Cambridge A Level Mathematics 9709 — 2016 Oct/Nov Paper 5 · Variant 1

1 A particle P of mass 0.3 kg moves in a circle with centre O on a smooth horizontal surface. P is attached to O by a light elastic string of modulus of elasticity 12 N and natural length l m. The speed of P is 4 m s−1, and the radius of the circle in which it moves is 2l m. Calculate l. [4]

Mark scheme: 1 T = 12 N B1 T = 12(2L–L)/L T = 0.3 x 42/r M1 Accn = v2/r 12 = 4.8/(2L) A1 ft candidates expression for T L = 0.2 A1 4

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2 0.6 m 0.6 m A B C A uniform wire is bent to form an object which has a semicircular arc with diameter AB of length 1.2 m, with a smaller semicircular arc with diameter BC of length 0.6 m. The end C of the smaller arc is at the centre of the larger arc (see diagram). The two semicircular arcs of the wire are in the same plane. (i) Show that the distance of the centre of mass of the object from the line ACB is 0.191 m, correct to 3 significant figures. [3] The object is freely suspended at A and hangs in equilibrium. (ii) Find the angle between ACB and the vertical. [4]

Mark scheme: 2 (i) CoM(large) = 0.6/(π/2) or B1 CoM(small) = 0.3/(π/2) (π x 0.6 + π x 0.3)D = M1 OR (2+1)D = 2(1.2/π) – 1(0.6/π) π x 0.6(1.2/π) – π x 0.3(0.6/π) Moments about ACB D = 0.191 m AG A1 3 (ii) (π x 0.6 + π x 0.3)H = M1 OR 3H = 2 x 0.6 + 1 x 0.9 π x 0.6 x 0.6 + π x 0.3 x 0.9 Moments about A H = 0.7 A1 tanθ = 0.191/0.7 M1 θ = 15.3° A1 4 2

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3 A small block B of mass 0.25 kg is released from rest at a point O on a smooth horizontal surface. After its release the velocity of B is v m s−1 when its displacement is x m from O. The force acting on B has magnitude 2 + 0.3x2 N and is directed horizontally away from O. (i) Show that vdv = 1.2x2 + 8. [2] dx (ii) Find the velocity of B when x = 1.5. [3] An extra force acts on B after x = 1.5. It is given that, when x > 1.5, vdv = 1.2x2 + 6 −3x. dx (iii) Find the magnitude of this extra force and state the direction in which it acts. [2]

Mark scheme: 3 (i) 0.25vdv/dx = 2 + 0.3x2 M1 vdv/dx = 1.2 x2 + 8 AG A1 2 (ii) ∫v d v = ∫ (1.2 x 2 + 8) dx M1 v2/2 = 0.4x3 + 8x ( + c) A1 Allow c = 0 without working v = 5.17 A1 3 (iii) 0.25vdv/dx = 0.3x2 + 1.5 – 0.75x M1 Force is 0.5 + 0.75x N towards O A1 2

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4 B 2a m C 0.9 m A a m D The diagram shows the cross-section ABCD through the centre of mass of a uniform solid prism. AB = 0.9 m, BC = 2a m, AD = a m and angle ABC = angle BAD = 90Å. (i) Calculate the distance of the centre of mass of the prism from AD. [2] (ii) Express the distance of the centre of mass of the prism from AB in terms of a. [2] The prism has weight 18 N and rests in equilibrium on a rough horizontal surface, with AD in contact with the surface. A horizontal force of magnitude 6 N is applied to the prism. This force acts through the centre of mass in the direction BC. (iii) Given that the prism is on the point of toppling, calculate a. [3]

Mark scheme: 4 (i) (0.9a + 0.9a/2)Y = M1 1.5Y = 1 x 0.45 + 0.5 x 0.6 0.9a x 0.45 + 0.45a x 0.9 x 2/3 Moments about AD Y = 0.5 m A1 2 (ii) (0.9a + 0.9a/2)X = M1 1.5X = 1 x a/2 + 0.5 x 4a/3 0.9a x a/2 + 0.45a x (a + a/3) X = 7a/9 A1 2 (iii) 0.5 x 6 = (a – 7a/9) x 18 M1 Ft [Yi and (a–Xii)] A1 a = 0.75 A1 3 1

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5 A small ball B of mass 0.4 kg moves in a horizontal circle with centre O and radius 0.6 m on a smooth horizontal surface. One end of a light inextensible string is attached to B; the other end of the string is attached to a fixed point 0.45 m vertically above O. (i) Given that the tension in the string is 5 N, calculate the speed of B. [3] (ii) Find the greatest possible tension in the string for the motion, and the corresponding angular speed of B. [4]

Mark scheme: 5 (i) θ(= tan–10.45/0.6 = 36.87..) = 36.9° B1 Or tanθ = 3/4 0.4v2/0.6 = 5cosθ M1 v = 2.45 ms–1 A1 3 Or 6 (ii) Tsinθ = 0.4g M1 2 T = 6.67 N A1 Accept 0.66, 6 , 20/3 3 0.4ω2 x 0.6 = 6.67cosθ M1 ω = 4.71 rad s–1 A1 4 Accept 4.72 rad s–1 2

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6 A O 0.9 m P The diagram shows a smooth narrow tube formed into a fixed vertical circle with centre O and radius 0.9 m. A light elastic string with modulus of elasticity 8 N and natural length 1.2 m has one end attached to the highest point A on the inside of the tube. The other end of the string is attached to a particle P of mass 0.2 kg. The particle is released from rest at the lowest point on the inside of the tube. By considering energy, calculate (i) the speed of P when it is at the same horizontal level as O, [4] (ii) the speed of P at the instant when the string becomes slack. [3]

Mark scheme: 6 (i) EE = 8(0.9π – 1.2)2/(2 x 1.2) B1 Initial EE = 8.83 J 8.83 = 0.2g x 0.9 + 0.2v2/2 + M1 8(0.9π/2 – 1.2)2/(2 x 1.2) A1 v = 8.29 m s–1 A1 4 (ii) θ = 1.2/0.9 = 4/3 rad (=76.4°) B1 8.83 = 0.2g x 0.9 + 0.2g x 0.9cosθ M1 0.2 x 8.292/2 = 0.2g x 0.9cosθ + 0.2v2/2 + 0.2v2/2 v = 8.13 m s–1 A1 3 2

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7 A particle P is projected with speed 35 m s−1 from a point O on a horizontal plane. In the subsequent motion, the horizontal and vertically upwards displacements of P from O are x m and y m respectively. The equation of the trajectory of P is 1 + k2 x2 y = kx − , 245 where k is a constant. P passes through the points A 14, a and B 42, 2a , where a is a constant. (i) Calculate the two possible values of k and hence show that the larger of the two possible angles of projection is 63.435Å, correct to 3 decimal places. [5] For the larger angle of projection, calculate (ii) the time after projection when P passes through A, [2] (iii) the speed and direction of motion of P when it passes through B. [4]

Mark scheme: 7 (i) a = 14k – 0.8(1 + k2) and M1 Creates 2 simultaneous equations 2a = 42k – 7.2(1 + k2) 42k – 7.2(1 + k2) = 2[14k – 0.8(1 + k2)] M1 Creates a single equation in k k = 1/2 and 2 B1 Both values θ = tan–1k M1 With 1 of the candidates value of k θ= 63.435 AG A1 5 (ii) t = 14/(35cos63.435) M1 t (= 0.89442..) = 0.894 s A1 2 (iii) Vv = 35sin63.4 – g[42/(35cos63.4)] M1 Vv = 4.495 tanα = 4.495/(35cos63.4) α = 15.9° above the horizontal A1 Accept 16(.0)° V 2 = 4.4952 + (35cos63.4)2 M1 V = 16.3 m s–1 A1 4 OR 2a = 48 M1 42 x 2 – 7.2(1 + 22) V 2 = 352 – 2g x 48 V = 16.3 m s–1 A1 cosα= 35cos63.435/16.3 M1 α = 15.9° A1 4

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