1. A uniform circulr disc hs mss m, centre O nd rdius. The line POQ is dimeter of the disc. A circulr hole of rdius is mde in the disc with the centre of the hole t the point R on PQ where QR = 5, s shown in the digrm bove. The resulting lmin is free to rotte bout fixed smooth horizontl xis L which psses through Q nd is perpendiculr to the plne of the lmin. () Show tht the moment of inerti of the lmin bout L is 69m. (7) The lmin is hnging t rest with P verticlly below Q when it is given n ngulr velocity Ω. π Given tht the lmin turns through n ngle before it first comes to instntneous rest, find Ω in terms of g nd. (6) (Totl 1 mrks). A uniform lmin ABC of mss m is in the shpe of n isosceles tringle with AB = AC = 5 nd BC = 8. () Show, using integrtion, tht the moment of inerti of the lmin bout n xis through A, prllel to BC, is 9 m. (6) Edexcel Internl Review 1
The foot of the perpendiculr from A to BC is D. The lmin is free to rotte in verticl plne bout fixed smooth horizontl xis which psses through D nd is perpendiculr to the plne of the lmin. The lmin is relesed from rest when DA mkes n ngle α with the downwrd verticl. It is given tht the moment of inerti of the lmin bout n xis through A, 8 perpendiculr to BC nd in the plne of the lmin, is m. Find the ngulr ccelertion of the lmin when DA mkes n ngle θ with the downwrd verticl. (8) Given tht α is smll, (c) fin d n pproximte vlue for the period of oscilltion of the lmin bout the verticl. () (Totl 16 mrks). A uniform circulr disc hs mss m, centre O nd rdius. It is free to rotte bout fixed smooth horizontl xis L which lies in the sme plne s the disc nd which is tngentil to the disc t the point A. The disc is hnging t rest in equilibrium with O verticlly below A when it is struck t O by prticle of mss m. Immeditely before the impct the prticle is moving perpendiculr to the plne of the disc with speed (g). The prticle dheres to the disc t O. () Find the ngulr speed of the disc immeditely fter the impct. (5) Find the mgnitude of the force exerted on the disc by the xis immeditely fter the impct. (6) (Totl 11 mrks) Edexcel Internl Review
. A uniform lmin of mss M is in the shpe of right-ngled tringle OAB. The ngle OAB is 90, OA = nd AB =, s shown in the digrm bove. () Prove, using integrtion, tht the moment of inerti of the lmin OAB bout the edge OA is M. (You my ssume without proof tht the moment of inerti of uniform rod of mss m nd length l bout n xis through one end nd perpendiculr to the rod is ml.) (6) The lmin OAB is free to rotte bout fixed smooth horizontl xis long the edge OA nd hngs t rest with B verticlly below A. The lmin is then given horizontl impulse of mgnitude J. The impulse is pplied to the lmin t the point B, in direction which is perpendiculr to the plne of the lmin. Given tht the lmin first comes to instntneous rest fter rotting through n ngle of 10, find n expression for J, in terms of M, nd g. (7) (Totl 1 mrks) 5. A pendulum consists of uniform rod AB, of length nd mss m, whose end A is rigidly ttched to the centre O of uniform squre lmin PQRS, of mss m nd side. The rod AB is perpendiculr to the plne of the lmin. The pendulum is free to rotte bout fixed smooth horizontl xis L which psses through B. The xis L is perpendiculr to AB nd prllel to the edge PQ of the squre. () Show tht the moment of inerti of the pendulum bout L is 75m. () Edexcel Internl Review
The pendulum is relesed from rest when BA mkes n ngle α with the downwrd verticl 7 through B, where tn α =. When BA mkes n ngle θ with the downwrd verticl through B, the mgnitude of the component, in the direction AB, of the force exerted by the xis L on the pendulum is X. Find n expression for X in terms of m, g nd θ. (9) Using the pproximtion θ sin θ, (c) find n estimte of the time for the pendulum to rotte through n ngle α from its initil rest position. (6) (Totl 19 mrks) 6. A uniform solid right circulr cylinder hs mss M, height h nd rdius. Find, using integrtion, its moment of inerti bout dimeter of one of its circulr ends. [You my ssume without proof tht the moment of inerti of uniform circulr disc, of mss m nd rdius, bout dimeter is 1 m.] (Totl 10 mrks) Edexcel Internl Review
7. y R O x A region R is bounded by the curve y = x (y > 0), the x-xis nd the line x = ( > 0), s shown in the digrm bove. A uniform solid S of mss M is formed by rotting R bout the x-xis through 60. Using integrtion, prove tht the moment of inerti of S bout the x-xis is M. (You my ssume without proof tht the moment of inerti of uniform disc, of mss m nd rdius r, bout n xis through its centre perpendiculr to its plne is 1 mr.) (Totl 7 mrks) Edexcel Internl Review 5
8. S L O A lmin S is formed from uniform disc, centre O nd rdius, by removing the disc of centre O nd rdius, s shown in the digrm bove. The mss of S is M. () Show tht the moment of inerti of S bout n xis through O nd perpendiculr to its plne is 5 M. () The lmin is free to rotte bout fixed smooth horizontl xis L. The xis L lies in the plne of S nd is tngent to its outer circumference, s shown in the digrm bove. Show tht the moment of inerti of S bout L is 1 M. () S is displced through smll ngle from its position of stble equilibrium nd, t time t = 0, it is relesed from rest. Using the eqution of motion of S, with suitble pproximtion, (c) find the time when S first psses through its position of stble equilibrium. (6) (Totl 1 mrks) Edexcel Internl Review 6
9. () Prove, using integrtion, tht the moment of inerti of uniform rod, of mss m nd length, bout n xis perpendiculr to the rod through one end is m. () Hence, or otherwise, find the moment of inerti of uniform squre lmin, of mss M nd side, bout n xis through one corner nd perpendiculr to the plne of the lmin. () (Totl 6 mrks) 10. A uniform circulr disc hs rdius nd mss m. Prove, using integrtion, tht the moment of inerti of the disc bout n xis through its centre nd perpendiculr to the plne of the disc is 1 m. (Totl 5 mrks) Edexcel Internl Review 7
11. P L Q R A thin uniform rod PQ hs mss m nd length. A thin uniform circulr disc, of mss m nd rdius, is ttched to the rod t Q in such wy tht the rod nd the dimeter QR of the disc re in stright line with PR = 5. The rod together with the disc form composite body, s shown in the digrm. The body is free to rotte bout fixed smooth horizontl xis L through P, perpendiculr to PQ nd in the plne of the disc. () Show tht the moment of inerti of the body bout L is 77m. (7) When PR is verticl, the body hs ngulr speed ω nd the centre of the disc strikes sttionry prticle of mss 1 m. Given tht the prticle dheres to the centre of the disc, find, in terms of ω, the ngulr speed of the body immeditely fter the impct. () (Totl 11 mrks) 1. A uniform lmin of mss m is in the shpe of n equilterl tringle ABC of perpendiculr height h. The lmin is free to rotte in verticl plne bout fixed smooth horizontl xis L through A nd perpendiculr to the lmin. () Show, by integrtion, tht the moment of inerti of the lmin bout L is 5mh. 9 (9) Edexcel Internl Review 8
The centre of mss of the lmin is G. The lmin is in equilibrium, with G below A, when it is 6g given n ngulr speed. 5h Find the ngle between AG nd the downwrd verticl, when the lmin first comes to rest. (5) (c) Find the gretest mgnitude of the ngulr ccelertion during the motion. () (Totl 17 mrks) 1. A uniform lmin of mss m is in the shpe of rectngle PQRS, where PQ = 8 nd QR = 6. () Find the moment of inerti of the lmin bout the edge PQ. () P Q S R The flp on letterbox is modelled s such lmin. The flp is free to rotte bout n xis long its horizontl edge PQ, s shown in the digrm bove. The flp is relesed from rest in horizontl position. It then swings down into verticl position. g Show tht the ngulr speed of the flp s it reches the verticl position is. () Edexcel Internl Review 9
(c) Find the mgnitude of the verticl component of the resultnt force of the xis PQ on the flp, s it reches the verticl position. () (Totl 7 mrks) 1. A O B A body consists of two uniform circulr discs, ech of mss m nd rdius, with uniform rod. The centres of the discs re fixed to the ends A nd B of the rod, which hs mss m nd length 8. The discs nd the rod re coplnr, s shown in the digrm bove. The body is free to rotte in verticl plne bout smooth fixed horizontl xis. The xis is perpendiculr to the plne of the discs nd psses through the point O of the rod, where AO =. () Show tht the moment of inerti of the body bout the xis is 5m. (6) The body is held t rest with AB horizontl nd is then relesed. When the body hs turned through n ngle of 0, the rod AB strikes smll fixed smooth peg P where OP =. Given tht the body rebounds from the peg with its ngulr speed hlved by the impct, show tht the mgnitude of the impulse exerted on the body by the peg t the impct is 5g 9m. 6 (10) (Totl 16 mrks) Edexcel Internl Review 10
15. () Prove, using integrtion, tht the moment of inerti of uniform circulr disc, of mss m nd rdius, bout n xis through its centre O perpendiculr to the plne of the disc is 1 m. () The line AB is dimeter of the disc nd P is the mid-point of OA. The disc is free to rotte bout fixed smooth horizontl xis L. The xis lies in the plne of the disc, psses through P nd is perpendiculr to OA. A prticle of mss m is ttched to the disc t A nd prticle of mss m is ttched to the disc t B. Show tht the moment of inerti of the loded disc bout L is 1 m. (6) At time t = 0, PB mkes smll ngle with the downwrd verticl through P nd the loded disc is relesed from rest. By obtining n eqution of motion for the disc nd using suitble pproximtion, (c) find the time when the loded disc first comes to instntneous rest. (8) (Totl 18 mrks) Edexcel Internl Review 11
16. () Show by integrtion tht the moment of inerti of uniform disc, of mss m nd rdius, bout n xis through the centre of disc nd perpendiculr to the plne of the disc is 1 m. () A 1 B A uniform rod AB hs mss m nd length. A uniform disc, of mss m nd rdius 1, is ttched to the rod with the centre of the disc lying on the rod distnce from A. The rod lies in the plne of the disc, s shown in the digrm bove. The disc nd rod together form pendulum which is free to rotte bout fixed smooth horizontl xis L which psses through A nd is perpendiculr to the plne of the pendulum. Show tht the moment of inerti of the pendulum bout L is 7 m. () The pendulum mkes smll oscilltions bout its position of stble equilibrium. (c) Show tht the motion of the pendulum is pproximtely simple hrmonic, nd find the period of the oscilltions. (6) (Totl 1 mrks) Edexcel Internl Review 1
1. () Mss of disc removed = m B1 1 ( ) ( ) m + m A1 1 ( ) (5 ) m + m A1 1 1 I = m( ) + m( ) ( m( ) + m(5 ) ) D = 69 m * A1 7 m.0 = mx m 1 x = (from O) A1 1 1 π 69m Ω = mg( )(1 cos ) A Ω= 11g A1 6 [1]. () 8x δa= δx A1 8x m δm= δx. or δm = 8x δ x.ρ 1 D δ 8x m m I = x. x (= x x) 1 9 A1 m I = x dx 9 0 m x = 9 0 9m = * A1 6 Edexcel Internl Review 1
9m 8m m I A = + = (perp xes rule) A1 6 I I m A = G + ( ) (prllel xes rule) D A1 I D = I G + m (prllel xes rule) A1 I D m 5m = m = A1 6 6 5m mg sinθ = θ 6 6g θ = sinθ 5 A1 8 (c) 6g For smll θ, θ = θ SHM 5 5 T = π = 5π A1 6g g [16]. () MI of disc bout L = 1 m() + m() = 5m A1 CAM: m g. = (5 m + m( ) ) ω A1 ft ω = g A1 5 Edexcel Internl Review 1
M ( A), 0= Iθ B1 θ = 0 R( ), X = mθ = 0 B1 R( ), Y mg = mθ A1 Y g = + 9 D mg = 9 A1 6 [11]. () Mxδ x δ m = A1 1 Mxδ x δ I = ( x) A1 I = 0 8Mx d x D 8M = x 0 = M * A1 6 Edexcel Internl Review 15
J. = M ω A1 1 M ω = Mg ( 1 + cos 60 ) A solving for J D g J = M A1 7 [1] 1 5. () m ( ) m + m( ) = 1 + B1 A1 1 1 m + m + 6m = 75m * A1 1 75m ω = mg(cosθ cosα) + mg(cosθ cosα) A 8 8 ω = g(cosθ ) = g( 5cosθ ) A1 15 5 75 X 6mg cosθ = mω + mω = 0mω A 8 X = 6mg cosθ + 0m g(5cosθ ) 75 D 50mg cosθ 56mg = 5 A1 9 Edexcel Internl Review 16
(c) mg sinθ mg sin θ = 75m θ A1 θ g = sinθ 15 A1 g θ, SHM 15 Time = π 15 g 1 15 π g A1 6 [19] 6. M Mδx δ m = π δx. = π h h A1 1 δ I = δm. + δm. x A1 M = ( + x )δx h A1 h M = I ( + x )dx h A1 0 h M = x x h + 0 M = ( + h ) M = ( + h ) 1 A1 10 [10] Edexcel Internl Review 17
7. V = π xdx 0 = π A1 M M δm = π xδx = x δ x π 1 M M δi = xδx. y = x δx A1 I = M = x dx M DA1 7 0 [7] M π π 8. () I O = () () ( ) ( ) A1 π 5M = * A1 1 5M I dimeter = (perp. xes) A1 5M I L = + M () (prllel xes) 1M = A1 (c) 1M M(L), Mg sin θ = θ A1 sin θ θ θ 8g = θ 1 DA1 Time = 1 8g D π 1 = 8g A1 6 [1] Edexcel Internl Review 18
9. () I = m = x dx 0 m x A1 0 * m = A1 I x = I y = mc I z = I x + I y = 8 mc ( r xes) A1 [6] 10. x δx δm = m π δi = m π. π δx A1. π x δx I = m 1 x dx = m 0 A1 5 Edexcel Internl Review 19
11. () P Q L R DISC: I dim = 1 1 m = I L = 1 m + m() 1 m A1 A1 ROD: I L = 65 = m = m B1 I TOTAL = 65 m + m = 77 m (*) A1 7 CAM: 77 m 77 1 w = m + m() w A1 A1 77 w = w A1 109 [11] Edexcel Internl Review 0
1. () L A x l δx h b l b = = x h bx x l = = h A1 m m ρ = = 1 bh bh bx m m m δm =. δx. = = xδx h bh bh h 1 l δt L δ m + δmx A1 = 1 1 (l + 1x )δm 1 x mx = 1x δx 1 + h 80 m = x δx 6 h 0 m = x δx 9 h 0 m I L = h x δ x 9 h 0 0 m h =. 9 h 5mh = (*) A1 9 9 A1 Edexcel Internl Review 1
.. G Energy: 5m. h θ 9 1 5mh. 9 1 cosθ = h A h G 1 h 6g 5h = mg (1 cosθ) 6g h = mg (1 cosθ) A1 A1 5h θ = π (60 ) A1 5 (c) h 5mh M(A): mg. sin60 = θ 9 (mx L r cc n when t rest) A1ft on θ θ = g A1 mx 5h [17] Edexcel Internl Review
1. () I PQ = m () = 1 m A1 PQ Y. Energy: 1 1 m θ = mg A1 ft θ = g (*) A1 (c) R( ): Y mg = m θ Y = mg + m g A1 = 5 mg A1 [9] 1. () I = 1 m + m() + 1 m + m(5), A1 A1 (+) 1 (m) () + m A1 = 5 m (*) A1 6 Edexcel Internl Review
P ω 1 5m ω = mg 5g 1 5g ω = = 5 6 5 + mg mg A, 1, 0 A1 I ω / P I = 5m I = = 5m 5 m = 9m 5g 6 1 5g 5 1 5g 5 5g 6 5g 5 A, 1, 0 ft A1 10 [16] 15. () (δ I) = (ρ)πrδ r.r m Using (ρ) = π Completion: I = m r = 1 m (*) A1 0 Edexcel Internl Review
Am ( ) L L* B( m) r Disc: Use of xis theorem to find I L* 1 1 I L* = ( m 1 ) = m A1 Use of prllel xis theorem 1 I L = m + m = 1 m A1 For loded disc: I = 1 m + m 1 + m = m (*) A1 cso 6 Edexcel Internl Review 5
(c) A mg P θ mg B Iθ = mg mg sinθ mg sinθ mg sinθ A1 A1 [A1 for signs, A1 terms ] m θ = mg sinθ For smll ngles θ sinθ m θ = mgθ θ g = θ 7 A1 ft g SHM with ω = 7 π 7 π 7 Time = ; = π or ω g g ; A1 8 [18] Edexcel Internl Review 6
16. () r MI of element = πρrδr r m = πρ I = m r r d 0 = m r 0 = 1 m A1 I = I rod + I disc = m, + 1 m + m B1, = m + m + 9m = 7 m A1 (c).. mg mg 7 m θ = mg sin θ mg θ g = = 9mg sin θ sin θ sin θ A, 1, 0 Edexcel Internl Review 7
Smll oscilltions sin θ θ θ g = θ which is SHM A1 T = π g A1 6 [1] Edexcel Internl Review 8
1. In prt () mny correct solutions were seen. Errors rose in the use of the prllel xes rule. Despite distnces being quite clerly given nd mrked on the digrm some cndidtes ttempted to use, perhps misreding the question. A few filed to use the prllel xes rule t ll. It ws surprising to see such bsic errors in wht ws firly stndrd problem involving chnge of xis for moment of inerti. In prt most cndidtes relised tht n energy eqution ws required. A surprisingly lrge number, however, filed to relise tht the position of the centre of mss for the lmin ws required in the clcultion of Potentil Energy. Another common error ws to use mg rther thn mg for the weight of the lmin. It ws surprising to see number misquoting the expression for Kinetic Energy sometimes omitting the 1 or filing to squre the Ω.. This question proved to be too difficult for mny. Few completely correct solutions were seen. Prt () ws very poorly nswered. In order to find the mss of strip, the rtio of bse to height of the tringle ws required. The height ws esily found using,,5 tringle yet number of cndidtes mde errors here with 5 rther thn seen quite often. Mny cndidtes were unble to use pproprite methods for clculting the moment of inerti of strip bout the required xis through A. Mny tried to use n xis through BC insted. A number used n incorrect density, filing to understnd tht m ws the mss for the whole tringle nd not hlf of it. Methods used were often very difficult to follow. There ws often little indiction of wht the cndidte ws trying to do. 8 For prt, very mny thought tht m ws the moment of inerti required in their solution. Very few relised tht they needed to use this, together with the nswer from prt (), nd the perpendiculr xes rule. Of those who did, few pprecited tht they lso needed to use the prllel xes rule, in order to find the moment of inerti bout the xis required. Using their moment of inerti, in n eqution of motion, to find the ngulr ccelertion lso proved to be stumbling block for mny, with rther thn often seen. A significnt number chose, insted, to differentite n energy eqution nd both methods were generlly used correctly by those who got this fr. In prt (c) mny cndidtes filed to write down the ctul SHM eqution for θ in terms of θ before writing down the periodic time, thereby losing ll the mrks. Edexcel Internl Review 9
. A firly high number of good ttempts to this question were seen. In prt () most cndidtes used the correct moment of inerti for n xis tngentil to the disc nd some did then use the perpendiculr xes rule followed by the prllel xes rule. Only few, erroneously, used n xis perpendiculr to the disc. Most of the errors in this prt then resulted from trying to use energy conservtion for n inelstic impct where energy ws not conserved. Conservtion of ngulr momentum ws required here nd those who used this were generlly successful in scoring most of the vilble mrks. Very few completely correct solutions were seen to prt. Some ttempted to use energy conservtion gin. Of those who correctly relised tht equtions of motion were required, there were very mny incomplete solutions with only very smll number correctly justifying the fct tht the horizontl component of the force required ws zero. Of those who correctly followed through verticl eqution of motion, quite number filed to include the mss of the prticle to mke the totl mss m.. The simplest wy to split the lmin into strips for prt () ws to use the hint in the question nd hve the strips prllel to AB. Those who used strips prllel to OA mde the question more complicted nd more difficult but were, nevertheless, often successful. The commonest errors in the second prt were either to get the position of the centre of mss wrong or to forget to multiply J by when pplying the rottionl impulse-momentum principle. 5. The wekness of cndidtes when nswering questions bout rottionl mechnics ws shown by the lrge number who only ttempted prt () of this question. This sometimes involved bltnt fudging. In prt, the ttempt t the energy eqution ws often resonble, but using Newton's second lw long the rod often contined dimensionl errors, such s multiplying the component of the weight by distnce. In the finl prt it ws necessry to obtin n expression for the ngulr ccelertion either by tking moments or by differentiting the energy eqution from prt. Without this strting point, it ws not possible to use the pproximtion θ sin θ nd so few mrks could be obtined. There re still number of cndidtes who lern formul for the period of compound pendulum nd this should be strongly discourged s the questions re usully designed to test understnding nd the bility to work from 1 st principles. 6. Most cndidtes tried to use n increment of volume nd then integrte, but substntil minority did not use the prllel xes theorem to chnge to n xis through one of the ends of the cylinder, thus only gining possible mrks out of 10. A surprising number of cndidtes ppered to be considering cone or sphere. Some got their constnts nd vribles mixed up, both in creting n expression for the moment of inerti of the increment nd in integrtion. Edexcel Internl Review 0
7. A number of cndidtes correctly ttempted to split the solid into discs with their centres on the x-xis nd mny rrived t the finl nswer. Some, however, ssumed tht the solid ws hemisphere when finding the density nd n expression for y. The other common error ws to 1 1 tke the moment of inerti of disc to be δ mx insted of δ my. 8. The simplest (incorrect) method to get the printed nswer in prt () is by dding vrious pieces, ech of mss M, which is wht significnt number of the cndidtes did. The simplest correct method is by considering the rtio of the msses of the lrge disc nd the smll disc nd hence the mss of the ring. Prt using the perpendiculr nd prllel xis theorems ws lmost universlly successful. In the third prt, few cndidtes seemed unwre of the use of the term eqution of motion used in its rottionl sense. Most, however, took moments bout the xis nd proceeded correctly. Common errors were to omit the minus sign, which ws penlised when the period of SHM ws needed, the from, nd thinking tht hlf of the period ws needed rther thn qurter of it. A few cndidtes ignored the instruction to use n eqution of motion nd strted from formul for the period of compound pendulum, thus gining no mrks in prt (c). 9. The first prt proved to be n esy strter nd ws generlly well nswered. Prt ws more demnding but nevertheless ws well done by mny, either by using the perpendiculr xes rule directly or by strting from the centre of the squre nd using both the perpendiculr xes nd the prllel xes rules. A few cndidtes tried using integrtion but usully without success. 10. Nerly ll cndidtes provided good proofs of this stndrd result nd set out their working clerly. Only the wekest cndidtes filed to mke good progress in this question. 11. This proved to be the most strightforwrd question on the pper with mny fully correct solutions. The mjority found the moment of inerti of the body correctly the error for the minority ws to use the moment of inerti of the disc bout n xis through the centre rther thn bout the dimeter. In prt, few cndidtes thought tht energy ws conserved, rther thn ngulr momentum. 1. Prt () proved to be beyond the mjority of cndidtes. The proof ws not well known nd most cndidtes could not find the moment of inerti of n elementry strip. Use of the prllel nd perpendiculr xis theorems ws needed but they were rrely seen. However, mrks were gined in nd (c) since the inerti ws given lthough cndidtes often lost mrks becuse they used mgh insted of mgh/ in the potentil energy term nd in the ngulr eqution of motion. Edexcel Internl Review 1
1. This proved to be very strightforwrd question for the vst mjority of cndidtes nd full mrks here were regulrly obtined. It ws very plesing to note the cler understnding shown of the generl principles involved. 1. Most could verify tht the moment of inerti of the given body ws s stted. Most could lso successfully find the ngulr speed of it just before it hit the peg. The lst prt of the question proved to be more chllenging with reltively few fully correct solutions seen. Mny confused liner nd ngulr speeds nd/or impulses, filing to tke the moment of the impulse. Others too filed to relise tht the ngulr velocity would chnge direction t the impulse nd so hd wrong sign in their eqution. The question s whole proved to be good discrimintor, highlighting those with cler understnding of the principles involved in ngulr motion. 15. Prt () often provided cndidtes with mrks, lthough some cndidtes did need to fudge little, nd the pproch of some cndidtes did involve considerble mount of work. In prt good cndidtes did set out their working very clerly but often it ws difficult to know wht cndidtes were doing. A very frequent solution, with no explntion, ws I = 1 m + m + 1 m + m followed by = m, the given nswer. For these cndidtes, who hd not considered the need to pply the perpendiculr xis theorem, the nswer should hve been m! In prt (c) it ws quite common for cndidtes to omit one of the prticles or mke sign error in the eqution of motion. If the resulting eqution, fter the pproximtion of θ for sinθ, did not represent SHM the finl three mrks were not vilble. As stted in the introduction, if T I = π ws used, without proof, then only mrks were vilble for correct nswer. mgh 16. No Report vilble for this question. Edexcel Internl Review