Chapter 8 Vector applications

Transcription

1 MC Qld- 5 Vector applications Chapter Vector applications Exercise A Force diagrams and the triangle of forces a F ( F + F ) (4 + ) i ( + 4 ) F (4 + ) + ( + 4 ) c d a ( + 4 ) tan θ (third quadrant) (4 + ).0796 θ 0 + tan (.0796) The third force has magnitude.5 N at an angle of 7 anticlockwise of i. The answer is E. e f g h C A B C 9 C N The answer is C. From the figure in part a, sin θ 4 5 θ sin (0.) 5. In actual situation: Angle The answer is E. 4 a i F i 5 and F 4 i + 9 R F + F i 5 4 i i + 4 ii F i + 6 and F 4 i 9 i R i i 9 6 i a R F + F R 4 i + + i + 4 (4 + ) i + ( + 4 ) The answer is C. iii F i and F 5i + R i 5i + ( 5 ) i i R ii R ( 6) + ( ) 45 5

2 Vector applications MC Qld- 5 iii R ( 5) + ( ) c i F ( F + F ) 9 i 4 ii F ( F + F ) 6 i + iii F ( F + F ) ( 5 ) i + 5 i a F i 5, F 4 i + 9 and F 6 i R F + F + F i i i i + R + 5 c tan θ θ 6.4 d F 4 ( F + F + F ) i ii a F i, F i + and F i R i i + + i R F + G i + + i i + 0i + 5 R ( 0) + ( 5 ) To find the angle with north ( R is west of north) θ 9.4 So R acts N 9.4 W R is a force of 6.4 N acting N 9.4 W. ii F R, would e a force of 6.4 N acting S 9.4 E. i R R W + F i i 5 ii F R 50 i + 5 c i R A + B 6 i 0 0 i a i ( )i R c tan θ 0 θ 0 d F 4 ( F + F + F ) F 4 ( )i ( )i d i 4 i + ii F R 4 i ln i, notation: F 0 cos 40 i 0 sin i 6.4 G 7 cos 45 i + 7 sin 45 In i, notation o o F 5cos60 i + 5sin i + o o G 45cos60 i + 45sin i i H cos 0 i + sin 0. i R F + G + H 7.66 i i i i +.6 R (.) + (.6).94.9 N Find the angle R makes with north ( R is west of north)

3 MC Qld- 54 Vector applications 7 a tan θ..6 θ 6.9 Direction is N 6.9 W R is a force of.9 N acting N 6.9 W. ii F R F is.9 N acting S 6.9 E. Z X + Y XY cos (0)(0) cos Z 0 N sin(0 θ) sin 60 Y Z 0 sin 60 sin (0 θ ) 0 sin 60 0 θ 60 θ 0 50 sin 45 sin (0 θ) 00 sin (0 θ) θ θ (0 ) 4 β 60 (5 + 4 ) 77 α 00, β 5, Y 7 N and Z? Z sin(0 α) Y sin(0 β ) Z 7 sin 0 sin 45 Z 7sin0 sin 45 Z 7.6 N c X Y Triangle is isosceles. α 60 or 0 α θ 0 β 0 (since isosceles triangle) Z sin(0 α) X sin(0 θ ) Z X sin 0 sin0 Z sin0 X sin0 X X Exercise B Newton s First Law of Motion a 9 C A + B AB cos (0 θ ) (000)(00) cos (0 θ) cos (0 θ) cos (0 θ ) cos (0 θ) θ 00. θ a α 5, Z 00 N, X 50 N and β? 0 α 45 sin (0 θ) sin 45 X Z T H 5 cos 0.5 N T V 5 sin 0.6 N c Acceleration 0 F T H.5 N a H 00 N W 40g N Angle to the ertical 5 The answer is D. The horizontal component of T is the same as the magnitude of the gien horizontal force. T H 00 N The answer is B. c The ertical component of T is the same as mg. T V 40g N The answer is C.

4 Vector applications MC Qld- 55 d c T V W T V W a 4 a T 00 + (40 g) Ν 494 Ν The answer is A. Acceleration 0 R 0 c F cos N d 50 g 5g N 45 N TV tan θ TH TV T H 45 tan N Left rope: T T H i + T V 7 i + 45 Right rope: T T H i + T V 7 i a R (T cos 0 T cos 60 ) i + (T sin 0 + T sin 60 g) R 0 i + 0 Equating the i -components: T cos 0 T cos 60 0 T T 0 T T 0 T T [] Equating the -components: T sin 0 + T sin 60 g 0 T + T g 0 T + T 6g 0 [] Sustitute T T into [] T + T 6g 0 4T 6g T 4g N Sustitute T 4g into [] T 4g 4 g N cos θ where θ is the angle etween either rope and 4 ceiling. θ 4.4 Each rope is angled 4.4 from the ceiling. e T left T right N f 6 a R 4000 sin θ 50g sin θ θ.75 x 4 cos θ 4 cos x Maximum separation is metres. A V 0 cos N down A H 0 sin N left c R (F 77.) i + (N 9.9 W) (F 77.) i + (N g) (F 77.) i + (N 4.9) d a 0 R 0 F F 77. N e N N 4.9 N

5 MC Qld- 56 Vector applications 7 a a R 0 i + 0 (Not moing) Therefore, R O N c W.5g sin 0 i.5g cos i.7 N d R (T 7.4) i + (N.7) 0 i + 0 T T 7.4 N e N.7 0 (From part d) N.7 N W.5g sin 0 i.5g cos i.7 N c Ri H (H cos 0 7.4) i 0 i H N H 7.4 i.5 sin i 4. N d R (N.7 4.) 0 N N 7 N N 7 e Consider the i component of R. Ri H cos0 W sin 0 0 H cos0 W sin 0 H W H W W H W W 9 The weight of the ook is mg N The normal must alance the weight, N 9 N Friction opposes motion, F 00 N F μn F μ N a F ma N The weight of the car is 900g 60 N The normal must alance the weight, N 60 N c F μn F μ N The weight of the log is mg N The normal must alance the weight, N 940 N Friction opposes motion, F 00 N F μn F μ N a W mg N Breaking W into i and components: W 5.sin 40 i 5.cos 40 The normal is the component of weight N 5.cos N c The frictional force is the i component of weight F 5.sin40 7. N d F μn F μ N W mg Breaking W into i and components: W mg sin 5 i mg cos5 The normal is the component of weight N mgcos5 The frictional force is the i component of weight F mgsin5 F μn F μ N mg sin5 mg cos5

6 Vector applications MC Qld sin5 cos5 tan5 0.7 W mg N Breaking W into i and components: W mgsin 40 i mgcos40 9.6sin 40 i 9.6cos 40 The normal is the component of weight N 9.6cos N F μn N Friction opposes the i component of weight. The resultant force down the plane 9.6sin N Exercise C Momentum a Momentum mass elocity Momentum 40 N s East Momentum mass elocity Momentum 600 N s South c Mass 50 g 0.5 kg Momentum mass elocity Momentum 7.5 N s North d Mass.5 tonnes 500 kg Momentum mass elocity Momentum 500 N s West e Velocity 40 km/h 400/6 m/s Momentum mass elocity 400/ Momentum 97 N s North f Mass.4 tonnes 400 kg Velocity 0 km/h 00/6 m/s Momentum mass elocity /6. Momentum N s North a Speed momentum mass Speed 0 m/s Mass 0 g 0.0 kg Speed momentum mass Speed m/s c Mass. tonnes 00 kg Speed momentum mass Speed 0 m/s a Mass momentum speed Mass 50 kg Speed 40 km/h 400/6 m/s Mass momentum speed / / Mass 90 kg 4 a oect A: momentum mass elocity 0 0 momentum 0 N s East oect B: momentum mass elocity 4 momentum N s East total momentum N s East Using results from 4a: Momentum oect A 0 N s East Momentum oect B N s West Total momentum 0 N s East + N s West 0 N s East N s East N s East c Using results from 4a: Momentum oect A 0 N s East Momentum oect B N s North Total momentum 0 N s East + N s North x θ 4.9 Total momentum N s at 4.9 North of East d Using results from 4a: Momentum oect A 0 N s East Momentum oect B N s 0 North of East Total momentum 0 N s East + N s 0 North of East using the cosine rule x cos x 7.

7 MC Qld- 5 Vector applications sin50 7. sin50 7. θ 6. Total momentum 7. N s at 6. North of East e Using results from 4a: Momentum oect A 0 N s East Momentum oect B N s North of West Total momentum 0 N s East + N s North of West Using Cosine rule x cos x 5 sin 45 5 sin45 5 θ. Total momentum 5 N s at. North of East 5 Mass 00 kg Initial elocity 6 m/s towards the wall Final elocity m/s away from the wall Initial momentum mass initial elocity N s towards the wall Final momentum N s away from the wall Change in momentum final momentum initial momentum (900 N s away from the wall) ( 00 N s towards the wall) (900 N s away from the wall) ( 00 N s towards the wall) 700 N s away from the wall 6 a Momentum mass elocity N s Momentum mass elocity N s Change in momentum N s 7 a Initial momentum mass elocity 0 0 N s Final momentum mass elocity 0 0 N s Change in momentum final momentum initial momentum N s Initial momentum mass elocity 7 56 N s Final momentum mass elocity 04 N s Change in momentum final momentum initial momentum N s c Mass 0.95 tonne 950 kg Initial elocity 44 km/h 440/6 m/s Final elocity 0 km/h 00/6 m/s Initial momentum mass elocity /6 6 N s Final momentum mass elocity /6 N s Change in momentum final momentum initial momentum N s d Initial elocity 50 km/h 500/6 m/s Final elocity km/h 0/6 m/s Initial momentum mass elocity / N s Final momentum mass elocity 400 0/6 000 N s Change in momentum final momentum initial momentum N s Laurie s momentum mass elocity 5 60 N s down Alan s momentum mass elocity N s right Initial momentum (60 N s down) + (700 N s right) x θ 44 As momentum is consered, the momentum of the collision is N s at 44 elow the horizontal in the same direction as Alan was initially moing Mass after collision kg Velocity momentum mass m/s The players will moe at 5. m/s at an angle of 44 elow the horizontal in the same direction as Alan was originally moing 9 a Before collision: Car A s momentum mass elocity N s North Car B s momentum mass elocity N s North Momentum of collision ( 560 N s North) + (560 N s North) 0 N s North

8 Vector applications MC Qld- 59 After collision: Car A s momentum mass elocity N s North Car B s momentum Momentum of collision Car A s momentum ( 0 N s North) ( 0 N s North) 40 N s North Car B s elocity momentum mass m/s Car B s elocity after the collision is 6.6 m/s North Before collision: Car A s momentum mass elocity N s North Car B s momentum mass elocity N s South Momentum of collision ( 560 N s North) + (560 N s South) ( 560 N s North) + ( 560 N s North) 000 N s North After collision: Car A s momentum mass elocity 0 0 N s North Car B s momentum Momentum of collision Car A s momentum ( 000 N s North) (0 N s North) 60 N s North Car B s elocity momentum mass m/s Car B s elocity after the collision is 6 m/s North c Before collision: Car A s momentum mass elocity N s North Car B s momentum mass elocity 0 40 N s South Momentum of collision (640 N s North) + ( 40 N s South) ( 640 N s South) + ( 40 N s North) 600 N s South After collision: Car A s momentum mass elocity N s South Car B s momentum Momentum of collision Car A s momentum ( 600 N s South) (640 N s South) 960 N s South Car B s elocity momentum mass m/s Car B s elocity after the collision is.4 m/s South 0 Laurie s momentum mass elocity 5 60 N s at 60 elow horizontal and to the left Alan s momentum mass elocity N s right Initial momentum (60 N s at 60 elow horizontal) + (700 N s right) Using cosine rule x cos x 690. sin sin θ 5.6 As momentum is consered, the momentum of the collision is 690. N s at 5.6 elow the horizontal in the direction of Alan s original moement Mass after collision kg Velocity momentum mass m/s The players will moe at.7 m/s at an angle of 5.6 elow the horizontal in the direction of Alan s initial moement. a Before collision: A s momentum mass elocity N s 0 B s momentum mass elocity N s Momentum of collision (0.4 N s 0 ) + (0 N s) 0.4 N s 0 After collision: A s momentum mass elocity N s 0 B s momentum Momentum of collision A s momentum (0.4 N s 0 ) (0. N s 0 ) Using cosine rule: x cos x 0. sin sin0 0. θ 7.9 Momentum 0. N s at 47.9 B s elocity momentum mass m/s B s elocity after the collision is. m/s 47.9 Before collision: A s momentum mass elocity N s 60

9 MC Qld- 60 Vector applications B s momentum mass elocity N s 0 Momentum of collision Using cosine rule: x cos0 0. x 0.59 sin sin θ 40.9 Momentum of collision 0.59 N s 40.9 After collision: A s momentum mass elocity N s 45 B s momentum Momentum of collision A s momentum (0.59 N s 40.9 ) (0. N s 45 ) Using cosine rule: x cos x 0.59 sin sin θ 0.5 Angle is Momentum 0.59 N s at 7.4 B s elocity momentum mass m/s B s elocity after the collision is 5.9 m/s 7.4 c Before collision: A s momentum mass elocity N s 0 B s momentum mass elocity N s 50 Momentum of collision Using cosine rule: x cos x 0.65 sin sin θ 40.9 Angle is Momentum of collision 0.65 N s 70.9 After collision: A s momentum mass elocity N s 0 B s momentum Momentum of collision A s momentum (0.65 N s 70.9 ) (0. N s 0 ) Using cosine rule: x cos x 0.67 sin sin θ 5.4 Angle is Momentum 0.67 N s at 65.4 B s elocity momentum mass m/s B s elocity after the collision is.67 m/s 65.4 d Before collision: A s momentum mass elocity N s 0 B s momentum mass elocity N s 0 Momentum of collision Using cosine rule: x cos x 0. sin sin 0 0. θ 46.9 Angle is Momentum of collision 0. N s 7.

10 Vector applications MC Qld- 6 After collision: A s momentum mass elocity N s 0 p p 00 + rel w 0 w + 0 i i + 00 B s momentum Momentum of collision A s momentum (0. N s 7. ) (0. N s 0 ) Using cosine rule: x cos x 0.4 sin sin θ 5.9 Angle is 60 ( ) 4 Momentum 0.4 N s at 4.5 B s elocity momentum mass m/s B s elocity after the collision is.4 m/s 4 a Before the rifle is fired, the ullet and rifle are still therefore their momentum is 0. After firing, mass of ullet 5g 0.05 kg. Momentum of ullet mass elocity N s As momentum efore firing 0, momentum after firing 0. Momentum of rifle 4.5 N s in opposite direction to ullet Speed momentum mass The rifle will moe at m/s Exercise D Relatie elocity 6 i rel w w 4 + rel w 6 i w θ 4.0 An oserer would see the oat moing at 6.5 m/s at angle of 4 South of East. 00 p rel w 0 i (from the West means towards the East) w θ The elocity of the plane relatie to the ground is 0.5 km/h at 5.7 East of North 5 5 i+ rel w w + rel w i w i ( ) θ 5.06 An oserer would see the oat moing at 7. m/s at angle of 5. North of West. 4 6 w θ 7.5 The ferry heads upstream at an angle of 7.5 to the ank.

11 MC Qld- 6 Vector applications 5 9 km/h North-West 6 km/h North-West p rel p + p rel Relatie to the water, the person is traelling at km/h. 6 a 40 a 0 i a rel a 40 0 i 0 i + 40 d a rel a rel a 5 5 i+ 0 i 5 5 ( 0 ) i θ m/s at 7.5 West of North 4 4 i a a rel 0 i a a rel a rel θ km/h at 6.6 West of North 0 a rel 5 a a rel m/s North e i 0 i i ( 0) + ( + ) θ m/s at 4.9 South of East cos0 i sin 0 a a rel i 5cos65 i+ 5sin 65 a c i+ a 0 i 5 5 a rel i 5cos65 i 5sin ( 5cos65 ) i ( 5sin 65 ) ( + 5cos65 ) + ( + 5sin 65 ) 7.65 ( + 5sin65 ) + 5cos65 θ m/s at 5 South of East

12 Vector applications MC Qld- 6 7 Riding north: 5 c To the cyclist, the wind appears to e coming from the northeast. If the magnitude of this elocity is a then w rel c w Riding south: 5 c i a a + w rel c c a a i 5 + a i+ 5 [] a To the cyclist, the wind appears to e coming from the southeast. If the magnitude of this elocity is then Riding south: 5 c i + 5 a a a a i ( ) + 5 [] To the cyclist, the wind appears to e coming from 0 south of east. If the magnitude of this elocity is then w rel c w i + + w rel c c i 5 + i+ 5 [] True speed and direction of wind: In oth cases, are the same. Equating the x components in w [] and [] results in, therefore a. a Equating the y components in [] and [] and replacing with a: a a 5 5 a 0 a 0 5 Su in [] i w 5i i The wind is lowing at 5 km / h towards the west (or from the east). Riding north: 5 c To the cyclist, the wind appears to e coming from the northeast. If the magnitude of this elocity is a then a a i w rel c + w c w rel c w rel c w cos0 i+ sin 0 i + w rel c c i 5 i ( ) [] True speed and direction of wind: In oth cases, are the same. Equating the x components in w [] and [] results in a a 6 Equating the y components in [] and [] and replacing a with : Su in [] i+ 5 ( ) w i+ ( 5) w ( ) 0( ) i+ 5 0 ( ) 0( ) i+ 5 i ( ) i ( ) + 5( )

13 MC Qld- 64 Vector applications ( ) ( ) 5 5 θ.9 The wind is lowing at 9.4 km / h towards North of East (or from South of West or 7 West of South) 9 Let the initial elocity of the ogger e 9 i To the ogger, the rain appears to e towards her at an angle of 0. If the magnitude of this elocity is a then asin0 i acos0 r rel + r r rel asin0 i acos0 + 9 i 9 asin0 i acos0 [] Return ourney: ( 9 69.sin0 ) i 69.cos0 r w.006 i 6.94 (.006) + ( 6.94) θ.54 The rain is falling at 6. km / h at.5 towards the ogger s original direction. Exercise E Using ectors in geometry CD r + t DB t + q CD DB, r + t t + q t r + q t ( r + q ) r + q BD t q BC r q BD n BC, t q n ( r q ) t q + n r n q ( n ) q + n r 9 i To the ogger, the rain appears to e coming towards her at an angle of 5. If the magnitude of this elocity is then sin 5 i cos5 r rel + r r rel sin 5 i cos5 9 i sin5 9 i cos5 [] True speed and direction of rain: In oth cases, are the same. Equating the y components in r [] and [] results in acos0 sin 5 a cos0 sin 5 Equating the x components in [] and [] and replacing with a: a cos0 9 asin0 sin5 9 cos5 a cos0 sin5 + asin0 cos5 cos0 sin5 a( + sin0 cos5 ) 0.59a a 69. Su in [] n n q + n r Let AD a Let CE AB AD a CB CE DB AD a EB CE DE DB EB a AC AB CB a ( a ) DE DE AC and therefore DE is parallel to AC.

14 Vector applications MC Qld a DC DB a CA CD + DA + u c BA DB + DA c + u d ( + u ) ( + u ) + u + u or + u Since u 0 c c ( + u ) ( + u ) u + u or c + u Since c or c AC AB 5 AB + BC + CA 0 a + + c 0 Since u 0 AC DB ( a + ) ( a + ) a + a a + a But a (rhomus) AC DB 0 AC and DB are perpendicular. Therefore the diagonals of a rhomus intersect at right angles. Clearly a + > a + as the sum of any two side lengths in a triangle is greater than the length of the third side. or a + a + a + a AD AB + BD AD AD (AB + BD) AD or AD AB AD + BD AD But BD AD 0 (perpendicular) AD AB AD AB AD cos a AD AB cos a [] CD CB + BD CD CD ( CB + BD ) CD CD CB CD + BD CD But BD CD 0 (perpendicular) CD CB CD CB CD cos c CD CB cos c [] But CB AB and a c (equilateral triangle) CD AB cos a AD from [] So, AD DC DB a + AC a + 0 Let AD a Let AB Let E e the midpoint of AC. AC a + AE ( a + ) BE + ( a + ) ( a ) ED ( a ) + a ( a ) BE E is the midpoint of BD. Therefore the diagonals of a parallelogram isect each other. Let AO a Let OB OC a AB a + BC ( a + ) + a a AB BC ( a + ) ( a ) a But a (radii of circle) AB BC a a 0 AB is perpendicular to BC.

15 MC Qld- 66 Vector applications a Therefore the angle sutended y the diameter of a circle is a right angle. Let AB a Let BC uur EF AB + BC a + FG a + HG a + EH a + uur EF HG and FG EH Therefore EFGH is a parallelogram since the opposite sides are equal and parallel. since a,, c are perpendicular. DH a + + c a since a c a + + c a + + c a since a c CG DH cos θ CG DH a a a a a θ cos Therefore the acute angle etween the maor diagonals is Let DA a Let AB Let CE c Let O e the midpoint of CG CG + a + c or CG a + c CO OG ( a + c) DH a + + c DO DA + AG OG a + c ( a + c) a + + c ( a + + c) OH OG + GH ( a + c) + a + + c ( a+ + c) DO OH DH Therefore the maor diagonals of a cue isect each other. CG DH ( a + c) ( a + + c) a a+ a + a c a c+ c a+ c + c c a + c + a c But a c 0, since a and c are perpendicular and a c, since it is a cue. CG DH a a + a + 0 CG a a + c a + ( ) + c c a+ c c ( a+ )( a+ ) a a+ a + c a + + a But a and are perpendicular, therefore a 0 c a + Chapter reiew f + f i i + i + 9 f ( f + f ) i 9 f + ( 9) 5 The answer is B. Resoling horizontally F 0cos The magnitude of F 4.6 N The solution is A Resoling the ectors ertically g Fcos5 g F cos5 and

16 Vector applications MC Qld- 67 g cos5 0. N The solution is B. 4 f ( f+ f) ( 0i + 40 ) f Find the angle with f f f 6400 f f cosθ f f θ 5.4 6g Tcos6.4 + Tsin 6.4 T ( cos6.4 + sin 6 4).6T T 6. T T 5.6 T 6. N T 5.6 N c θ 6.4 [rope ] 90 θ 6.6 [rope ] d tan θ from part, 5 f is 9.4 N at 5.4 to f then 5 Using triangle ABD, x sin θ AB 6 a (600)(500) cos (0 θ) cos (0 θ) cos (0 θ ) θ 4.4 θ.6 7 x AB sin θ x 5 5x Let θ angle etween T and horizontal. Resoling ertically o ( θ) W Tsin 90 + T 6g Tcosθ + Tsin θ [] Resoling horizontally Tcos 90 o ( θ) Tcosθ T T cosθ T T [] cosθ T T, therefore [] ecomes T T cosθ θ 6.4 [] Sustitute for θ and T in [] W mg N Breaking W into i and components: W mgsin 40 i mgcos40 9.sin 40 i 9.cos40 The normal is the component of weight N 9.cos N F μn N Friction opposes the i component of weight. The resultant force down the plane 9.sin N

17 MC Qld- 6 Vector applications mass 5 kg speed 00/0 m/s momentum mass speed Ns The solution is D 9 Mass 5 g 0.05 kg Initial elocity 40 m/s Final elocity 0 m/s Initial momentum mass initial elocity Ns Final momentum mass final elocity Ns Change in momentum The solution is B 0 Mass of car 70 kg Velocity of car 00 km/h 00/.6 m/s Momentum of car 70 00/ Ns Mass of oect 45 g kg Momentum of oect momentum of car 500 Ns Speed of oect momentum of oect / mass 500 / m/s The solution is C North moing car: Mass 50 kg Speed 70 km/hr 70/.6 m/s Momentum mass speed 50 70/ East moing car Mass 00 kg Speed 0 km/h 0/.6 m/s Momentum mass speed 00 0/.6.9 Magnitude The solution is D Mass 50 g 0.5 kg Velocity 0 m/s (speed is same efore and after, only direction changes) Momentum mass elocity Change in momentum final momentum initial momentum x + cos60 9 x As the sides of the triangle are all Ns, the triangle is equilateral and θ 60. The change of momentum is Ns perpendicular to the wall. Player A: Mass 00 kg Velocity m/s Momentum mass elocity 00 Ns Player B: Mass 5 kg Velocity 9 m/s Momentum mass elocity 765 Ns After collision: Mass kg Momentum is consered x θ 4.7 Speed momentum mass After the collision, they moe at 6 m/s at 4.7 to A s original motion. 4 a rel a i + ( i + 4 i + ) The solution is B 5 The car is traelling north. For the other car to appear as if it s traelling south-east, the actual elocity of the car must hae an easterly component. The only possile option is C 6 5 w rel w i + rel w i 5 w x θ The oat would moe at. km/h at 65.6 to the ank. 7 Riding east: 5 i c To the cyclist, the wind appears to e coming from 60 E of N. If the magnitude of this elocity is a then acos0 i asin 0 w rel c w a a i c + w rel c

18 Vector applications MC Qld- 69 Riding west: 5 i c a a i 5i + a 5 i [] a To the cyclist, the wind appears to e coming from 60 W of N. If the magnitude of this elocity is then w rel c w cos0 i sin 0 i + w rel c c i 5 i ( ) i 5 + [] True speed and direction of wind: In oth cases, are the same. Equating the y components w a in [] and [] results in, therefore a. Equating the x components in [] and [] and replacing with a: a a a 0 0 a 0 0 Su in [] 5 i w i The wind is lowing at.7 km / h towards the south (or from the north). AP p a PB p PB AP p ( p a) p p a p a+ p a+ The solution is D. 9 Let AB a Let AD DB a AC a+ AC DB ( a + ) ( a ) a But a, since ABCD is a square. So, AC DB a a 0 Therefore the diagonals of a square intersect at right angles. AO AC a + DO AO AD ( a+ ) ( a ) OB AO + AB ( a+ ) + a ( a ) DO Therefore the midpoint of AC is also the midpoint of BD The diagonals of a square isect each other at 90 0 ( u + ) ( u ) u u u + u u Modelling and prolem soling Using g 9. m/s Weight down plane 960sin N Water falling at 5 l/min and l of water weighs kg Mass of trailer is increasing at 5 kg/min If the mass of the trailer is M, then M t (where t is in minutes) The weight of the trailer down the plane 9.M sin5 9.(00 + 5t)sin5 Once the weight down the plane is oer 000 N, Arnie won t e ale to support it ( t ) sin t 9.sin t t Arnie will e ale to support the trailer for 7.77 mins (7 mins 46 secs).

19 MC Qld- 70 Vector applications This can e thought of as how far down the each will the sweep drag Jodie. Jodie swims at km/h. To coer 0m (0.km) it will take hr. During this time, the sweep is dragging her down the each at 4 km/h. The distance dragged km. She should walk 60 m up the each efore she starts to swim. a The weight of the all N down Motion is upwards, therefore resistance to motion is downwards. Resistance.0 N down Resultant force N down When the all is at the maximum height, its elocity is 0, therefore the resistance is 0. The only force acting on the all will e weight.96 N down. c The weight of the all.96 N down Motion is downwards, therefore resistance to motion is upwards. Resistance.0 N up Resultant force N down 4 a Velocity at ottom of hill 4. km/h m/s Initial elocity 0 m/s Time 0 s Mass 500 kg Weight sin0 N Normal cos0 N Retardation 0g N Δelocity Acceleration time 0. m/s R Force mass acceleration N R Force Weight Friction Retardation Friction Weight R Force Retardation sin0 00 0g N Friction μ Normal μ cos0 0.0 Handrake turning off means that friction will e the only force to slow the car down Friction μ Normal N Friction mass acceleration Friction Acceleration Mass u + as s 500 s 9. m 5 a A rhomus is a quadrilateral with 4 equal sides. OA OC ( ) ( ) AB 6 a BC ( ) + ( 5 6) + ( ) OA OC AB BC therefore the quadrilateral is a rhomus. (4,, 4) (, 5, ) ac a c cos θ cos θ cos θ θ 09.5 c i If two ectors are perpendicular, their dot product is 0. OA (p i + q + r k ) 0 (4 i k ) (p i + q + r k ) 0 4p + q + 4r 0 [] OC (p i + q + r k ) 0 ( i + 5 k ) (p i + q + r k ) 0 p + 5q r 0 [] []: 4p + 0q r 0 [] ] + []: 0p + q + 0r 0 q 0 Su q in [] 4p + 4r 0 p r As it is a unit ector, p + q + r Su q 0, p > 0 and p r, ii p p p r As the unit ector is perpendicular to OA and OC, it is perpendicular to the ase of the pyramid. The magnitude of the height can e thought of as the resolution of OD in the direction of the unit ector. This is calculated y finding the dot product of the ectors. OD (, 0, ) As ZY is parallel to OX, ZY i + 5 As YX is parallel to OZ ut in the opposite direction, YX 6 i. c OY y ZX x z x+ 6 i 9i+ 5 5 i+ 5 6i i+ 5

20 Vector applications MC Qld- 7 d OY OY ZX 5 e ZX OY ZX cos θ 06 4 cosθ θ θ 9. f OZ is on the x-axis. Vector is the unit ector perpendicular to OZ. The ector resolute is the component of OX 5 4 g As X is on ZY and ZY is parallel to OX, x 6i+ k i k i+ 5k The ector oining P to X is found as follows: PX x p i+ 5 ( 6+ k) i+ 5k ( ) ( 5 5 ) k i+ k PX is perpendicular to ZX. This means that the dot product is 0. 0 ( k) ( ) + (5 5k) k + 5 5k 4 6k h k x 6+ i i+ 0 5 X has co-ordinates (,0 5 ) ZX i+ 5 ZX 4 XP ZX p x i+ 0 i i 5 ( 9 ) ( 5 ) + 5 Area of triangle ase perp. height ZX XP The triangle has an area of units.