Cambridge A Level Mathematics 9709 — 2018 May/June Paper 4 · Variant 3

1 v m s−1 16 0 t (s) 0 40 600 The diagram shows the velocity-time graph for a train which travels from rest at one station to rest at the next station. The graph consists of three straight line segments. The distance between the two stations is 9040 m. (i) Find the acceleration of the train during the first 40 s. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Find the length of time for which the train is travelling at constant speed. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (iii) Find the distance travelled by the train while it is decelerating. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 1(i) Total: 1 1(ii) [ 1 2 9040 (600 ) 16 T = + × ] M1 Equating area of the trapezium to the total distance or using s = ½ (u + v)t or equivalent Time is 530 (s) A1 Total: 2 1(iii) [ 1 2 (600 530 40) 16 s = × − − × ] M1 Use of triangular area, or equivalent Distance is 240 (m) A1 Total: 2

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2 A small ball is projected vertically downwards with speed 5 m s−1 from a point A at a height of 7.2 m above horizontal ground. The ball hits the ground with speed V m s−1 and rebounds vertically upwards with speed 12V m s−1. The highest point the ball reaches after rebounding is B. Find V and hence find the total time taken for the ball to reach the ground from A and rebound to B. [5] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 [ 2 2 5 2 7.2 V g = + × × ] 13 V = A1 [13 = 5 + gt t = ….. ] 0.8 (s) M1 Use of uvast to find time for A to reach ground [0 = 6.5 – gt t = ….. ] 0.65 (s) M1 Use of uvast to find time from ground to B Total time is 1.45 (s) A1 Total: 5

More questions on Kinematics of motion in a straight line

3 8 N 12 N 30Å 60Å 18 N Coplanar forces of magnitudes 8 N, 12 N and 18 N act at a point in the directions shown in the diagram. Find the magnitude and direction of the single additional force acting at the same point which will produce equilibrium. [6] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 3 M1 For resolving forces in any one direction E.g. 18 12sin60 8sin30 X = + ° − ° 14 + 6√3 A1 One correct equation or expression E.g. 8cos30 12cos60 Y = ° + ° 6 + 4√3 A1 Second correct equation or expression (X and Y may denote components of resultant of given 3 forces or may be components of the fourth force that would produce equilibrium) [(14 + 6√3)2 + (6 + 4√3)2] or [tan–1 (6 + 4√3)/( 14 + 6√3)] M1 Use of Pythagoras or appropriate trig to find magnitude or angle Magnitude is 27.6 (N) A1 Not for resultant Direction is 27.9° below ‘negative x-axis’ A1 Not for 27.9° only; direction must be clearly specified Total: 6

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4 P A 1.6 kg B 0.8 kg 1 Two particles A and B, of masses 0.8 kg and 1.6 kg respectively, are connected by a light inextensible string. Particle A is placed on a smooth plane inclined at an angle 1 to the horizontal, where sin 1 = 35. The string passes over a small smooth pulley P fixed at the top of the plane, and B hangs freely (see diagram). The section AP of the string is parallel to a line of greatest slope of the plane. The particles are released from rest with both sections of the string taut. Use an energy method to find the speed of the particles after each particle has moved a distance of 0.5 m, assuming that A has not yet reached the pulley. 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Mark scheme: 4 [ 1 2 × 0.8 × v2] or [ 1 2 × 1.6 × v2] Gain in KE 2 2 1 1 2 2 0.8 1.6 v v = × × + × × A1 Total KE [Gain in PEA= 0.8 g × 0.5 × sinθ ] or [Loss in PEB= 1.6 g × 0.5] M1 For PE change of either particle (irrespective of sign) Loss in PE = 1.6 g × 0.5 – 0.8 g × 0.5 × 0.6 A1 Change of PE [ 2 1.2 8 2.4 v = − ] M1 Energy equation originating from 4 terms Speed is 2.16 (m s–1) A1 Total: 6 SC for using Newton II equations and v2 = u2 + 2as (max 2/6) [16 – T = 1.6a and T – 8sinθ = 0.8a] → a = 4.67 (ms–2) B1 [v2 = 2 × 14 3 × 0.5] → speed is 2.16 (ms–1) B1 Alternative method 1 for Question 4 [ 1 2 × 0.8 × v2] or [0.8 g × 0.5 × sinθ ] M1 For KE gain or PE gain of particle A 1 2 × 0.8 × v2 + 0.8 g × 0.5 × 0.6 A1 Total energy gain for particle A [16 – T =1.6a and T – 8sinθ = 0.8a → T = ....] 8.53 M1 Forms and solves Newton II equations to find tension T WDT = 128 15 × 0.5 A1 Finds WDTension [ 1 2 × 0.8 × v2 + 0.8 g × 0.5 × 0.6 = 128 15 × 0.5] M1 Energy equation (3 terms) Question Answer Marks Guidance 4 Speed is 2.16 (m s–1) A1 Total: 6 Alternative method 2 for Question 4 [ 1 2 × 1.6 × v2] or [1.6 g × 0.5] M1 For KE gain or PE loss of particle B 1.6 g × 0.5 – 1 2 × 1.6 × v2 A1 Energy change for particle B [16 – T = 1.6a and T – 8sinθ = 0.8a → T = ....] 8.53 M1 Forms and solves Newton II equations to find tension T WDT = 128 15 × 0.5 A1 Finds WDTension 1.6 g × 0.5 – 1 2 × 1.6 × v2 = 128 15 × 0.5] M1 Energy equation (3 terms) Speed is 2.16 (m s–1) A1 Total: 6

More questions on Kinematics of motion in a straight line

5 A particle of mass 3 kg is on a rough plane inclined at an angle of 20Å to the horizontal. A force of magnitude P N acting parallel to a line of greatest slope of the plane is used to keep the particle in equilibrium. The coefficient of friction between the particle and the plane is 0.35. Show that the least possible value of P is 0.394, correct to 3 significant figures, and find the greatest possible value of P. 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Mark scheme: 5 [ 0.35 3 cos20 F g = × ° ] M1 For use of F R µ = [ 1 3 sin 20 P F g + = ° ] M1 Attempted resolving equation for minimum case 1 0.394 P = (AG) A1 Correct given answer from correct work [ 2 3 sin 20 P F g = + ° ] M1 Attempted resolving equation for maximum case 2 20.1(N) = P A1 Total: 6

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6 A car of mass 1400 kg travelling at a speed of v m s−1 experiences a resistive force of magnitude 40v N. The greatest possible constant speed of the car along a straight level road is 56 m s−1. (i) Find, in kW, the greatest possible power of the car’s engine. 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(ii) Find the greatest possible acceleration of the car at an instant when its speed on a straight level road is 32 m s−1. 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(iii) The car travels down a hill inclined at an angle of 1Å to the horizontal at a constant speed of 50 m s−1. The power of the car’s engine is 60 kW. Find the value of 1. 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Mark scheme: 6(i) [ 40 56 56 × ] M1 For equating Power Velocity to Resistance, or equivalent Power is 125 (kW) A1 Total: 2 6(ii) Driving force is 125 440 32 B1ft Follow through their power from (i) [ 125 440 40 32 1400 32 a − × = ] M1 For 3-term Newton II equation 2 1.89 (m s ) − = a A1 Total: 3 Question Answer Marks Guidance 6(iii) [ 60 000 1400 sin 40 50 0 50 g θ + − × = ] M1 For 3-term Newton II equation A1 Correct equation [ 800 sin 14 000 θ° = ] M1 3.3 θ = A1 Total: 4

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7 A particle P moves in a straight line starting from a point O. The velocity v m s−1 of P at time t s is given by v = 12t −4t2 for 0 ≤t ≤2, v = 16 −4t for 2 ≤t ≤4. (i) Find the maximum velocity of P during the first 2 s. 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(ii) Determine, with justification, whether there is any instantaneous change in the acceleration of P when t = 2. 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(iii) Sketch the velocity-time graph for 0 ≤t ≤4. [3] v (m s−1) t (s) 0 2 4 (iv) Find the distance travelled by P in the interval 0 ≤t ≤4. 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Mark scheme: 7(i) [ d 12 8 d v t t = − ] or e.g. [–4[(t – 1.5)2 – 2.25]] M1 For attempted differentiation of 2 12 4 t t − (or for alternative e.g. completing the square) [Maximum v when 2 1.5 12 1.5 4 1.5 t v = ⇒ = × −× ] M1 For finding and using t Maximum velocity is 9 (m s–1) A1 Total: 3 7(ii) [ d 12 8 d v t t = − = –4] M1 Finding acceleration for 0 ⩽ t ⩽ 2 when t = 2 Acceleration for 2 ⩽ t ⩽ 4 is –4 No instantaneous change A1 Both values correct, with correct statement Total: 2 Question Answer Marks Guidance 7(iii) B1 Quadratic shape (with max) for 0 ⩽ t ⩽ 2 B1 Line with negative gradient from (2, …) to (4,0) B1 All correct, smooth join and key values indicated Total: 3 7(iv) Area of triangle is 8 B1 (May be obtained by integrating 16 – 4t or use of uvast) [ 2 2 3 4 3 (12 4 ) d 6 t t t t t − = − ∫ ] M1 Integration attempt for 0 ⩽ t ⩽ 2 [ 2 3 2 3 4 4 3 3 6 2 2 6 0 0 × − × − × + × ] DM1 Use of limits 0 and 2; condone absence of zero terms Area under curve is 40 3 or 13.3 A1 Distance travelled is 64 (m) 3 or 21.3 (m) A1 Total: 5

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