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1 1. 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). 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) Edexcel Internl Review 1

2 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). Prticles P nd Q hve mss m nd m respectively. Prticle P is ttched to one end of light inextensible string nd Q is ttched to the other end. The string psses over circulr pulley which cn freely rotte in verticl plne bout fixed horizontl xis through its centre O. The pulley is modelled s uniform circulr disc of mss m nd rdius. The pulley is sufficiently rough to prevent the string slipping. The system is t rest with the string tut. A third prticle R of mss m flls freely under grvity from rest for distnce before striking nd dhering to Q. Immeditely before R strikes Q, prticles P nd Q re t rest with the string tut. 1 g () Show tht, immeditely fter R strikes Q, the ngulr speed of the pulley is. (5) When R strikes Q, there is n impulse in the string ttched to Q. Find the mgnitude of this impulse. () Given tht P does not hit the pulley, (c) find the distnce tht P moves upwrds before first coming to instntneous rest. (6) (Totl 1 mrks) Edexcel Internl Review

3 . A uniform circulr disc hs mss m nd rdius. The disc cn rotte freely bout n xis tht is in the sme plne s the disc nd tngentil to the disc t point A on its circumference. The disc hngs t rest in equilibrium with its centre O verticlly below A. A prticle P of mss m is moving horizontlly nd perpendiculr to the disc with speed (kg), where k is constnt. The prticle then strikes the disc t O nd dheres to it t O. Given tht the disc rottes through n ngle of 90 before first coming to instntneous rest, find the vlue of k. (Totl 10 mrks) 5. 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) Edexcel Internl Review

4 6. A uniform rod AB, of mss m nd length, is free to rotte in verticl plne bout fixed smooth horizontl xis through A. The rod is hnging in equilibrium with B below A when it is hit by prticle of mss m moving horizontlly with speed v in verticl plne perpendiculr to the xis. The prticle strikes the rod t B nd immeditely dheres to it. () Show tht the ngulr speed of the rod immeditely fter the impct is v. 8 (5) Given tht the rod rottes through 10 before first coming to instntneous rest, find v in terms of nd g. (6) (c) find, in terms of m nd g, the mgnitude of the verticl component of the force cting on the rod t A immeditely fter the impct. (5) (Totl 16 mrks) 7. A rod AB hs mss m nd length. It is free to rotte bout fixed smooth horizontl xis through the point O of the rod, where AO =. The rod is hnging in equilibrium with B below O when it is struck by prticle P, of mss m, moving horizontlly with speed v. When P strikes the rod, it dheres to it. Immeditely fter striking the rod, P hs speed v. Find the distnce from O of the point where P strikes the rod. (Totl 7 mrks) Edexcel Internl Review

5 1. () MI of disc bout L = 1 m() + m() = 5m CAM: m g. = (5 m + m( ) ) ω ft ω = g A1 5 M ( A), 0= Iθ B1 θ = 0 R( ), X = mθ = 0 B1 R( ), Y mg = mθ Y g = + 9 DM1 mg = 9 A1 6 [11]. () Mxδ x δ m = 1 Mxδ x δ I = ( x) I = 0 8Mx d x DM1 8M = x 0 Edexcel Internl Review 5

6 = M * A1 6 J. = M ω 1 M ω = Mg ( 1 + cos 60 ) M1 A solving for J DM1 g J = M A1 7 [1]. () u = g B1 CAM bout 0: m g = m ω + m 1 ω + m ω M1 A g 6 = ω 1 g w = A1 5 For Q: I = mw mu I = 6mw mw = mw = m g = m 8 g A1 Edexcel Internl Review 6

7 (c) PE gin of = KE loss of + KE loss of + KE loss of + PE loss of P P Q pulley Q mgd = m w + m w + m w + mgd M1 A mgd = m w 1 g gd =. = [1]. MI of disc bt dimeter = 1 m MI of disc bout xis = 1 m + m = 5 m M1 CAM: m.. kg = ( 5 m + m )ω ω = Energy : 1. 9m 9. kg k = 9 16 kg 81 A1 = mg A1 A1 [10] 5. () P Q L R DISC: I dim = 1 1 m = I L = 1 m + m() 1 m ROD: I L = 65 = m = m B1 Edexcel Internl Review 7

8 I TOTAL = 65 m + m = 77 m (*) 7 CAM: 77 m 77 1 w = m + m() w A1 77 w = w A1 109 [11] 6. () R B mg 60 mg A mg m v B I A = { m + m() } 16m mv () = I A ω = ω ft v ω = 8 (*) no wrong working seen A1 cso 5 Gin in PE = mg (1 + cos60 ) Attempt t 1 I ω = gin in PE M1 1 16m v = mg (1 + cos60 ) A1 ft 8 c Finding v v = 1 g 6 (c) Accelertion of C of G = ( ω ) B1 R mg = mrω ; = m( ω ) Substitution of ω (nd v) nd finding R = M1 R = g 16 A1 5 [16] Edexcel Internl Review 8

9 7. O v P MI of rod bout O is 1 m() + m = 7 m B1 Moment of momentum: mv x = ( 7 m + mx )ω xω = v A1 ft M1 m xω x = ( 7 m + mx )ω 7 x = 7 x = 7 [7] Edexcel Internl Review 9

10 1. 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.. Mny cndidtes tried to use conservtion of energy (or the equivlent route using Newton s second lw etc) in prt () in spite of the fct tht collision hd occurred. Even those who used conservtion of ngulr momentum often missed out term nd so hd to do some interesting fudges to get the printed nswer. In prt most cndidtes did not relise tht the liner momentum of both R nd Q hd to be considered nd there were very few correct solutions seen. In the third prt most relised tht n energy method gve the esiest solution but did not lwys include ll of the prticles nd the pulley. Some tried to write down equtions of motion for the prticles nd the pulley to find n ccelertion. Few however successfully reched the end of this long method.. Able cndidtes scored well here, though the question ws found to be more chllenging for others, with number filing to relise tht they would need both n energy eqution (for the motion fter the impct) nd momentum eqution (for the impct itself). A number thought tht they would only need the energy eqution. Some rithmeticl slips lso occurred in the processing of the equtions. 5. 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. Edexcel Internl Review 10

11 6. Prt () ws generlly well nswered lthough, s usul, the given nswer sometimes emerged from incorrect work. Mny good cndidtes gined full mrks in prt but there were two firly common errors: (i) omitting term in the potentil energy element nd, more importntly, (ii) using 1 mv insted of 1 Iω for the kinetic energy element. The finl prt proved discrimintor even mong the better cndidtes; there ws much confused thinking, with correct eqution of motion being very rre. 7. No Report vilble for this question. Edexcel Internl Review 11

PhysicsAndMathsTutor.com

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