Cambridge A Level Mathematics 9709 — 2018 Oct/Nov Paper 4 · Variant 1

1 A particle of mass 0.2 kg moving in a straight line experiences a constant resistance force of 1.5 N. When the particle is moving at speed 2.5 m s−1, a constant force of magnitude F N is applied to it in the direction in which it is moving. Given that the speed of the particle 5 seconds later is 4.5 m s−1, find the value of F. [4] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 a = 0.4 A1 F – 1.5 = 0.2a M1 For use of Newton’s second law F = 1.58 A1 4

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2 A high-speed train of mass 490 000 kg is moving along a straight horizontal track at a constant speed of 85 m s−1. The engines are supplying 4080 kW of power. (i) Show that the resistance force is 48 000 N. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) The train comes to a hill inclined at an angle 1Å above the horizontal, where sin 1Å = 200.1 Given that the resistance force is unchanged, find the power required for the train to keep moving at the same constant speed of 85 m s−1. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2(i) Resistance = Driving force = 4080000 85 = 48 000 N 1 2(ii) DF = 85 P B1 DF = P v DF – 48 000 – 490 000 g × 1 200 = 0 M1 For applying Newton’s second law (3 terms) P = 72 500 × 85 = 6.16 MW A1 3

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3 A van of mass 2500 kg descends a hill of length 0.4 km inclined at 4Å to the horizontal. There is a constant resistance to motion of 600 N and the speed of the van increases from 20 m s−1 to 30 m s−1 as it descends the hill. Find the work done by the van’s engine as it descends the hill. [5] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 3 [KE gained ( )( ) 2 2 1 2500 30 20 625000J 2 = × × − = PE lost ( ) 2500 400sin4 697564.7J = × = g A1 Both KE and PE correct [WD by engine 2 1 2500 400sin 4 2500 20 2 + × + × × g 2 1 600 400 2500 30 2 = × + × × ] M1 Using work-energy equation in the form WD by engine + PE lost = WD against F + KE gain Work done by engine + PE lost = 600 × 400 + 625 000 A1 Work-energy equation all correct Work done = 167 000 J (167 435.2…) A1 5

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4 Two particles A and B, of masses m kg and 0.3 kg respectively, are attached to the ends of a light inextensible string. The string passes over a fixed smooth pulley and the particles hang freely below it. The system is released from rest, with both particles 0.8 m above horizontal ground. Particle A reaches the ground with a speed of 0.6 m s−1. (i) Find the tension in the string during the motion before A reaches the ground. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Find the value of m. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(i) a = 0.225 A1 T – 0.3 g = 0.3a M1 For using Newton’s second law for the 0.3 kg particle T = 3.07 N (3.0675 N) A1 4 Question Answer Marks Guidance 4(ii) mg – T = ma, m(10 – 0.225) = 3.0675 M1 For using Newton’s second law applied to the m kg particle m = 0.314 kg (0.31381…) A1 2

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5 25 N 15 N 30Å 40Å A B C 30 N Coplanar forces, of magnitudes 15 N, 25 N and 30 N, act at a point B on the line ABC in the directions shown in the diagram. (i) Find the magnitude and direction of the resultant force. [6] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) The force of magnitude 15 N is now replaced by a force of magnitude F N acting in the same direction. The new resultant force has zero component in the direction BC. Find the value of F, and find also the magnitude and direction of the new resultant force. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 5(i) M1 For resolving forces horizontally or vertically o.e. 25 cos 30 – 15 cos 40 (= 10.1599…) A1 25 sin 30 + 15 sin 40 – 30 (= –7.8581…) A1 M1 For using a method for either magnitude or direction Magnitude = ( ) 2 2 10.15... 7.858... + = 12.8 N A1 Magnitude = 12.844… Angle 37.7° below the horizontal in the direction BA A1 6 Question Answer Marks Guidance 5(ii) F cos 40 = 25 cos 30 M1 For equating forces in the direction BC to zero F = 28.3 A1 F = 28.2628… New resultant force = 28.26…sin 40 + 25 sin 30 – 30 = 0.667 N upwards B1 3

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6 A particle is projected from a point P with initial speed u m s−1 up a line of greatest slope PQR of a rough inclined plane. The distances PQ and QR are both equal to 0.8 m. The particle takes 0.6 s to travel from P to Q and 1 s to travel from Q to R. (i) Show that the deceleration of the particle is 2 m s−2 and hence find u, giving your answer as an 3 exact fraction. [6] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Given that the plane is inclined at 3Å to the horizontal, find the value of the coefficient of friction between the particle and the plane. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 6(i) M1 For using constant acceleration equations such as 2 1 2 = + s ut at or equivalent complete methods to find expressions for PQ or QR or PR For PQ 0.8 = 0.6u + 0.18a A1 For PR 1.6 = 1.6u + 1.28a A1 or for QR 0.8 = (u + a × 0.6) × 1 + 0.5a M1 Solving simultaneously two relevant equations in u and a Deceleration = 2 3 ms–2 A1 AG 23 15 = u B1 6 Question Answer Marks Guidance 6(ii) R = mg cos 3 B1 F = µmg cos 3 M1 For use of F = µR 2 sin3 cos3 3 µ   − − × = × −     mg mg m M1 For using Newton’s second law (3 terms) µ = 0.0144 (0.014350…) A1 4

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7 A particle moves in a straight line starting from rest from a point O. The acceleration of the particle at time t s after leaving O is a m s−2, where a = 5.4 −1.62t. (i) Find the positive value of t at which the velocity of the particle is zero, giving your answer as an exact fraction. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Find the velocity of the particle at t = 10 and sketch the velocity-time graph for the first ten seconds of the motion. [3] ........................................................................................................................................................ ........................................................................................................................................................ (iii) Find the total distance travelled during the first ten seconds of the motion. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 7(i) 5.4 1.62 d = ∫ − v t t ( ) 2 5.4 0.81 = − + v t t C A1 2 5.4 0.81 0 − = t t M1 For solving v = 0 2 20 6 3 3 = = t s A1 4 7(ii) v(10) = –27 ms–1 B1 Inverted parabola B1 v = 0 at t = 0, negative at t = 10 and through 2 6 ,0 3       B1 3 Question Answer Marks Guidance 7(iii) ( ) 2 5.4 0.81 = ∫ − s t t dt M1 For using integration of v to find s ( ) 2 3 2.7 0.27 = − + s t t C A1 At t = 6 2 3 , displacement = 40 M1 For evaluating the integral at the time when v = 0 At t = 10 displacement = 0 M1 For evaluating the integral at time t = 10 Total distance = 80 m A1 5

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