Cambridge A Level Mathematics - Further 9231 — 2023 May/June Paper 3 · Variant 3

1 A particle P of mass m is attached to one end of a light inextensible string of length a. The other end of the string is attached to a fixed point O. The particle P is held at the point A, where OA makes an angle a with the downward vertical through O, and with the string taut. The particle P is projected perpendicular to OA in an upwards direction with speed 3ag . It then starts to move along a circular path in a vertical plane. The string goes slack when P is at B, where OB makes an angle i with the upward vertical. Given that cos a = 45 , find the value of cosi. [4] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 1   2 1 1 .3 cos cos 2 2      m ag mv mga M1 Energy equation, 4 terms, dimensionally correct, mass must be present, allow sign errors, allow sin in both terms on RHS 2 cosmv mg a B1 N2L, may include tension initially but not awarded until tension = 0 used 3 1 4 . cos cos 2 2 5            mag m ag mga 3 7 cos 2 10  M1 Dependent on tension = 0 and on an energy equation, eliminate 2v . 7 cos 15  A1 If no m in energy equation and no further errors, award SCB2 for correct final answer 4

More questions on Circular motion

2 One end of a light elastic string, of natural length a and modulus of elasticity mmg , is attached to a fixed point O. The string lies on a smooth horizontal surface. A particle P of mass m is attached to the other end of the string. The particle P is projected in the direction OP. When the length of the string is 4 3 a , the speed of P is 2ag . When the length of the string is 53 a , the speed of P is 12 2ag . Find the value of m. [4] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 2 2 2 2 2 1 1 2 1 2 2 4 3 3                               mg v m v a a a M1 Kinetic energy = elastic potential energy, 4 terms, dimensionally correct, allow sign errors. 2 2 1 1 1 2 1 2 2 2 2 3 3                               mg m ag ag a a a A1 With v substituted. Solve [ 2 3 3 ] 4 9        v g a M1 Solve to find value for  dependent on energy equation with 3 or 4 terms 9 2  A1 SCB2 for 9 2  mg if given  not used 4

More questions on Hooke's law

3 A 3a D 5a x C B E 6a A uniform lamina is in the form of a triangle ABC, with AC = 8a, BC = 6a and angle ACB = 90°. The point D on AC is such that AD = 3a. The point E on CB is such that CE = x (see diagram). The triangle CDE is removed from the lamina. (a) Find, in terms of a and x, the distance of the centre of mass of the remaining object ADEB from AC. 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The object ADEB is on the point of toppling about the point E when the object is in the vertical plane with its edge EB on a smooth horizontal surface. (b) Find x in terms of a. 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Mark scheme: 3(a) [Mass is proportional to area] Area Centre of mass from AC ABC 1.6 .8 2 a a ( 2 24 )  a 2a DEC 1 .5 2 x a 1 3 x ADEB 2 5 24 2  a xa x B1 All correct for ABC and DEC. Moments [about AC] 2 2 5 1 5 24 24 2 2 3 2            x a xa a a x ax M1 All moment terms present, dimensionally correct, allow sign error. A1 All correct moments about AC.   2 2 288 5 3 48 5    a x x a x A1 AEF 4 Question Answer Marks Guidance 3(b) On the point of toppling about E:   2 2 288 5 , 3 48 5     a x x x x a x B1 FT FT their expression for x from part (a). Rearrange to 3-term quadratic: 2 2 10 144 288 0    x ax a M1 Allow 3-term inequality.    2 5 12 12 0,    x a x a 12 5  x a A1 Single correct answer, no inequality, CWO. 3

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4 2u u A B 30° m m Two identical smooth uniform spheres A and B each have mass m. The two spheres are moving on a smooth horizontal surface when they collide with speeds u and 2u respectively. Immediately before the collision, A’s direction of motion makes an angle of 30° with the line of centres, and B’s direction of motion is perpendicular to the line of centres (see diagram). After the collision, A and B are moving in the same direction. The coefficient of restitution between the spheres is e. (a) Find the value of e. 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(b) Find the loss in the total kinetic energy of the spheres as a result of the collision. 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Mark scheme: 4(a) Let speeds of A and B along line of centres after collision be VA and VB cos30    A B V V u (1) M1 Allow sign errors, allow missing m. cos30     A B V V eu (2) M1 Signs on LHS must be consistent with (1). Speeds perpendicular to line of centres after collision are sin30 u and 2u Moving in same direction, so sin30 2   A B V V u u (3) B1 SOI 4  B A V V Use 4  B A V V in (1): 5 cos30   A V u From (2): 3 cos30   A V eu then Combine to find equation in e only. M1 A complete method to find equation in e only 3 5 e  A1 Question Answer Marks Guidance 4(a) Alternative method for question 4(a) Let speeds of A and B along line of centres after collision be VA and VB cos30    A B V V u (1) M1 Allow sign errors, allow missing m. cos30     A B V V eu (2) M1 Signs on LHS must be consistent with (1). Speeds perpendicular to line of centres after collision are sin30 u and 2u Moving in same direction, so sin30 2   A B V V u u (3) B1 SOI 4  B A V V Solve (1) and (2):   1 1 cos30 2    A V u e ,   1 1 cos30 2    B V u e Substitute in (3) to find equation in e only . M1 Note: 4 3, 3 10 10   A B u u V V 3 5 e  A1 5 4(b) KE after =   2 2 2 2 1 1 ( 2 2 2 2                  A B u m V m u V ) B1 Correct expression for KE for one of the spheres, after collision, with both components. KE for A after = 2 7 50 mu or KE for B after = 2 56 25 mu or KE loss for A = 2 9 25 mu or KE gain for B = 2 6 25 mu B1 Implied by total KE after = 2 119 50 mu . Total loss in KE = 2 3 25 mu B1 Term   2 1 2 2 m u may be omitted from KE of B before and after. 3

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5 One end of a light elastic string, of natural length 12a and modulus of elasticity kmg, is attached to a fixed point O. The other end of the string is attached to a particle of mass m. The particle moves with constant speed 32 3ag in a horizontal circle with centre at a distance 12a below O. The string is inclined at an angle i to the downward vertical through O. (a) Find, in terms of a, the extension of the string. 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(b) Find the value of k. 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Mark scheme: 5(a) cos   T mg 2 27 sin 4 mv ag T m r r            B1 12 tan  r a used M1 Divide: 27 tan , 4 12tan     so 3 tan 4  M1 Finds value for tan OE. Reduces to equation in  or x, no k. 9 ,  r a extension of string = 3a A1 Alternative method for question 5(a) Let L be stretched length of string. cos   T mg B1 Or 12a T mg L   2 27 sin 4 mv ag T m r r            B1 sin  r L used M1 Use 12 cos a L and   0.5 2 2 144 sin   L a L and eliminate T. M1 2 2 2 [ 144 81 ] 15 L a a L a    , extension of string = 3a A1 5 Question Answer Marks Guidance 5(b) Hooke’s law:   12 12  kmg L a T a B1 Eliminate T:   12 12 12   kmg L a mgL a a M1 5 12    L k L a A1 3

More questions on Circular motion

6 A particle of mass m kg falls vertically under gravity, from rest. At time t s, P has fallen x m and has velocity v m s -1 . The only forces acting on P are its weight and a resistance of magnitude kmgv N, where k is a constant. (a) Find an expression for v in terms of t, g and k. 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(b) Given that k = 0.05, find, in metres, how far P has fallen when its speed is 12 m s -1 . 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Mark scheme: 6(a)   1   dv m mg kv dt B1 Mass must be seen at this point or earlier. [SUVAT does not apply.]   1 ln 1     kv gt A k M1 Separate variables and integrate to logarithm. A1 Correct, with constant of integration. 0, 0 [ 0] t v A    M1 Use initial condition to evaluate their constant.   1 1    kgt v e k A1 Any correct form with v as subject. Final A0 if numerical value of g present. 5 Question Answer Marks Guidance 6(b) 0.05 k  and so   0.5 20 1    t dx e dt *M1 Attempt to integrate if expression contains a term of the form ct be . Integrate:   0.5 20 2     t x t e B A1 1 1           kgt x t e B k gk   0, 0 40 t x B    DM1 Use initial condition to evaluate their constant. When 12,  v from part (a), 0.5 1 0.05 12 0.4, 2ln0.4 t e t      M1 1.83… 40ln 0.4 40 0.4 40     x 12.7  A1 5 40ln 24 2  Alternative method for question 6(b)   1   dv v g kv dx leading to 1 1 1          dv kgdx kv *M1 Separate variables and write in integrable form   1 ln 1     v kv kgx B k DM1 A1 Dependent on previous M1. Attempt to integrate.   0, 0 0 v x B    and 0.05, 12 1 2 20ln0.4 0.5 k v x     M1 Dependent on both previous M1s. Use initial condition to evaluate their constant and use 12  v 12.7  x A1 5 40ln 24 2  5

More questions on Linear motion under a variable force

7 The points O and P are on a horizontal plane, a distance 8 m apart. A ball is thrown from O with speed u m s -1 at an angle i above the horizontal, where tan i = 43 . At the same instant, a model aircraft is launched with speed 5 m s -1 parallel to the horizontal plane from a point 4 m vertically above P. The model aircraft moves in the same vertical plane as the ball and in the same horizontal direction as the ball. The model aircraft moves horizontally with a constant speed of 5 m s -1 . After T s, the ball and the model aircraft collide. (a) Find the value of T. 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(b) Find the direction in which the ball is moving immediately before the collision. 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Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................

Mark scheme: 7(a) For aircraft, 5 d T For ball, 2 1 4 sin 10 2 u T T       B1 To point of collision. For ball, cos 8 5 8 u T d T       B1 Eliminate u:   2 2 5 4 5 4 1 4 10 , 5 2 4 T u T T u T       and   5 5 8 3   T u T     2 3 4 5 4 5 8    T T *M1 Dependent on LHS of second B1 being 4, expression involving only T 2 3 4 4 0    T T DM1 Dependent on previous M1. Obtain and solve 3-term quadratic. 2  T A1 Single correct answer. 6 Note 8  d used leads to 2 3 T  B1B1B0M1M1A0 Question Answer Marks Guidance 7(b)   1 sin 10 tan cos     u T u M1 OE Accept ‘tan = ……’ 1 8 tan 9  A1 OE Direction is 41.6° below the horizontal A1 CAO Note: 8  d used leads to 20.9 above the horizontal. 3

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