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1 . A prticle of mss 0.5 kg is ttched to one end of light elstic spring of nturl length 0.9 m nd modulus of elsticity λ newtons. The other end of the spring is ttched to fixed point O on rough plne which is inclined t n ngle θ to the horizontl, where sin θ =. The 5 coefficient of friction between the prticle nd the plne is 0.5. The prticle is held on the plne t point which is.5 m down the line of gretest slope from O, s shown in the digrm bove. The prticle is relesed from rest nd first comes to rest gin fter moving 0.7 m up the plne. Find the vlue of λ. (Totl 9 mrks). A light elstic string, of nturl length nd modulus of elsticity 6mg, hs one end ttched to fixed point A. A prticle P of mss m is ttched to the other end of the string nd hngs in equilibrium t the point O, verticlly below A. () Find the distnce AO. () The prticle is now rised to point C verticlly below A, where AC >, nd is relesed from rest. Show tht P moves with simple hrmonic motion of period π. g (5) It is given tht OC =. 4 (c) Find the gretest speed of P during the motion. () Edexcel Internl Review

2 The point D is verticlly bove O nd OD =. The string is cut s P psses through D, 8 moving upwrds. (d) Find the gretest height of P bove O in the subsequent motion. (4) (Totl 5 mrks). A prticle P of weight 40 N is ttched to one end of light elstic string of nturl length 0.5 m. The other end of the string is ttched to fixed point O. A horizontl force of mgnitude 0 N is pplied to P, s shown in the digrm bove. The prticle P is in equilibrium nd the elstic energy stored in the string is 0 J. Clculte the length OP. (Totl 0 mrks) 4. A light elstic string hs nturl length nd modulus of elsticity mg. A prticle P of mss m is ttched to one end of the string. The other end of the string is ttched to fixed point A. The prticle is relesed from rest t A nd flls verticlly. When P hs fllen distnce + x, where x > 0, the speed of P is v. gx () Show tht v = g( + x). (4) Find the gretest speed ttined by P s it flls. (4) Edexcel Internl Review

3 After relese, P next comes to instntneous rest t point D. (c) Find the mgnitude of the ccelertion of P t D. (6) (Totl 4 mrks) 5. A light elstic string hs nturl length 8 m nd modulus of elsticity 80 N. The ends of the string re ttched to fixed points P nd Q which re on the sme horizontl level nd m prt. A prticle is ttched to the mid-point of the string nd hngs in equilibrium t point 4.5 m below PQ. () Clculte the weight of the prticle. (6) Clculte the elstic energy in the string when the prticle is in this position. () (Totl 9 mrks) 6. A nd B re two points on smooth horizontl floor, where AB = 5 m. A prticle P hs mss 0.5 kg. One end of light elstic spring, of nturl length m nd modulus of elsticity 6 N, is ttched to P nd the other end is ttched to A. The ends of nother light elstic spring, of nturl length m nd modulus of elsticity N, re ttched to P nd B, s shown in the digrm bove. () Find the extensions in the two springs when the prticle is t rest in equilibrium. (5) Edexcel Internl Review

4 Initilly P is t rest in equilibrium. It is then set in motion nd strts to move towrds B. In the subsequent motion P does not rech A or B. Show tht P oscilltes with simple hrmonic motion bout the equilibrium position. (4) (c) Given tht the initil speed of P is 0 m s, find the proportion of time in ech complete oscilltion for which P stys within 0.5 m of the equilibrium position. (7) (Totl 6 mrks) 7. A prticle P of mss m is ttched to one end of light elstic string, of nturl length nd modulus of elsticity mg. The other end of the string is ttched to fixed point O. The prticle P is held in equilibrium by horizontl force of mgnitude 4 mg pplied to P. This force cts in the verticl plne contining the string, s shown in the digrm bove. Find () the tension in the string, the elstic energy stored in the string. (5) (4) (Totl 9 mrks) Edexcel Internl Review 4

5 8. One end A of light elstic string, of nturl length nd modulus of elsticity 6mg, is fixed t point on smooth plne inclined t 0 to the horizontl. A smll bll B of mss m is ttched to the other end of the string. Initilly B is held t rest with the string lying long line of gretest slope of the plne, with B below A nd AB =. The bll is relesed nd comes to instntneous rest t point C on the plne, s shown in the digrm bove. Find () the length AC, (5) the gretest speed ttined by B s it moves from its initil position to C. (7) (Totl mrks) 9. A light elstic string of nturl length 0.4 m hs one end A ttched to fixed point. The other end of the string is ttched to prticle P of mss kg. When P hngs in equilibrium verticlly below A, the length of the string is 0.56 m. () Find the modulus of elsticity of the string. () A horizontl force is pplied to P so tht it is held in equilibrium with the string mking n ngle θ with the downwrd verticl. The length of the string is now 0.7 m. Find the ngle θ. () (Totl 6 mrks) Edexcel Internl Review 5

6 0. A prticle P of mss m lies on smooth plne inclined t n ngle 0º to the horizontl. The prticle is ttched to one end of light elstic string, of nturl length nd modulus of elsticity mg. The other end of the string is ttched to fixed point O on the plne. The prticle P is in equilibrium t the point A on the plne nd the extension of the string is. 4 The prticle P is now projected from A down line of gretest slope of the plne with speed V. It comes to instntneous rest fter moving distnce. By using the principle of conservtion of energy, () find V in terms of nd g, (6) find, in terms of nd g, the speed of P when the string first becomes slck. (4) (Totl 0 mrks). A l B l P A light elstic string, of nturl length l nd modulus of elsticity λ, hs its ends ttched to two points A nd B, where AB = l nd AB is horizontl. A prticle P of mss m is ttched to the mid-point of the string. Given tht P rests in equilibrium t distnce l below AB, s shown in the digrm bove, 5mg () show tht λ =. 6 (9) Edexcel Internl Review 6

7 The prticle is pulled verticlly downwrds from its equilibrium position until the totl length of the elstic string is 7.8l. The prticle is relesed from rest. Show tht P comes to instntneous rest on the line AB. (6) (Totl 5 mrks). A prticle P of mss m is ttched to one end of light elstic string, of nturl length nd modulus of elsticity.6mg. The other end of the string is fixed t point O on rough horizontl tble. The prticle is projected long the surfce of the tble from O with speed (g). At its furthest point from O, the prticle is t the point A, where OA = 4. () Find, in terms of m, g nd, the elstic energy stored in the string when P is t A. () Using the work-energy principle, or otherwise, find the coefficient of friction between P nd the tble. (6) (Totl 9 mrks). A prticle P of mss 0.5 kg is ttched to one end of light elstic string. The string hs nturl length 0.8 m nd modulus of elsticity λ N. The other end of the string is ttched to fixed point A. In its equilibrium position, P is 0.85 m verticlly below A. () Show tht λ = 9.. () The prticle is now displced to point B, 0.95 m verticlly below A, nd relesed from rest. Prove tht, while the string remins stretched, P moves with simple hrmonic motion of period 7 π s. (6) (c) Clculte the speed of P t the instnt when the string first becomes slck. () Edexcel Internl Review 7

8 The prticle first comes to instntneous rest t the point C. (d) Find, to significnt figures, the time tken for P to move from B to C. (5) (Totl 6 mrks) 4. Two light elstic strings ech hve nturl length 0.75 m nd modulus of elsticity 49 N. A prticle P of mss kg is ttched to one end of ech string. The other ends of the strings re ttched to fixed points A nd B, where AB is horizontl nd AB =.5 m..5 m A P B The prticle is held t the mid-point of AB. The prticle is relesed from rest, s shown in the figure bove. () Find the speed of P when it hs fllen distnce of m. (6) Given insted tht P hngs in equilibrium verticlly below the mid-point of AB, with APB = α, show tht tnα + 5 sinα = 5. (6) (Totl mrks) Edexcel Internl Review 8

9 5. A 60 P F N A prticle P of mss 0.8 kg is ttched to one end of light elstic string, of nturl length. m nd modulus of elsticity 4 N. The other end of the string is ttched to fixed point A. A horizontl force of mgnitude F newtons is pplied to P. The prticle P is in equilibrium with the string mking n ngle 60 with the downwrd verticl, s shown in the figure bove. Clculte () the vlue of F, () (c) the extension of the string, the elstic energy stored in the string. () () (Totl 8 mrks) 6. A light elstic string of nturl length l hs one end ttched to fixed point A. A prticle P of mss m is ttched to the other end of the string nd hngs in equilibrium t the point O, where AO = 5 l. 4 () Find the modulus of elsticity of the string. () Edexcel Internl Review 9

10 The prticle P is then pulled down nd relesed from rest. At time t the length of the string is 5l + x. 4 Prove tht, while the string is tut, d x 4gx = dt l (5) When P is relesed, AP 4 7 = l. The point B is distnce l verticlly below A. (c) Find the speed of P t B. (4) (d) Describe briefly the motion of P fter it hs pssed through B for the first time until it next psses through O. () (Totl mrks) 7. A light elstic string hs nturl length l nd modulus of elsticity 4mg. One end of the string is ttched to fixed point A nd the other end to fixed point B, where A nd B lie on smooth horizontl tble, with AB = 4l. A prticle P of mss m is ttched to the mid-point of the string. 5l The prticle is relesed from rest t the point of the line AB which is from B. The speed of P t the mid-point of AB is V. () Find V in terms of g nd l. (7) Explin why V is the mximum speed of P. () (Totl 9 mrks) Edexcel Internl Review 0

11 8. A light spring of nturl length L hs one end ttched to fixed point A. A prticle P of mss m is ttched to the other end of the spring. The prticle is moving verticlly. As it psses through the point B below A, where AB = L, its speed is (gl). The prticle comes to instntneous rest t point C, 4L below A. () Show tht the modulus of elsticity of the spring is 8mg. 9 (4) At the point D the tension in the spring is mg. Show tht P performs simple hrmonic motion with centre D. (5) (c) Find, in terms of L nd g, (i) the period of the simple hrmonic motion, (ii) the mximum speed of P. (5) (Totl 4 mrks) 9. A B P Edexcel Internl Review

12 Two light elstic strings ech hve nturl length nd modulus of elsticity λ. A prticle P of mss m is ttched to one end of ech string. The other ends of the strings re ttched to points A nd B, where AB is horizontl nd AB =. The prticle is held t the mid-point of AB nd relesed from rest. It comes to rest for the first time in the subsequent motion when PA nd PB mke ngles α with AB, where tn α = 4, s shown in the digrm bove. Find λ in terms of m nd g. (Totl 7 mrks) 0. A 60 L P A prticle P of mss m is ttched to one end of light string. The other end of the string is ttched to fixed point A. The prticle moves in horizontl circle with constnt ngulr speed ω nd with the string inclined t n ngle of 60 to the verticl, s shown in the digrm bove. The length of the string is L. () Show tht the tension in the string is mg. () Find ω in terms of g nd L. (4) Edexcel Internl Review

13 The string is elstic nd hs nturl length 5 L. (c) Find the modulus of elsticity of the string. () (Totl 7 mrks). A prticle P of mss m is ttched to one end of light elstic string of length nd modulus of elsticity mg. The other end of the string is fixed t the point A which is t height bove smooth horizontl tble. The prticle is held on the tble with the string mking n ngle β with the horizontl, where tn β = 4. () Find the elstic energy stored in the string in this position. () The prticle is now relesed. Assuming tht P remins on the tble, find the speed of P when the string is verticl. () By finding the verticl component of the tension in the string when P is on the tble nd AP mkes n ngle θ with the horizontl, (c) show tht the ssumption tht P remins in contct with the tble is justified. (5) (Totl mrks). A prticle P of mss m is held t point A on rough horizontl plne. The coefficient of friction between P nd the plne is. The prticle is ttched to one end of light elstic string, of nturl length nd modulus of elsticity 4mg. The other end of the string is ttched to fixed point O on the plne, where OA =. The prticle P is relesed from rest nd comes to rest t point B, where OB <. Using the work-energy principle, or otherwise, clculte the distnce AB. (Totl 6 mrks) Edexcel Internl Review

14 . 0 cm A prticle of mss 5 kg is ttched to one end of two light elstic strings. The other ends of the strings re ttched to hook on bem. The prticle hngs in equilibrium t distnce 0 cm below the hook with both strings verticl, s shown in the digrm bove. One string hs nturl length 00 cm nd modulus of elsticity 75 N. The other string hs nturl length 90 cm nd modulus of elsticity λ newtons. Find the vlue of λ. (Totl 5 mrks) 4. A light elstic string hs nturl length 4 m nd modulus of elsticity 58.8 N. A prticle P of mss 0.5 kg is ttched to one end of the string. The other end of the string is ttched to fixed point A. The prticle is relesed from rest t A nd flls verticlly. () Find the distnce trvelled by P before it comes to instntneous rest for the first time. (7) The prticle is now held t point 7 m verticlly below A nd relesed from rest. Find the speed of the prticle when the string first becomes slck. (5) (Totl mrks) Edexcel Internl Review 4

15 5. A light elstic string AB hs one end A ttched to fixed point on ceiling. A prticle P of mss 0. kg is ttched to B. When P hngs in equilibrium with AB verticl, AB = 00 cm. The prticle P is replced by nother prticle Q of mss 0.5 kg. When Q hngs in equilibrium with AB verticl, AB = 0 cm. Find () the nturl length of the string, the modulus of elsticity of the string. (5) () (Totl 7 mrks) 6. In test your strength gme t n musement prk, competitors hit one end of smll lever with hmmer, cusing the other end of the lever to strike bll which then moves in verticl tube whose totl height is djustble. The bll is ttched to one end of n elstic spring of nturl length m nd modulus of elsticity 0 N. The mss of the bll is kg. The other end of the spring is ttched to the top of the tube. The bll is modelled s prticle, the spring s light nd the tube is ssumed to be smooth. The height of the tube is first set t m. A competitor gives the bll n initil speed of 0 m s. () Find the height to which the bll rises before coming to rest. (6) The tube is now djusted by reducing its height to.5 m. The spring nd the bll remin unchnged. Find the initil speed which the bll must now hve if it is to rise by the sme distnce s in prt (). (5) (Totl mrks) Edexcel Internl Review 5

16 . λ 0.6 λ 0. 7 EPE lost = = λ A ( ) R R = mg cosθ 4 = 0.5g = 0.4g 5 F = µ R= g A P.E. gined = E.P.E. lost work done ginst friction λ 0.6 λ g 0.7 sinθ = g 0.7 A A λ = λ = λ = N or.7 A [9] Edexcel Internl Review 6

17 . () ( ) R T = mg B 6mge Hooke's lw: T = 6mge mg = e= AO = 4 A ( ) ( ) H.L. 6mg x mg x T = = Bft Eqn. of motion -mg + T = mx mg ( x) -mg + = mx mgx = mx g x= x A period π g * A 5 Edexcel Internl Review 7

18 (c) v = ω ( x ) v 4 g mx = 0 A vmx = 4 (g) A (d) g x = v = g = 64 v = u + s g 0= gh 64 A h = 8 9 Totl height bove O = + = A 4 [5] Edexcel Internl Review 8

19 . T cos θ = 40 ttempt t both equtions A T sin θ = 0 A leding to T = 50 A x E = λ = 0 B λx HL T = = 50 leding to x = 0.4 A OP = =0.9 (m) Aft [0] mgx 4 4. () mv + = mg( + x) leding to v g( + x) A (, 0) gx = * cso A 4 Gretest speed is when the ccelertion is zero x mgx T = λ = = mg x = A g 8g v = g + = v = ( 6g) ccept exct equivlents A 4 Edexcel Internl Review 9

20 Alterntive v = g ( + x) gx Differentiting with respect to x dv gx v = g dx dv = 0 x = A dx g 8g v = g + = v = ( 6g) ccept exct equivlents A gx v = 0 g + x = (c) ( ) 0 x 4x 4 = (x )(x + ) = 0 x = A At D, λ mx = mg ft their Aft x = g A 6 Alterntive pproch using SHM for nd (c) If SHM is used mrk nd (c) together plcing the mrks in the gird s shown. Estblishment of equilibrium position x mge T = λ = = mg e = b ba NL, using y for displcement from equilibrium position ( y + e) mg g my = mg = y b ba ω = g Edexcel Internl Review 0

21 Speed t end of free fll u = g c Using A for mplitude nd v = ω ( x ) u = g when g 4 y = g = A c 9 4 A = ca 4 g ω c ca Mximum speed A = = ( 6g) 4 g Mximum ccelertion Aω = = g ca [4] 5. () Resolving verticlly: Tcosθ =W A,,0 Hooke s Lw: 80.5 T = 4 A W = 84N A EPE = 80.5, =45 (or wrt 45) Aft,A 4 (lterntive = 45) [9] Edexcel Internl Review

22 6. () Hooke s lw: Equilibrium ( d ) ( 4 ) 6 d = AA d =. A so extensions re.m nd 0.8m. A If the prticle is displced distnce x towrds B then Aft ( + x) ( 0.8 x) ( x) 6. m x = = 0 Aft 0 x = 40x or x = ( SHM) A m (c) π T = Bft 40 0 = theirω Bft x = sin ωt their, their ω = sin 40t A 4 π π 40t = ( t = ) Proportion 4t T 4π = = π A [6] Edexcel Internl Review

23 7. () ( ) 4 T sin θ = A mg ( ) T cosθ = mg A Leding to 4 = T mg + ( mg) 5 T = mg A 5 HL x 5 mge T = λ mg = ft their T Aft 5 e = 9 x mg 5 5 E = λ = = mg A [9] 8. () Let x be the distnce from the initil position of B to C GPE lost = EPE gined 6mgx mgx sin 0º = A = A Leding to x = 6 7 AC = 6 A 5 Edexcel Internl Review

24 The gretest speed is ttined when the ccelertion of B is zero, tht is where the forces on B re equl. 6mge T = mg sin 0º = e = A 6mg CE mv + = mg sin 0º A = A Leding to g 6g v = = A 7 4 Alterntive pproch to using clculus with energy. Let distnce moved by B be x CE 6mg mv + x = mgx sin 0º A = A 6g v = gx x For mximum v d d ( v g v ) = v = g x = 0 dx dx x = A 6g g v = g = 4 g v = A 7 4 Edexcel Internl Review 4

25 Alterntive pproch to using clculus with Newton s second lw. As before, the centre of the oscilltion is when extension is A NL mg sin 0 T = m.ẋ 6mg + x.. mg = m x A 6 g g x ω = 6 A.. x = 6g g vmx = ω = = A 7 4 [] λ e 9. () T or = mg (even T = m is, A0, A0 sp cse) l λ 0.6 = g A 0.4 λ = 49 N or 5g A mg R( ) T cos θ = mg or cos θ = T cosθ = 9.6 or 4g.cosθ = g or mg. cosθ = mg (ft on their λ)aft 0.4 π cosθ = θ = 60 (or rdins) A Specil cse T sin θ = mg giving θ = 0 is A0 A0 unless there is evidence tht they think θ is with horizontl then A A0 [6] 0. () Energy eqution with t lest three terms, including K.E term mv +... mg mg , + mg..sin 0, = A, A, A V = g da 6 Edexcel Internl Review 5

26 Using point where velocity is zero nd point where string becomes slck: mw = mg 9.., mg..sin 0 A, A 6 4 g w = A 4 8 Alterntive (using point of projection nd point where string, AA becomes slck): mg mw mv, = mg A 6 8 g So w = 8 In prt () D requires EE, PE nd KE to hve been included in the energy eqution. If sign errors led to V = g, the lst two mrks re M0 A0 In prts () nd A mrks need to hve the correct signs In prt for need one KE term in energy eqution of t lest terms with distnce to indicte first method, nd two KE terms in energy 4 eqution of t lest 4 terms with distnce 4 to indicte second method. SHM pproch in prt. (Condone this method only if SHM is proved) Using v = ω ( x ) with ω g = nd x = ±. 4 g Using = to give w =. 8 [0] Edexcel Internl Review 6

27 . () A.5l B T l P mg T AP = ((.5l) + (l) ) =.5l 4 cos α = 5 A B λ(.5l.5l) λ Hooke s Lw T = =.5l A 5mg T cos α = mg T = 8 A λ 4 λ 5mg = mg = 5 8 λ = 5mg * cso A 9 6 A.5l B.9l h P h = ((.9l) (.5l) ) =. 6l A 5mg (.4l) Energy mv + mg h = 6.5l ft their h Aft=A Leding to v = 0 * cso A 6 [5].6mg.6mg. () E.P.E. = x = A = 0. mg A Edexcel Internl Review 7

28 4 Friction = µmg work done by friction = µ mg A Work-energy: m.g = µmgd + 0. mg ( relevnt terms) Aft Solving to find µ : µ = 0.6 A 6 st : llow for ttempt to find work done by frictionl force (i.e. not just finding friction). nd : relevnt terms, i.e. energy or work terms! A f.t. on their work done by friction [9]. () A x T = λ = λ (0.05) = 0.5g 0.8 (0.8)(0.5g) = 9. (*) A T = (x ) 0.8 mg T = m ( term equn) g 0.8 x = 96 x SHM with period (x ) = 0.5 x (or equivlent) A π π π = = s (*) ω 4 7 A cso 6 st must hve extn s x + k with k 0 (but llow if e.g. x + 0.5), or must justify lter For lst four mrks, must be using x (not ) A (c) v = 4 {(0.) (0.05) } Aft =.(4 ). m s ( s.f.) Accept 7 /0 A Using x = 0 is M0 Edexcel Internl Review 8

29 ... (d) Time T under grvity = (= 0.7 s) Bft g Complete method for time T from B to slck. π + t, where 0.05 = 0.sin 4t A [ e.g. 8 OR T, where 0.05 = 0. cos 4 T ] T = 0.496s A Totl time = T + T = 0.7 s A 5 must be using distnce for when string goes slck. Using x = 0. (i.e. ssumed end of the oscilltion) is M0 [6] 4. () A 0.75 m B m P AP = ( ) =.5 A Conservtion of energy v + = g for ech incorrect term A (, 0) 0.75 Leding to v.8 (ms ) ccept.8 A 6 Edexcel Internl Review 9

30 A 0.75 m B y T T P g R ( ) Tcosα = g A Hooke s Lw 0.75 y = sinα T = A 0.75 sinα = 49 sinα 9.8 = 49 Eliminting T cosα sinα tn α = 5 ( sin α) 5 = tn α + 5sin α cso A 6 [] 5. () F = T sin 60 i T cos 60 = 0.8g both [or Z F cos60 = 0.8g cos 0 ] (M) F = 0.8g tn 60 4(N) ccept.6 A T = 0.8g sin 0 HL 5.68 = (= 5.68) llow in () 4 x. x 0.78 (cm) ccept A c) E = 4 x 6. (J) ccept 6.5 Aft. [8] Edexcel Internl Review 0

31 6. A l B 4l O x P () HL T = mg = λ l 4 l λ = 4mg A NL mg T m x 4mg ( l + x) 4 mg = m x l A d x 4g = x cso A 5 dt l (c) v ω ( x ) = 4g l l l 6 4 A Leding to v = (gl) A 4 or energy, gl 4mg. l 6 = mv + mg. l for the first A in (c) 4 (d) P first moves freely under grvity, B then (prt) SHM. B [] Edexcel Internl Review

32 7. () A M X B l l l 4 4mg( l) 4mg( l) 40 Elstic energy when P is t X: E = + mgl = l l 9 A 4 mgl 4mh( l) 4mg( l) 4 mv + = + A=Aft l l l V 8 + 4gl = gl + gl 9 9 V 8gl = 9 solving for V 5 8gl V = 9 or exct equivlents A 7 The mximum speed occurs when = 0 B At M the prticle is in equilibrium (the sum of the forces is zero) = 0 B [9] Alterntive using Newton s second lw. () A T M X T B l x 4 mi ( l + x) 4mg ( l x) HL T =, T = x l l 8mg NL m x = T T = x l This is SHM, centre M l =, ω 8g = l v = ω ( x ) v 8g l = x l 9 Depends on showing SHM At M, x = 0, V 8gl 8gl =, V = 9 9 or exct equivlents A A, Aft, A 7 Edexcel Internl Review

33 The prticle is performing SHM bout the mid-point of AB. B The mximum speed occurs t the centre of the oscilltion (when x = 0) B [9] 8. () KE loss + PE loss = EPE Gin λ(l). mgl + mgl = L A( e.e.) 8mg (*) = λ 9 A 4 mg T = m x A 8mg mg ( x + e) = mx 9L A 8g x = x 9 L Hence SHM bout D A c.s.o. 5 (c) (i) Period = π 9L = π = π Aft w 8g g 8mg 9L (ii) mg = e e = 9L 8 9L 5L 9 = L = 8 8 5L 8g v mx = 9w = 8 9L 5 = gl 4 B A 5 [4] Edexcel Internl Review

34 9. Extn t bottom = = cosα λ ( ) Energy: mg tn α = mg = λ (0.67 or better) Second M0 if treted s equilibrium Third for solving for λ A A A ft A [7] 0. T mg () ( ) T cos60 = mg T = mg * B ( ) T sin60 = mrω r = Lsin60 g ω = L A B A 4 (c) Applying Hooke s Lw: mg = ( L) 5 λ (L 5 L); λ mg ;A [7] Edexcel Internl Review 4

35 . () R P T Length of string (L) = mg EPE = = 49 mg 6 0 ( L ) B A Energy eqution: ½ mv + v = mg 49 = ( mg 6 4 5g or equivlent [llow.46 g, ) C Aft ] A (c) T V = T sin θ [implied by R + T sin θ = mg] B mg T V in terms of θ [ ( ) sinθ] sinθ mg( AP ) mg x or in terms of x or AP [or (. or ] AP ( + x) (i) T = ½ mg( sinθ) or R = ½ mgsinθ A Complete method to show R > 0 A OR (ii) mgx T = ½ mg( sinθ); mg( ); AP + x Complete method to show T V < mg or tht T V mg not poss OR A A (iii) T = ½ mg( sinθ) A s θ increses T V decreses; T V < T Vmx = 0 7 mg < mg A 5 [In ll cses: For A ll working correct nd rg. convincing] [] Edexcel Internl Review 5

36 . 0 B X R T P mg d Attempt to relte Fd to EPE 4 mg( ) mg d = [A ft only on omission of g in F] F A F = R = mg 4 mg x B B A ft Finl nswer: d = 4 A 6 [6]. T T T = 5g λ = 49 A λ = 4 A 5 B [5] 4. () 58. x 48 = (x + 4) A A x x 8 = 0 A (x + 4)(x ) = 0, x = Distnce fllen = 6 m A 7 0.5v = A A 4 v = 4. m s A 5 [] Edexcel Internl Review 6

37 5. λ( 00 l) l λ( 0 l) l = 0.g A = 0.5g A 5(00 l) = (0 l) l = 85 cm A 5 λ = 0.g 85 5 = 6.66 N A [7] 6. () Energy: 0 = 9.8 h + 0h + 9.6h 00 = 0 0 h A A h = 9.6 ± ( = (or.80) m A 6 V = A A V =.7 ( s.f.) or ( s.f.) m s A 5 [] Edexcel Internl Review 7

38 . This ws probbly the lest well done of ll the questions nd correct solutions were reltively rre. The most common mistke ws the ssumption tht there ws no finl EPE but this ws often combined with other errors to give huge vriety of different wrong nswers. A surprisingly lrge proportion of cndidtes treted this s n equilibrium question, either strting with T = μr + mg sin θ or slipping n EPE term in s well for good mesure. Others relised tht it ws n energy question but forgot to include the work done ginst friction; these ttempts either used only the frictionl force in their eqution or ignored it completely, offering s their solution Initil EPE = mgh. Another common error ws to include the GPE term twice, once s energy nd gin s prt of the Work done expression, showing lck of understnding of the origin of the mgh formul. Very mny cndidtes scored only the mrks for finding friction, while those who thought tht this ws simple conversion of EPE into GPE hd no need to find the friction nd so didn t even ern these. Some cndidtes who included ll necessry terms fell t the ccurcy hurdle. Inexplicbly, finl extension/ compression of 0. ws not uncommon nd other errors rose from inpproprite use of the vrious lengths mentioned,.5, 0.9 nd 0.7. There were lso ll the usul sign errors generted by mistkes in identifying gins nd losses. A few cndidtes produced perfect solution but lost the finl mrk by giving their nswer s The mjority gined full mrks for prt () but some used m insted of m for the mss nd others forgot to complete the question by dding to their extension. However, prt ws different story. This ws stndrd question nd should hve been routine for most but ws very poorly done. Some cndidtes mesured the extension from the nturl length, which cn produce correct result provided the pproprite substitution is employed. Probbly they were not fully wre of wht they were doing nd so stopped work when they did not rech the required eqution. Mny hd inconsistent msses, with mg nd m x ppering in the sme line of working, lthough some relised nd bcktrcked successfully. The omission of mg in the eqution of motion led to the correct nswer when the extension ws incorrectly mesured from the nturl length so cndidtes ssumed tht they hd nswered correctly. Poor nottion ws often seen; cc. or ws used for ccelertion (s well s its length ppliction) nd e ws used for the vrible extension long with n ccelertion x. There were mny sign errors seen in the equtions nd lthough some cndidtes relised they hd mde errors nd either corrected their work or fiddled the result, others seemed to think tht the eqution x = ω x proved S.H.M. Good ttempts were presented for prts (c) nd (d) lthough some thought tht the mplitude ws /4; ω = ws nother common error. Some forgot to complete their g work in prt (d) by dding nd to obtin the finl nswer. 8 8 Edexcel Internl Review 8

39 . This ws probbly the best nswered question on the pper. Most cndidtes obtined T (not lwys shown explicitly) by resolving in two directions. There were few who drew tringle of forces. Good cndidtes were ble to see the tringle without working nd wrote down immeditely tht T = 50. A few unfortuntely thought they were deling with 40g nd 0 even though ws it clerly stted in the question tht the weight ws 40 N nd not tht the mss ws λx 40 kg. The most common mistke ws in the rithmetic 50 = so 0.5 forgot to complete their solution by dding 0.5 to their extension to find OP. λ x = 00! Very few 4. Most cndidtes mnged to rrive t the required result in prt (), though some unnecessrily split the motion into two prts, considering freefll initilly to find the kinetic energy when the string becme tut nd then proceeding to consider the tut string nd others would clerly hve filed hd not the nswer been provided. Prts nd (c) were often difficult to disentngle. Some cndidtes took n SHM pproch to the Exminer. The min fult ws not when to strt considering SHM but not estblishing correct eqution to prove tht the motion ws SHM; no credit is given for mking ssumptions of this nture. A fully correct solution using SHM ws rre, the equtions frequently being unstisfctory due to using x for the distnce from the equilibrium point nd confusing it with x s defined in the question to be the extension of the string. For the non-shm solutions, in prt mny cndidtes ssumed tht the mximum speed occurred when x = 0 rther thn when = 0. In prt (c) most substituted v = 0 in the result from prt (). Some did not expect to obtin qudrtic nd so stopped working (or rn out of time?). Of those who obtined solution for their qudrtic eqution, some would then try incorrectly to use their vlue for x s the mplitude in SHM insted of using n eqution of motion nd Hooke s lw. Mny equtions of motion omitted the weight of the prticle. from the strt of 5. This ws strightforwrd opening question but mny cndidtes spoiled their solutions with creless errors. Most mnged to use Pythgors to clculte the length of the extended string. However some ppered to hve misinterpreted the informtion given in the question nd took the nturl length s m insted of 8 m. Hooke s lw ws well known but problems rose s cndidtes were confused between the full string nd hlf strings, with some thinking the tensions in these were different. Significnt numbers gve the mss s their finl nswer insted of the weight s demnded. Prt gve rise to fewer problems. The formul for the elstic potentil energy ws known by ll but smll number of cndidtes nd mny who hd n incorrect extension in prt () either completely recovered to gin full mrks or gined of the mrks by follow through. Edexcel Internl Review 9

40 6. Prt () ws usully hndled confidently with Hooke s Lw in the equilibrium position pplied to ech spring nd the resulting tensions equted. There were some unfortunte creless errors such s totl extension = 5 = which ppered fr too often. Mistkes in the extensions led to difficulties in the lter prts of the question but some ble cndidtes didn t think to check their extensions when their confidently pplied method in filed to work s expected. The work seen in prt suggested tht mny cndidtes re not wre tht to prove SHM they need n eqution of motion tht reduces to the form x = ω x ; some even tried to give written explntion of the motion, with no equtions provided t ll. Even if the question hd not told them tht the equilibrium position ws the centre of the oscilltion, cndidtes should hve known this nd mesured their displcement from this point; mny chose some other point insted. There were lso numerous sign errors in the eqution of motion, often corrected but not lwys in vlid mnner. Lck of success in prt ment mny cndidtes hd no suitble vlue for ω to use in prt (c). Some were content to invent number nd continue, thereby mking the method nd follow through mrks vilble, others gve up, though this my hve been due to lck of time. Most who worked on this prt relised tht the mplitude could be obtined using mx v mx = ω nd hence the time required using x = sinωt. Solutions using x = cosωt were seen occsionlly but the extr work needed to obtin the necessry time ws often omitted. The instruction to find the proportion of the time ws usully ignored completely resulting in the loss of three mrks s the period ws not clculted since the necessity for it ws not pprent. 7. Most cndidtes resolved horizontlly nd verticlly in () nd then found the tngent of their ngle between the string nd the horizontl or verticl before proceeding to obtin the tension. It ws not uncommon to see solutions which strted with the Pythgors eqution in line of the mrk scheme. This is risky pproch s errors in this eqution cuse mny mrks to be lost if the eqution is not derived from the resolving equtions. Hooke s lw nd the formul for elstic potentil energy were well known nd frequently pplied correctly in to rech correct nswer. However, omission of some or ll of the letters m, g nd in the finl nswer ws firly common. 8. Mny cndidtes used unduly complicted methods for both prts of this question. Some tried to use S.H.M. but very few ttempted to estblish the motion s being S.H.M. before quoting nd using the stndrd formule. In prt () there ws confusion between equilibrium nd rest positions. Some cndidtes used Hooke s lw nd resolved prllel to the plne, finding the equilibrium extension nd climed tht the prticle ws t rest t this point. They then proceeded to use the sme extension (now correctly) in to obtin the gretest speed. Energy methods often hd missing term or grvittionl potentil energy term which ws inconsistent with the positions involved. Alterntive methods using Newton s second lw nd clculus were resonbly populr but mny were incomplete. 9. Prt ws generlly well done lthough few left m in their finl nswer. Most nswered prt correctly but severl used T sin θ insted of T cos θ s the resolved prt of T in the verticl direction, nd others used cos θ = 0.56/0.7 which ws not pproprite. Edexcel Internl Review 40

41 0. There were some excellent solutions but lso gret mny inccurte nd muddled ttempts. Although the theory ws well known, it ws very common for one or more energy terms to be left out (most commonly the initil EPE or the GPE). Confusion between energy losses nd gins often led to wrong signs, difficulty which is esily delt with by equting initil energy to finl energy insted. Mny cndidtes seemed uncler bout the reltive positions they were considering nd there ws gret del of inccurcy in identifying extensions, heights nd positions of zero KE. In prt () it ws usully possible to follow the cndidte s line of resoning nd identify the mistkes but mny of the ttempts t used inconsistent initil energy vlues from vriety of loctions nd filed to follow the Advice to cndidtes tht their methods should be mde cler to the exminer. The clerest solutions strted with very well lbelled digrm showing ll the significnt points nd identifying the extension ssocited with ech. Sttements such s Energy t A =..., energy t B = helped to void muddled vlues nd usully led to correct or nerly correct equtions. By contrst, gret mny ttempts t contined no sttement t ll but simply n eqution of unidentifible terms.. Prt () ws well done. A correct use of Hooke s Lw, combined with verticl resolution nd the geometry of the sitution, ws the stndrd pproch. A few mde the error of not relising tht there were two tensions for the resolution but only one for Hooke s Lw. Mny good rguments were seen in prt. There were two min methods; using conservtion of energy with three terms nd showing v = 0, nd finding the elstic potentil energy loss nd showing tht it ws equl to the gin in grvittionl potentil energy t the level of AB. There ws tendency for cndidtes to drop the l in their lengths. As energy is involved, this gve dimensionlly incorrect solutions but cndidtes who did this could usully gin 4 of the 6 mrks if their solution ws otherwise correct.. Some excellent solutions were seen here with mny gining full mrks. Of those who did not, some mde smll slip in the processing (or occsionlly in the signs of the work energy eqution, though such mistkes were rrer thn might hve been expected); weker cndidtes mde more fundmentl errors, e.g. equting energy with force.. Prt () ws lmost universlly correct. Prt cused considerble problems: mny ssumed tht they could find the ccelertion s (unspecified), nd tht if they showed tht this ws equl to 96x they hd succeeded in showing tht the motion ws SHM. Such cndidtes filed to relise tht ny ccelertion given s n unspecified needs its direction clerly specified. Hence, without the use of n expression for x s equl to 96x, they could mke no progress. Weker cndidtes lso filed to see tht the eqution of motion hd to include the weight s well. In prts (c) nd (d) common mistke ws effectively to ssume tht the prticle cme to rest t the end of n oscilltion within the simple hrmonic motion. Nevertheless, more ble cndidtes were ble to complete the question ccurtely nd overll, this proved to be good discriminting question for the finl one on the pper. Edexcel Internl Review 4

42 4. Prt () ws well done. The only common error ws considering the elstic potentil energy in only one prt of the string insted of in both prts. Most cndidtes relised tht energy ws involved nd the few who ttempted using Newton s Second Lw lmost ll filed to consider generl point of the motion nd so gined no credit. Nerly ll cndidtes could strt prt by resolving verticlly nd writing down some form of Hooke s Lw. The mnipultions required to obtin the required trigonometric reltion, however, were demnding nd even strong cndidtes often needed two or three ttempts to complete this nd the time spent on this ws sometimes reflected in n inbility to complete the pper. This ws prticulr the cse if cndidtes ttempted to use or gin informtion by writing down n eqution of energy. This leds to very complicted lgebr nd is not prcticl method of solving questions of this type t this level. (Correctly pplied it leds to qurtic not solvble by elementry methods.) For those who were successful in prt (), writing T in terms of, sy, the ngle mde by ech prt of the string with the verticl proved the criticl step. If they obtined T = 0. 75, or 0.75 sin its equivlent, the mjority of cndidtes hd the necessry trigonometric skills to complete the question. 5. This provided firly strightforwrd strt nd most scored well but there were fewer full mrks thn expected. The nswer to () ws often given s.58, losing mrk, the vlue of F ws frequently used mistkenly for T in nd ccurcy errors were firly common in (c) when sig fig nswer to ws used to clculte sig fig vlue for the energy. 6. Aprt from (), which ws lmost lwys correct, this ws not well done. With the definition ofclerly defined in the question, the proof in should hve been strightforwrd. For mny it ws, but there were gin huge number of fiddled ttempts. Often, cndidtes strted out correctly with mg T = ± m, only to cross out the mg subsequently, when they relised tht T = λx / l = m led to the required expression, nd therefore lose the mrk for this eqution. Prt (c) ws not well done either. Most chose to use the formul v =ω ( x ) but vlues for nd x were not often correct. Use of energy ws populr s well but mny omitted either the EPE or the GPE. Prt (d) ws gret discrimintor nd only the best cndidtes were ble to describe the trnsition from SHM to free motion under grvity to SHM clerly nd ccurtely. Edexcel Internl Review 4

43 7. This proved the hrdest question on the pper nd mny scored no mrks. The use of energy ws seen more frequently thn pproches using Hooke s Lw nd n eqution of motion, but energy ws not well hndled. There re four elstic energies involved in the question; two different energies in the initil position nd two identicl energies t the mid-point. It ws common to see three of these four energies omitted. When it ws relised tht there were two energies t the mid-point, it ws not unusul for cndidtes to think these cncelled out. Those using Hooke s Lw gin often only hd one string in tension nd no credit cn be gined using SHM methods unless vrible extension is used. With the SHM method, cndidtes often hd difficulty finding the extension of the two strings but, on blnce, this method tended to be used more successfully thn energy. It is worth recording tht few cndidtes produced completely correct solutions, using Newton s Second Lw, mesuring the displcements from A, B or the initil position of P. They then used methods of solving second order differentil equtions to complete the solution without reference to SHM or ny further mechnicl principles. This module is probbly now tken by high proportion of cndidtes who hve studied Further Pure Mthemtics nd such solutions my be more commonly seen thn in the pst. Questions requiring explntion remin unpopulr with cndidtes. In this cse, the comments expected were tht the mximum speed occurs when the ccelertion is zero nd the mid-point is the position of equilibrium nd, hence, the ccelertion there is zero. 8. Most relised tht the first prt required the use of energy nd were ble to obtin the required result. There ws disppointing response to the stndrd proof in prt ; mny cndidtes simply ignored the weight nd scored no mrks. Prt (c) (i) ws well done but there were few correct solutions to (ii), where most were unble to find the mplitude of the oscilltion. 9. Even mong ble cndidtes, this ws rrely recognised s question bout energy, with most ttempts ssuming tht the lowest point reched would be the equilibrium position. Consequent use of Tsinα = mg nd Hooke s lw lost 5 of the 7 vilble mrks leving most cndidtes ble to score only the mrks for finding the extension. This ws often correct but, s in question, there were mny numericl nd lgebric slips. Some did not even relise tht the extension could be found, nd persisted with n extension x. Those who did correctly use n energy method to solve the problem often produced concise nd ccurte solutions but significnt number forgot to include the EPE from both strings. 0. Although the nswer ws given in prt () it is still good to report tht ll but hndful of cndidtes gined the mrk. Prt ws generlly well nswered, lthough T = mrω, Tsin60 = mlω, nd errors in eliminting r, were occsionlly seen. Probbly the most common error in this question occurred in prt (c), where L, insted of of Hooke s lw. 5 L, ws often seen in the denomintor Edexcel Internl Review 4

44 . Although in prt () some cndidtes did not find the correct vlue for the extension of the string, nd in prt the elstic energy in the string when it ws verticl ws sometimes omitted, these two prts were generlly nswered well. There were some very impressive solutions, showing good mnipultive nd deductive skills, by the most ble cndidtes in prt (c), but this proved to be the most chllenging prt of the pper nd the modl score ws one mrk. Mny cndidtes did not follow the instruction, nd the hint, to consider the verticl component of the string t generl point. The most common mistke ws to ssume tht the mximum vlue of T being t the point of relese, nd tht it remins on the tble t tht point, ws sufficient justifiction for the prticle to remin on the tble during the motion. The fct tht, during the motion, s T decresed so sinθ incresed nd so the behviour of Tsinθ ws not obvious, ws not pprecited by the mjority of cndidtes.. Good cndidtes found this strightforwrd, finding AB immeditely from single ppliction of the work-energy eqution. However, lrge number of cndidtes did not hve cler strtegy for the problem nd mrks of, or were common. Common errors were: using, not the extension, in the energy term; omitting the frictionl force, or equting it to the tension, or including the tension when considering the work done, 4mg e.g. (T F)x = ; nd not relising tht the string ws slck t B, so introducing n extr elstic energy term. Cndidtes who used two-stge pproch, first finding the velocity t the instnce the string becme slck, were rrely successful.. This proved to be strightforwrd strter for most nd full mrks ws often seen. The most common error ws to ssume tht the tensions in the two strings re equl but there were other odd errors in use of Hooke s Lw nd/or rithmetic. 4. Cndidtes who considered the energy chnges from the strt to the finish were generlly much more successful thn those who considered intermedite points. A significnt number tried to use SHM formule but were given no credit for it unless they hd first proved tht the motion ws simple hrmonic. Some lso tried to use differentil eqution but usully without success. Mny thought, in prt, tht the string went slck t the equilibrium position nd few lost the lst mrk for giving their finl nswer to more thn SF. This ws deemed inpproprite s the nswer involved tking g s 9.8. Edexcel Internl Review 44

45 5. No Report vilble for this question. 6. No Report vilble for this question. Edexcel Internl Review 45