Cambridge A Level Mathematics 9709 — 2010 Oct/Nov Paper 5 · Variant 3

1 0.9 m D C 0.9 m A B 1.8 m ABCD is a uniform lamina with AB = 1.8 m, AD = DC = 0.9 m, and AD perpendicular to AB and DC (see diagram). (i) Find the distance of the centre of mass of the lamina from AB and the distance from AD. [4] The lamina is freely suspended at A and hangs in equilibrium. (ii) Calculate the angle between AB and the vertical. [2]

Mark scheme: 1 (i) 2mx0.45 + mx0.3 = 3mv M1 Table of values idea v = 0.4m (from AB) A1 2mx0.45 + mx(0.9+0.3) = 3mh M1 Table of values idea h = 0.7m (from AD) A1 [4] (ii) tanα = 0.4/0.7 M1 α = 29.7° A1ft Accept 0.519 radians [2]

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2 A particle P is projected with speed 26 m s−1 at an angle of 30◦above the horizontal from a point O on a horizontal plane. (i) For the instant when the vertical component of the velocity of P is 5 m s−1 downwards, find the direction of motion of P and the height of P above the plane. [4] (ii) P strikes the plane at the point A. Calculate the time taken by P to travel from O to A and the distance OA. [3]

Mark scheme: 2 (i) tanα = 5/(26cos30°) M1 α = 12.5° (0.219rad) below the horizontal A1 Accept 77.5°/1.35rad with downward vertical 52 = (26sin30°)2 – 2gs M1 s = 7.2m A1 [4] (ii) –(26sin30°) = (26sin30°) – gT M1 Or time to greatest height if later doubled T = 2.6s A1 OA = (26cos30°) × 2.6 = 58.5m A1 Or B1 for OA = 262sin(2 × 30°)/10 = [3] 58.5

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3 A a° P 5 rad s–1 0.3 m B Q Particles P and Q have masses 0.8 kg and 0.4 kg respectively. P is attached to a fixed point A by a light inextensible string which is inclined at an angle α◦to the vertical. Q is attached to a fixed point B, which is vertically below A, by a light inextensible string of length 0.3 m. The string BQ is horizontal. P and Q are joined to each other by a light inextensible string which is vertical. The particles rotate in horizontal circles of radius 0.3 m about the axis through A and B with constant angular speed 5 rad s−1 (see diagram). (i) By considering the motion of Q, find the tensions in the strings PQ and BQ. [3] (ii) Find the tension in the string AP and the value of α. [5]

Mark scheme: 3 (i) TPQ = (0.4g) = 4N B1 TBQ = 0.4 × 52 × 0.3 M1 Uses F = mω2r TBQ = 3N A1 [3] (ii) Tcosα = 0.8g + 4 M1 Attempts to find either component of T Tsinα = 0.8x52x0.3 A1 Both components correct T2 = 122 + 62 M1 Or any equivalent method to find T TAP = 13.4N ( = 6 5 N) A1 α ° = tan–1 (6/12) = tan–1 (1/2) = 26.6° B1ft OR Tcosα = 0.8g + 4 M1 Attempts to find either component of T Tsinα = 0.8x52x0.3 A1 Both components correct tanα = 6/12 M1 α = 26.6 A1 TAP = 13.4N B1ft [5] GCE A LEVEL – October/November 2010 9709 53

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4 B 1.2 m A 30° A uniform rod AB has weight 15 N and length 1.2 m. The end A of the rod is in contact with a rough plane inclined at 30◦to the horizontal, and the rod is perpendicular to the plane. The rod is held in equilibrium in this position by means of a horizontal force applied at B, acting in the vertical plane containing the rod (see diagram). (i) Show that the magnitude of the force applied at B is 4.33 N, correct to 3 significant figures. [3] (ii) Find the magnitude of the frictional force exerted by the plane on the rod. [2] (iii) Given that the rod is in limiting equilibrium, calculate the coefficient of friction between the rod and the plane. [3]

Mark scheme: 4 (i) M1 Moments about A Fx1.2sin60° = 15 × 0.6cos60° A1 F = 4.33N AG A1 [3] (ii) Fcos30° + Fr = 15cos60° M1 Resolving parallel to the plane Fr = 3.75N A1 OR 15 × 0.6cos60° = 1.2Fr M1 Moments about B Fr = 3.75N A1 OR Fcos30° × 0.6 = Fr x 0.6 M1 Moments about centre of rod Fr = 3.75N A1 [2] (iii) R = 15cos30° + 4.33cos60° M1 R = 15.2 A1 R = 15.155… Accept 15.1 µ (= 3.75/15.2) = 0.247 B1ft From their F and R found but not R=W [3] λ ( 2 2 )/

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5 2.4 m A B 0.5 m P A light elastic string has natural length 2 m and modulus of elasticity λ N. The ends of the string are attached to fixed points A and B which are at the same horizontal level and 2.4 m apart. A particle P of mass 0.6 kg is attached to the mid-point of the string and hangs in equilibrium at a point 0.5 m below AB (see diagram). (i) Show that λ = 26. [4] P is projected vertically downwards from the equilibrium position, and comes to instantaneous rest at a point 0.9 m below AB. (ii) Calculate the speed of projection of P. [5] [Question 6 is printed on the next page.]

Mark scheme: 5 (i) T = λ ( 2.1 2 + 5.0 2 – 1)/1 B1 T = 0.3λ or T = 0.3x26 2xTx0.5/1.3 = 6 B1 T = 0.3λ = 7.8 M1 λ = 26 AG A1 [4] (ii) EE1 = 2x26x0.32/2x1 M1 (= 2.34) Use of EPE formula, either EE2 = 2x26( 2.1 2 + 9.0 2 – 1) 2/2x1 A1 (= 6.5) Both expressions correct M1 Conservation of energy (including KE/GPE/EPE) 0.6v2/2 + 0.6x10x(0.9 – 0.5) = 6.5 – 2.34 A1 V = 2.42ms–1 A1 [5] GCE A LEVEL – October/November 2010 9709 53

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6 2 m s–1 P 30° A particle P of mass 0.2 kg is projected with velocity 2 m s−1 upwards along a line of greatest slope on a plane inclined at 30◦to the horizontal (see diagram). Air resistance of magnitude 0.5v N opposes the motion of P, where v m s−1 is the velocity of P at time t s after projection. The coefficient of friction 1 between P and the plane is 2 √3. The particle P reaches a position of instantaneous rest when t = T. dv (i) Show that, while P is moving up the plane, dt = −2.5(3 + v). [3] (ii) Calculate T. [4] (iii) Calculate the speed of P when t = 2T. [5]

Mark scheme: 6 (i) M1 N2L with 3 force terms 0.2dv/dt = –0.5v – 0.2gsin30° – dv/dt = –2.5v – 5 – (5 3 )/(2 3 ) A1 0.2gcos30°/(2 3 ) dv/dt = –2.5(3 + v) AG A1 [3] M1 Separates variables and integrates (ii) ∫ dv /(3 + v ) = −5.2 ∫dt ln(3 + v) = –2.5t (+ c) A1 t = 0, v = 2, hence c = ln5 Or equivalent use of limits ln3 = 2.5T + ln5 M1 [ln(3 + v)] 02 = [–2.5] T0 T = 0.204 A1 T = 0.4ln(5/3) [4] (iii) 0.2dv/dt = 0.2gsin30° – 0.2gcos30°/(2 3 ) – dv/dt = 5 – 2.5v – (5 3 )/(2 3 ) M1 0.5v A1 ∫ dv/(1 - v) = 2.5 ∫dt –ln(1 – v) = 2.5t (+ c) t = 0, v = 0, hence c = 0 B1 Or equivalent –ln(1 – v) = 2.5x0.4ln(5/3) M1 Uses t = T v = 0.4ms–1 A1 [5]

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