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Cambridge AS & A Level Mathematics (9709), Paper 5, October/November 2019. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
9709_w19_qp_539709_w19_ms_539709/53/O/N/19
Each question as it's printed, and the mark scheme's points for it, as text. Diagrams, tables and graphs are in the PDFs above.
Question
B 3N 0.1 m A A uniform solid cone has weight 5 N and base radius 0.1 m. AB is a diameter of the base of the cone. The cone is held in equilibrium, with A in contact with a rough horizontal surface and AB vertical, by a force applied at B. This force has magnitude 3 N and acts parallel to the axis of the cone (see diagram). Calculate the height of the cone. [3]
Mark scheme
1 h B1 Conservation of momentum at 4 5× h M1 Take moments about A = 3 × 0.2 4 (h = ) 0.48 m A1 3
Question
A particle is projected from a point on horizontal ground with speed 15 m s−1 at an angle of 1Å above the horizontal. The particle strikes the ground 2 s after projection. (i) Find 1. [2] (ii) Calculate the time after projection at which the direction of motion of the particle is 20Å below the horizontal. [4]
Mark scheme
2(i) –15sinθ = 15sinθ – 2g M1 Use v = u + at vertically (θ =) 41.8 A1 2 2(ii) v M1 v = vertical velocity Vertically: = ± tan 20 15cosθ v = ( ± ) 4.07 A1 –4.07 = 15sin41.8 – gt M1 Use v = u + at vertically (t =) 1.41 s A1 4
Question
A smooth horizontal surface has two fixed points O and A which are 0.8 m apart. A particle P of mass 0.25 kg is projected with velocity 3 m s−1 horizontally from A in the direction away from O. The velocity of P is v m s−1 when the displacement of P from O is x m. A force of magnitude kv2 x−2 N opposes the motion of P. dv (i) Show that v = −4kv2 x−2 . [1] dx (ii) Express v in terms of k and x. [5]
Mark scheme
3(i) dv dv B1 AG 0.25v = − kv 2 x −2 → v = −4kv 2 x −2 dx dx 1 3(ii) dv M1 Attempt to integrate ∫ v = −4k ∫ x dx −2 4k A1 ln v = (+c) x x = 0.8, v = 3 hence c = ln3 – 5k A1 Finds c 4k M1 ln v = + ln 3 − 5k x 4k −5 k A1 v=3 x 5
Question
A small ball B is projected with speed 30 m s−1 at an angle of 60Å above the horizontal from a point O. At time t s after projection the horizontal and vertically upwards displacements of B from O are x m and y m respectively. (i) Express x and y in terms of t and hence find the equation of the trajectory of the ball. [4] (ii) Find the value of x for which OB makes an angle of 45Å above the horizontal. [3]
Mark scheme
4(i) x = 30cos60t B1 Use horizontal motion gt 2 B1 gt 2 y = 30sin60t – Use s = ut + vertically 2 2 30sin 60 x 5x2 M1 Attempt to eliminate t y= − 30cos60 (30cos 60)2 x2 A1 y = 1.73x – 0.0222 x2 or y = 3 x − 45 4 4(ii) y M1 x = y or tan 45 = x x M1 x common to all three terms 1 = 1.73 – 0.0222x or 1 = 3 − 45 x = 32.9 A1 3
Question
A particle P of mass 0.3 kg is attached to one end of a light elastic string of natural length 0.6 m and modulus of elasticity 9 N. The other end of the string is attached to a fixed point O on a smooth plane inclined at 30Å to the horizontal. OA is a line of greatest slope of the plane with A below the level of O and OA = 0.8 m. The particle P is released from rest at A. (i) Find the initial acceleration of P. [4] (ii) Find the greatest speed of P. [5]
Mark scheme
5(i) 9 × (0.8 − 0.6) M1 λx 0.4 T= Use T = . Note OP = 0.6 l sin 30 T=3N A1 0.3a = 3 – 0.3gsin30 M1 Use Newton’s Second Law along the slope a = 5 m s −1 A1 4 5(ii) 9e M1 Note the maximum speed is at the equilibrium position 0.3 g sin 30 = 0.6 e = 0.1 A1 9 × (0.8 − 0.6) 2 9 × 0.12 B1 EPE = or 2 × 0.6 2 × 0.6 0.3v 2 9 × (0.8 − 0.6) 2 9 × 0.12 M1 Set up a 4 term energy equation = − − 0.3 g × 0.1sin 30 2 2 × 0.6 2 × 0.6 v = 0.707 m s−1 A1 5
Question
A and B are two fixed points on a vertical axis with A 0.6 m above B. A particle P of mass 0.3 kg is attached to A by a light inextensible string of length 0.5 m. The particle P is attached to B by a light elastic string with modulus of elasticity 46 N. The particle P moves with constant angular speed 8 rad s−1 in a horizontal circle with centre at the mid-point of AB. (i) Find the speed of P. [2] (ii) Calculate the tension in the string BP and hence find the natural length of this string. [7]
Mark scheme
6(i) 0.32 + r 2 = 0.52 hence r = 0.4 B1 Use Pythagoras’s theorem 8 × 0.4 = 3.2 m s−1 B1 Use v = rω 2 6(ii) 3 3 B1 Resolve vertically A × − B × = 0.3 g 5 5 4 4 0.3 × 3.22 M1A1 Use Newton’s Second Law horizontally A × +B × = 0.3 × 82 × 0.4 or 5 5 0.4 M1 Attempt to solve for B B = 2.3 N A1 46(0.5 − L ) M1 λx 2.3 = Use T = and attempt to solve L l 10 A1 L = 0.476 m or 21 7
Question
C 0.9 m A 0.4 m B ABC is the cross-section through the centre of mass of a uniform prism which rests with AB on a rough horizontal surface. AB = 0.4 m and C is 0.9 m above the surface (see diagram). The prism is on the point of toppling about its edge through B. (i) Show that angle BAC = 48.4Å, correct to 3 significant figures. [3] A force of magnitude 18 N acting in the plane of the cross-section and perpendicular to AC is now applied to the prism at C. The prism is on the point of rotating about its edge through A. (ii) Calculate the weight of the prism. [3] (iii) Given also that the prism is on the point of slipping, calculate the coefficient of friction between the prism and the surface. [4] Additional Page If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
Mark scheme
7(i) BG = 0.3 m B1 G is the CoM vertically above B. M is the mid-point of AB and E is v the point vertically below C on AB extended. CE 0.9 M1 Use of similar triangles and trigonometry of a right angled ME = 3 × 0.2 = 0.6 and tan A = = triangle AE 0.8 A = 48.4° A1 AG 3 7(ii) 0.9 B1 Use trigonometry of a right angled triangle AC = = 1.20(41...) sin 48.4 18 × 1.2041 = 0.4W M1 Moments about A W = 54.2 N A1 3 7(iii) H = 18sinA = 18sin48.4 (= 13.46) B1 Resolve horizontally V = 54.2 – 18cos48.4 (= 42.25) B1ft Resolve vertically 13.46 M1 Use F = µR µ= 42.25 µ = 0.319 A1 Accept 0.32 4
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Mathematics 9709/53 for October/November 2019 is out of 50 marks, with 1 hour 15 minutes to answer it.
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