Cambridge A Level Mathematics - Further 9231 — 2023 Oct/Nov Paper 3 · Variant 2

1 One end of a light inextensible string of length a is attached to a fixed point O. The other end of the string is attached to a particle of mass m. The string is taut and makes an angle i with the downward vertical through O, where cos i = 23 . The particle moves in a horizontal circle with speed v. Find v in terms of a and g. [4] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1 ↑ T cosθ= mg B1 mv 2 B1 → T sinθ= a sinθ Eliminate T and substitute for 𝜃 M1 5 A1 v = ag 6 4

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2 A particle P of mass 0.5 kg moves in a straight line. At time t s the velocity of P is v m s -1 and its displacement from a fixed point O on the line is x m. The only forces acting on P are a force of 150 magnitude 2 N in the direction of increasing displacement and a resistive force of magnitude ( x + 1) 450 3 N . When t = 0 , x = 0 and v = 20 . ( x + 1) Ax + B Find v in terms of x, giving your answer in the form v = , where A and B are constants to be ( x + 1) determined. [6] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 2 dv 150 450 M1 Allow sign errors. 0.5v = − dx ( x + 1) 2 ( x + 1) 3 300 450 M1A1 Correct powers, allow sign errors. Integrate: 0.5v 2 = − + + A x + 1 ( x + 1) 2 x = 0 , v = 20 ; A= 50 M1 Use initial condition. 2 100( x 2 − 4 x + 4) A1 AEF Rearrange: v = ( x + 1) 2 2 100( x − 2) 2 10( x − 2) A1 v = so v =  ( x + 1) 2 ( x + 1) 20 − 10 x From initial condition, sign must be negative, v = Signs dealt with convincingly. x + 1 6

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3 B D E A C 2a A uniform lamina is in the form of an isosceles triangle ABC in which AC = 2a and angle ABC = 90° . The point D on AB is such that the ratio DB : AB = 1: k. The point E on CB is such that DE is parallel to AC. The triangle DBE is removed from the lamina (see diagram). (a) Find, in terms of k, the distance of the centre of mass of the remaining lamina ADEC from the midpoint of AC. 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When the lamina ADEC is freely suspended from the vertex A, the edge AC makes an angle i with the downward vertical, where tan i = 185 . (b) Find the value of k. 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Mark scheme: 3(a) Mass is proportional to area B1 At least two areas correct, at least one distance correct Centre of mass Area from AC 1 1 ABC  2 a  a 2 3a 1  a  2 2  a  BDE 2  a −   2  k  3  k  1  a  2 2 ADEC −  2   + a x 2  k   1  a  2 2  1 2  2  a    a  2 M1A1 Moments equation, dimensionally correct, correct number x  −  2   + a  = a  a −  a −       of terms.   2  k  3  3  k    k    2 A1 3 a k + k − 2 a k − 3k + 2 ( ) ( ) x = Allow unsimplified single fraction . k 2 − 1 3k ( k + 1) 3k ( ) 4 3(b) x  5  B1 With their x from part (a). tan= =   a  18  So 18 k 2 + k − 2 = 5 3k 2 + 3k M1 Obtain a polynomial in k only, e.g. k 3 − 13k + 12 = 0 , ( ) ( ) 2 may be implied. k + k − 12 = 0 ( k + 4 )( k − 3 ) = 0, A1 CWO k = 3 only 3

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4 u B A i d m m d d Two smooth vertical walls meet at right angles. The smooth sphere A, with mass m, is at rest on a smooth horizontal surface and is at a distance d from each wall. An identical smooth sphere B is moving on the horizontal surface with speed u at an angle i with the line of centres when the spheres collide (see diagram). After the collision, the spheres take the same time to reach a wall. The coefficient of restitution between the spheres is 1.2 (a) Find the value of tani. 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(b) Find the percentage loss in the total kinetic energy of the spheres as a result of this collision. 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Mark scheme: 4(a) After collision, A has velocity vA towards wall on right and B has component B1 of velocity towards lower wall of u sin . Same distance and time, so vA = u sin . Along line of centres: M1 Both, consistent signs, must be cos. PCLM: mv A + mvB = mu cos NEL: v A − v B = eu cos 2sin = (1 + e ) cos M1 Eliminating vA and vB to find an equation in . Condone common factor of u. 3 1 Note: v A = u cos and vB = u cos. 4 4 tan= 3 A1 4 4 4(b) 1 2 2 2 B1 Both components of velocity of B needed. Final KE = m ( v A + ( u sin) + vB ) 2 1 2 1 2 2 2 1 2  9 9 1  M1 vA, vB, substituted, ft their final KE with both = mu Loss = mu − m  1 − − −  components of velocity of B included. ( v A + ( u sin) + vB ) 2 2 2  25 25 25  3 2 A1 Loss = mu 25 Percentage loss = 24% 3

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5 vB B i a O α A vA A bead of mass m moves on a smooth circular wire, with centre O and radius a, in a vertical plane. The a with the downward vertical bead has speed vA when it is at the point A where OA makes an angle through O, and cos a = 35 . Subsequently the bead has speed vB at the point B, where OB makes an angle i with the upward vertical through O. Angle AOB is a right angle (see diagram). The reaction of the wire on the bead at B is in the direction OB and has magnitude equal to 16 of the magnitude of the reaction when the bead is at A. (a) Find, in terms of m and g, the magnitude of the reaction at B. 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(b) Given that v A = kag , find the value of k. 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Mark scheme: 5(a) mv A2 B1 At A: R A − mg cos= a mv B2 B1 At B: − RB + mg cos = a 1 2 1 2 M1 All terms present, allow sign errors, cos/sin mix. Energy: mv A − mvB = mga ( cos+ cos) 2 2 A1 AEF 1 M1 Any equivalent working, leading to a dimensionally Eliminate velocities, use RB = R A and substitute angle values 6 correct equation in RB and mg. 1 1 a ( RA − mg cos) − a ( − RB + mg cos) = mga ( cos+ cos) 2 2 3 4 7 6 RB − mg + RB − mg = 2mg  5 5 5 3 A1 RB = mg 5 6 5(b) 3 3 mv A2 M1 From first equation in (a), 6  mg − mg = 5 5 a v A2 = 3ag , so A1 CAO k = 3 2

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6 A particle P is projected with speed u at an angle a above the horizontal from a point O on a horizontal plane and moves freely under gravity. The horizontal and vertical displacements of P from O at a subsequent time t are denoted by x and y respectively. (a) Derive the equation of the trajectory of P in the form gx 2 2 a . 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During its flight, P must clear an obstacle of height h m that is at a horizontal distance of 32 m from the point of projection. When u = 40 2 m s -1 , P just clears the obstacle. When u = 40 m s -1 , P only achieves 80% of the height required to clear the obstacle. (b) Find the two possible values of h. 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Mark scheme: 6(a) → x = u cost 1 2 B1 Both correct.  y = u sint − gt 2 2 M1 Eliminate. x 1  x  Eliminate t: y = u sin − g   ucos 2  ucosα  gx 2 2 A1 y = x tan − sec  AG 2 2u 3 6(b) 4 g 32 2 2 16 2 B1 h = 32tan − sec  [= 32tan− sec ] 2 5 2  40 5 g 32 2 2 8 2 B1 h = 32tan− sec  [= 32tan− sec ] 2 5 2 40 2 ( ) 8 2 5 16 2 M1 Equate expressions for h and obtain a 3-term quadratic in 32tan− sec = (32tan− sec ) tan. 5 4 5 8 2 2 32t − 1 + t = 40t − 4 1 + t ( ) ( ) 5 3t 2 − 10t + 3 = 0 1 A1 Both correct t = 3, 3 h = 80 M1 For using their value of t to work out one value of h. 80 h = A1 Both correct 9 6

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7 5a A B 3a 4a P F A particle P of mass m is attached to one end of a light rod of length 3a. The other end of the rod is able to pivot smoothly about the fixed point A. The particle is also attached to one end of a light spring of natural length a and modulus of elasticity kmg. The other end of the spring is attached to a fixed point B. The points A and B are in a horizontal line, a distance 5a apart, and these two points and the rod are in a vertical plane. Initially, P is held in equilibrium by a vertical force F with the stretched length of the spring equal to 4a (see diagram). The particle is released from rest in this position and has a speed of 65 2ag when the rod becomes horizontal. (a) Find the value of k. 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(b) Find F in terms of m and g. 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(c) Find, in terms of m and g, the tension in the rod immediately before it is released. 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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) 2 B1 1  6  Gain in KE = m 2ag and Gain in GPE = mg  3a sin   2  5  1 B1 kmg 2 2 2 Loss in EPE = ( 3a ) − a ( ) a 1 M1 KE, GPE and at least one EPE present, allow sign errors, 2 kmg 2 2 dimensionally correct. 1  6 2 ( 3a ) − a Energy equation: m  2 ag + mg  3a sin= ( ) 2  5  a A1 All correct. 36 12 A1 mg + mg = 4kmg 25 5 24 k = 25 5 7(b) 3a 72 M1 Hooke’s law. In lower position, tension T in spring = kmg  = 3kmg = mg a 25 Perpendicular to rod, T = ( F + mg )cos 72 A1 So, ( F + mg ) cos= mg , 25 19 F = mg 5 Alternative method for question 7(b) 3a 72 M1 In lower position, tension T in spring = kmg  = 3kmg = mg Hooke’s law. a 25 Eliminate T. T  cos= Tsin T  sin+ Tcos= F + mg 19 A1 F = mg 5 2 7(c) Let tension in rod = T ' M1 Parallel to rod, T  = ( F + mg )sin 96 A1 T  = mg 25 Alternative method for question 7(c) T  cos= Tsin M1 At least one equation seen with their T and/or F. T  sin+ Tcos= F + mg 96 A1 T  = mg 25 2

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