Cambridge A Level Mathematics - Further 9231 — 2022 May/June Paper 3 · Variant 3

1 A uniform lamina OABC is a trapezium whose vertices can be represented by coordinates in the x-y plane. The coordinates of the vertices are O (0, 0), A (15, 0), B (9, 4) and C (3, 4). Find the x-coordinate of the centre of mass of the lamina. [4] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 1 Area Distance from Oy Triangle OCD 6 2 Rectangle DEBC 24 6 Triangle BAE 12 11 Trapezium OCBA 42 x where D is point with coordinates (3, 0) and E is a point with coordinates (9, 0). Other options possible for RHS of moments equation, for example: (1) OAC: 30  6 and ABC: 12  9 (2) OBC: 12 4 and OAB: 30 8 (3) Subtraction: 60 7.5 6 1 12 13     Parts that would give correct total area 42 B1 Moments about Oy 42 6 2 24 6 12 11 x       (=288) A1 Correct equation. 288 42 x  6.86  A1 4

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2 A particle P of mass m is attached to one end of a light elastic string of natural length a and modulus of elasticity 43 mg . The other end of the string is attached to a fixed point O on a rough horizontal surface. The particle is at rest on the surface with the string at its natural length. The coefficient of friction between P and the surface is 1.3 The particle is projected along the surface in the direction OP with a speed of 12 ga . Find the greatest extension of the string during the subsequent motion. [5] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 2 2 4 1 3 . 2 mg x a 1 3mgx B1 Work term correct Loss in KE = gain in EPE + work done against friction 2 2 4 1 1 1 3 2 2 3 mg mv x mgx a    M1 Energy equation with 3 terms, allow sign error. 2 2 1 1 1 3 2 4 3 g ga x gx a    2 2 16 8 3 0 x ax a       4 4 3 0 x a x a    M1 Obtain and attempt to solve a 3-term quadratic equation. 1 4 x a  A1 5

More questions on Hooke's law

3 A particle P is projected with speed 25 ms -1 at an angle i above the horizontal from a point O on a horizontal plane and moves freely under gravity. After 2 s the speed of P is 15 ms -1 . (a) Find the value of sini. 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(b) Find the range of the flight. 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Mark scheme: 3(a) Components of velocity : 25cos 25sin 2g      B1 Speed =     2 2 25cos 25sin 2g     M1 A1 Expression for speed or square of speed.     2 2 2 25cos 25sin 2 15 g      2 625 100 sin 4 225 g g     M1 Attempt to solve and find value for sin 800 4 sin 1000 5   A1 5 3(b) Time of flight 2 25sin g         = 4 (s) B1 Range = 2 25sin 25cos g     M1 Any equivalent method. Range = 60 (m) A1 CWO Alternative method for question 3(b) 2 4 1 3 45 y x x   B1 Equation of trajectory.. Substitute 0 y  and solve M1 60 (m) A1 3

More questions on Motion of a projectile

4 One end of a light inextensible string of length a is attached to a fixed point O. A particle of mass m is attached to the other end of the string and is held with the string taut at the point A. At A the string makes an angle i with the upward vertical through O. The particle is projected perpendicular to the string in a downward direction from A with a speed u. It moves along a circular path in the vertical plane. When the string makes an angle a with the downward vertical through O, the speed of the particle is 2u and the magnitude of the tension in the string is 10 times its magnitude at A. It is given that u = 23 ga . (a) Find, in terms of m and g, the magnitude of the tension in the string at A. 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(b) Find the value of cosa. 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Mark scheme: 4(a)   2 2 1 1 cos cos 2 2 mv mu mga      M1 Energy equation with all necessary terms, GPE terms must be resolved, allow sin/cos mix, allow sign error.     2 2 1 1 2 cos cos 2 2 m u mu mga      A1 2u may be substituted later. Implied by   3 2 cos cos 2 3 ag ga      At A, 2 cos m T mg u a    B1 N2L Also, 2 10 cos 4 m T mg u a    B1 N2L and use of tension (10T). Use all three (two N2L and energy) equations to find T in terms of m and g only. M1 Might see   3 2 9 cos cos 3 m T mg ga a         cos cos 1       10cos cos 4     1 3 T mg  A1 6 4(b) Substitute back, 1 4 2 10 cos 3 3 m mg mg ga a      M1 Any appropriate method to obtain cos. 2 cos 3  A1 2

More questions on Circular motion

5 A particle P of mass 4 kg is moving in a horizontal straight line. At time t s the velocity of P is v ms -1 and the displacement of P from a fixed point O on the line is x m. The only force acting on P is a resistive force of magnitude ( 4e -x + 12) e -x N . When t = 0, x = 0 and v = 4 . 1 + 3 e x (a) Show by integration that v = x . 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(b) Find an expression for x in terms of t. 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Mark scheme: 5(a)   4 4e 12 e x x dv v dx     B1   2 2 1 1 e 3e 2 2 x x v A      M1 Expression of the correct form. 9 4, 0, 2 v x A    A1   2 2 2 e 6e 9 3 e x x x v         1 3e 3 e e x x x v      A1 AG Must see the factorisation. Condone lack of justification for taking positive square root. 4 5(b) d 1 3e d e x x x t   so e 1 d 3e 1 x x dx t         1 ln 3e 1 3 x t B    M1* A1 Integration to obtain ln term Correct answer with constant of integration 1 0, 0, ln4 3 t x B    3e 1 3 ln 4 x t   DM1 Find the constant and substitute into their general solution. 3 4 1 ln e 3 3 t x         A1 OE 4

More questions on Linear motion under a variable force

6 C b u i 60° A B AB and BC are two fixed smooth vertical barriers on a smooth horizontal surface, with angle ABC = 60° . A particle of mass m is moving with speed u on the surface. The particle strikes AB at an angle i with AB. It then strikes BC and rebounds at an angle b with BC (see diagram). The coefficient of restitution between the particle and each barrier is e and tan i = 2 . The kinetic energy of the particle after the first collision is 40% of its kinetic energy before the first collision. (a) Find the value of e. 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(b) Find the size of angle b. 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Mark scheme: 6(a) Energy loss: 2 2 2 2 1 2 1 2 , 2 5 2 5 mv mu v u    B1 Energy loss. cos cos v u    sin sin v eu    B1 Both. Combine to form equation in e only 2 2 1 4 5 5 5 e    M1     2 2 2 cos sin v u eu     1 2 e  A1 4 6(b) tan tan e    , so tan 1, 45      B1 For 2nd collision   cos cos 180 60 w v        sin sin 180 60 w ev      M1 Both. May be implied by the A1.   tan tan 120 e their     M1 Divide to find . 61.8   A1 4

More questions on Momentum

7 B D A i a E a C A uniform cylinder with a rough surface and of radius a is fixed with its axis horizontal. Two identical uniform rods AB and BC, each of weight W and length 2a, are rigidly joined at B with AB perpendicular to BC. The rods rest on the cylinder in a vertical plane perpendicular to the axis of the cylinder with AB at an angle i to the horizontal. D and E are the midpoints of AB and BC respectively and also the points of contact of the rods with the cylinder (see diagram). The rods are about to slip in a clockwise direction. The coefficient of friction between each rod and the cylinder is n. The normal reaction between AB and the cylinder is R and the normal reaction between BC and the cylinder is N. (a) Find the ratio R : N in terms of n. 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(b) Given that n = 13 , find the value of tani. 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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) , AB BC F R F N Moments about B, (sin cos Na Ra Wa      ) Moments about centre, (cos sin AB BC F a F a Wa      ) Moments about D,   cos sin BC F a Na Wa      Moments about E,   cos sin AB Ra F a Wa      B1 One moments equation about any point involving all relevant forces, resolved if necessary (AEF). Parallel to AB, sin sin AB N F W W      Perpendicular to AB, cos cos BC F R W W      B1 Two resolutions: all relevant terms, different frictional forces [Vertical: cos cos sin sin BC AB R F N F W W          Horizontal: sin cos sin cos BC AB F F R N        ] Alternative approach using two moments equations can earn the B1B1       1 1 1 2 N R N R        M1 Combine appropriate equations.     1 1 1 1 1 1 2 2 N R                    1 1 1 1 2 2 2 2 N R                  M1 Collect terms to obtain ratio/fraction in terms of only (CWO), any equivalent simplified form. : 1 :1 R N     A1 6 Question Answer Marks Guidance 7(b) Divide resolution equations: tan N R N R       M1 Must include  terms. Use 1 3 2 : tan 7 3 N R N N    M1 FT their answer to part (a). 1 tan 7  A1 3

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