of the curvature of the earth, receivers far away from transmitters cannot receive radio waves

transmitter

receiver

poor reception earth

438

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16.2 Electromagnetic spectrum (SB p.287)

Propagation of radio waves (method 1) repeater

transmitter

build a repeater

receiver

earth

439

Radio wave

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16.2 Electromagnetic spectrum (SB p.287)

Radio wave

Propagation of radio waves (method 2) ionosphere

earth 440

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reflection on the ionosphere

16.2 Electromagnetic spectrum (SB p.288)

Experiment 16B Interference of microwaves

Expt. VCD

441

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Microwave

16.2 Electromagnetic spectrum (SB p.289)

Microwave

Experiment 16B Results 3 cm microwave transmitter connect to a power supply

442

constructive interference and destructive interference occur

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microwave receiver microammeter

Class Practice 3 ： 3 cm microwave transmitter To power supply

microwave receiver microammeter

Chris used the above experimental set-up to study the interference of microwaves. He placed the receiver at the position of central maximum. He then moved the receiver towards A along the line AB. While doing so, the reading on the microammeter first dropped __________ rose (rose / dropped) and then __________ (rose / dropped) again. This variation continued when he further moved the receiver towards A. If Chris record the first maximum at N where S1N = 52 cm, the distance Ans S2N should be __________ cm. 55 Manhattan Press (H.K.) Ltd. © 2001 443 wer

16.2 Electromagnetic spectrum (SB p.290)

Properties of microwave

Microwave

Since the wavelengths of microwaves are short, they hardly diffract. This is the basic working principle of radar

radar aerial

444

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Microwave

16.2 Electromagnetic spectrum (SB p.290)

Radar detection

P1 is transmitted pulse P2 is reflected pulse

screen of radar

radar aerial

445

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16.2 Electromagnetic spectrum (SB p.291)

Police use radar to check speeding on roads

446

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Microwave

Microwave

16.2 Electromagnetic spectrum (SB p.291)

Satellite communication communication satellite

earth station receiver

earth 447

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16.2 Electromagnetic spectrum (SB p.292)

Microwave

Microwave oven microwaves of frequency 2.45 GHz

When water molecules inside the food are struck by microwaves, →molecules oscillate vigorously →internal energy of foods increases →temperature of food increases gradually →the food is cooked 448

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16.2 Electromagnetic spectrum (SB p.293)

Applications of infrared radiation

Infrared radiation

Due to the longer wavelength, infrared radiation is less scattered by fine particles than the visible light, and so passes through haze easily ordinary photograph

449

infrared photograph

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more clear

16.2 Electromagnetic spectrum (SB p.293)

Applications of infrared radiation

Infrared radiation

Infrared radiation emitted form plants can be detected by infrared photographs. This gives details of the vegetation (in red)

taken at Shatin

450

taken from an earth resource satellite

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16.2 Electromagnetic spectrum (SB p.294)

Applications of infrared radiation check the teeth

Infrared radiation

of a killer whale remote controls auto-focus camera

451

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Infrared radiation

16.2 Electromagnetic spectrum (SB p.294)

Applications of infrared radiation infrared night-vision equipment

452

display of infrared signals

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Visible spectrum Experiment 16C Visible spectrum and infrared radiation

16.2 Electromagnetic spectrum (SB p.296)

Expt. VCD

453

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16.2 Electromagnetic spectrum (SB p.296)

Experiment 16C Dispersion

w

gh i l e hit

red orange yellow green blue indigo violet

t

prism

454

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Visible spectrum

Applications of ultraviolet radiation - sunbathe

16.2 Electromagnetic spectrum (SB p.298)

used in producing vitamin D

455

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Ultraviolet spectrum

16.2 Electromagnetic spectrum (SB p.298)

Ultraviolet spectrum

Harm of sunbathe exposure to too much radiation may cause skin cancer

456

ultraviolet radiation

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ozone layer

Ultraviolet spectrum

16.2 Electromagnetic spectrum (SB p.299)

Applications of ultraviolet radiation check the genuineness of banknotes

457

reveal fluorescent security marks

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16.2 Electromagnetic spectrum (SB p.299)

X-ray

Applications of X-ray X-ray photograph

458

reveal the contents of luggage

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X-ray diffraction pattern from crystalline sodium chloride (NaCl)

16.2 Electromagnetic spectrum (SB p.300)

Gamma ray radioactive substances should be kept in lead-shielded box and handled with forceps

459

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Gamma ray

16.2 Electromagnetic spectrum (SB p.300)

Application of gamma ray radiotherapy

460

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Gamma ray

Chapter 17 Sound 17.1 Production and Propagation of Sound 17.2 Wave Nature of Sound 17.3 Properties of Sound 17.4 Ultrasonics Manhattan Press (H.K.) Ltd. © 2001

Section 17.1

• Production and Propagation of Sound Manhattan Press (H.K.) Ltd. © 2001

17.1 Production and propagation of sound (SB p.311)

Production of sound a piano

a chorus a double bass

463

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17.1 Production and propagation of sound (SB p.312)

Sounds are produced by vibrations of particles in a medium copper strips in a harmonica vibrate

vocal cords vibrate

guitar strings vibrate

drumskin vibrates

464

cymbals strike one another and then vibrate Manhattan Press (H.K.) Ltd. © 2001

17.1 Production and propagation of sound (SB p.312)

Sound wave are longitudinal waves direction of sound wave displacement of air molecules

loudspeaker

wavewavelength length molecules vibrate back and forth Sound waves propagates through the oscillation of air molecules

465

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17.1 Production and propagation of sound (SB p.313)

Loudspeaker is stationary

no vibration

466

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17.1 Production and propagation of sound (SB p.313)

Loudspeaker pushs forwards air molecules are pushed forwards

467

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17.1 Production and propagation of sound (SB p.313)

Loudspeaker drags backwards air molecules are dragged backwards

468

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17.1 Production and propagation of sound (SB p.313)

Vibrations of a tuning fork

A tuning fork vibrates and gives sound when it is struck rarefactions compressions

vibrating tuning fork

469

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17.1 Production and propagation of sound (SB p.314)

Does sound propagate in the following medium? solid

 liquid

Sound must propagate through medium

 gas





vacuum

470

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17.1 Production and propagation of sound (SB p.314)

Speed of propagation of sound about 200 m s -1

to 300 m s-1

gas about 1500 m s -1 to 3000 m s-1

liquid about 5000 m s-1 to 6000 m s-1

solid

471

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17.1 Production and propagation of sound (SB p.315)

Class Practice 1 ： During a thunderstorm, Edmond hears the thunderclap 6 s after he has seen the flash of lightning. If sound travels in air at a speed of 350 m s-1 , find the distance between the thundercloud and Edmond. Distance travelled Speed × Time taken

＝ ˍˍˍˍˍˍˍˍˍˍ 350 × 6

＝ ˍˍˍˍˍˍˍˍˍˍ 2 100 m

＝ ˍˍˍˍˍˍˍˍˍˍ 472

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Ans wer

Section 17.2

Wave Nature of Sound • Reflection • Refraction • Diffraction

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17.2 Wave nature of sound (SB p.316)

Echo — reflection of sound

474

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Reflection

17.2 Wave nature of sound (SB p.317)

Refraction of sound — like the refraction of light

air water direction of propagation of sound

475

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Refraction

Refraction

17.2 Wave nature of sound (SB p.317)

Sound and temperature sound speed high temperature

low temperature

476

Manhattan Press (H.K.) Ltd. © 2001

fast

slow

Refraction

17.2 Wave nature of sound (SB p.317)

The direction of propagation of sound curves downwards at night People can hear the voice farther away easily at night higher temperature direction of propagation of sound

warmer air

cooler air

lower temperature

477

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Refraction

17.2 Wave nature of sound (SB p.317)

The direction of propagation of sound curves upwards in daytime In daytime, people feel more difficult to hear the voice farther away, but people at the top of the hill can hear the voice from the people at the bottom of the hill easily lower temperature

cooler air

direction of propagation of sound

warmer air

higher temperature

478

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Diffraction Diffraction — person in the room can hear the sound of television outside …...

17.2 Wave nature of sound (SB p.318)

diffracted sound reflected sound

479

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17.2 Wave nature of sound (SB p.318)

Diffraction — person can hear around corner …...

480

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Diffraction

Diffraction

17.2 Wave nature of sound (SB p.318)

Intro. VCD

Experiment 17A Interference of sound waves CRO signal generator

Expt. VCD loudspeaker

microphone 481

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Diffraction

17.2 Wave nature of sound (SB p.318)

Experiment 17A

Results

constructive interference and destructive interference occur 482

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Section 17.3

Properties of Sound • Pitch and frequency • Loudness and intensity

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17.3 Properties of sound (SB p.319)

Diffraction

Musical notes and noises • Both musical notes and noises are sound waves • Waveforms of musical notes are regular, so pleasant to hear • Waveforms of noises are irregular, so unpleasant to hear

484

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17.3 Properties of sound (SB p.320)

Experiment 17B notes

Expt. VCD

Musical CRO

signal generator

pitch?

loudspeaker loudness?

485

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quality?

17.3 Properties of sound (SB p.321)

Pitch and frequency

The higher the pitch, the higher the frequency

frequencies of tuning forks increase gradually, producing sounds of eight different frequencies 486

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17.3 Properties of sound (SB p.322)

Loudness and intensity

Intensity of sound • A sound carries larger amount of energy has a higher intensity and larger amplitude • Different sensations of sounds of different intensities are called loudness

487

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17.3 Properties of sound (SB p.322)

Frequency original trace

increased frequency

decreased frequency

488

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Loudness and intensity

17.3 Properties of sound (SB p.322)

Intensity of sound

original trace

increased intensity

decreased intensity

489

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Loudness and intensity

17.3 Properties of sound (SB p.322)

Loudness and intensity

Audible frequency range of human • 20 Hz to 20 kHz • Most sensitive frequency range: 500 Hz to 5 kHz

490

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17.3 Properties of sound (SB p.324)

Quality

491

Different musical instruments generate different qualities of sound which depend on the waveforms of the notes

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Quality

17.3 Properties of sound (SB p.324)

Quality Musical instruments produce fundamental frequency note and overtones Fundamental frequency note: determines the pitch Overtones: other higher frequencies that superpose with the fundamental one Quality: different overtones of different amplitudes give characteristic quality

492

fundamental frequency

overtone (twice the fundamental frequency

resultant waveform

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Quality

Section 17.4

Ultrasonics • Applications of ultrasonics Manhattan Press (H.K.) Ltd. © 2001

17.4 Ultrasonics (SB p.326)

Ultrasonics (or ultrasound) • Frequency: above 20 kHz • Out of the audible range of human • Bats and dolphins can emit and detect ultrasound, this help them to decide the positions of obstacles and prey and communicate with each other

494

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Applications of ultrasonics

17.4 Ultrasonics (SB p.327)

Applications of ultrasonics

use of ultrasonics in cleaning spectacles ultrasonic imaging

detecting cracks in metals 222

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Applications of ultrasonics

17.4 Ultrasonics (SB p.328)

Sonar — applications on the surface of the sea

use sonar to detect depth of the sea

transmitter ultrasonic waves waves reflected from bottom

496

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Applications of ultrasonics

17.4 Ultrasonics (SB p.328)

Sonar — applications beneath the sea a submarine uses sonar to detect other objects

reflected sonar from the submarine on the left

497

sonar transmitted from the submarine on the right

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The End

498

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