1. A gaden game is played with a small ball B of mass m attached to one end of a light inextensible sting of length 13l. The othe end of the sting is fixed to a point A on a vetical pole as shown in the diagam above. The ball is hit and moves with constant speed in a hoizontal cicle of adius 5l and cente C, whee C is vetically below A. Modelling the ball as a paticle, find (a) the tension in the sting, (3) (b) the speed of the ball. (4) (Total 7 maks). A bend of a ace tack is modelled as an ac of a hoizontal cicle of adius 10 m. The tack is not banked at the bend. The maximum speed at which a motocycle can be idden ound the bend without slipping sideways is 8 m s 1. The motocycle and its ide ae modelled as a paticle and ai esistance is assumed to be negligible. (a) Show that the coefficient of fiction between the motocycle and the tack is. 3 (6) Edexcel Intenal Review 1
The bend is now econstucted so that the tack is banked at an angle α to the hoizontal. The maximum speed at which the motocycle can now be idden ound the bend without slipping sideways is 35 m s 1. The adius of the bend and the coefficient of fiction between the motocycle and the tack ae unchanged. (b) Find the value of tan α. (8) (Total 14 maks) 3. A paticle P of mass m moves on the smooth inne suface of a hemispheical bowl of adius. The bowl is fixed with its im hoizontal as shown in the diagam above. The paticle moves 3g with constant angula speed in a hoizontal cicle at depth d below the cente of the bowl. (a) Find, in tems of m and g, the magnitude of the nomal eaction of the bowl on P. (4) (b) Find d in tems of. (4) (Total 8 maks) Edexcel Intenal Review
4. A ough disc otates about its cente in a hoizontal plane with constant angula speed 80 evolutions pe minute. A paticle P lies on the disc at a distance 8 cm fom the cente of the disc. The coefficient of fiction between P and the disc is μ. Given that P emains at est elative to the disc, find the least possible value of μ. (Total 7 maks) 5. A B The diagam above shows a paticle B, of mass m, attached to one end of a light elastic sting. The othe end of the sting is attached to a fixed point A, at a distance h vetically above a smooth hoizontal table. The paticle moves on the table in a hoizontal cicle with cente O, whee O is vetically below A. The sting makes a constant angle θ with the downwad vetical and B moves with constant angula speed ω about OA. h O (a) g Show that ω. h (8) The elastic sting has natual length h and modulus of elasticity mg. Given that 3 tan θ =, 4 (b) find ω in tems of g and h. (5) (Total 13 maks) Edexcel Intenal Review 3
6. A light inextensible sting of length l has one end attached to a fixed point A. The othe end is attached to a paticle P of mass m. The paticle moves with constant speed v in a hoizontal cicle with the sting taut. The cente of the cicle is vetically below A and the adius of the cicle is. Show that g = v (l ). (Total 9 maks) 7. A h R B One end of a light inextensible sting is attached to a fixed point A. The othe end of the sting is attached to a fixed point B, vetically below A, whee AB = h. A small smooth ing R of mass m is theaded on the sting. The ing R moves in a hoizontal cicle with cente B, as shown in the diagam above. The uppe section of the sting makes a constant angle θ with the downwad vetical and R moves with constant angula speed ω. The ing is modelled as a paticle. g 1+ sinθ (a) Show that ω =. h sinθ (7) (b) g Deduce that ω >. h () Edexcel Intenal Review 4
3g Given that ω =, h (c) find, in tems of m and g, the tension in the sting. (4) (Total 13 maks) 8. 3a C 4a V A hollow cone, of base adius 3a and height 4a, is fixed with its axis vetical and vetex V downwads, as shown in the figue above. A paticle moves in a hoizontal cicle with cente C, 8g on the smooth inne suface of the cone with constant angula speed. 9a Find the height of C above V. (Total 11 maks) Edexcel Intenal Review 5
9. A O a 30 P A paticle P of mass m is attached to one end of a light inextensible sting of length a. The othe end of the sting is fixed to a point A which is vetically above the point O on a smooth hoizontal table. The paticle P emains in contact with the suface of the table and moves in a kg cicle with cente O and with angula speed, whee k is a constant. Thoughout the 3a motion the sting emains taut and APO = 30, as shown in the figue above. kmg (a) Show that the tension in the sting is. 3 (3) (b) Find, in tems of m, g and k, the nomal eaction between P and the table. (3) (c) Deduce the ange of possible values of k. () g The angula speed of P is changed to. The paticle P now moves in a hoizontal cicle a above the table. The cente of this cicle is X. (d) Show that X is the mid-point of OA. (7) (Total 15 maks) Edexcel Intenal Review 6
10. A paticle P of mass m moves on the smooth inne suface of a spheical bowl of intenal adius. The paticle moves with constant angula speed in a hoizontal cicle, which is at a depth 1 below the cente of the bowl. (a) Find the nomal eaction of the bowl on P. (4) (b) Find the time fo P to complete one evolution of its cicula path. (6) (Total 10 maks) 11. A paticle P of mass 0.5 kg is attached to one end of a light inextensible sting of length 1.5 m. The othe end of the sting is attached to a fixed point A. The paticle is moving, with the sting taut, in a hoizontal cicle with cente O vetically below A. The paticle is moving with constant angula speed.7 ad s 1. Find (a) the tension in the sting, (4) (b) the angle, to the neaest degee, that AP makes with the downwad vetical. (3) (Total 7 maks) 1. A ough disc otates in a hoizontal plane with constant angula velocity ω about a fixed vetical axis. A paticle P of mass m lies on the disc at a distance 4 3 a fom the axis. The coefficient of fiction between P and the disc is 3 5. Given that P emains at est elative to the disc, (a) pove that ω 9g. 0a (4) Edexcel Intenal Review 7
The paticle is now connected to the axis by a hoizontal light elastic sting of natual length a and modulus of elasticity mg. The disc again otates with constant angula velocity ω about the axis and P emains at est elative to the disc at a distance 3 4 a fom the axis. (b) Find the geatest and least possible values of ω. (7) (Total 11 maks) 13. A 60 L P A paticle P of mass m is attached to one end of a light sting. The othe end of the sting is attached to a fixed point A. The paticle moves in a hoizontal cicle with constant angula speed ω and with the sting inclined at an angle of 60 to the vetical, as shown in the diagam above. The length of the sting is L. (a) Show that the tension in the sting is mg. (1) (b) Find ω in tems of g and L. (4) Edexcel Intenal Review 8
The sting is elastic and has natual length 5 3 L. (c) Find the modulus of elasticity of the sting. () (Total 7 maks) 14. A ca moves ound a bend which is banked at a constant angle of 10 to the hoizontal. When the ca is tavelling at a constant speed of 18 m s 1, thee is no sideways fictional foce on the ca. The ca is modelled as a paticle moving in a hoizontal cicle of adius metes. Calculate the value of. (Total 6 maks) 15. A l 3 l P l B A paticle P of mass m is attached to the ends of two light inextensible stings AP and BP each of length l. The ends A and B ae attached to fixed points, with A vetically above B and AB = 3 l, as shown in the diagam above. The paticle P moves in a hoizontal cicle with constant angula speed ω. The cente of the cicle is the mid-point of AB and both stings emain taut. Edexcel Intenal Review 9
(a) (b) Show that the tension in AP is 6 1 m(3lω + 4g). Find, in tems of m, l, ω and g, an expession fo the tension in BP. (7) () (c) Deduce that ω 4 g. 3l () (Total 11 maks) 16. A 5l B 3l A light inextensible sting of length 8l has its ends fixed to two points A and B, whee A is vetically above B. A small smooth ing of mass m is theaded on the sting. The ing is moving with constant speed in a hoizontal cicle with cente B and adius 3l, as shown in the diagam above. Find (a) (b) the tension in the sting, the speed of the ing. (3) (5) (c) State biefly in what way you solution might no longe be valid if the ing wee fimly attached to the sting. (1) (Total 9 maks) Edexcel Intenal Review 10
17. A a 3a O P A paticle P of mass m is attached to one end of a light inextensible sting of length 3a. The othe end of the sting is attached to a fixed point A which is a vetical distance a above a smooth hoizontal table. The paticle moves on the table in a cicle whose cente O is vetically below A, as shown in the diagam above. The sting is taut and the speed of P is (ag). Find (a) the tension in the sting, (6) (b) the nomal eaction of the table on P. (4) (Total 10 maks) Edexcel Intenal Review 11
1. (a) 1 cos α = 13 B1 R ( ) T cosα = mg M1 1 T = mg 13 13 T = mg oe 1 A1 3 (b) Eqn of motion v Tsinα = m 5 l M1 A1 13mg 5 v = m 1 13 5 l M1 dep 5gl v = 1 5 gl gl 5gl v = accept 5 o o any othe equiv) 3 1 1 A1 4 [7] Edexcel Intenal Review 1
. (a) R = mg B1 Use of limiting fiction, F = μr B1 m8 µ R = M1 A1 10 8 µ = = * cao M1 A1 6 10 9.8 3 (b) R cos α μr sin α = mg M1 A1 mv µ R cos α + R sinα = M1 A1 v µ cosα + sinα = cosα µ sinα g Eliminating R M1 cosα + 3sinα = 3cosα sinα leading to 5 4 Substituting values M1 7 tan α = awt 0. M1 A1 8 1 [14] Edexcel Intenal Review 13
3. (a) R sin θ = mxω M1 A1 x 3g R = mx M1 3mg R = A1 (b) R cos θ = mg M1 A1 3mg d = mg M1 d = 3 A1 [8] 4. 80 π 1 8π ω = ad s = 8.377... 60 3 B1 Accept v = π 0.67 ms as equivalent 75 Fo least value of μ ( ) ( ) R = mg B1 µ mg = mω M1 A1 = A1 0.08 8π µ = 0.57 accept 0.573 M1 A1 9.8 3 [7] Edexcel Intenal Review 14
5. (a) N T w l h mg T cos θ + N = Mg (1) M1A1 T sin θ = mω () M1A1 sin θ = fom () T = mlω l sub into (1) ml cos θ ω + N = mg M1 N = mg mhω A1 g Since in contact with table N...0 ω,, * h M1A1 cso 8 (b) : h: l = 3 :4:5 extension = 4 h B1 mg h mg T = = M1A1 h 4 T = mlω = mh g ω ω = M1A1 5 4 5h 5 [13] Edexcel Intenal Review 15
6. A l T P mg T cos θ = mg mv T sin θ = tan θ = ( l ) o equivalent M1A1 M1A1 M1A1 v tan θ = Eliminating T M1 g ( l v = ) g Eliminating θ M1 g = v (l ) * cso A1 9 [9] 7. (a) T cos θ = mg B1 T + T sin θ = mω (3 tems) M1A1 = h tan θ B1 mg mω h sinθ (1 + sinθ ) = cosθ cosθ (eliminate ) M1 g 1+ sinθ ω = (*) h sinθ (solve fo ω ) M1A1 7 Allow fist B1M1A1 if assume diffeent tensions (so next M1 is effectively fo eliminating and T. Edexcel Intenal Review 16
(b) g 1 g g ω = + 1 > (sinθ < 1) ω > (*) h sinθ h h M1A1 M1 equies a valid attempt to deive an inequality fo ω. (Hence putting sin θ = 1 immediately into expession of ω [assuming this is the citical value] is M0.) (c) 3g g 1+ sinθ 1 = sinθ = h h sinθ M1 A1 3 T cos θ = mg T = mg o 1.15mg (awt) 3 M1A1 4 [13] 8. R h mg tan α = 4 3 o equivalent B1 tan α = h o 3a = B1 h 4a 3 R( ) R sin α = mg R = mg 5 R( ) R cos α = mω 8g 10mg = m R = 9a 9a M1 A1 M1 A1 9a 5 100mg tan α = mg = Eliminating R M1 A1 8 3 9a 3 9a 3 = = a 4 8 3a 4 h = = = a tanα 3 A1 M1 A1 [11] Edexcel Intenal Review 17
kg 9. (a) NL U T cos 30 = m (a cos 30 ) M1 A1 3a kmg T = 3 * cso A1 3 (b) i R = mg T sin30 M1 A1 = mg 1 k A1 3 3 (c) (R 0) k 3 ignoe k > 0, accept k < 3 M1 A1 (d) A T a X mg NL U g T cos θ = m (a cos θ) a M1 A1 (T = 4mg) i T sin θ = mg M1 Eliminating T M1 AX = a sin θ = 1 AO = a sin 30 = a AX = a A1 1 AO, as equied cso B1, A1 7 [15] Edexcel Intenal Review 18
10. P x mg R 1 (a) 1 1 sinθ = = ( θ = 30 ) B1 Rsinθ = mg M1 A1 R = mg A1 4 (b) Rcosθ = mxω M1 A1 = m( cosθ)ω A1 g ω = 1 π T = = π ω g 1 o exact equivalent M1 A1 6 A1 Note: x = 3 [10] 11. (a) 1.5 T 0.5g = 1.5 sinθ B1 Tsinθ = mw M1A1 Tsinθ = 0.5 1.5sinθ.7 T = 5.4675 N A1 4 (AWRT 5.5 N) Edexcel Intenal Review 19
(b) Tcosθ = 5 M1 A1 0.5g cosθ = 5.4675 θ = 6 A1 3 (neaest degee) [7] 1. (a) ( ), R = mg B1 4a m ω (seen and used) 3 B1 4a m ω 3 mg 3 5 M1 ω 9g (*) 0a A1 c.s.o 4 (b) mg a mg T = = a 3 3 B1 3 mg 4a ( ), mg + m ωmax 5 3 = 3 M1 A1 f.t 19g 0a = ω max A1 3 mg 4a ( ), mg + m ωmin 5 3 = 3 M1 A1 f.t g 0 a min A1 7 If only one answe, must be clea whethe max o min fo final A1 [11] Edexcel Intenal Review 0
13. T mg (a) ( ) T cos60 = mg T = mg * B1 1 (b) ( ) T sin60 = mω = Lsin60 g ω = L M1A1 B1 A1 4 (c) Applying Hooke s Law: mg = 3 ( L) 5 λ (L 5 3 L); λ 3mg M1;A1 [7] 14. R 10 mg ( ) R cos10 = mg M1 A1 ( ) R sin10 = mv [A1 ft on sin/cos mix] M1 A1 ft 18 Solving fo : = g tan10 M1 = 190 (m) [Accept 187, 188] A1 6 [Resolving along slope: mg sin10 = ft, then M1 A1] m(18) cos10, MA1A1 [6] Edexcel Intenal Review 1
A θ T l P 15. (a) l mg θ S B ( ) (T S) cosθ = mg M1 A1 ( ) (T + S) sinθ = mω M1 A1 ft = m(l sin θ)ω A1 Finding T in tems of l, m, ω and g M1 T = 1 6 m(3lω + 4g) (*) A1 7 (b) Finding S: S = 1 6 m(3lω 4g) M1 A1 any coect fom (c) Setting S 0; ω 4 g (*) (no wong woking seen) 3l M1 A1 [11] 16. (a) 4l 5l T T mg 3, 4, 5 B1 R( ) T sin θ = mg T = 5mg 4 M1 A1 3 Edexcel Intenal Review
(b) R ( ) T + T cos θ = mv 3l M1 A 8 5mg = 5 4 mv 3l M1 v = 6 gl A1 5 (c) Could not assume tensions same B1 1 [9] 17. A a 3a T N 0 P mg OP = a 8 R( ): T sin θ = mv a 8 B1 M1 A1 T 8a 3a = m 4ga a 8 (sin θ) B1 ft 3mg T = R( ): T cos θ + N = mg M1 A1 6 M1 A1 N = mg 3 mg 3 1 = 1 mg M1 A1 4 [10] Edexcel Intenal Review 3
1. Fo the ovewhelming majoity this povided a vey staightfowad intoduction leading to full maks. Mistakes made by the weakest students, both hee and elsewhee, evealed thei lack of undestanding of diffeent aeas of the syllabus and thei eliance on standad equations. In paticula, some seemed intent on using SHM wheeve they felt a familia fomula seemed applicable.. Almost all candidates achieved full maks in pat (a) but a few lost the last mak fo using decimals though thei woking. The given answe was a faction and once accuacy has been lost though the use of decimals it cannot be egained. Pat (b) poved to be much moe of a challenge. The majoity seemed to be tying to esolve hoizontally and vetically as they poduced a coect hoizontal equation (M1A1) but thei second equation was often R = mg cosθ (M0A0). Some even used R = mg, as they had in pat (a). Othes poduced coect hoizontal and vetical equations (M1A1M1A1) but then included R = mg cosθ o R = mg to aid elimination of R. This made thei solution invalid and no futhe maks could be gained. A few candidates attempted to esolve paallel and pependicula to the plane but did not ealise that both of these equations needed a component of the acceleation. 3. Some candidates found this question to be vey staightfowad and gave vey neat concise solutions to both pats. Howeve, that was not so in the majoity of cases. Many poduced hoizontal and vetical equations with coect components of the eaction. Howeve, not all knew what to do with these equations and no futhe wok was shown. Fequently candidates equated the hoizontal component of the eaction to mω (whee was the given adius of the 3 bowl). Most of these candidates then found tanθ = and linked back to the eaction and d (in tems of ) via sinθ and cosθ. As they thought that was the hoizontal adius (athe than the adius of the bowl) they also used θ = to aive at what appeaed to be a coect esult. Few d maks could be awaded, howeve, fo wok which was based on such a seious initial eo. 3mg Some candidates simply wote down F = mω =, which is a coect equation, although it was often not clea that the candidates eally undestood this. Many could not poduce any equations to enable d to be found o even evealed thei lack of undestanding by poceeding to 3mg teat as a component of the equied eaction. 4. Candidates seemed to have geat difficulty changing fom evolutions pe minute to adians pe second. This seemed to be wong at least as often as it was coect. A supising numbe of solutions involved inequalities, not always the coect way ound. Unfotunately, some gave thei final answe as an inequality and so failed to answe the question as set. Many did not notice that the distance was given in centimetes and so used 8 instead of 0.08 in thei calculation. Those who knew how to tackle this question poduced succinct solutions. Edexcel Intenal Review 4
5. It was disappointing to note how many candidates missed the point of the mechanics in pat (a) of this question. Cetainly having a eaction foce fom contact with the table was an iitation to them so much so that they eithe ignoed it all togethe o put it in and then said N = 0. The actual condition of finding an expession fo N and using N 0 was lost on too many. Most pefeed pat (b) whee they could simply find some set values and plug in esults. The ones who did this pat incoectly wee usually those who had insisted that T cos θ = mg in pat (a) and hence it also applied in pat (b). Most pefeed pat (b) whee they could simply find some set values and plug in esults. The ones who did this pat incoectly wee usually those who had insisted that T cos θ = mg in pat (a) and hence it also applied in pat (b). They used it to find an incoect value of T to use in T sin θ = mω. The elating of 3:4:5 to a given side of h was often dubious with h being missed out of some lengths and some caeless tigonomety was also seen. Candidates often stopped at ω and so lost the final mak. 6. This poved to be the easiest question on the pape and full maks wee vey common. Those who failed to complete the question nealy always had the esolutions coect but failed to spot the tigonometic elation between the lengths. Methods of solution using a centifugal foce wee vey uncommon. 7. Pats (a) and (c) wee geneally well done. In pat (b), a fully justified deivation of the given answe was only aely seen. Most assumed that they could put the maximum value of sin θ = 1 diectly into the expession obtained in (a) without any moe discussion. 8. This question was a vey good disciminato. Thee wee many excellent concise solutions leading quickly to full maks. Howeve thee was also a substantial minoity of students who failed to get stated. Inadequate diagam often led to candidates confusing an angle with its complementay angle and this led to eos in esolution. A common eo was to assume that the line of action of the eaction went though the cente of the cicula end of the cone. In esolving, the vetical equation was moe often incoect than the hoizontal equation. R =mg, R=mg cos α and R=mg sin α wee all common eos. An unexpected featue of the esponses was, fo the fist time fo some yeas in any numbes, to see some students using centifugal o, even, centipetal foces. Such methods wee not envisaged when this set of mechanics specifications wee designed but, if used coectly, ae accepted. Edexcel Intenal Review 5
9. Fo a significant numbe of elatively weak candidates this question ensued a espectable mak. If conical pendulum methods wee well known, it poved vey staightfowad and lage numbes of candidates gained full maks. Howeve, the show that in pat (a) thew many, who wee not entiely happy with this topic, completely off tack. It was vey clea to many that mω a gave the ight numeical answe, so they decided that this must be the ight way to do the question. It was vey common to see a solution which had stated with T cos30 = mω a continue with the cos 30 scibbled out. This then had futhe implications in pat (c) whee the candidates used the method which had woked in pat (a) and failed to esolve hee also. This betayed a complete lack of undestanding and justifiably led to the loss of a significant numbe of maks. Thee was futhe evidence in this question of inadequate showing by able students when answes ae pinted. Calculation of length OA as a was almost tivial but needed to be shown not assumed. Pefect solutions which ended up saying OX = a/, so X is the mid-point of OA wee not uncommon. 10. Although the geneal pinciples involved in this question wee usually well undestood, many eos of detail wee seen. Many assumed to be the adius of the cicle athe than the sphee. This led to an incoect angle in (a) and a significant ovesimplification of the question in (b). A common eo in (b), even among those who ealised in (a) that was the adius of the sphee, was to fail to calculate the adius of the cicle. Almost all of those who could find the angula velocity knew how to convet this to the time fo a complete evolution. 11. This poved to be a good stat fo most candidates although a few used = 1.5 and a significant numbe failed to give the answe in pat (b) to the neaest degee. 1. Pat (a) was easonably well done by most but the second pat was challenging. The use of centipetal foce continues to cause much confusion. The most fequent eos in (b) wee (i) using only one of the equations T ± F= mω, and then failing to indicate whethe thei answe was the maximum o minimum. (ii) putting F = 0 to find the minimum value and (iii) being unable to find o making an eo with the extension of the elastic sting. In (b) some candidates dew the diagam as a conical pendulum and wasted a geat deal of time tying to solve the unsolvable. 13. Although the answe was given in pat (a) it is still good to epot that all but a handful of candidates gained the mak. Pat (b) was geneally well answeed, although T = mω, Tsin60 = mlω, and eos in eliminating, wee occasionally seen. Pobably the most common eo in this question occued in pat (c), whee L, instead of of Hooke s law. 3 5 L, was often seen in the denominato Edexcel Intenal Review 6
14. This question was well answeed by the majoity of candidates, who fequently scoed 5 o 6 maks, the loss of the final mak fo giving an answe to moe than 3 significant figues. Some candidates did use equations showing seious eos in the mechanics, like R = mg cos10 and v mg sin10 = m ; invaiably these eos meant that a maximum of maks was available. 15. All but the weakest candidates, who consideed the stings sepaately, made some pogess in this question. Some wee handicapped by mixing up sine and cosine, by not eliminating successfully, and by aithmetic eos in finding tigonometic atios. In pats (a) and (c) given esults had to be deived and, as usual, they often emeged fom incoect woking. In pat (c) it was quite common to see T 1 (top sting) 0, o T 1 + T 0 used as the condition fo deiving the esult. 16. The majoity scoed well on pat (a) although many calculated an angle athe than using the 3-4-5 tiangle. Thee wee some who esolved incoectly, eithe using the wong tig. atio o wose, tying to esolve along the sting, fogetting that thee is an acceleation component in that diection. In the second pat a lage numbe of candidates included only one tension, usually the one in the uppe pat of the sting, in thei equation of motion but then went on, in the final pat, to coectly state that, if the paticle wee attached to the sting, the tensions in the two potions of the sting would be diffeent. 17. No Repot available fo this question. Edexcel Intenal Review 7