Cambridge A Level Mathematics 9709 — 2019 Feb/March Paper 5 · Variant 2

1 A particle is projected with speed 24 m s−1 at an angle of 30Å above the horizontal. Find the speed and direction of motion of the particle at the instant 4 s after projection. [5] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 ′x = 24cos30 ( = 12 3 ) B1 Use horizontal motion ′y = 24sin30 – 4g ( = –28) B1 Use vertical motion 2 V = ( ) 2 24 30 cos + ( ) 2 24 30 4 − sin g = (12 3 )2 + (–28)2 OR tanα = (24sin30 – 4g)/(24cos30) = –28/(12 3 )2 M1 Where V is the required speed and α is the angle below the horizontal V = 34.9 m s–1 A1 α = 53.4° below the horizontal A1 5

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2 1 m 2 m 3 m A uniform object is made by joining together three solid cubes with edges 3 m, 2 m and 1 m. The object has an axis of symmetry, with the cubes stacked vertically and the cube of edge 2 m between the other two cubes (see diagram). (i) Calculate the distance of the centre of mass of the object above the base of the largest cube. 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The smallest cube is now removed from the object. It is replaced by a heavier uniform cube with 1 m edges which is made of a different material. The centre of mass of the object is now at the base of the 2 m cube. (ii) Find the ratio of the masses of the two cubes of edge 1 m. 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Mark scheme: 2(i) Total volume (= 27 + 8 + 1) = 36 B1 36x = 27×1.5 + 8×4 + 1×5.5 M1 Take moments about base of largest cube x ( = 13/6 ) = 2.17 m A1 3 2(ii) Mass of new cube = 35 + m B1 Where m is the mass of the new cube (35 + m) × 3 =27×1.5 + 8×4 + 5.5m (leads to m = 13) M1 Take moments about base of largest cube 13:1 or 1:13 A1 Accept 13 3

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3 A small ball is projected from a point O on horizontal ground. At time t s after projection the horizontal and vertically upwards displacements of the ball from O are x m and y m respectively, where x = 4t and y = 6t −5t2. (i) Find the equation of the trajectory of the ball. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Hence or otherwise calculate the angle of projection of the ball and its initial speed. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 3(i) x = 4t and y = 6t – 5 2t y [= 6x/4 – 5 2 ( / 4) x ] = 1.5x –5 2 x /16 or 1.5x – 0.3125 2 x A1 2 3(ii) tanθ = 1.5 M1 Use the trajectory equation from the formula sheet θ = 56.3° A1 2 2 V cos 56.3 =16 M1 Again use the trajectory equation V = 7.21 m s–1 A1 OR Vcosθ = 4 and Vsinθ =6 M1 Initial horizontal and vertical velocities 2 2 V cos θ + 2 2 2 4 θ = V sin + 2 6 OR tanθ = 6/4 M1 Use Pythagoras's theorem or trigonometry of a right angled triangle V = 7.21 m s–1 A1 θ = 56.3° A1 4

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4 A 60Å O P A particle P of mass 0.3 kg is attached to a fixed point A by a light elastic string of natural length 0.8 m and modulus of elasticity 16 N. The particle P moves in a horizontal circle which has centre O. It is given that AO is vertical and that angle OAP is 60Å (see diagram). Calculate the speed of P. [6] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 4 Tcos60 = 0.3g M1 Resolve vertically T = 6 N A1 T = 16e/0.8 ( = 6 ) leads to e = 0.3 M1 Use T =λx/L r = (0.8 + 0.3)sin60 ( = 1.1sin60) A1 Tsin60 = 0.3 2 v /(1.1sin60) M1 Use N2L horizontally v = 4.06 m s–1 A1 6

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5 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 24 N. The other end of the string is attached to a fixed point O. The particle P is released from rest at the point 0.4 m vertically below O. (i) Find the greatest speed of P. 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(ii) Calculate the greatest distance of P below O. 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Mark scheme: 5(i) 0.3g = 24e/0.6 M1 Note greatest speed occurs at the equilibrium position. Use T=λx/L e = 0.075 m A1 Fall = 0.275 m PE Change = 0.3g × 0.275 B1 0.3 2 v /2 = 0.3g × 0.275 – 24 × 2 0.075 /(2 × 0.6) M1 Set up a 3 term energy equation v = 2.18 m s–1 A1 5 Question Answer Marks Guidance 5(ii) 0.3g(0.2 + E) = 24 2 E /(2 × 0.6) M1 Set up an energy equation. Note v = 0 at the greatest distance 20 2 E – 3E – 0.6 = 0 M1 Attempt to solve a 3 term quadratic equation E = 0.264 and so greatest distance is 0.864 m A1 3

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6 2r 5r 2r Fig. 1 Fig. 1 shows the cross-section of a solid cylinder through which a cylindrical hole has been drilled to make a uniform prism. The radius of the cylinder is 5r and the radius of the hole is r. The centre of the hole is a distance 2r from the centre of the cylinder. (i) Find, in terms of r, the distance of the centre of mass of the prism from the centre of the cylinder. 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P N 30Å Fig. 2 The prism has weight W N and is placed with its curved surface on a rough horizontal plane. The axis of symmetry of the cross-section makes an angle of 30Å with the vertical. A horizontal force of magnitude P N acting in the plane of the cross-section through the centre of mass is applied to the cylinder at the highest point of this cross-section (see Fig. 2). The prism rests in limiting equilibrium. (ii) Find the coefficient of friction between the prism and the plane. 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Mark scheme: 6(i) Area of hole = π 2 r and Area of original circle = 25π 2 r Area of cross-section = 24π 2r A1 2 πr (2r) = 24π 2r (d) M1 Take moments about the centre of the cylinder d = r/12 ( = 0.083333....r) A1 4 6(ii) P(2 × 5r) = W(r/12)cos60 M1 Take moments about the point of contact with the plane P = Wcos60/120 = W/240 = 0.00417W ( = F ) A1 µ = (Wcos60/120)/W M1 Use F = µR Note R = W by resolving vertically µ = 1/240 = 0.00417 A1 4

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7 A particle P is projected horizontally from a point O on a rough horizontal surface. The coefficient of friction between the particle and the surface is 0.2. A horizontal force of magnitude 0.06t N directed away from O acts on P, where t s is the time after projection. P comes to rest when t = 4. (i) The particle begins to move again when t = 8. Show that the mass of P is 0.24 kg. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ dv (ii) Show that, for 0 ≤t ≤4, = 0.25t −2, and find the speed of projection of P. [5] dt ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (iii) Find the distance from O at which P comes to rest. 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Mark scheme: 7(i) 0.2mg = 0.06 × 8 M1 Resolve along the plane m = 0.24 kg AG A1 2 7(ii) m ୢ௩ ୢ௧ = 0.06t – 0.2mg or 0.24 ୢ௩ ୢ௧ = 0.06t – 0.2 × 0.24g M1 Use N2L along the plane ୢ௩ ୢ௧ = 0.25t – 2 AG A1 dv ∫ = ( ) 0.25 2 d t t ∫ − M1 Attempt to integrate v = 0.25 2 / 2 t – 2t + c , Put v = 0 and t = 4 ( leads to c = 6 ) M1 Attempt to find c Initial velocity = 6 m s–1 A1 5 7(iii) x = (∫0.25 2t /2 – 2t + 6)dt M1 Attempt to integrate x = 0.25 3t /6 – 2t + 6t ( + k ) A1ft ft candidates c from part (ii) Finds or assumes k = 0 and substitutes t = 4 OR uses limits of 0 and 4 M1 OP = 32/3 = 10 2 3 = 10.7 m A1 4

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