Cambridge A Level Mathematics 9709 — 2019 May/June Paper 4 · Variant 1

1 78 N 50 N ! 1 112 N Given that tan ! = 12 and tan 1 = 43, show that the coplanar forces shown in the diagram are in 5 equilibrium. [3] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 (Y =) 78 × 12/13 + 50 × 4/5 – 112 = 78 sin 67.4 + 50 sin 53.1 – 112 or vertically (3 terms) [X = 30 – 30 = 0 Y = 72 + 40 – 112 = 0] A1 Correct expressions horizontally and vertically X = 0 and Y = 0 A1 From convincing exact calculations Alternative method for question 1 112 50 78 sin59.5 sin157.4 sin143.1 = = M1 Attempt to use Lami, one pair of terms A1 All terms correct 112 50 78 130 56 / 65 5 /13 3 / 5 = = = A1 Exact values seen and used and shown to be = 130 cos [180 – (θ + α)] = 33/65 and sin [180 – (θ + α)] = 56/65 3

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2 A particle P is projected vertically upwards with speed 25 m s−1 from a point 3 m above horizontal ground. (i) Find the time taken for P to reach its greatest height. 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(ii) Find the length of time for which P is higher than 23 m above the ground. 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(iii) P is higher than h m above the ground for 1 second. Find h. 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Mark scheme: 2(i) [0 = 25 – 10t] or other complete method for finding t to highest point t = 2.5 A1 3 Question Answer Mark Guidance 2(ii) [20 = 25t – ½ gt2] M1 Applying s = ut + ½at2 with s = 20, u = 25 [t = 1 and t = 4] M1 Solve a 3-term quadratic for t, factorising or formula Required time = 4 – 1 = 3 seconds A1 Alternative method for question 2(ii) [v2 = 252 + 2 × (-10) × 20 → v = ± 15] M1 Using v2 = u2 + 2as with u = 25, s = 20 and a = –g [-15 = 15 – 10T] or equivalent M1 Use v at s = 20 to find the time, T, taken to reach the maximum height and to return to s = 20 Required time = 1.5 + 1.5 = 3 seconds A1 3 2(iii) Max height reached at 2.5 s, hence reaches h after 2 s h – 3 = 25 × 2 – 5 × 22 M1 Using their t from 2(i) – 0.5 in s = ut + ½at2 Allow finding h without taking note of the additional 3 m h = 33 m A1 Alternative method for question 2(iii) Maximum height = ½ × (25 + 0) × 2.5 [= 31.25] o.e. In 0.5 s it falls distance ½ × 10 × 0.52 [= 1.25] M1 For attempting to find both the maximum height and the distance fallen in 0.5 seconds h = 31.25 – 1.25 + 3 = 33 m A1 2

More questions on Kinematics of motion in a straight line

3 A lorry has mass 12 000 kg. (i) The lorry moves at a constant speed of 5 m s−1 up a hill inclined at an angle of 1 to the horizontal, where sin 1 = 0.08. At this speed, the magnitude of the resistance to motion on the lorry is 1500 N. Show that the power of the lorry’s engine is 55.5 kW. 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When the speed of the lorry is v m s−1 the magnitude of the resistance to motion is kv2 N, where k is a constant. (ii) Show that k = 60. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (iii) The lorry now moves at a constant speed on a straight level road. Given that its engine is still working at 55.5 kW, find the lorry’s speed. 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Mark scheme: 3(i) Power = DF × 5 M1 Using P = Fv (their 2 term DF × 5) Power = 11 100 × 5 = 55.5 kW A1 AG 3 3(ii) k × 52 = 1500, k = 60 B1 AG 1 3(iii) DF = 60v2 B1 Using DF = resistance = 60v2 55500 = DF × v = 60v2 × v = 60v3 M1 P = Fv used and attempt to solve a 2-term cubic equation for v v = 9.74 ms-1 A1 3

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4 A particle of mass 1.3 kg rests on a rough plane inclined at an angle 1 to the horizontal, where tan 1 = 12 . The coefficient of friction between the particle and the plane is -. 5 (i) A force of magnitude 20 N parallel to a line of greatest slope of the plane is applied to the particle and the particle is on the point of moving up the plane. Show that - = 1.6. 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The force of magnitude 20 N is now removed. (ii) Find the acceleration of the particle. 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(iii) Find the work done against friction during the first 2 s of motion. 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Mark scheme: 4(i) R = 13 cos 67.4 = 13 (5/13) [R = 5] B1 Resolve forces perpendicular to plane. Allow 67.4 used F + 13 sin 67.4 = F + 13(12/13) = 20 [F = 8] B1 Resolve forces parallel to plane. Allow 67.4 used M1 Use F = µR µ = 8/5 = 1.6 A1 AG Must be from exact working here 4 Question Answer Mark Guidance 4(ii) 13 sin 67.4 – F = 1.3a F = µR = 8 → [4 = 1.3a] M1 For applying Newton’s second law along the plane and also using F = µR (3 terms) a = 3.08 ms-2 A1 Allow a = 40/13 2 4(iii) s = 0 + 0.5 × (40/13) × 22 [= 80/13 = 6.15] M1 Use s = ut + ½at2 with u = 0 and their a ≠ ±g to find the distance moved in the first 2 seconds WD = 8 × 6.15 M1 WD = F × d WD = 49.2 J A1 Allow WD = 640/13 J Alternative method for question 4(iii) s = 0 + 0.5 × (40/13) × 22 [= 80/13 = 6.15] M1 [v = (40/13) × 2] and [WD = 1.3g(80/13)(12/13) – ½ × 1.3 × (80/13)2] M1 Finding v after 2 seconds and using WD = PE loss – KE gain WD = 49.2 J A1 Allow WD = 640/13 J 3

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5 A particle P moves in a straight line from a fixed point O. The velocity v m s−1 of P at time t s is given by v = t2 −8t + 12 for 0 ≤t ≤8. (i) Find the minimum velocity of P. 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(ii) Find the total distance travelled by P in the interval 0 ≤t ≤8. 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Mark scheme: 5(i) a = 2t - 8 M1 Differentiate to find a a = 0 → t = 4 M1 Set a = 0 and solve for t Minimum v = –4 ms-1 A1 Full marks available for correct use of a v-t graph or correct use of “t = -b/2a” Alternative method for question 5(i) v = (t – 4)2 – 4 M1 Attempt to complete the square for v [t = 4] M1 Choose the t value which gives minimum v Minimum v = –4 ms–1 A1 3 5(ii) v = 0 when (t – 2)(t – 6) = 0 M1 Find values of t when v = 0, factorise or formula t = 2 or t = 6 A1 [s = ⅓ t3 – 4t2 + 12t (+c)] M1 Integrate v to find s A1 Correct integration 0 ≤ t ≤ 2 s1 = 8/3 – 16 + 24 = 32/3 2 ≤ t ≤ 6 s2 = (216/3 – 144 + 72) – (8/3 – 16 + 24)= -32/3 6 ≤ t ≤ 8 s3 = (512/3 – 4 × 82 + 12 × 8) – (216/3 – 144 + 72) = 32/3 M1 Attempt to find s1, s2 and s3 Look for consideration of the need for 3 intervals Allow use of symmetry when finding s1, and s3 A1 2 correct values of displacement Total distance = 32 m A1 All correct 7

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6 P A 0.2 kg B 0.4 kg 0.5 m 1Å Two particles A and B, of masses 0.4 kg and 0.2 kg respectively, are connected by a light inextensible string. Particle A is held on a smooth plane inclined at an angle of 1Å to the horizontal. The string passes over a small smooth pulley P fixed at the top of the plane, and B hangs freely 0.5 m above horizontal ground (see diagram). The particles are released from rest with both sections of the string taut. (i) Given that the system is in equilibrium, find 1. 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(ii) It is given instead that 1 = 20. In the subsequent motion particle A does not reach P and B remains at rest after reaching the ground. (a) Find the tension in the string and the acceleration of the system. 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(b) Find the speed of A at the instant B reaches the ground. 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[Question 6 continues on the next page.] (c) Use an energy method to find the total distance A moves up the plane before coming to instantaneous rest. 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Mark scheme: 6(i) Particle A: T = 4 sin θ Particle B: T = 2 M1 Eliminate T and solve for θ θ = 30 A1 3 6(ii)(a) A: T – 4 sin 20 = 0.4a B: 2 – T = 0.2a System: 2 – 4 sin 20 = (0.4 + 0.2)a M1 Apply Newton’s second law to A or to B or to the system A1 Two correct equations M1 Solve for a or T T = 1.79 and a = 1.05 A1 Both correct 4 6(ii)(b) v2 = 2 × 1.053 × 0.5 = 1.053 M1 Attempt to find v using their a ≠ ±g v = 1.03 ms–1 A1 2 Question Answer Mark Guidance 6(ii)(c) Loss in KE = ½ × 0.4 × 1.053 = 0.2106 Gain in PE = 0.4 × 10 × d sin 20 M1 Attempt KE loss or PE gain for particle A only after particle B hits the ground. A1ft Both correct, d is distance moved up the plane after B hits ground ½ × 0.4 × 1.053 = 0.4 × 10 × d sin 20 M1 Apply KE loss = PE gain A1 FT Correct energy equation Total dist A moves up plane = 0.5 + d = 0.654 m A1 5

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