Chapter 8 — Vector applicationsthefinneymathslab.weebly.com/uploads/8/1/0/4/... · Vector...

M C 1 1 Q l d - 8 152 V e c t o r a p p l i c a t i o n s

Exercise 8A — Force diagrams and the triangle of forces 1 a

b

c

d

e

f

g

h

i

j

2 a R

% = 1F

% + 2F

%

R%

= 4 i%

+ 3 j%

+ 3 3i%

+ 4 3 j%

= (4 + 3 3 ) i%

+ (3 + 4 3 ) j%

The answer is C.

b 3F%

= −( 1F%

+ 2F%

)

= −(4 + 3 3 ) i%

− (3 + 4 3 ) j%

F3 = 2 2(4 3 3) (3 4 3)+ + +

= 16 24 3 27 9 24 3 48+ + + + +

= 100 48 3+ ≈ 13.53

tan θ = (3 4 3)(4 3 3)

− +− +

(third quadrant)

= 1.0796 θ = 180° + tan−1(1.0796) = 180° + 47° = 227° The third force has magnitude 13.5 N at an angle of 227°

anticlockwise of i%

. The answer is E.

3 a

C2 = A2 − B2

= 52 − 42

= 25 − 16 = 9 C = 9 C = 3 N The answer is C. b From the figure in part a, sin θ = 4

5

θ = sin−1(0.8) = 53.1° In actual situation: Angle = 180° − 53.1° = 126.9° ≈ 127° The answer is E. 4 a i 1F

% = 13 i

% − 5 j

% and 2F

% = −4 i

% + 9 j

%

R%

= 1F%

+ 2F%

= 13 i%

− 5 j%

− 4 i%

+ 9 j%

= 9 i%

+ 4 j%

ii 1F%

= 8 i%

+ 6 j%

and 2F%

= −14 i%

− 9 j%

R%

= 8 i%

+ 6 j%

−14 i%

− 9 j%

= −6 i%

− 3 j%

iii 1F%

= 2 2i%

− 3 j%

and 2F%

= − 3 5i%

+ 2 j%

R%

= 2 2i%

− 3 j%

− 3 5i%

+ 2 j%

= ( 2 2 3 5− ) i%

− j%

b i R%

= 2 29 4+

= 97

ii R%

= 2 2( 6) ( 3)− + −

= 45 = 3 5

Chapter 8 — Vector applications

V e c t o r a p p l i c a t i o n s M C 1 1 Q l d - 8 153

iii R%

= 2 2(2 2 3 5) ( 1)− + −

= 8 12 10 45 1− + +

= 54 12 10− c i 3F

% = −( 1F

%+ 2F%

)

= −9 i%

− 4 j%

ii 3F%

= −( 1F%

+ 2F%

)

= 6 i%

+ 3 j%

iii 3F%

= −( 1F%

+ 2F%

)

= (3 5 2 2)− i%

+ j%

5 i a 1F%

= 3 i%

− 5 j%

, 2F%

= 4 i%

+ 9 j%

and 3F%

= −6 i%

− 2 j%

R%

= 1F%

+ 2F%

+ 3F%

= 3 i%

− 5 j%

+ 4 i%

+ 9 j%

− 6 i%

− 2 j%

= i%

+ 2 j%

b R%

= 2 21 2+

= 5 c tan θ = 2

1

= 2 θ = 63.4° d 4F

% = − ( 1F

%+ 2F%

+ 3F%

)

= − i%

− 2 j%

ii a 1F%

= i%

− j%

, 2F%

= −2 i%

+ 3 j%

and 3F%

= 2i%

− 2 j%

R%

= i%

− j%

− 2 i%

+ 3 j%

+ 2i%

− 2 j%

= ( 2 1)i−%

b R%

= 2 1−

c tan θ = 0 θ = 0° d 4F

% = − ( 1F

%+ 2F%

+ 3F%

)

4F%

= − ( 2 1)i−%

= (1 2)i−%

6 a i

In , i j

% %notation

25cos60 25sin 60

25 25 32 2

F i j

i j

= +

= +

o o

% % %

% %

45cos60 45sin 60

45 45 32 2

G i j

i j

= − +

−= +

o o

% % %

% %

25 25 3 45 45 32 2 2 220 70 32 210 35 3

R F G

i j i j

i j

i j

= +

−= + + +

−= +

= − +

% % %

% %% %

% %

% %

( ) ( )2210 35 3

377561.44

R = − +

==

%

To find the angle with north ( R%

is west of north)

10tan35 30.1659.4

θ

θ

=

== °

So R%

acts N 9.4° W R

%is a force of 61.4 N acting N 9.4° W.

ii 3F%

= − R%

, would be a force of 61.4 N acting S 9.4° E.

b i R%

= W%

+ F%

R

% = −80 j

% + 50 i

% + 75 j

%

= 50 i%

− 5 j%

ii 3F%

= − R%

= −50 i%

+ 5 j%

c i R%

= A%

+ B%

= 26 i

% − 80 j

% − 30 i

% + 92 j

%

= −4 i%

+ 12 j%

ii 3F%

= − R%

= 4 i%

− 12 j%

d i

ln i

%, j%

notation:

F%

= −10 cos 40° i%

− 10 sin 40° j%

= −7.66 i%

− 6.43 j%

G%

= −7 cos 45° i%

+ 7 sin 45° j%

= −4.95 i%

+ 4.95 j%

H%

= 12 cos 20° i%

+ 12 sin 20° j%

= 11.28 i%

+ 4.10 j%

R%

= F%

+ G%

+ H%

= −7.66 i

% − 6.43 j

% − 4.95 i

% + 4.95 j

% + 11.28 i

%

+ 4.10 j%

= −1.33 i%

+ 2.62 j%

R%

= 2 2( 1.33) (2.62)− +

= 2.94 ≈ 2.9 N Find the angle R

%makes with north ( R

%is west of north)


tan θ = 1.332.62

θ = 26.9° Direction is N 26.9° W R

%is a force of 2.9 N acting N 26.9° W.

ii 3F%

= − R%

3F%

is 2.9 N acting S 26.9° E.

7 a

Z2 = X2 + Y2 − 2XY cos 60° = 102 + 102 − 2(10)(10) cos 60° = 100 Z = 10 N

b sin(180 )Y

θ° − = sin 60Z

°

sin (180° − θ ) = 10 sin 6010

°

= sin 60° 180° − θ = 60° θ = 120°

8

C2 = A2 + B2 − 2AB cos (180° − θ ) 17002 = 10002 + 12002 − 2(1000)(1200) cos (180° − θ) 2 890 000 = 1 000 000 + 1 440 000 − 2 400 000 cos (180° − θ) 2 400 000 cos (180° − θ ) = − 450 000 cos (180° − θ) = −0.1875 180° − θ = 100.8° θ = 180° − 100.8° = 79.2°

9

a α = 135°, Z%

= 200 N, X%

= 150 N and β = ?

180 − α = 45°

sin (180 )X

θ° −

%

= sin 45Z

°

%

sin (180° − θ) = 150 sin 45200

°

sin (180° − θ) = 0.5303 180° − θ = 32° θ = (180 − 32)° = 148° ⇒ β = 360° − (135° + 148°) = 77° b α = 100°, β = 135°, Y

% = 27 N and Z

% = ?

sin(180 )

Zα° −

% = sin(180 )

Yβ° −

%

sin 80

Z°

% = 27sin 45°

Z%

= 27sin80sin 45

°°

Z%

= 37.6 N

c X%

= Y%

⇒ Triangle is isosceles.

α = 60° or 180° − α = 120°

⇒ 180° − θ = 180° − β = 180 1202

° − °

= 30° (since isosceles triangle)

sin(180 )

Zα° −

% = sin(180 )

Xθ° −

%

sin 120

Z°

% = sin30

X°

%

Z%

= sin120

sin30X°

°%

= 12

32

X%

= 3 X%

Exercise 8B — Newton’s First Law of Motion 1 a

TH = 25 cos 20° = 23.5 N

b TV = 25 sin 20° = 8.6 N c Acceleration = 0 F = TH = 23.5 N

2 a H = 300 N W = 40g N Angle to the vertical = 25° The answer is D.

b The horizontal component of T%

is the same as the magnitude of the given horizontal force.

TH = 300 N The answer is B. c The vertical component of T

% is the same as mg.

TV = 40g N The answer is C.


d

T%

= 2 2300 (40 )g+

= 2 2300 392+ = 493.6 Ν ≈ 494 Ν The answer is A.

3 a

b Acceleration = 0 ⇒ R

%= 0

c F = 2 × 20 000 cos 10° = 39 392.3 N

4 a

b R

% = (T1cos 30° − T2 cos 60°) i

% +

(T1 sin 30° + T2 sin 60° − 8g) j%

R%

= 0 i%

+ 0 j%

Equating the i%

-components: T1 cos 30° − T2 cos 60° = 0

32

T1 − 12 T2 = 0

3 T1 − T2 = 0 T2 = 3 T1 [1] Equating the j

%-components:

T1 sin 30° + T2 sin 60° − 8g = 0

12 T1 + 3

2 T2 − 8g = 0

T1 + 3 T2 − 16g = 0 [2] Substitute T2 = 3 T1 into [2] T1 + 3T1 − 16g = 0 4T1 = 16g T1 = 4g N Substitute T1 = 4g into [1] T2 = 3 × 4g = 4 3 g N

5 a

b cos θ = 3

4 where θ is the angle between either rope and ceiling.

θ = 41.4° Each rope is angled 41.4° from the ceiling.

c 2 TV = W

TV = 2

W

= 502g

= 25g N = 245 N

d

tan θ = V

H

TT

TH = V

tanT

θ

= 245tan 41.4°

≈ 278 N Left rope: T

% = −TH i

% + TV j

%

= −278 i%

+ 245 j%

Right rope: T%

= TH i%

+ TV j%

= 278 i%

+ 245 j%

e Tleft = Tright = 2 2278 245+ = 370.6 N

f

jR

%% = 2 × 4000 sin θ − 50g = 0

8000 sin θ − 245 = 0 sinθ = 245

8000

θ = 1.75° x = 4 cos θ = 4 cos 1.75° = 3.998 ≈ 4 2x = 8 Maximum separation is 8 metres.

6 a

b AV = 120 cos 40° = 91.9 N down AH = 120 sin 40° = 77.1 N left c R

% = (F − 77.1) i

% + (N − 91.9 − W) j

%

= (F − 77.1) i%

+ (N − 91.9 − 40g) j%

= (F − 77.1) i%

+ (N − 483.9) j%

d a%

= 0 ⇒ R

% = 0

F − 77.1 = 0 F = 77.1 N e N − 483.9 = 0 N = 483.9 N


7 a

b R%

= 0 i%

+ 0 j%

(Not moving) Therefore, R = O N

c W%

= −1.5g sin 30° i%

− 1.5g cos 30° j%

≈ −7.4 i%

− 12.7 j%

N

d R%

= (T − 7.4) i%

+ (N − 12.7) j%

= 0 i%

+ 0 j%

T − 7.4 = 0 T = 7.4 N e N − 12.7 = 0 (From part d) N = 12.7 N

8 a

b W

% = −1.5g sin 30° i

% − 1.5g cos 30° j

%

= −7.4 i%

− 12.7 j%

N

c iR%

= (H cos 30° − 7.4) i%

= 0 i%

H × 32

− 7.4 = 0

H = 7.4 × 23

≈ 8.5 N ⇒ H

% = 7.4 i

% − 8.5 sin 30° j

%

= 7.4 i%

− 4.3 j%

N

d jR%

= (N − 12.7 − 4.3) j%

= 0 j%

N − 12.7 − 4.3 = 0 N = 17 N ⇒ N

% = 17 j

%

e Consider the i%

component of R.

cos30 sin30 0

cos30 sin30

3 12 2

3

33

3 333

iR H W

H W

H W

H W

WH

W

W

= ° − ° =

° = °

× = ×

=

=

= ×

=

% % %

% %

% %

% %

%%

%

%

9 The weight of the book is mg = 40 × 9.8 = 392 N The normal must balance the weight, N = 392 N Friction opposes motion, F = 300 N

3003920.77

F NFN

μ

μ

=

=

=

=

10 a F = ma = 1 900 × 4 = 7 600 N

b The weight of the car is 1 900g = 18 620 N The normal must balance the weight, N = 18 620 N c

7600186200.41

F NFN

μ

μ

=

=

=

=

11 The weight of the log is mg = 300 × 9.8 = 2 940 N The normal must balance the weight, N = 2 940 N Friction opposes motion, F = 200 N

200

29400.07

F NFN

μ

μ

=

=

=

=

12 a

6 9.858.8 N

W mg== ×=

%

b Breaking W%

into i%

and j%

components:

58.8sin 40 58.8cos40W i j= − ° − °% % %

The normal is the j%

component of weight

N = 58.8cos40° = 45.04 N c The frictional force is the i

%component of weight

F = 58.8sin40° = 37.8 N d

37.845.040.84

F NFN

μ

μ

=

=

=

=

13

W mg=%

Breaking W

%into i

%and j

%components:

sin 35 cos35W mg i mg j= − ° − °% % %


component of weight

N = mgcos35° The frictional force is the i

%component of weight

F = mgsin35°

sin35cos35

F NFNmgmg

μ

μ

=

=

°=°


sin35cos35tan350.7

°=°

= °=

14

2 9.819.6 N

W mg== ×=

%

Breaking W%

into i%

and j%

components:

sin 40 cos40

19.6sin 40 19.6cos40

W mg i mg j

i j

= − ° − °

= − ° − °% % %

% %


component of weight

N = 19.6cos 40° = 15.01 N F = μN = 0.2 × 15.01 = 3 N Friction opposes the i

%component of weight.

The resultant force down the plane = 19.6sin40° −3 = 12.6 – 3 = 9.6 N

Exercise 8C — Momentum 1 a Momentum = mass × velocity = 2.5 × 16 = 40 Momentum = 40 N s East

b Momentum = mass × velocity = 4 × 150 = 600 Momentum = 600 N s South c Mass = 250 g = 0.25 kg Momentum = mass × velocity = 0.25 × 30 = 7.5 Momentum = 7.5 N s North d Mass = 2.5 tonnes = 2 500 kg Momentum = mass × velocity = 2 500 × 13 = 32 500 Momentum = 32 500 N s West e Velocity = 40 km/h = 400/36 m/s Momentum = mass × velocity = 88 × 400/36 = 977.8 Momentum = 978 N s North f Mass = 3.4 tonnes = 3 400 kg Velocity = 120 km/h = 1200/36 m/s Momentum = mass × velocity = 3 400 × 1200/36 = 113 333.3 Momentum = 113 333 N s North

2 a Speed = momentum ÷ mass = 2 000 ÷ 25 = 80 Speed = 80 m/s

b Mass = 80 g = 0.08 kg Speed = momentum ÷ mass = 3 000 ÷ 0.08 = 37 500 Speed = 37 500 m/s c Mass = 2.1 tonnes = 2 100 kg Speed = momentum ÷ mass = 42 000 ÷ 2 100 = 20 Speed = 20 m/s

3 a Mass = momentum ÷ speed = 2 500 ÷ 50 = 50 Mass = 50 kg

b Speed = 40 km/h = 400/36 m/s Mass = momentum ÷ speed = 22 000 ÷ 400/36 = 22 000 × 36/400 = 1980 Mass = 1980 kg

4 a object A: momentum = mass × velocity = 3 × 10 = 30 momentum = 30 N s East object B: momentum = mass × velocity = 2 × 4 = 8 momentum = 8 N s East total momentum = 38 N s East

b Using results from 4a: Momentum object A = 30 N s East Momentum object B = 8 N s West Total momentum = 30 N s East + 8 N s West = 30 N s East – 8 N s East = 22 N s East c Using results from 4a: Momentum object A = 30 N s East Momentum object B = 8 N s North Total momentum = 30 N s East + 8 N s North

2 230 8

96431

x = +

==

830tan

14.9

θθ

=

= °

Total momentum = 31 N s at 14.9° North of East d Using results from 4a: Momentum object A = 30 N s East Momentum object B = 8 N s 30° North of East Total momentum= 30 N s East + 8 N s 30° North of

East

using the cosine rule 2 2 230 8 2 30 8cos150

1379.737.1

x

x

= + − × × °==


sin sin1508 37.1

8sin15037.1sin

6.2

θ

θθ

°

°

=

=

= °

Total momentum = 37.1 N s at 6.2° North of East e Using results from 4a: Momentum object A = 30 N s East Momentum object B = 8 N s North of West Total momentum = 30 N s East + 8 N s North of

West

Using Cosine rule 2 2 230 8 2 30 8cos45

624.5925

x

x

= + − × × °==

sin sin 458 25

8sin 4525sin

13.1

θ

θθ

°

°

=

=

= °

Total momentum = 25 N s at 13.1° North of East 5 Mass = 300 kg Initial velocity = 6 m/s towards the wall Final velocity = 3 m/s away from the wall Initial momentum = mass × initial velocity = 300 × 6 = 1 800 N s towards the wall Final momentum = 300 × 3 = 900 N s away from the wall Change in momentum = final momentum – initial momentum

= (900 N s away from the wall) – (1 800 N s towards the wall)

= (900 N s away from the wall) – (−1 800 N s towards the wall)

= 2 700 N s away from the wall 6 a Momentum = mass × velocity = 100 × 10 = 1 000 N s

b Momentum = mass × velocity = 100 × 5 = 500 N s Change in momentum = 500 – 1 000 = −500 N s

7 a Initial momentum = mass × velocity = 10 × 12 = 120 N s Final momentum = mass × velocity = 10 × 8 = 80 N s Change in momentum = final momentum – initial momentum = 80 – 120 = –40 N s

b Initial momentum = mass × velocity = 8 × 7 = 56 N s Final momentum = mass × velocity = 8 × 13 = 104 N s Change in momentum = final momentum – initial momentum = 104 – 56 = 48 N s

c Mass = 0.95 tonne = 950 kg Initial velocity = 44 km/h = 440/36 m/s Final velocity = 80 km/h = 800/36 m/s Initial momentum = mass × velocity = 950 × 440/36 = 11 611 N s Final momentum = mass × velocity = 950 × 800/36 = 21 111 N s Change in momentum = final momentum – initial momentum = 21 111 – 11 611 = 9 500 N s d Initial velocity = 50 km/h = 500/36 m/s Final velocity = 18 km/h = 180/36 m/s Initial momentum = mass × velocity = 400 × 500/36 = 5 555.56 N s Final momentum = mass × velocity = 400 × 180/36 = 2 000 N s Change in momentum = final momentum – initial momentum = 2 000 – 5 555.56 = −3 556 N s

8 Laurie’s momentum = mass × velocity = 85 × 8 = 680 N s down Alan’s momentum = mass × velocity = 100 × 7 = 700 N s right

Initial momentum = (680 N s down) + (700 N s right)

2 2680 700

952400975.9

x = +

==

680700tan

44

θθ

=

= °

As momentum is conserved, the momentum of the collision is 975.9 N s at 44° below the horizontal in the same direction as Alan was initially moving

Mass after collision = 85 + 100 = 185 kg Velocity = momentum ÷ mass = 975.9 ÷ 185 = 5.3 m/s The players will move at 5.3 m/s at an angle of 44° below

the horizontal in the same direction as Alan was originally moving

9 a Before collision: Car A’s momentum = mass × velocity = 320 × 8 = 2 560 N s North Car B’s momentum = mass × velocity = 280 × 2 = 560 N s North Momentum of collision = (2 560 N s North) + (560 N s North) = 3 120 N s North


After collision: Car A’s momentum = mass × velocity = 320 × 4 = 1 280 N s North Car B’s momentum = Momentum of collision – Car A’s momentum = (3 120 N s North) – (1 280 N s North) = 1 840 N s North Car B’s velocity = momentum ÷ mass = 1840 ÷ 280 =6.6 m/s Car B’s velocity after the collision is 6.6 m/s North

b Before collision: Car A’s momentum = mass × velocity = 320 × 8 =2 560 N s North Car B’s momentum = mass × velocity = 280 × 2 = 560 N s South Momentum of collision = (2 560 N s North) + (560 N s South) = (2 560 N s North) + (−560 N s North) = 2 000 N s North After collision: Car A’s momentum = mass × velocity = 320 × 1 = 320 N s North Car B’s momentum = Momentum of collision – Car A’s momentum = (2 000 N s North) – (320 N s North) = 1 680 N s North Car B’s velocity = momentum ÷ mass = 1 680 ÷ 280 =6 m/s Car B’s velocity after the collision is 6 m/s North c Before collision: Car A’s momentum = mass × velocity = 320 × 2 = 640 N s North Car B’s momentum = mass × velocity = 280 × 8 = 2 240 N s South Momentum of collision = (640 N s North) + (2 240 N s South) =(−640 N s South) + (2 240 N s North) = 1 600 N s South After collision: Car A’s momentum = mass × velocity = 320 × 2 = 640 N s South Car B’s momentum = Momentum of collision – Car A’s momentum = (1 600 N s South) – (640 N s South) = 960 N s South Car B’s velocity = momentum ÷ mass = 960 ÷ 280 =3.4 m/s Car B’s velocity after the collision is 3.4 m/s South

10 Laurie’s momentum = mass × velocity = 85 × 8 = 680 N s at 60° below horizontal

and to the left Alan’s momentum = mass × velocity = 100 × 7 = 700 N s right

Initial momentum = (680 N s at 60° below horizontal) + (700 N s right)

Using cosine rule 2 2 2700 680 2 700 680cos60

476400690.2

x

x

= + − × × °==

sin sin60680 690.2

θ °=

680sin60sin 690.258.6

θ

θ

°=

= °

As momentum is conserved, the momentum of the collision is 690.2 N s at 58.6° below the horizontal in the direction of Alan’s original movement

Mass after collision = 85 + 100 = 185 kg Velocity = momentum ÷ mass = 690.2 ÷ 185 = 3.7 m/s The players will move at 3.7 m/s at an angle of 58.6° below

the horizontal in the direction of Alan’s initial movement. 11 a Before collision: A’s momentum = mass × velocity = 0.1 × 4 = 0.4 N s 30° B’s momentum = mass × velocity = 0.1 × 0 = 0 N s Momentum of collision = (0.4 N s 30°) + (0 N s) = 0.4 N s 30° After collision: A’s momentum = mass × velocity = 0.1 × 2 = 0.2 N s 10°

B’s momentum = Momentum of collision – A’s momentum = (0.4 N s 30°) – (0.2 N s 10°) Using cosine rule: 2 2 20.4 0.2 2 0.4 0.2cos20

0.04960.22

x

x

= + − × × °==

sin sin200.2 0.22

0.2sin20sin 0.2217.9

θ

θ

θ

°=

=

= °

Momentum = 0.22 N s at 47.9° B’s velocity = momentum ÷ mass = 0.22 ÷ 0.1 =2.2 m/s B’s velocity after the collision is 2.2 m/s 47.9°

b Before collision: A’s momentum = mass × velocity = 0.1 × 4 = 0.4 N s 60°


B’s momentum = mass × velocity = 0.1 × 2 = 0.2 N s 0°

Momentum of collision

Using cosine rule: 2 2 20.4 0.2 2 0.4 0.2cos120

0.280.529

x

x

= + − × × °==

sin sin1200.4 0.529

0.4sin1200.529sin

40.9

θ

θθ

°

°

=

=

= °

Momentum of collision = 0.529 N s 40.9° After collision: A’s momentum = mass × velocity = 0.1 × 3 = 0.3 N s −45°

B’s momentum = Momentum of collision –A’s momentum = (0.529 N s 40.9°) – (0.3 N s −45°) Using cosine rule: 2 2 20.529 0.3 2 0.529 0.3cos85.9

0.3470.59

x

x

= + − × × °==

sin sin 85.90.3 0.59

0.3sin85.90.59sin

30.5

θ

θθ

°=

=

= °

Angle is 40.9° + 30.5° = 71.4° Momentum = 0.59 N s at 71.4° B’s velocity = momentum ÷ mass = 0.59 ÷ 0.1 = 5.9 m/s B’s velocity after the collision is 5.9 m/s 71.4° c Before collision: A’s momentum = mass × velocity = 0.1 × 3 = 0.3 N s 30° B’s momentum = mass × velocity = 0.1 × 2 = 0.2 N s 150° Momentum of collision


0.070.265

x

x

= + − × × °==

sin sin 600.2 0.265

0.2sin 600.265sin

40.9

θ

θθ

°

°

=

=

= °

Angle is 30° + 40.9° = 70.9° Momentum of collision = 0.265 N s 70.9° After collision: A’s momentum = mass × velocity = 0.1 × 1 = 0.1 N s 80°

B’s momentum = Momentum of collision –A’s momentum = (0.265 N s 70.9°) – (0.1 N s 80°) Using cosine rule: 2 2 20.265 0.1 2 0.265 0.1cos9.1

0.027890.167

x

x

= + − × × °==

sin sin 9.10.1 0.167

0.1sin 9.10.167sin

5.4

θ

θθ

°

°

=

=

= °

Angle is 70.9° − 5.4° = 65.4° Momentum = 0.167 N s at 65.4° B’s velocity = momentum ÷ mass = 0.167 ÷ 0.1 = 1.67 m/s B’s velocity after the collision is 1.67 m/s 65.4° d Before collision: A’s momentum = mass × velocity = 0.1 × 3 = 0.3 N s 30° B’s momentum = mass × velocity = 0.1 × 4 = 0.4 N s −120° Momentum of collision


0.0420.21

x

x

= + − × × °==

sin sin 300.3 0.21

0.4sin 300.21sin

46.9

θ

θθ

°

°

=

=

= °

Angle is 46.9° − 120° = −73.1° Momentum of collision = 0.21 N s −73.1°


After collision: A’s momentum = mass × velocity = 0.1 × 2 = 0.2 N s 0°

B’s momentum = Momentum of collision – A’s momentum = (0.21 N s −73.1°) – (0.2 N s 0°) Using cosine rule: 2 2 20.21 0.2 2 0.21 0.2cos73.1

0.05970.24

x

x

= + − × × °==

sin sin 73.10.2 0.24

0.2sin 73.10.24sin

52.9

θ

θθ

°

°

=

=

= °

Angle is 360° − (73.1° + 52.9°) = 234° Momentum = 0.24 N s at 234.5° B’s velocity = momentum ÷ mass = 0.24 ÷ 0.1 = 2.4 m/s B’s velocity after the collision is 2.4 m/s 234°

12 a Before the rifle is fired, the bullet and rifle are still therefore their momentum is 0.

After firing, mass of bullet = 25g = 0.025 kg. Momentum of bullet = mass × velocity = 0.025 × 180 = 4.5 N s As momentum before firing = 0, momentum after

firing = 0. Momentum of rifle = 4.5 N s in opposite direction to

bullet b Speed = momentum ÷ mass = 4.5 ÷ 1.5 = 3 The rifle will move at 3 m/s

Exercise 8D — Relative velocity 1

~ rel b wv = 16

~i

~ wv = −4

~j

~bv =

~ rel b wv +

~ wv

= 16~i + −4

~j

( )22

~16 4

4 1716.49

bv = + −

==

416tan

14.03

θθ

−=

= − °

An observer would see the boat moving at 16.5 m/s at angle of 14° South of East.

2 ~ rel p wv = 300

~j

~ wv = 30

~i (from the West means towards the East)

~ pv =

~ rel p wv +

~ wv

= 300 ~j + 30

~i

= 30~i + 300

~j

2 2

~30 300

301.496b

v = +

=

30030tan

84.289

θθ

=

= °

90° − 84.3° = 5.7° The velocity of the plane relative to the ground is 301.5 km/h

at 5.7° East of North 3

~ rel b wv = 15 15

2 2~ ~i j− +

~ wv = 3

~j

~bv =

~ rel b wv +

~ wv

= 15 152 2~ ~

i j− + +3~j

= ( )15 152 2~ ~

3i j− + +

( ) ( )2 215 152 2~

3

17.25b

v −= + +

=

15

2152

3tan

52.06

θ

θ

−

+=

= − °

An observer would see the boat moving at 17.3 m/s at angle of 52.1° North of West.

4 |~ wv | = 6

|~bv | 2 220 6

19.08= −=

19.086tan

72.5

θθ

=

= °

The ferry heads upstream at an angle of 72.5° to the bank.


5 ~bv = 19 km/h North-West

~ rel p bv = −6 km/h North-West

~ pv =

~ rel p bv +

~bv

= −6 + 19 = 13 Relative to the water, the person is travelling at 13 km/h. 6 a

~ av =40

~j

~bv = 20

~i

~ rel a bv =

~ av −

~bv

= 40~j − 20

~i

=−20~i + 40

~j

|~ rel a bv | 2 220 40

44.7= +=

4020tan

63.4

θθ

−=

= − °

90 °− 63.4° = 26.6° 44.7 km/h at 26.6° West of North b

~ rel a bv = 20

~j

~bv = −15

~j

~ av =

~ rel a bv +

~bv

= 20~j + −15

~j

= 5~j

5 m/s North

c ~ av = 25 25

2 2~ ~i j+

~bv = 20

~i

~ rel a bv =

~ av −

~bv

= 25 252 2~ ~

i j+ − 20~i

= ( )25 252 2~ ~

20 i j− +

|~ rel a bv | ( ) ( )2 2

25 252 2

20

17.83

= − +

=

25

225

2

tan20

82.51

θ

θ

=−

= − °

90° − 82.5° = 7.5° 17.8 m/s at 7.5° West of North d

~ av = 4 4

2 2~ ~i j−

~ rel a bv = −30

~i

~bv =

~ av −

~ rel a bv

( )4 42 2~ ~~

4 42 2~ ~

30

30

i j i

i j

= − − −

= + −

|~bv | ( ) ( )2 2

4 42 2

30

32.95

−= + +

=

42

42

tan30

4.9

θ

θ

−

=+

= − °

32.95 m/s at 4.9° South of East e

~ av

~ ~

3 3 32 2~ ~

3cos30 3sin30i j

i j

= ° − °

= −

~bv =

~ ~5cos65 5sin 65i j− ° + °

~ rel a bv =

~ av −

~bv

( ) ( )

3 3 32 2~ ~~ ~

3 3 32 2~ ~

5cos65 5sin 65

5cos65 5sin 65

i j i j

i j

⎛ ⎞= − − − ° + °⎜ ⎟⎝ ⎠

= + ° − + °

|~ rel a bv | ( ) ( )

2 23 3 32 25cos65 5sin 65

7.65

= + ° + + °

=

( )3

23 3

2

5sin 65tan

5cos6552

θ

θ

− + °=

+ °= − °

7.7 m/s at 52° South of East


7 Riding north:

~cv = 15

~j

To the cyclist, the wind appears to be coming from the north-east. If the magnitude of this velocity is a then

~ rel w cv =

2 2~ ~

a ai j− −

~ wv =

~ rel w cv +

~cv

( )2 2~ ~ ~

2 2~ ~

15

15 [1]

a a

a a

i j j

i j

−

−

= − +

= + −

Riding south:

~cv = −15

~j

To the cyclist, the wind appears to be coming from the south-east. If the magnitude of this velocity is b then

~ rel w cv =

2 2~ ~

b bi j− +

~ wv =

~ rel w cv +

~cv

( )2 2~ ~ ~

2 2~ ~

15

15 [2]

b b

b b

i j j

i j

−

−

= + −

= + −

True speed and direction of wind: In both cases,

~ wv are the same. Equating the x components in

[1] and [2] results in 2 2a b− −= , therefore a = b.

Equating the y components in [1] and [2] and replacing b with a:

2 2

22

30 22

15 15

30

15 2

a a

a

a

− = −

=

=

=

Sub in [1]

~ wv ( )

( )

15 2 15 22 2~ ~

~ ~

~

15

15 15 15

15

i j

i j

i

−= + −

= − + −

= −

The wind is blowing at 15 km / h towards the west (or from the east).

8 Riding north:

~cv = 15

~j

To the cyclist, the wind appears to be coming from the north-east. If the magnitude of this velocity is a then

~ rel w cv =

2 2~ ~

a ai j− −

~ wv =

~ rel w cv +

~cv

( )2 2~ ~ ~

2 2~ ~

15

15 [1]

a a

a a

i j j

i j

−

−

= − +

= + −

Riding south:

~cv = −15

~j

To the cyclist, the wind appears to be coming from 30° south of east. If the magnitude of this velocity is b then

~ rel w cv

~ ~

32 2~ ~

cos30 sin30

b b

b i b j

i j−

= − ° + °

= +

~ wv =

~ rel w cv +

~cv

( )

32 2~ ~ ~

32 2~ ~

15

15 [2]

b b

b b

i j j

i j

−

−

= + −

= + −


~ wv are the same. Equating the x components in

[1] and [2] results in

3226

2

ba

ba

−− =

=

Equating the y components in [1] and [2] and replacing a with b:

615 1522 2330 2 2

1 330 2

601 3

b b

bb

b

b

⎛ ⎞⎜ ⎟⎝ ⎠

− = −

= +

+=

=+

Sub in [2]

~ wv ( ) ( )( )3 60 601

2 21 3 1 3~ ~15i j−

+ += + −

( )( ) ( )

( ) ( )

( ) ( )( )( ) ( )

30 3 301 3 1 3~ ~

30 3 1 3 30 1 3

1 3 1 3~ ~

30 3 1 3 30 1 3

2 2~ ~

~ ~

~ ~

15

15

15

15 3 1 3 15 1 3 15

15 3 1 3 15 2 3

i j

i j

i j

i j

i j

−+ +

− − −

− −

− − −

− −

= + −

⎛ ⎞= + −⎜ ⎟⎝ ⎠⎛ ⎞= + −⎜ ⎟⎝ ⎠

= − + − − −

= − − −

|~ wv | ( ) ( )2 2

15 3 1 3 15 2 3

19.439

⎡ ⎤ ⎡ ⎤= − + −⎣ ⎦ ⎣ ⎦=


( )( )

15 2 3tan

15 3 1 3

11.9

θ

θ

− −=

−

= °

The wind is blowing at 19.4 km / h towards 12° North of East (or from 12° South of West or 78° West of

South)

9 Let the initial velocity of the jogger be

~ jv = 9

~i

To the jogger, the rain appears to be towards her at an angle of 10°. If the magnitude of this velocity is a then

~ rel r jv =

~ ~sin10 cos10a i a j− ° − °

~ rv =

~ rel r jv +

~ jv

( )

~ ~~

~ ~

sin10 cos10 9

9 sin10 cos10 [1]

a i a j i

a i a j

= − ° − ° +

= − ° − °

Return journey:

~ jv = −9

~i

To the jogger, the rain appears to be coming towards her at an angle of 5°. If the magnitude of this velocity is b then

~ rel r jv

~ ~sin5 cos5b i b j= ° − °

~ rv =

~ rel r jv +

~ jv

( )

~ ~~

~ ~

sin 5 cos5 9

sin5 9 cos5 [2]

b i b j i

b i b j

= ° − ° −

= ° − − °

True speed and direction of rain: In both cases,

~ rv are the same. Equating the y components in

[1] and [2] results in

cos10sin 5

cos10 sin5a

a bb °

°

− ° = − °=

Equating the x components in [1] and [2] and replacing b with a:

( )

cos10cos5cos10cos5

cos10 sin 5cos5

9 sin10 sin5 9

18 sin5 sin10

sin10

0.259869.282

a

a

a

a

a

aa

°°

°°

° °°

− ° = ° −

= ° + °

= + °

==

Sub in [1]

~ rv ( )

~ ~

~ ~

9 69.282sin10 69.282cos10

3.0306 68.2294

i j

i j

= − ° − °

= − −

|~ wv | ( ) ( )2 23.0306 68.2294

68.297

= − + −

=

3.0306tan

68.22942.54

θ

θ

−=−

= °

The rain is falling at 68.3 km / h at 2.5° towards the jogger’s original direction.

Exercise 8E — Using vectors in geometry 1 CD

uuur = − r

% + t

%

DBuuur

= − t%

+ q%

CDuuur

= DBuuur

, ⇒ − r

% + t

% = − t

% + q

%

2 t%

= r%

+ q%

t%

= 12 ( r

% + q

%)

= 12 r%

+ 12 q%

2 BDuuur

= t%

− q%

BCuuur

= r%

− q%

BDuuur

= 1n

BCuuur

,

⇒ t%

− q%

= 1n

( r%

− q%

)

t%

= q%

+ 1n

r%

− 1n

q%

= (1 − 1n

) q%

+ 1n

r%

= 1nn−⎛ ⎞

⎜ ⎟⎝ ⎠

q%

+ 1n

r%

3

Let ADuuur

= a%

Let CEuuur

= b%

ABuuur

= 2 ADuuur

= 2 a

%

CBuuur

= 2 CEuuur

= 2 b

%

DBuuur

= ADuuur

= a

%

EBuuur

= CEuuur

= b

%

DEuuur

= DBuuur

− EBuuur

= a

% − b

%

ACuuur

= ABuuur

− CBuuur

= 2 a

% − 2 b

%

= 2( a%

− b%

)

= 2 DEuuur

DE

uuur = 1

2 ACuuur

and therefore DE is parallel to AC.


4

a DCuuur

= − DBuuur

a

% = − v

%

b CAuuur

= CDuuur

+ DAuuur

b

% = v

% + u

%

c BAuuur

= − DBuuur

+ DAuuur

c

% = − v

% + u

%

d b b⋅% %

= ( ) ( )v u v u+ ⋅ +% % % %

= v2 + 2 u v⋅

% % + u2

or b2 = v2 + u2 Since u v⋅% %

= 0 c c⋅

% % = ( ) ( )v u v u− + ⋅ − +

% % % %

= v2 − 2 u v⋅% %

+ u2

or c2 = v2 + u2 Since u v⋅% %

= 0 Since b2 = c2

or b = c ⇒ AC = AB

5 ABuuur

+ BCuuur

+ CAuuur

= 0 ⇒ a

% + b

% + c

% = 0

6

ADuuur

= ABuuur

+ BDuuur

AD AD⋅

uuur uuur = (AB BD) AD+ ⋅

uuur uuur uuur

or 2|AD| = AB AD⋅uuur uuur

+ BD AD⋅uuur uuur

But BD AD⋅uuur uuur

= 0 (perpendicular)

⇒ 2|AD| = AB AD⋅uuur uuur

= |AB| |AD| cos a |AD| = |AB| cos a [1]

CDuuur

= CBuuur

+ BDuuur

CD CD⋅

uuur uuur = ( CB BD+

uuur uuur) CD⋅uuur

2|CD| = CB CD⋅uuur uuur

+ BD CD⋅uuur uuur

But BD CD⋅uuur uuur

= 0 (perpendicular)

⇒ 2|CD| = CB CD⋅uuur uuur

= |CB| |CD| cos c |CD| = |CB| cos c [2] But |CB| = |AB| and a = c (equilateral triangle) |CD| = |AB| cos a = |AD| from [1] So, |AD| = |DC|

7

DBuuur

= − a%

+ b%

ACuuur

= a%

+ b%

AC DB⋅uuur uuur

= ( ) ( )a b a b+ ⋅ − +% % % %

= 2 2a a b a b b− + ⋅ − ⋅ +% % % %

= b2 − a2

But a = b (rhombus) ⇒ AC DB⋅

uuur uuur = b2 − b2

= 0 AC

uuur and DB

uuur are perpendicular.

Therefore the diagonals of a rhombus intersect at right angles.

8

Clearly | | | | | |a b a b+ > +% % % %

as the sum of any two side lengths in a triangle is greater than the length of the third side.

or

| | | |a b+% %

= | |a b+% %

⇒ | | | | | |a b a b+ ≥ +

% % % %

9

Let ADuuur

= a%

Let ABuuur

= b%

Let E be the midpoint of AC. AC

uuur = a

% + b

%

⇒ AEuuur

= 12 ( )a b+% %

BEuuur

= 12 ( )b a b− + +% %

= 12 ( )a b−% %

EDuuur

= 12 ( )a b a− − +% % %

= 12 ( )a b−% %

= BEuuur

E is the midpoint of BD. Therefore the diagonals of a parallelogram bisect each

other.

10

Let AOuuur

= a%

⇒ OCuuur

= a%

Let OBuuur

= b%

ABuuur

= a%

+ b%

BCuuur

= −( a%

+ b%

) + 2 a%

= a

% − b

%

AB BC⋅uuur uuur

= ( ) ( )a b a b+ ⋅ −% % % %

= 2a%

− 2b%

But a = b (radii of circle) AB BC⋅

uuur uuur = a2 − a2

= 0 AB

uuur is perpendicular to BC

uuur.


Therefore the angle subtended by the diameter of a circle is a right angle.

11

Let ABuuur

= a%

Let BCuuur

= b%

EFuur

= 1 12 2AB BC+uuur uuur

= 12 a%

+ 12 b%

FGuuur

= 12 a−%

+ 12 b%

HGuuur

= 12 a%

+ 12 b%

EHuuur

= 12 a−%

+ 12 b%

⇒ EFuur

= HGuuur

and FG

uuur = EHuuur

Therefore EFGH is a parallelogram since the opposite sides

are equal and parallel.

12 a

Let DAuuur

= a%

Let ABuuur

= b%

Let CEuuur

= c%

Let O be the midpoint of CGuuur

CG

uuur = b a c− + +

% % %

or CGuuur

= a b c− +% % %

⇒ COuuur

= OGuuur

= 12 ( )a b c− +

% % %

DHuuur

= a b c+ +% % %

DOuuur

= DA AG OG+ −uuur uuur uuur

= 12 ( )a c a b c+ − − +

% % % % %

= 1 1 12 2 2a b c+ +% % %

= 12 ( )a b c+ +

% % %

OHuuur

= OGuuur

+ GHuuur

= 1

2 ( )a b c b− + +% % % %

= 1 1 12 2 2a b c+ +% % %

= 12 ( )a b c+ +

% % %

DOuuur

= OHuuur

= 12

DHuuur

Therefore the major diagonals of a cube bisect each other. b CG DH⋅

uuur uuur = ( ) ( )a b c a b c− + ⋅ + +

% % % % % %

= a a a b a c b a b b b c c a c b c c⋅ + ⋅ + ⋅ − ⋅ − ⋅ − ⋅ + ⋅ + ⋅ + ⋅% % % % % % % % % % % % % % % % % %

= 2 2 2 2a b c a c− + + ⋅% %

But a c⋅

% % = 0, since a

% and c

% are perpendicular and

| | | | | |a b c= =% % %

, since it is a cube.

⇒ CG DH⋅uuur uuur

= 2 2 2 0a a a− + + = a2

|CG|uuur

= | |a b c− +% % %

= 2 2 2( )a b c+ − +

since a%

, b%

, c%

are perpendicular.

= 2 2 2a b c+ +

= 23a since a = b = c |DH|

uuur = | |a b c+ +

% % %

= 2 2 2a b c+ +

= 23a since a = b = c

∴ cos θ = CG DH|CG| |DH|

⋅uuur uuur

uuur uuur

= 2

2 23 3

a

a a

= 2

23aa

= 13

θ = cos−113

⎛ ⎞⎜ ⎟⎝ ⎠

= 70.529 Therefore the acute angle between the major diagonals is

70.53°.

13

c a b= +

% % %

( )( )

2 2 2

2

2

c c a b a ba a a b b b

c a b a b

⋅ = + += ⋅ + ⋅ + ⋅

= + + ⋅

% % % % % %

% % % % % %

% %

But a%

and b%

are perpendicular, therefore 0a b⋅ =% %

and

2 2 2c a b= +

Chapter review 1 1 2f f+

% % = −3 i

% + 7 j

% + 5 i

% + 2 j

%

= 2 i%

+ 9 j%

3f%

= − 1 2( )f f+% %

= −2 i%

− 9 j%

3f%

= ( )2 22 ( 9)− + −

= 85 The answer is B. 2 Resolving horizontally 2 20cos30

3402

20 334.6

F = × °

= ×

==

The magnitude of F = 34.6 N The solution is A 3 Resolving the vectors vertically

2 2 cos25

22cos25

g FgF

= °

=°


cos2510.8 N

g=

=

The solution is B. 4 ( )

( )3 1 2

80 40

f f f

i j

= − +

= − +% % %

% %

( ) ( )2 23 80 40

800089.4

f = − + −

=≈

%

Find the angle with 1f%

1 3 6400f f⋅ = −% %

1 3

1 3

cos

640080 8000153.4

f f

f fθ

θ

⋅=

−=×

= °

% %

% %

3f%

is 89.4 N at 153.4° to 1f%

5

4002 = 6002 + 5002 − 2 (600)(500) cos (180° − θ) 160 000 = 610 000 − 600 000 cos (180° − θ)

cos (180° − θ ) = 450 000600 000

= 0.75 180° − θ = 41.4° θ = 138.6°

6 a

b Let θ = angle between T2 and horizontal.

Resolving vertically

( )1 2

1 2

sin 90 sin

6 cos sin [1]

W T T

g T T

θ θ

θ θ

= − +

= +

o

Resolving horizontally

( )1 2

1 2

2 1

cos 90 cos

sin cossin [2]cos

T T

T T

T T

θ θ

θ θθθ

− =

=

=

o

T2 = 2T1, therefore [2] becomes

1 1sin2cos

2 tan63.43 [3]

T T θθ

θθ

=

== °

Substitute for θ and T2 in [1]

( )1 1

1

1

1

2 1

6 cos63.24 2 sin 63.24cos63.24 2sin 63 24

2.23626.3252.6

g T TT

TTT T

= ° += ° + °====

T1 = 26.3 N T2 = 52.6 N c θ = 63.4° [rope 2] 90 − θ = 26.6° [rope 1]

d

tan θ = 21 from part b,

then sinθ = 25

Using triangle ABD,

sin θ = ABx

AB = sin

xθ

= 25

x

= 52

x

7

4 9.839.2 N

W mg== ×=

%

Breaking W%

into i%

and j%

components:

sin 40 cos40

39.2sin 40 39.2cos40

W mg i mg j

i j

= − ° − °

= − ° − °% % %

% %


component of weight

N = 39.2cos40° = 30.03 N F = μN = 0.25 × 30.03 = 7.5 N Friction opposes the i

%component of weight.

The resultant force down the plane = 39.2sin40° −7.5 = 25.2 – 7.5 = 17.7 N


8 mass = 85 kg speed = 100/10 m/s momentum = mass × speed = 85 × 10 = 850 Ns The solution is D 9 Mass = 85 g = 0.085 kg Initial velocity = 40 m/s Final velocity = –30 m/s Initial momentum = mass × initial velocity = 0.085 × 40 = 3.4 Ns Final momentum = mass × final velocity = 0.085 × −30 = −2.55 Ns Change in momentum = −2.55 – 3.4 = −5.95 The solution is B 10 Mass of car = 1170 kg Velocity of car = 100 km/h = 100/3.6 m/s Momentum of car = 1170 × 100/3.6 = 32 500 Ns Mass of object = 45 g = 0.045 kg Momentum of object = momentum of car = 32 500 Ns Speed of object = momentum of object / mass = 32 500 / 0.045 = 722 222.22 m/s The solution is C 11 North moving car: Mass = 1 250 kg Speed = 70 km/hr = 70/3.6 m/s Momentum = mass × speed = 1 250 × 70/3.6 =24 305.555 East moving car Mass = 1 300 kg Speed = 80 km/h = 80/3.6 m/s Momentum = mass × speed = 1 300 × 80/3.6 = 28 888.889

Magnitude 2 224305.000 28888.88937753.52

= +=

The solution is D 12 Mass = 150 g = 0.15 kg Velocity = 20 m/s (speed is same before and after, only

direction changes) Momentum = mass × velocity = 0.15 × 20 = 3

Change in momentum = final momentum – initial momentum 2 2 23 3 2 3 3cos60

93

x

x

= + − × × °==

As the 3 sides of the triangle are all 3 Ns, the triangle is equilateral and θ = 60°.

The change of momentum is 3 Ns perpendicular to the wall.

13 Player A: Mass = 100 kg Velocity = 8 m/s Momentum = mass × velocity = 800 Ns Player B: Mass = 85 kg Velocity = 9 m/s Momentum = mass × velocity = 765 Ns After collision: Mass = 100 + 85 = 185 kg Momentum is conserved

2 2800 7651106.8988

x = +=

765800tan

43.7

θθ

=

= °

Speed = momentum ÷ mass = 1106.8988 ÷ 185 = 5.98 After the collision, they move at 6 m/s at 43.7° to A’s

original motion. 14

~ rel a bv =

~ av −

~bv

= 2~i + 3

~j − (−2

~i + 2

~j )

= 4~i +

~j

The solution is B 15 The car is travelling north. For the other car to appear as if

it’s travelling south-east, the actual velocity of the car must have an easterly component. The only possible option is C

16 ~ wv = −5

~j

~ rel b wv = 11

~i

~bv =

~ rel b wv +

~ wv

= 11~i − 5

~j

x 2 211 512.08

= +=

115tan

65.55

θθ

=

= °

The boat would move at 12.1 km/h at 65.6° to the bank. 17 Riding east:

~cv = 15

~i

To the cyclist, the wind appears to be coming from 60° E of N. If the magnitude of this velocity is a then

~ rel w cv

~ ~

32 2~ ~

cos30 sin30

a a

a i a j

i j−

= − ° − °

= −

~ wv =

~ rel w cv +

~cv


( )

32 2~ ~~

32 2~ ~

15

15 [1]

a a

a a

i j i

i j

−= − +

= − −

Riding west:

~cv = −15

~i

To the cyclist, the wind appears to be coming from 60° W of N. If the magnitude of this velocity is b then

~ rel w cv

~ ~

32 2~ ~

cos30 sin30

b b

b i b j

i j

= ° − °

= −

~ wv =

~ rel w cv +

~cv

( )

32 2~ ~~

32 2~ ~

15

15 [2]

b b

b b

i j i

i j

= − −

= − + −


~ wv are the same. Equating the y components

in [1] and [2] results in 2 2a b− −= , therefore a = b.

Equating the x components in [1] and [2] and replacing b with a:

3 32 2

2 32

303

3303 3

15 15

30

10 3

a a

a

a

− = − +

=

=

= ⋅

=

Sub in [1]

~ wv ( )

( )

10 3 3 10 32 2~ ~

302 ~ ~

~

~

15

15 5 3

5 3

8.66

i j

i j

j

j

= − −

= − −

= −

= −

The wind is blowing at 8.7 km / h towards the south (or from the north).

18 AP p a= −uuur

%%

PB b p= −uuur

% %

( )

2 13 3

PB 2AP

2

2 2

3 2

b p p a

b p p a

p a b

p a b

=

− = −

− = −

= +

= +

uuur uuur

% %% %

% %% %

% %%

% %%

The solution is D.

19

Let ABuuur

= a%

Let ADuuur

= b%

DBuuur

= a b−% %

ACuuur

= a b+% %

AC DB⋅uuur uuur

= ( ) ( )a b a b+ ⋅ −% % % %

= a2 − b2

But a = b, since ABCD is a square. So, AC DB⋅

uuur uuur = a2 − a2

= 0 Therefore the diagonals of a square intersect at right angles.

( )

1AO AC212

a b

=

= +

uuur uuur

% %

( )

( )

DO AO AD1212

a b b

a b

= −

= + −

= −

uuur uuur uuur

% % %

% %

( )

( )

OB AO AB12

12DO

a b a

a b

= − +

= − + +

= −

=

uuur uuur uuur

% % %

% %uuur

Therefore the midpoint of AC is also the midpoint of BD The diagonals of a square bisect each other at 90° 20 ( ) ( )u v u v+ ⋅ −

% % % %

= u u u v u v v v⋅ − ⋅ + ⋅ − ⋅% % % % % % % %

= 2 2| | | |u v−

Modelling and problem solving 1

Using g = 9.8 m/s2 Weight down plane = 1 960sin15° = 507.285 N Water falling at 25 l/min and 1 l of water weighs 1 kg Mass of trailer is increasing at 25 kg/min If the mass of the trailer is M, then M = 200 + 25t (where t is

in minutes) The weight of the trailer down the plane = 9.8M sin15° =9.8(200 + 25t)sin15° Once the weight down the plane is over 1 000 N, Arnie

won’t be able to support it.

( )1000

9.8sin15

194.25525

1000 9.8 200 25 sin15

200 25

394.25525 194.255

7.77

t

t

tt

°

= + °

+ =

===

=

Arnie will be able to support the trailer for 7.77 mins (7 mins 46 secs).


2 This can be thought of as how far down the beach will the sweep drag Jodie.

Jodie swims at 8km/h. To cover 120m (=0.12km) it will take 0.12

8 0.015= hr. During this time, the sweep is dragging her

down the beach at 4 km/h. The distance dragged = 0.015 × 4 = 0.06 km.

She should walk 60 m up the beach before she starts to swim.

3 a The weight of the ball = 0.2 × 9.8 = 1.96 N down Motion is upwards, therefore resistance to motion is

downwards. Resistance = 1.0 N down Resultant force = 1.96 + 1.0 = 2.96 N down b When the ball is at the maximum height, its velocity is 0,

therefore the resistance is 0. The only force acting on the ball will be weight = 1.96 N

down. c The weight of the ball = 1.96 N down

Motion is downwards, therefore resistance to motion is upwards. Resistance = 1.0 N up

Resultant force = 1.96 – 1.0 = 0.96 N down 4 a Velocity at bottom of hill = 43.2 km/h = 12 m/s Initial velocity = 0 m/s Time = 10 s Mass = 1 500 kg Weight = 1 500 × 9.8 sin10° N Normal = 1 500 × 9.8 cos10° N Retardation = 20g N

Acceleration

1210

2

velocitytime

1.2 m/s

Δ=

=

=

R Force = mass × acceleration = 1 500 × 1.2 = 1 800 N R Force = Weight – Friction − Retardation Friction = Weight – R Force − Retardation = 1 500 × 9.8 sin10° − 1 800 – 20g = 556.6282 N Friction = μ × Normal μ 556.6282

1500 9.8cos100.038

=× °

=

b Handbrake turning off means that friction will be the only force to slow the car down

Friction = μ × Normal = 0.038 × 1 500 × 9.8 = 558.6 N Friction = mass × acceleration

Acceleration

558.61500

FrictionMass

−

=

=

v2 = u2 + 2as 0 = 122 + −2 558.6

1500 s

s = 193.3 m 5 a A rhombus is a quadrilateral with 4 equal sides.

|OA| 2 2 24 1 4

33

= + +

=

|OC| ( ) ( )2 222 5 2

33

= − + + −

=

|AB| ( ) ( ) ( )2 2 22 4 6 1 2 4

33

= − + − + −

=

|BC| ( ) ( ) ( )2 2 22 2 5 6 2 2

33

= − − + − + − −

=

|OA| = |OC| = |AB| = |BC| therefore the quadrilateral is a rhombus.

b ~ ~a c⋅ = (4, 1, 4) ⋅ (−2, 5, −2)

= −8 + 5 + −8 = −11 =|

~a | × |

~c | cos θ

= 33 33× cos θ cos θ = 11

33−

θ = 109.5° c i If two vectors are perpendicular, their dot product is 0. OA (p

~i + q

~j + r

~k ) = 0

(4~i +

~j + 4

~k ) (p

~i + q

~j + r

~k ) = 0

4p + q + 4r = 0 [1] OC (p

~i + q

~j + r

~k ) = 0

(−2~i + 5

~j − 2

~k ) (p

~i + q

~j + r

~k ) = 0

−2p + 5q − 2r = 0 [2] 2 × [2]: −4p + 10q – 2r = 0 [3] 1] + [3]: 0p + 11q + 0r = 0 q = 0 Sub q in [1] 4p + 4r = 0 p = −r

As it is a unit vector, 2 2 2 1p q r+ + =

Sub q = 0, p > 0 and p = r, 2

12

12

2 1

2 1

p

pp

r

=

==

= −

ii As the unit vector is perpendicular to OA and OC, it is perpendicular to the base of the pyramid. The magnitude of the height can be thought of as the resolution of OD in the direction of the unit vector. This is calculated by finding the dot product of the vectors.

OD ⋅ ( 12

, 0, − 12

) 3 12 3 29 1

3 2 3 2

2103 2 2

5 23

−= −

= +

= ⋅

=

6 a

b As ZY

uuuris parallel to OX

uuur, ZYuuur

= 3~i + 5

~j

As YXuuur

is parallel to OZuuur

but in the opposite direction, YXuuur

= −6~i .

c ~

~ ~

~ ~

OY

6

9 5

y

x i

i j

=

= +

= +

uuur

( )~~

~ ~~

~ ~

ZX

3 5 6

3 5

x z

i j i

i j

= −

⎛ ⎞= + −⎜ ⎟⎝ ⎠

= − +

uuur


d | OYuuur

| 2 29 5

106

= +

=

| ZXuuur

| ( )2 23 5

34

= − +

=

OYuuur

⋅ ZXuuur

= 9 × −3 + 5 × 5 = −2 =| OY

uuur| | ZXuuur

| cos θ 2 = 106 34 cosθ θ = 91.9° e 5

9tan

29.1

θθ

=

= °

f OZ is on the x-axis. Vector ~j is the unit vector

perpendicular to OZ. The vector resolute is the

~j component of OX = 5

~j

g As X is on ZY and ZY is parallel to OX,

( )~ ~ ~ ~

~ ~

6 3 5

6 3 5

x i k i j

k i k j

⎛ ⎞= + +⎜ ⎟⎝ ⎠

= + +

The vector joining P to X is found as follows:

( )

( ) ( )

~ ~

~ ~~ ~

~ ~

PX

3 5 6 3 5

3 3 5 5

x p

i j k i k j

k i k j

= −

⎛ ⎞ ⎛ ⎞= + − + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= − − + −

uuur

PXuuur

is perpendicular to ZXuuur

. This means that the dot product is 0.

0 = (−3 – 3k) (−3) + (5 – 5k)5 = 9 + 9k + 25 – 25k = 34 – 16k

k = 3416

= 178

( )17 17

8 8~ ~ ~3 58 8~ ~

6 3 5

12 10

x i j

i j

= + × + ×

= +

X has co-ordinates ( )3 58 812 ,10

h ~ ~

ZX 3 5i j= − +uuur

| ZXuuur

| 34= XP

uuur =

~p −

~x

3 58 8~ ~~ ~

3 58 8~ ~

12 10 3 5

9 5

i j i j

i j

⎛ ⎞ ⎛ ⎞= + − +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= +

ZX uuur

( ) ( )2 23 58 8

385232

9 5= +

=

Area of triangle = 12 base × perp. height

= 12 |ZX| × |XP|

382512 32

6502512 16

25512 42558

78

34

31

=

=

= ×

=

=

The triangle has an area of 7831 units2.

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