Cambridge A Level Mathematics 9709 — 2020 Oct/Nov Paper 4 · Variant 1

1 A particle B of mass 5 kg is at rest on a smooth horizontal table. A particle A of mass 2.5 kg moves on the table with a speed of 6 m s−1 and collides directly with B. In the collision the two particles coalesce. (a) Find the speed of the combined particle after the collision. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the loss of kinetic energy of the system due to the collision. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 1(a) 6 × 2.5 = 2.5v + 5v M1 Apply conservation of momentum, 3 terms implied v = 2 ms–1 A1 2 1(b) Use KE = ½ mv2 either before or after collision M1 Allow this for either particle KE(before) = 0.5 × 2.5 × 62 KE(after) = 0.5 × 7.5 × 22 A1 FT Both correct FT on v Loss of KE = 30 J A1 3

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2 A car of mass 1400 kg is moving along a straight horizontal road against a resistance of magnitude 350 N. (a) Find, in kW, the rate at which the engine of the car is working when it is travelling at a constant speed of 20 m s−1. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the acceleration of the car when its speed is 20 m s−1 and the engine is working at 15 kW. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2(a) P = 350 × 20 M1 Using P = Fv P = 7 kW A1 2 2(b) 15 000 = DF × 20 [DF = 750] B1 Using P = Fv DF – 350 = 1400a M1 Use Newton’s 2nd law, 3 terms a = 2 7 ms–2 A1 a = 0.286 3

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3 8 N P N 1Å 30Å 45Å 12 N 10 N Coplanar forces of magnitudes 8 N, 12 N, 10 N and P N act at a point in the directions shown in the diagram. The system is in equilibrium. Find P and 1. [6] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 3 Resolve forces either horizontally or vertically M1 Correct number of relevant terms P cos θ = 12 + 8 cos 30 – 10 cos 45 [= 11.857] A1 P sin θ = 10 sin 45 – 8 sin 30 [= 3.071] A1 ( ) 2 2 11.857 3.071 P = + M1 OE. Use of correct method for finding P 1 3.071 tan 11.857 −  θ =     M1 OE. Use of correct method for finding θ P = 12.2 and θ = 14.5 A1 Both correct 6

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4 A particle P moves in a straight line. It starts from rest at a point O on the line and at time t s after leaving O it has acceleration a m s−2, where a = 6t −18. Find the distance P moves before it comes to instantaneous rest. [6] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 4 *M1 Attempt to integrate a [s = t3 – 9t2 (+ C)] #M1 Attempt to integrate v v = 3t2 – 18t s = t3 – 9t2 A1 Both integrals correct v = 0, 3t2 – 18t = 0 [t = 6] *DM1 Attempt to find t when v = 0 s = 63 – 9 × 62 – [0] #DM1 Substitute limits correctly into s s = 108 m A1 Answer must be positive 6

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5 0.8 kg 0.2 kg 0.5 m Two particles of masses 0.8 kg and 0.2 kg are connected by a light inextensible string that passes over a fixed smooth pulley. The system is released from rest with both particles 0.5 m above a horizontal floor (see diagram). In the subsequent motion the 0.2 kg particle does not reach the pulley. (a) Show that the magnitude of the acceleration of the particles is 6 m s−2 and find the tension in the string. 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(b) When the 0.8 kg particle reaches the floor it comes to rest. Find the greatest height of the 0.2 kg particle above the floor. 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Mark scheme: 5(a) 0.8g – T = 0.8a, T – 0.2g = 0.2a, For system: 0.8g – 0.2g = (0.8 + 0.2)a A1 Any 2 correct equations Attempt to solve for either a or T M1 a = 6 ms–2 and T = 3.2 N A1 AG. Both correct 4 5(b) v2 = 2 × 6 × 0.5 M1 Attempt to find v or v2 as 0.8 kg particle reaches the ground using a from 5(a) 0 = 6 – 20s M1 Attempt to find the extra height reached by 0.2 kg particle using v2 from previous M1 mark Greatest height = 0.5 + 0.5 + 0.3 = 1.3 m A1 3

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6 A car of mass 1500 kg is pulling a trailer of mass 750 kg up a straight hill of length 800 m inclined at an angle of sin−1 0.08 to the horizontal. The resistances to the motion of the car and trailer are 400 N and 200 N respectively. The car and trailer are connected by a light rigid tow-bar. The car and trailer have speed 30 m s−1 at the bottom of the hill and 20 m s−1 at the top of the hill. (a) Use an energy method to find the constant driving force as the car and trailer travel up the hill. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ After reaching the top of the hill the system consisting of the car and trailer travels along a straight level road. The driving force of the car’s engine is 2400 N and the resistances to motion are unchanged. (b) Find the acceleration of the system and the tension in the tow-bar. 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Mark scheme: 6(a) KE (initial) = ½ × 1500 × 302 + ½ × 750 × 302 PE gain = 2250 × 10 × 800 × 0.08 B1 WD against friction = 600 × 800 B1 ½ × 2250 × 302 + DF × 800 = 600 × 800 + ½ × 2250 × 202 + 2250 × 10 × 800 × 0.08 M1 Use energy equation. DF = 1700 N A1 DF = 1696.875 N 5 Question Answer Marks Guidance 6(b) 2400 – 600 = 2250a or T – 200 = 750a and 2400 – 400 – T = 1500a M1 Apply Newton’s second law to the system or to each of the car and trailer separately A1 Two correct equations Attempting to solve for a or for T M1 T = 800 N and a = 0.8 ms–2 A1 4

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7 0.2 kg A 1 m B 1 m 30Å C Three points A, B and C lie on a line of greatest slope of a plane inclined at an angle of 30Å to the horizontal, with AB = 1 m and BC = 1 m, as shown in the diagram. A particle of mass 0.2 kg is released from rest at A and slides down the plane. The part of the plane from A to B is smooth. The part of the plane from B to C is rough, with coefficient of friction - between the plane and the particle. (a) Given that - = 1 3, find the speed of the particle at C. 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(b) Given instead that the particle comes to rest at C, find the exact value of -. 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Mark scheme: 7(a) 2 1 0.2 10 0.5 0.2 2 × × = × × B v M1 Attempt PE loss = KE gain from A to B 2 10 = B v A1 Alternative method for the first 3 marks 0.2 × 10 × sin 30 = 0.2a, a = 5 (M1) Attempt to find acceleration a for motion from A to B 2 2 0 2 5 1 = + × × B v (M1) Use v2 = u2 + 2as in attempt to find speed at B 2 10 = B v (A1) Question Answer Marks Guidance 7(a) THEN, either this method for the next 5 marks R = 0.2 × 10 × cos 30 = √3 B1 3 3 0.2 10 1.5 2 2 F = × × × = M1 For using F = µR where R must be a component of 0.2g PE loss = 0.2 × 10 × 0.5 = 1 WD against F = 1.5 × 1 M1 Attempt to find either PE loss or WD against F from B to C 2 1 1 0.2 10 0.2 10 0.5 1.5 1 0.2 2 2 × + × × = × + C v M1 Apply work-energy equation for motion from B to C as KE at B + PE at B = WD against F + KE at C with vB ≠ 0 vc = 5 = 2.24 ms–1 A1 OR, this method for the next 5 marks R = 0.2 × 10 × cos 30 = √3 (B1) 3 3 0.2 10 1.5 2 2 F = × × × = (M1) For using F = µR where R must be a component of 0.2g 0.2 × 10 sin 30 – 1.5 = 0.2a a = –2.5 (M1) Attempt to find acceleration a for motion from B to C 2 10 2 2.5 1 = + × − × cv (M1) Use v2 = u2 + 2as in attempt to find vc using vB ≠ 0 vc = 5 = 2.24 ms–1 (A1) 8 Question Answer Marks Guidance 7(a) Alternative method for question 7(a) PE loss = 0.2 × 10 × 2 sin 30 = 2 M1 Attempt PE loss for motion from A to C KE gain 2 1 0.2 2 = × × C v M1 Attempt KE gain for motion from A to C Both PE loss and KE gain correct A1 R = 0.2 × 10 × cos 30 = √3 B1 3 3 0.2 10 1.5 2 2 F = × × × = M1 For using F = µR where R must be a component of 0.2g WD against F = 1.5 × 1 M1 Attempt WD against F 2 1 0.2 10 1 1.5 1 0.2 2 × × = × + × × C v M1 Attempt work-energy equation for motion from A to C vc = 5 = 2.24 ms–1 A1 8 Question Answer Marks Guidance 7(b) 0 = 10 + 2a [a = –5] M1 Attempt to find a for motion from B to C, using 2 10 = B v , 0 = C v 0.2 × 10 × sin 30 – F = 0.2 × -5 M1 Attempt Newton’s 2nd law for motion from B to C 2 3 = μ M1 Use F = µR where R is a component of 0.2g but R = 0.2g is M0 2 3 μ = A1 Any correct exact form such as 2/3√3 Alternative method for question 7(b) PE loss = 0.2 × 10 × 1 sin 30 = 1 M1 Attempt PE loss for motion from B to C 1 + ½ × 0.2 × 10 = F × 1 M1 Work-Energy equation for motion from B to C in the form PE at B + KE at B = WD against F using 2 10 = B v , 0 = C v 3 F = μ M1 Use F = µR leading to an equation in µ where R is a component of 0.2g 2 3 μ = A1 Any correct exact form such as 2/3√3 Question Answer Marks Guidance 7(b) Alternative method for question 7(b) PE loss = 0.2 × 10 × 2 sin 30 = 2 M1 Attempt PE loss for motion from A to C 2 = F × 1 M1 Work-Energy equation for motion from B to C 3 F = μ M1 Use F = µR leading to an equation in µ where R is a component of 0.2g 2 3 μ = A1 Any correct exact form such as 2/3√3 4

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