A Level Fur ther Mathematics for OCR A

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1 Brighter Thinking A Level Fur ther Mthemtics for OCR A Mechnics Student Book (AS/A Level) Jess Brker, Nthn Brker, Michele Conwy nd Jnet Such This resource hs een sumitted to OCR s endorsement process

2 Contents Contents Introduction... iv How to use this ook... v 1 Work, energy nd power 1 1 Section 1: The work done y force... Section : Kinetic energy nd the work energy principle...5 Section 3: Potentil energy, mechnicl energy nd conservtion of mechnicl energy...8 Section 4: Work done y force t n ngle to the direction of motion Section 5: Power Mixed prctice 1... Dimensionl nlysis 6 Section 1: Defining nd clculting dimensions...7 Section : Units nd dimensions of sums, differences nd ngles Section 3: Finding dimensions from units nd derivtives, conversion of units nd predicting formule...36 Section 4: Summry of dimensions nd units Mixed prctice Momentum nd collisions 1 45 Section 1: Momentum nd impulse...46 Section : Collisions nd the principle of conservtion of momentum Section 3: Restitution, kinetic energy nd impulsive tension...59 Mixed prctice Circulr motion 1 78 Section 1: Liner speed vs ngulr speed...79 Section : Accelertion in horizontl circulr motion Section 3: Horizontl circulr motion in 3D...87 Mixed prctice Centres of mss 1 10 Section 1: Centre of mss of system of point msses Section : Centres of mss of stndrd shpes Section 3: Centres of mss of composite odies Mixed prctice Focus on Proof Focus on Prolem-solving Focus on Modelling Work, energy nd power 136 Section 1: Work done y vrile force f(x) Section : Hooke s lw, work done ginst elsticity nd elstic potentil energy Section 3: Prolem solving involving work, energy nd power Section 4: Using vectors to clculte work done, kinetic energy nd power Mixed prctice Liner motion under vrile force 169 Section 1: Working with ccelertion, velocity nd displcement Section : Vrile force Mixed prctice Momentum nd collisions 193 Section 1: Vrile force nd vector nottion Section : Olique impcts nd the impulse momentum tringle...01 Section 3: Olique collisions of two spheres nd impulsive tensions in strings Mixed prctice Circulr motion 3 Section 1: Conservtion of mechnicl energy...4 Section : Components of ccelertion ( generl model)...35 Section 3: Prolem-solving situtions...40 Mixed prctice Centres of mss 56 Section 1: Centres of mss y integrtion...57 Section : Equilirium of rigid ody Mixed prctice Focus on Proof...89 Focus on Prolem-solving...91 Focus on Modelling...94 Cross topic review exercise...96 Pper 1 Prctice questions Pper Prctice questions Formule Answers Glossry Index...35 Acknowledgements...39 Cross-topic review exercise iii

3 1 Work, energy nd power 1 In this chpter you will lern how to: clculte the work done y force clculte kinetic energy use the work energy principle equte grvittionl potentil energy to work done ginst grvity perform clcultions using power. Before you strt GCSE A Level Mthemtics Student Book 1 A Level Mthemtics Student Book 1 You should know how to convert units of distnce, speed nd time. You should know how to clculte the weight of n oject from its mss, nd know the unit of weight. You should e le to use Newton s second lw of motion: F = m A Level Mthemtics Student Book You should e le to resolve force into components t right ngles to ech other. 4 A force of 8 N cts on prticle t n ngle of 0 to the positive horizontl direction. Wht re the horizontl nd verticl components of the force? The reltionship etween work nd energy You hve lredy studied the effect of force or system of forces in A Level Mthemtics. In this chpter, you will lern the definition of the work done y force, which is quntity tht is mesured in joules, the sme units tht re used for energy. You will lern out propulsive nd resistive forces. You will lern out the reltionship etween work done nd two different types of energy: kinetic nd grvittionl potentil energy. You will lso lern out power which is the rte of doing work. Work done ginst these forces decreses the kinetic energy of the cr. 1 Convert metres to kilometres. Clculte the weight of cr of mss 1150 kg, stting the unit with your nswer. 3 A resultnt force of 50 N cts on n oject of mss.5 kg. Clculte the ccelertion of the oject. 1

4 A Level Further Mthemtics for OCR A Mechnics Student Book Ides of work, energy nd power re crucil in engineering, enling engineers to design mchines to do useful work. Hydroelectric power sttions work y converting the work done y flling wter first into kinetic energy s the hydroelectric turines rotte nd then into electricity. Section 1: The work done y force Work is done y force when the oject it is pplied to moves. Some forces promote movement while others resist it. For exmple, when you cycle into reeze, your pedlling promotes movement ut the reeze cts ginst your movement. Forces tht promote movement re clled propulsive forces nd those tht resist movement re known s resistive forces. Other propulsive forces include the tension in rope eing used to drg n oject cross the ground nd the driving force of vehicle engine. The driving force of n engine is often descried s its trctive force. Other resistive forces include vehicle rking, nd resistnce from still ir. Key point 1.1 For force cting in the direction of motion: work done = force distnce work done is mesured in joules (J) 1 joule = 1 newton 1 metre, i.e. 1 J = 1 N m For exmple force of 5 newtons cting on n oject tht moves 15 metres in the direction of the force does 5 15 = 75 joules of work. Douling the force to 10 newtons over the sme distnce would doule the mount of work done to 150 joules. Likewise, douling the distnce moved to 30 metres with n unchnged force of 5 newtons would doule the mount of work done to 150 joules. WORKED EXAMPLE 1.1 Fst forwrd In Chpter 6, you will lern out elstic potentil energy nd its conversion to kinetic energy. A ox is pushed 5 m cross horizontl floor y horizontl force of 5 N. Clculte the work done y the force. Work done = force distnce = 5 5 = 15 J Use the definition of work done. Stte units of work done (J) with your nswer.

5 1 Work, energy nd power 1 WORKED EXAMPLE 1. A truck driver driving long horizontl rod pplies rking force of 75 kn for 5 m. Clculte the work done y the rkes, giving your nswer in kj. 75 kn = N Convert 75 kn to N s you need to work in stndrd units. Work done y rkes = rking force distnce Tip = = J = 1880 kj (3 s.f.) Use the definition of work done. Chnge J to kj. Work done y the rkes ginst movement is equivlent to work done y the truck ginst the rkes. WORKED EXAMPLE 1.3 A 50 kg crte is lifted 1 m y rope nd pulley system. Clculte the work done ginst grvity. Work done = force distnce work done ginst grvity = weight height gined Weight of crte = = 490 N Work done ginst grvity = weight height gined = J Apply the definition of work done to the grvittionl force. Force ecomes weight nd distnce ecomes height gined. Clculte the weight of the crte, sed on the usul pproximtion for the ccelertion due to grvity of 9.8 m s. Use the definition of work done. 3

6 A Level Further Mthemtics for OCR A Mechnics Student Book Key point 1. Fst forwrd When mss, m, is rised or lowered through height h: work done y or ginst grvity = weight height = mg h WORKED EXAMPLE 1.4 Lter on in this chpter you will lern the equivlence of work done y or ginst grvity nd grvittionl potentil energy. A competitor of mss 75 kg dives from 10-metre-high diving ord into pool. Air resistnce verges 1 newtons s he descends 10 metres through the ir. Resistnce from the wter then verges 3000 newtons s he descends metres further. Clculte: the totl work done y grvity s he descends 1 metres the totl work done ginst ir nd wter resistnce during this descent. work done y grvity = 75g 1 Use mgh to clculte the work done y 880 joules grvity. work done ginst ir resistnce = 1 10 = 10 joules work done ginst wter resistnce = 3000 = 6000 joules Totl work done ginst resistnces = 610 joules WORKED EXAMPLE 1.5 Use force distnce to clculte the work done y ech of the resistnces. A vn of mss 150 kg trvels long stright rod. The driving force of the vehicle engine is 500 newtons nd resistnce to motion is 0 newtons, on verge. The vn trvels 1.5 km from one delivery to the next, descending 8 metres in height. Find: the work done y the vehicle engine the work done y grvity c the work done ginst resistnce. 1.5 km = 1500 metres work done y vehicle engine = = joules work done y grvity = 150g joules c work done ginst resistnce = Convert distnce to SI units. = joules Use force distnce to clculte the work done y the vehicle engine. Use mgh to clculte the work done y grvity. Use force distnce to clculte the work done ginst resistnce. 4

7 1 Work, energy nd power 1 EXERCISE 1A 1 A prcel is drgged 5 metres cross horizontl floor y horizontl rope. The tension in the rope is 1 N. Clculte the work done y the tension in the rope. Susn clims verticl rock 3 m high. Susn s mss is 65 kg. Clculte the work done y Susn ginst grvity. 3 Sunil descends verticl ldder. His mss is 8 kg nd the work done y grvity is 150 J. Find the height Sunil descends. 4 A ll of mss 100 g is dropped from window. Clculte the work done y grvity s it flls verticlly to the ground 6 m elow. 5 A puck slides 50 metres cross n ice rink, ginst resistive force of.5 N. Clculte the work done ginst resistnce. 6 A cyclist trvelling on horizontl ground nd pplies driving force of 5 N ginst hedwind of 10 N nd resistnce from friction of 5 N. The cyclist trvels 1. km. Find: the work done y the cyclist the totl work done ginst wind nd friction. 7 A fish sket is rised from the se floor to fishing ot t se level, 18 metres ove. The mss of the sket is 15 kg. The resistnce to motion from the sewter is 50 N. Clculte the totl work done, ginst grvity nd wter resistnce, in rising the fish sket. 8 A driving force of 400 N does 50 kj of work moving vn long horizontl rod from A to B. Resistnce to motion verges 185 N. Clculte the work done ginst resistnce s the vn moves from A to B. Section : Kinetic energy nd the work energy principle Kinetic energy is the energy n oject hs ecuse it is moving. Key point 1.3 An oject of mss m moving with speed v hs kinetic energy 1 mv. If mss is mesured in kg nd speed is mesured in m s 1, kinetic energy is mesured in joules. WORKED EXAMPLE 1.6 Tip If speed is not given in m s 1, you should convert to m s 1 efore you strt the rest of your clcultions. A prticle of mss 1.5 kg is moving with kinetic energy 48 joules. Clculte the speed of the prticle. Kinetic energy = 1 mv Use the formul for kinetic energy. 48 = v v = 64 so v = +8 m s -1 Sustitute nd rerrnge to find speed As mss ws given in kg nd kinetic energy in joules, speed is in m s 1. Since speed is positive quntity, v = 8 m s -1. 5

8 A Level Further Mthemtics for OCR A Mechnics Student Book WORKED EXAMPLE 1.7 A cyclist slows down from 5 km h 1 to 10 km h 1. The comined mss of the cyclist nd her icycle is 95 kg. Clculte the loss of kinetic energy. Let u e the strting speed nd v e the finl speed: u = m s 1 nd 36. v = 10 =.78 m s 1 (3 s.f.) 36. Loss of kinetic energy = 1 mv 1 mu Loss of kinetic energy = = 190 J (3 s.f.) WORKED EXAMPLE 1.8 Clculte the increse in kinetic energy when ot of mss tonnes chnges velocity from 3i + 4j m s 1 to 4.5i + 4.5j m s 1. Give your nswer in kj. (Strting speed) = = 5 (Finl speed) = = 40.5 Gin in kinetic energy = 1 m(v u ) Use Pythgors theorem to convert the velocity vectors to speeds. You need the squre of the speed, not the velocity vector, for the kinetic energy formul. You cn write 1 mv 1 mu in fctorised form. tonnes = 000 kg Convert tonnes to 000 kg. Gin in kinetic energy = (40.5 5) 15.5 kj The work energy principle is n essentil ide in mechnics tht enles us to clculte nd compre the work necessry to cuse chnge in kinetic energy. Key point 1.4 The work done y force cuses n equl increse in its kinetic energy. work done = 1 mv 1 mu To convert km h 1 to m s 1 you must multiply y the conversion fctor , which simplifies to division y 3.6. Loss of kinetic energy = initil kinetic energy finl kinetic energy Divide y 1000 to convert joules to kj. 6

9 1 Work, energy nd power 1 WORKED EXAMPLE 1.9 A prticle of mss 1.6 kg t rest on smooth horizontl plne is cted on y constnt horizontl force of 8 N. Find the speed of the prticle fter it hs trvelled 5 metres. Work done = 8 5 = 40 joules Stephen is driving his cr long horizontl rod t 55 km h 1 when he notices roken-down vehicle, just off the rod, 150 m hed. Stephen nd his cr hve mss of 105 kg nd the totl resistnce to motion is ssumed constnt t 500 N. Stephen elieves he should slow down nd tht he cn slow down sufficiently without pplying the rkes. Clculte Stephen s speed, in km h 1, s he reches the roken down vehicle, tking ccount of the resistnce to motion. Work done ginst resistnce = resistive force distnce = = J Assume tht Stephen llows the resistnce to motion to slow his cr down over 150 m. There is no driving or rking force. Clculte the work done ginst resistnce. 55 km h 1 = m s Convert 55 km h 1 to m s 1. Loss of kinetic energy = 1 m (u v ) Write down the expression for loss of kinetic energy. ( ) = u v u v = v v = 9.34 m s 1 (3 s.f.) v km h 1 v = 33.6 km h 1 (3 s.f.) Work done = force distnce 40 = v Work energy principle: work done = 1 mv v = 50 v = 7.07 m s 1 (3 s.f.) WORKED EXAMPLE 1.10 Work energy principle: work done ginst resistnce = loss of KE Rerrnge nd solve for v. Convert ck to km h 1. 7

10 A Level Further Mthemtics for OCR A Mechnics Student Book EXERCISE 1B 1 Clculte the kinetic energy of cyclist nd her icycle hving comined mss of 70 kg, trvelling t 1 m s 1. Give your nswer in kj. Clculte the mss of n thlete who is running t 8.5 m s 1, with kinetic energy 3500 J. 3 Clculte the speed of us of mss 0 tonnes with kinetic energy 1100 kj. Give your nswer in km h 1. 4 A ox of mss 5 kg is pulled from A to B cross smooth horizontl floor y horizontl force of mgnitude 10 N. At point A, the ox hs speed 1.5 m s 1 nd t point B the ox hs speed.8 m s 1. Ignoring ll other resistive forces, find: the increse in kinetic energy of the ox the work done y the force c the distnce AB. 5 Clculte the loss of kinetic energy when ot of mss 3.5 tonnes reduces in velocity from (3i + 4j) m s 1 to (.5i + 3j) m s 1. 6 A cr driver rkes on horizontl rod nd slows down from 0 m s 1 to 1 m s 1. The mss of the cr nd its occupnts is 1150 kg. Find the loss in kinetic energy. Given tht the work done ginst resistnce to motion is 50 kj, find the work done y the rkes. 7 A child of mss 35 kg descends smooth slide, fter propelling herself onto the top t 1.6 m s 1. Ignoring ir resistnce, clculte her speed t the ottom of the slide, which is.1 metres lower down thn the top. 8 A ullet of mss 10 grms psses horizontlly through trget of thickness 5 cm. The speed of the ullet is reduced from 40 m s 1 to 90 m s 1. Clculte the mgnitude of the verge resistive force exerted on the ullet. 9 A trin with mss 100 tonnes is trvelling t 108 km h 1 on horizontl trcks, when the driver sees speed reduction sign. The trin s speed must e reduced to 75 km h 1 over 500 m. Resistnce to motion is pproximtely 8 kn. Clculte the rking force required, in kn. 10 A pckge of mss 500 grms slides down prcel chute of length 3.5 metres, strting from rest. The ottom of the chute is. metres elow the top. The speed of the pckge t the ottom of the chute is 4.5 m s 1. Find the resistnce to motion on the chute. 11 Use the eqution of motion, F = m, together with the formul, v = u + s, to derive the reltion: Fs = 1 m(v u ). 1 Eddy cycles up hill. His mss, together with his icycle, is 9 kg. His driving force is 15 N nd resistnce from friction is 45 N. Eddy trvels 350 metres long the rod which rises through verticl height of 3 metres. His strting speed is 8. m s 1. Find his finl speed. Section 3: Potentil energy, mechnicl energy nd conservtion of mechnicl energy Consider n oject of mss m flling freely under grvity from height h 1 to height h, with strting speed u nd finl speed v. h 1 h 1 h u v h ground level 8

11 1 Work, energy nd power 1 Since the only externl force cting on the oject is grvity the work energy principle ecomes: work done y grvity = increse in kinetic energy mv mu mg( h1 h) = Rerrnging this gives: mv mu mgh1 mgh = mu mv mgh1 + = mgh + Ech side of this eqution is the sum of two terms, one of which is kinetic energy. The other term is grvittionl potentil energy. Grvittionl potentil energy equls the work done y grvity cusing n oject of weight mg to fll through height h if no other externl force cts on the oject. Key point 1.5 Grvittionl potentil energy (GPE) = mgh, where h is the height ove ground (zero) level. The sum of kinetic energy nd grvittionl potentil energy is conserved if no externl force cts on the oject. The sum is usully clled mechnicl energy. Key point 1.6 If the only force cting on n oject is its weight then mechnicl energy is conserved: GPE + KE = mgh + 1 mv = constnt, where h is the verticl height ove the zero level. This digrm my help you to understnd the formul for conservtion of mechnicl energy more esily. As n oject descends in height it speeds up, so grvittionl potentil energy is converted into kinetic energy. As n oject scends in height it slows down, so kinetic energy is converted into grvittionl potentil energy. mechnicl energy is conserved 1 mu mgh 1 oject speeds up s it descends potentil energy converted to kinetic energy 1 mv mgh oject slows down s it scends kinetic energy converted to potentil energy 1 mw mgh 3 ground level Tip Whilst you cn choose ny height s your ground (zero) level it is usully est to choose the lowest height reched y the moving oject. 9

12 A Level Further Mthemtics for OCR A Mechnics Student Book WORKED EXAMPLE 1.11 Fisl throws ll of mss 15 grms verticlly upwrds from ground level with speed of 1.5 m s 1. Assuming no externl forces pply: Clculte the speed of the ll fter it hs risen 5 metres. Clculte the mximum height gined y the ll. Strting KE = = 9.77 J (3 s.f.) Strting KE = mgh + KE t 5 m = 0.15g 5 + KE t 5 m KE t 5 m = 3.64 J (3 s.f.) 1 mv = Speed = = 7.63 m s 1 (3 s.f.) Strting KE = mgh mx 9.77 = 0.15g h mx h mx = 7.98 metres (3 s.f.) Using energy to solve prolems The principle of conservtion of mechnicl energy pplies to the sitution where the only force cting on n oject is its weight. Use conservtion of mechnicl energy over the first 5 metres of the scent. Tke the grvittionl potentil energy t ground level to e zero. Clculte the kinetic energy of the ll. Clculte the speed of the ll. You hve lredy used the work energy principle to solve prolems involving other propulsive nd resistive forces. You re now redy to comine the work energy principle nd the principle of conservtion of energy: (GPE 1 + KE 1 ) + work done y driving forces work done ginst resistive forces = (GPE + KE ) When using this formul you must rememer tht you do not need to include work done y grvity or work done ginst grvity s these re lredy ccounted for. Use conservtion of mechnicl energy over the whole scent (finl kinetic energy is zero). All the kinetic energy will hve een converted into grvittionl potentil energy. Clculte the mximum height gined. 10

13 1 Work, energy nd power 1 WORKED EXAMPLE 1.1 A 1.5 kg pckge is ttched to one end of n inextensile string. The string nd pckge re eing rised y the ction of pulley. The tension in the string is 18 N. Find the height gined y the pckge s it increses in speed from 1. m s 1 to 3. m s 1. Increse in KE = ( ) = 6.6 joules Propulsive work done resistive work done = 6.6 joules 18 h - 1.5g h = h = = metres g WORKED EXAMPLE m mgh 1 + KE 1 + work done y Helen - work done ginst resistnce = mgh + KE KE 1 = KE = 0 Work done ginst resistnce = resistive force distnce = = J vlley floor Clculte the increse in kinetic energy. Work energy principle: totl work done = increse in KE work done = tension h mg h Helen cycles from rest t the top of sloping trck, 35 m ove the vlley floor. She pedls downhill, then continues long horizontl trck, efore scending 10 m on n uphill trck nd stopping. The totl distnce she trvels is 500 m, nd the verge resistnce to motion is 55 N. The comined mss of Helen nd her icycle is 6 kg. Clculte the totl work done y Helen, nd the verge driving force she pplies. 10 m Use the work energy principle nd the principle of conservtion of energy. Helen hs no kinetic energy t the strt of her ride nd none t the end. You do not need to consider her motion throughout her ride: just t the strt nd t the end. Clculte the work Helen does ginst resistnce. The distnce used is the totl distnce long the rod. 6g (35-10) + work done y Helen = Let oth kinetic initil nd finl energy e zero nd rerrnge: mgh 1 mgh + work done y Helen work done ginst resistnce = 0 mgh 1 mgh is the loss of potentil energy of Helen nd her icycle. Continues on next pge... 11

14 A Level Further Mthemtics for OCR A Mechnics Student Book Work done y Helen J Helen s verge driving force work doney Helen = distnce = = 4.6 N Conservtion of energy is n importnt principle throughout Physics. Work done y moving oject ginst resistnce, which is lost mechnicl energy, is converted to other forms of energy such s het nd noise. This mens tht totl energy is still conserved. Did you know? The mechnicl equivlent of het ws first proposed y Jmes Joule nd explins the reltionship etween mechnicl energy nd het energy. EXERCISE 1C 1 Clculte the increse in potentil energy when mss of 5 kg is rised 5 m. Clculte the loss of potentil energy when mss of tonnes is lowered 10 m. 3 A oy of mss 68 kg gins 3600 J of potentil energy when climing verticl rope. Clculte the height he gins. 4 A toy trin loses 1.5 J of potentil energy when it descends spirl trck losing 50 cm in height. Find the mss of the toy trin. 5 Richrd strikes golf ll off n elevted tee. The golf ll hs mss 45 grms nd Richrd imprts n initil speed of 35 m s 1 to the ll. c d Find the initil kinetic energy of the golf ll. The ll lnds on the green 5 metres elow the tee. Clculte the loss of potentil energy of the ll when it lnds on the green. Clculte the kinetic energy of the ll when it lnds on the green. Clculte the speed of the golf ll when it lnds on the green. 6 Anit dives off highord into diving pool. When Anit leves the ord she hs speed of 7.5 m s 1 nd she is 8 metres ove the wter surfce. Anit s mss is 60 kg. c d Find Anit s kinetic energy s she leves the ord. Clculte Anit s kinetic energy s she enters the wter. Clculte Anit s speed s she enters the wter. Use Helen s work done, together with her distnce trvelled (500 m) to clculte her verge driving force. Wht modelling ssumptions hve you mde to simplify your clcultions? 7 Wing serves 58 grm tennis ll with speed of 9.5 m s 1 from height 3 metres ove the level of the tennis court. Assuming there re no resistive forces cting on the ll, clculte: 1 the kinetic energy of the ll s Wing serves it

15 1 Work, energy nd power 1 the potentil energy lost y the ll s it descends to the level of the court c d the kinetic energy of the ll s it strikes the court the speed of the tennis ll s it strikes the court. 8 Preeti descends slide strting from rest. Her mss is 58 kg. Overll, her chnge in verticl height ove the ground is 5.8 m nd her speed t the ottom of the slide is 6.5 m s 1. Clculte the work done ginst resistnce during her descent. 9 A pckge of mss 0.8 kg is projected down smooth sloping prcel chute with speed of. m s 1. The height of the ottom of the chute is 7.5 m verticlly elow the top. Assuming there re no externl resistive forces, clculte: the loss of potentil energy of the pckge the speed of the pckge t the ottom of the chute. 10 Krol slides on his sledge down stright trck of length 10 m, descending 3 m. The comined mss of Krol nd his sledge is 7 kg. Krol s strting speed is.8 m s 1 nd his speed t the end of his descent is 10.5 m s 1. Clculte the verge resistnce to motion, R N, during Krol s descent. 11 Lorett nd her icycle hve comined mss of 75 kg. Lorett cycles up stright hill AB, ccelerting from rest t A to 4 m s 1 t B. The level of point A is 5.5 m elow the level of B. Find: the increse in kinetic energy of Lorett nd her icycle s she cycles from A to B the increse in potentil energy of Lorett nd her icycle. During her ride, the resistnce to motion is constnt t 60 N prllel to the rod surfce nd Lorett does 8500 J of work. c Clculte the distnce from A to B. Section 4: Work done y force t n ngle to the direction of motion The prolems you hve worked with so fr hve hd ll forces in the direction of movement, either promoting motion or resisting it directly. But in mny cses the forces cusing motion re not in the direction of motion. Exmples re: mn drgging sledge long horizontl ground y pulling on rope tht is ngled upwrds child descending slide under grvity; the child s weight cts verticlly downwrds ut she trvels down the slide t n ngle to the verticl. If force is pplied t n ngle θ to the direction of motion s shown, the resolved component of the force tht does work is F cos θ. The resolved component tht is perpendiculr to the direction of motion, F sin θ, does no work. force, F θ F cos θ oject moving resolved component of F this wy in the direction of motion 13

16 A Level Further Mthemtics for OCR A Mechnics Student Book Key point 1.7 If force is cting t n ngle θ to the direction of movement: work done = force cos θ distnce Rewind Resolving forces is covered in A Level Mthemtics Student Book, Chpter 1. Fst forwrd In chpter 6 you will lern tht the formul for work done y force t n ngle to movement is the sclr product of the force vector nd the displcement vector. WORKED EXAMPLE 1.14 Jml is drgging his son on sledge on horizontl ground. He is pulling rope, ttched to the sledge, t n ngle of 30 to the horizontl. The tension in the rope is 50 N. Find the work Jml does drgging the sledge 8 m. Work done = force cos θ distnce Work done = 50 cos = 3680 J (3 s.f) WORKED EXAMPLE 1.15 Use the definition of work done y force cting t n ngle to the direction of motion. A girl of mss 0 kg descends stright smooth slide, strting from rest. The slide is 5 m in length nd inclined t 10 to the horizontl. Use work nd energy to clculte the speed of the girl t the ottom of the slide. Let θ e the ngle etween the verticl nd the slope. θ = 80 Work done y girl s weight = weight cos θ distnce = 0g cos joules The slope mkes n ngle of 10 with the horizontl ut 80 with the verticl. Clculte the work done y grvity. Use the work energy principle: 170. = 1 0 work done y grvity = increse of kinetic v energy. v = 4.13 m s 1 (3 s.f.) The girl is trvelling t 4.13 m s 1 t the ottom of the slide. 14

17 1 Work, energy nd power 1 A ox of mss 500 g is projected with speed.5 m s 1 up smooth inclined plne. The plne slopes t 0 to the horizontl. Clculte how fr the ox trvels up the plne y considering conservtion of energy. The slope mkes n ngle of 0 with the horizontl ut 70 with the verticl. GPE 1 = 0 At the end of the movement: GPE = 0.5g cos 70 x 1.678x J KE 1 = GPE = 1.678x x = m (3 s.f.) The ox trvels m up the plne. EXERCISE 1D WORKED EXAMPLE 1.16 The slope mkes n ngle of 0 with the horizontl ut 70 with the verticl. Let grvittionl potentil energy t strt of movement e zero (GPE 1 = 0). Let x m e the distnce the ox trvels up the plne. As the ox trvels x metres up the plne, its grvittionl potentil energy increses. This is equivlent to the work done ginst the weight of the ox: weight cos θ distnce Compre mechnicl energy when the ox is projected with mechnicl energy when it comes to rest: GPE 1 + KE 1 = GPE + KE KE = 0 (when the ox comes to rest). A force of 5 newtons is cting t constnt 0 to the line of movement of prticle tht moves 5 metres. Clculte the work done y the force. A force of 1.8 kn is cting t constnt 5 to the line of movement of prticle tht moves 1. metres. Clculte the work done y the force. Give your nswer in kj. A prticle is cted on y force of 1 newtons cting t constnt ngle of 15 to its direction of movement. The force does 850 joules of work. Find the distnce moved y the prticle. A prticle is cted on y force of F newtons cting t constnt ngle of 8 to its direction of movement. The force does 150 joules of work moving the prticle 55 metres. Find the vlue of F. A prticle is cted on y force of 10.5 newtons cting t constnt ngle to its direction of movement. The force does 750 joules of work moving the prticle through 15 metres. Find the constnt ngle. Clculte the increse in potentil energy when mss of 500 grms is moved 10 m up plne inclined t 5 to the horizontl. Clculte the loss of potentil energy when mss of 1.5 kg descends 0 m long plne inclined t 35 to the horizontl. 8 A cr is towed t constnt speed long horizontl stright rod. The tow rope is t 30 to the horizontl nd the tension in the tow rope is 180 N. The work done y the force is 4800 N. Clculte the distnce moved y the cr. 15

18 A Level Further Mthemtics for OCR A Mechnics Student Book 9 A lock of mss 3 kg is relesed from rest on smooth plne inclined t 18 to the horizontl, nd descends 50 m down the plne. Clculte: the loss of potentil energy of the lock the gin in kinetic energy of the lock. 10 A lock of mss 0 kg is drgged up smooth slope from rest t X to Y. The distnce XY is 8 m nd the slope is inclined t 6 to the horizontl. The rope used to drg the lock is prllel to the slope nd hs tension of 30 N. Find: the work done y the tension in the rope the chnge in potentil energy of the lock c the speed of the lock t Y. Section 5: Power Power is the rte of doing work. Averge power is defined s the totl work done y force divided y the time tken. Key point 1.8 When the force pplied is constnt: Averge power = work done time tken Did you know? Often you consider power in reltion to driving force ut it pplies eqully to ny constnt force cting on n oject. Power is mesured in wtts (W). 1 joule per second is equl to 1 wtt. WORKED EXAMPLE 1.17 A crne lifts.5 tonne concrete lock 35 m in 50 seconds. Clculte the verge power rting of the crne during the lift, giving your nswer in kw. Work done ginst grvity = weight height gined = 500g 35 = g J Power = 87500g 50 =3430 W or 3.43 kw Jmes Wtt ( ) ws Scottish engineer nd scientist. The unit of power is nmed fter him. Clculte the work done y the crne lifting the lock ginst grvity. Use the definition of power = work done time tken. 16

19 1 Work, energy nd power 1 WORKED EXAMPLE 1.18 The engine rkes on truck hve power rting of 5 kw. Clculte the totl work done in 5 seconds y the rking force t this verge power rting, giving your nswer in kj. Work done y rkes = = J or 150 kj WORKED EXAMPLE 1.19 Rerrnge the definition of power to mke work done the suject: work done = power time Convert your nswer to kj. A pump is used to rise wter from well. In one minute, 1500 litres of wter is rised 8 metres efore eing ejected into tnk t speed of 7.5 m s 1. The density of wter is 1000 kg m 3. Clculte the gin of potentil energy of the wter per second. Clculte the gin of kinetic energy of the wter ejected per second. c Clculte the power of the pump, in wtts. There re 1000 litres in 1 m 3, with mss of 1000 kg. Work out the mss of 1500 litres of wter litres of wter hs mss of 1500 kg. Gin of PE of wter per second = 1500g 8 60 = 1960 joules Gin of KE of wter per second = 60 = joules c Work done y pump = = 660 joules (3 s.f.) Therefore power = 660 (3 s.f.) You cn mke use of n lterntive formul for power when solving prolems. From the definitions of power nd work done: work done Power = time tken = force distnce = force distnce = force speed time tken time This definition llows you to work out power t specific point in time if you know the force nd the speed. This is often referred to s 'instntneous power' nd cn e used to work out power when either the force or the velocity vries over time. Use mgh to clculte the gin in potentil energy of the wter per second. Use 1 mv to clculte the gin of kinetic energy of the wter per second. Use conservtion of mechnicl energy: work done y the pump = gin in GPE + gin in KE The gin of totl mechnicl energy per second is the power of the pump. 17

20 A Level Further Mthemtics for OCR A Mechnics Student Book Key point 1.9 power = trctive force speed WORKED EXAMPLE 1.0 Chris is cycling t constnt speed of 8 m s 1 on horizontl rod with power output of 00 W. Clculte the totl resistnce to Chris nd his icycle in newtons. 00 = trctive force 8 trctive force = 5 N Trctive force totl resistive force = 0 trctive force = totl resistive force totl resistive force = 5 N WORKED EXAMPLE 1.1 power = trctive force speed As Chris is trvelling t constnt speed the resultnt force is zero. Juli is riding her motorike long horizontl rod with constnt speed 90 km h 1, t n engine power of 15 kw. Juli decides to overtke nd increses to full power, 0 kw. Assuming the resistnce to motion is unchnged, clculte Juli s ccelertion. Juli nd her motorike hve comined mss of 465 kg. 90 km h 1 = 5 m s 1 Convert Juli s speed to m s 1. Power = trctive force speed trctive force = power speed For the motorike: trctive force = = 600 N Trctive force resistnce to motion = 0 trctive force = resistnce to motion resistnce to motion = 600 N When Juli increses to full power: trctive force = power speed = = 800 N resultnt force = trctive force resistnce to motion = = 00 N Using F = m 00 = 465 ccelertion Accelertion = m s Use the definition of power, ut rerrnge to mke trctive force the suject. Clculte the trctive force of Juli nd her motorike when she is trvelling t constnt speed. As Juli is cruising t constnt speed the resultnt force is zero. Hence clculte the resistive force. When Juli increses the power, the trctive force increses so tht it is greter thn the resistive force, nd she ccelertes. Clculte the resultnt force. Use F = m to clculte Juli s initil ccelertion. 18

21 1 Work, energy nd power 1 WORKED EXAMPLE 1. A cr of mss 150 kg is trvelling long stright horizontl rod ginst resistnce to motion of kvkvn, where v is the speed of the cr nd k is constnt. When the engine is producing power of 13.5 kw, the cr hs speed 1.5 m s 1 nd is ccelerting t 0.45 m s. Find the vlue of k. The mximum constnt speed of the cr on this rod is 8 m s 1. Find the engine s mximum power, giving your nswer in kw. 1 kv N Let T e the trctive force of the cr engine. Resistnce vries with v 1. Resultnt force = T kv 1 Resultnt force = T resistnce T = power speed T = = 1080 N Resultnt force = mss ccelertion 1080 k = k = or 146 units Rerrnge the formul: power = force speed Clculte T. Use F = m. Rerrnge to find the vlue of k. T kv 1 = 0 Let the new driving force e T. As the cr is now moving with constnt speed the resultnt force is now zero. mximum power T = speed mximum power = 0 8 Mximum power = = 1.6 kw (3 s.f.) R 150 g 0.45 ms 1.5 ms 1 T Rerrnge to find the mximum power rting of the cr engine. 19

22 A Level Further Mthemtics for OCR A Mechnics Student Book EXERCISE 1E 1 A 38-tonne truck is le to rke from 65 km h 1 to rest in 0 seconds. Find the verge power rting of the rkes. A crne lifts -tonne concrete lock 5 m in 10 seconds. Find the verge power of the crne. 3 A lift of mss 800 kg cn ccommodte up to 1 people, ssumed to hve comined mss no more thn 1500 kg. Clculte the verge power required y the lift motor to rise the mximum lod through 60 m in 45 s. 4 A cr engine hs mximum driving force of 750 N when trvelling t 80 km h 1. Clculte the verge power of the engine. 5 A trin engine hs verge power rting of.5 MW. Clculte the trctive force when the trin is trvelling t 16 km h 1. 6 Find the verge power exerted y climer of mss 78 kg when climing verticl distnce of 35 m in 3 minutes. 7 A ot is trvelling t constnt speed of 16 km h 1. The ot hs mss 10.5 tonnes nd the engine is working t its mximum power output of 8 kw. Clculte the work done when the ot is displced (5i + 6j) km. 8 Find the verge power of n engine tht lifts 500 gs of flour 5 m in 1 hour. Ech g of flour hs mss 50 kg. 9 A pump is used to rise wter from well tht is 1 metres deep. Wter is rised t rte of 50 kg per second, nd is ejected into pipe t speed of 5.5 m s 1. c Clculte the gin of potentil energy of the wter per second. Clculte the gin of kinetic energy of the wter ejected per second. Clculte the power of the pump, in wtts. 10 Victori is cycling on level ground. Victori nd her cycle hve comined mss of 61.5 kg nd she is working t rte of 380 W. Given tht Victori is ccelerting t 0.65 m s, find the sum of the resistive forces cting on Victori nd her cycle t the instnt when her speed is 7.5 m s Stn is driving his 35-tonne truck on horizontl rod. Stn ccelertes from 50 km h 1 to 65 km h 1, which is his mximum speed t 500 kw power output. Find the mximum ccelertion of the truck, ssuming tht totl resistnce is constnt. 1 Vince is driving his vn ginst constnt resistnce to motion of 5500 N. The vn hs mss.8 tonnes nd engine power 15 kw. Vince s ccelertion t the instnt when his speed is u ms 1 is 0.9 m s. Clculte u. 13 The resistnce to motion of cr is kv 3 N, where v m s 1 is the speed of the cr nd k is constnt. The power of the cr s engine is 1 kw, nd the cr hs constnt speed of 7.5 m s 1 long horizontl rod. Show tht k A rocket, Athen, of mss 500 kg is moving in stright line in spce, without ny resistnce to motion. Athen s rocket motor is working t constnt rte of 500 kw nd her mss is ssumed to e constnt. Athen s speed increses from 90 m s 1 to 140 m s 1 in time t seconds. Clculte the vlue of t. Clculte Athen s ccelertion when her speed is 10 m s 1. 0

23 1 Work, energy nd power 1 Checklist of lerning nd understnding In this chpter you hve lerned the reltionship etween mechnicl work nd mechnicl energy. You hve lerned the work energy principle nd the principle of conservtion of energy. You hve used the definition of power to solve prolems. Work done = force distnce Work done ginst grvity = weight height gined (mgh) Kinetic energy = 1 mv Grvittionl potentil energy: GPE = mgh Work-energy principle: work done = chnge in energy If force is cting t n ngle θ to the direction of motion: Work done = force cos θ distnce work done Averge power = time tken Power = trctive force speed 1

24 A Level Further Mthemtics for OCR A Mechnics Student Book Mixed prctice 1 1 A womn drgs suitcse t constnt speed in stright line long horizontl ground y mens of plstic tether ttched to the suitcse. The tether mkes n ngle of 5 with the horizontl nd the tension in the tether is 50 N. Clculte the work done in moving the suitcse 150 m. A cr is pulled t constnt speed long horizontl stright rod y force of 00 N inclined t 35 to the horizontl. Given tht the work done y the force is 5000 J, clculte the distnce moved y the cr. [ OCR, A Level Further Mthemtics, 479, June 008] 3 Find the verge power exerted y rock climer of mss 80 kg when climing verticl distnce of 60 m in 4 minutes. 4 A lock is eing pushed in stright line long horizontl ground y force of 18 N inclined t 15 elow the horizontl. The lock moves distnce of 6 m in 5 s with constnt speed. Find: the work done y the force the power with which the force is working. [ OCR, A Level Further Mthemtics, 479/01, Jnury 013] 5 A nd B re two points on line of gretest slope of smooth inclined plne, with B verticl distnce of 8 m elow the level of A. A prticle of mss 0.75 kg is projected down the plne from A with speed of m s 1. Find: the loss in potentil energy of the prticle s it moves from A to B the speed of the prticle when it reches B. [ OCR, A Level Further Mthemtics, 479/01, June 013] 6 The power developed y the engine of cr s it trvels t constnt speed of 3 m s 1 on horizontl rod is 0 kw. Clculte the resistnce to the motion of the cr. The cr, of mss 1500 kg, now trvels down stright rod inclined t to the horizontl. The resistnce to the motion of the cr is unchnged. Find the power produced y the engine of the cr when the cr hs speed 3 m s 1 nd is ccelerting t 0.1 m s. [ OCR, A Level Further Mthemtics, 479/01, June 013] 7 A cr of mss 1600 kg moves long stright horizontl rod. The resistnce to the motion of the cr hs constnt mgnitude 800 N nd the cr s engine is working t constnt rte of 0 kw. Find the ccelertion of the cr t n instnt when the cr s speed is 0 m s 1. The cr now moves up hill inclined t 4 to the horizontl. The cr s engine continues to work t 0 kw nd the mgnitude of the resistnce to motion remins t 800 N. Find the gretest stedy speed t which the cr cn move up the hill. [ OCR, A Level Further Mthemtics, 479, June 01] 8 A stone of mss 50 kg strts from rest nd is drgged 35 m up slope inclined t 7 to the horizontl y rope inclined t 5 to the slope. The tension in the rope is 10 N nd the resistnce to the motion of the lock is 0 N. The lock is initilly t rest. Clculte:

25 1 Work, energy nd power 1 the work done y the tension in the rope c the chnge in the potentil energy of the lock the speed of the lock fter it hs moved 35 m up the slope. 9 A cr of mss 150 kg trvels long stright rod inclined t to the horizontl. The resistnce to the motion of the cr is kv N, where v m s 1 is the speed of the cr nd k is constnt. The cr trvels t constnt speed of 5 m s 1 up the slope nd the engine of the cr works t constnt rte of 1 kw. Clculte the vlue of k. Clculte the constnt speed of the cr on horizontl rod. [ OCR, A Level Further Mthemtics, 479, June 011] 10 A cr of mss 1500 kg trvels long stright horizontl rod. The resistnce to the motion of the cr is kv 1 N, where v m s 1 is the speed of the cr nd k is constnt. At the instnt when the engine produces power of W, the cr hs speed 15 m s 1 nd is ccelerting t 0.4 m s. Find the vlue of k. It is given tht the gretest stedy speed of the cr on this rod is 30 m s 1. Find the gretest power tht the engine cn produce. [ OCR, A Level Further Mthemtics, 479/01, Jnury 013] 11 The resistnce to the motion of cr is kv 3 N, where v m s 1 is the cr s speed nd k is constnt. The power exerted y the cr s engine is W, nd the cr hs constnt speed 5 m s 1 long horizontl rod. Show tht k = 4.8. With the engine operting t much lower power, the cr descends hill of inclintion α, where sin α = At n instnt when the speed of the cr is 16 m s 1, its ccelertion is 0.3 m s. Given tht the mss of the cr is 700 kg, clculte the power of the engine. [ OCR, A Level Further Mthemtics, 479, Jnury 011] 1 The mximum power produced y the engine of smll eroplne of mss tonnes is 18 kw. Air resistnce opposes the motion directly nd the lift force is perpendiculr to the direction of motion. The mgnitude of the ir resistnce is proportionl to the squre of the speed nd the mximum stedy speed in level flight is 80 m s 1. Clculte the mgnitude of the ir resistnce when the speed is 60 m s 1. The eroplne is climing t constnt ngle of to the horizontl. Find the mximum ccelertion t n instnt when the speed of the eroplne is 60 m s 1. [ OCR, A Level Further Mthemtics, 479, June 010] 13 The resistnce to the motion of cr of mss 600 kg is kv N, where v m s 1 is the cr s speed nd k is constnt. The cr scends hill of inclintion α, where sin α = 1. The power exerted y the cr s 10 engine is W nd the cr hs constnt speed 0 m s 1. Show tht k = 0.6. The power exerted y the cr s engine is incresed to W. 3

26 A Level Further Mthemtics for OCR A Mechnics Student Book Clculte the mximum speed of the cr while scending the hill. The cr now trvels on horizontl ground nd the power remins W. c Clculte the ccelertion of the cr t n instnt when its speed is 3 m s 1. [ OCR, A Level Further Mthemtics, 479/01, June 008] 14 A cr of mss 100 kg hs mximum speed of 30 m s 1 when trvelling on horizontl rod. The cr experiences resistnce of kv N, where v m s 1 is the speed of the cr nd k is constnt. The mximum power of the cr s engine is W. Show tht k = 50. c Find the mximum possile ccelertion of the cr when it is trvelling t 0 m s 1 on horizontl rod. The cr clims hill, which is inclined t n ngle of 10 to the horizontl, t constnt speed of 15 m s 1. Clculte the power of the cr s engine. [ OCR, A Level Further Mthemtics, 479/01, Jnury 008] 15 A spce shuttle of mss 400 kg is moving in stright line in spce. There is no resistnce to motion, nd the mss of the shuttle is ssumed to e constnt. With its motor working t constnt rte of 650 kw the shuttle's speed increses from 10 m s 1 to 160 m s 1 in time t seconds. Clculte the vlue of t. Clculte the ccelertion of the shuttle t the instnt when its speed is 150 m s A cr of mss 1100 kg hs mximum power of W. The resistive forces hve constnt mgnitude of 1400 N. c Clculte the mximum stedy speed of the cr on the level. The cr is moving on hill of constnt inclintion α to the horizontl, where sin α = Clculte the mximum stedy speed of the cr when scending the hill. Clculte the ccelertion of the cr when it is descending the hill t speed of 10 m s 1 working t hlf the mximum power. [ OCR, A Level Further Mthemtics, 479, June 009] 17 A cr of mss 800 kg experiences resistnce of mgnitude kv N, where k is constnt nd v m s 1 is the cr s speed. The cr s engine is working t constnt rte of P W. At n instnt when the cr is trvelling on horizontl rod with speed 0 m s 1 its ccelertion is 0.75 m s. At n instnt when the cr is scending hill of constnt slope 1 to the horizontl with speed 10 m s 1 its ccelertion is 0.5 m s. Show tht k = 0.900, correct to 3 deciml plces, nd find P. The power is incresed to 1.5P W. Clculte the mximum stedy speed of the cr on horizontl rod. [ OCR, A Level Further Mthemtics, 479, Jnury 009] 18 A cyclist nd her icycle hve comined mss of 70 kg. The cyclist scends stright hill AB of constnt slope, strting from rest t A nd reching speed of 4 m s 1 t B. The level of B is 6 m ove the level of A. For the cyclist s motion from A to B, find: the increse in kinetic energy 4

27 1 Work, energy nd power 1 the increse in grvittionl potentil energy. c During the scent the resistnce to motion is constnt nd hs mgnitude 60 N. The work done y the cyclist in moving from A to B is 8000 J. Clculte the distnce AB. [ OCR, A Level Further Mthemtics, 479/01, June 007] 19 A cr of mss 800 kg is moving t constnt speed of 0 m s 1 on stright rod down hill inclined t n ngle α to the horizontl. The engine of the cr works t constnt rte of 10 kw nd there is resistnce to motion of 1300 N. Show tht sin α = The cr now trvels up the sme hill nd its engine now works t constnt rte of 0 kw. The resistnce to motion remins 1300 N. The cr strts from rest nd its speed is 8 m s 1 fter it hs trvelled distnce of.1 m. Clculte the time tken y the cr to trvel this distnce. [ OCR, A Level Further Mthemtics, 479/01, June 014] 0 A cr of mss 1500 kg trvels up line of gretest slope of stright rod inclined t 5 to the horizontl. The power of the cr s engine is constnt nd equl to 5 kw nd the resistnce to the motion of the cr is constnt nd equl to 750 N. The cr psses through point A with speed 10 m s 1. Find the ccelertion of the cr t A. The cr lter psses through point B with speed 0 m s 1. The cr tkes 8 s to trvel from A to B. Find the distnce AB. [ OCR, A Level Further Mthemtics, 479, Jnury 01] 1 A cr of mss 700 kg is moving long horizontl rod ginst constnt resistnce to motion of 400 N. At n instnt when the cr is trvelling t 1 m s 1 its ccelertion is 0.5 m s. c Find the driving force of the cr t this instnt. Find the power t this instnt. The mximum stedy speed of the cr on horizontl rod is 35 m s 1. Find the mximum power of the cr. The cr now moves t mximum power ginst the sme resistnce up slope of constnt ngle θ to the horizontl. The mximum stedy speed up the slope is 1 m s 1. d Find θ. [ OCR, A Level Further Mthemtics, 479, Jnury 010] A prticle of mss 500 grms moves long the x-xis under the ction of propulsive force F. The prticle s displcement depends on time, t seconds, s follows: x = 3t +t metres. Find the power of force F when t = 5 seconds. speed (m s 1 ) 3 A vn of mss 1500 kg trvels long horizontl rod ginst constnt resistive force of 5 N. The vn trvels with constnt ccelertion from rest, t time t = 0 seconds, to 15 ms 1 t time t = 30 seconds. It then trvels t constnt speed for 10 seconds efore decelerting to rest over 5 15 seconds. The speed-time grph illustrtes the motion. O time (s) 5

28 Dimensionl nlysis In this chpter you will lern how to: understnd the concept of dimensions use the lnguge nd symols of dimensionl nlysis understnd the connections etween units nd dimensions check the vlidity of formul y using dimensionl considertions predict formule y using dimensionl nlysis. Before you strt A Level Mthemtics Student Book 1 GCSE GCSE GCSE GCSE A Level Mthemtics Student Book 1 A Level Mthemtics Student Book You should e le to work with indices nd surds. You should e le to rerrnge formule. You should e le to solve simultneous equtions. You should e le to express direct nd indirect proportion in mthemticl terms. You should know common re nd volume formule. You should e fmilir with the SI units of mss (kg), length (m) nd time (s). You should know the definition of rdin. 1 Simplify: 5 8 rr 9 r (r 4 ) c d 7 Mke x the suject of the formul: 3 xzt y = t Solve the equtions: x+ 3y= 3x y= 16 4 P is inversely proportionl to r. If P = when r =, wht is the vlue of P when r = 6? 5 Wht, in terms of π, is: the volume the surfce re of sphere of rdius 3 cm? 6 Wht re the SI units of velocity? 7 Wht is the ngle, in rdins, of sector of circle of rdius 4 cm nd rc length 5 cm? Continues on next pge... 6

29 Dimensionl nlysis A Level Mthemtics Student Book 1 A Level Mthemtics Student Book 1 Chpter 1 You should know the definitions nd units of velocity nd ccelertion. You should know the definition nd units of force. You should know the definitions of kinetic energy 1 mv nd potentil energy (mgh). Wht is dimensionl nlysis? In dimensionl nlysis you look t the type of unit you re deling with rther thn the specific units. You use it s mthemticl wy of checking tht equtions nd formule re correct, nd re comining like quntities, nd lso to predict nd estlish formule. Section 1: Defining nd clculting dimensions The dimension of given quntity descries wht sort of quntity you re mesuring, so ny distnce or length, whtever its units, hs the dimension of length nd hs the symol L. You cn use squre rckets to men the dimension of so [metres] = L The dimension of distnce is L. The dimeter of pin, the rdius of circle, the length of running trck, the distnce from London to Hong Kong re ll distnces which would e mesured in different units ut which re ll mesurements of length or distnce nd hve the dimension L. The other common dimensions tht you use in Mechnics re M for mss nd T for time. The mss of spider, the mss of n elephnt, the mss of sphere might ll e mesured in different units there is even unit of mss in Americ clled slug ut re ll mesurements of mss with the dimension M. Similrly, time whether mesured in seconds, dys or centuries hs the dimension T. In some rnches of science, other dimensions re used, for exmple, the dimensions of temperture, electric current, mount of light nd mount of mtter. 8 A prticle moving in stright line with constnt velocity trvels 10 m in seconds, wht is its velocity? Stte the units. 9 A prticle moving in stright line with constnt ccelertion increses its velocity from 8 m s 1 to 15 m s 1 in 10 seconds. Wht is the ccelertion? Stte the units. 10 A mss of 3 kg is cted on y constnt force of 1 N. Wht is its ccelertion? Stte the units. 11 A mss of 3 kg is held t height of metres verticlly ove the ground. The prticle is relesed from rest. By equting its loss in potentil energy to its gin in kinetic energy, find its speed t the instnt when it hits the ground. Did you know? There is connection etween dimensionl nlysis nd the greenhouse effect. The concept of dimensionl nlysis is often ttriuted to Joseph Fourier, fmous French Mthemticin nd Physicist. He is est known for the Fourier series, which is widely used in Mthemtics nd Physics, nd for his work on het flow. Fourier is widely recognised s eing the first scientist to suggest tht the Erth s tmosphere would ct s n insultion lyer the ide now known s the greenhouse effect. 7

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