Cambridge A Level Mathematics 9709 — 2021 Feb/March Paper 4 · Variant 2

1 Two particles P and Q of masses 0.2 kg and 0.3 kg respectively are free to move in a horizontal straight line on a smooth horizontal plane. P is projected towards Q with speed 0.5 m s−1. At the same instant Q is projected towards P with speed 1 m s−1. Q comes to rest in the resulting collision. Find the speed of P after the collision. [3] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 ±0.2 × 0.5 or ±0.3 × 1 B1 For initial momentum for either particle. Allow kg or g. 0.2 × 0.5 + 0.3 × (−1) = 0.2 × v + 0 M1 For conservation of momentum. Dimensions correct. Allow if 3 relevant momentum terms are seen regardless of sign. Speed = 1 m s–1 A1 Allow if final answer given as v = 1 or speed = 1 from an equation whose solution is v = –1 3

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2 A car of mass 1400 kg is travelling at constant speed up a straight hill inclined at ! to the horizontal, where sin ! = 0.1. There is a constant resistance force of magnitude 600 N. The power of the car’s engine is 22 500 W. (a) Show that the speed of the car is 11.25 m s−1. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ The car, moving with speed 11.25 m s−1, comes to a section of the hill which is inclined at 2Å to the horizontal. (b) Given that the power and resistance force do not change, find the initial acceleration of the car up this section of the hill. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2(a) Driving force = DF 22500 v = B1 DF – 1400g × 0.1 – 600 = 0 M1 Apply Newton’s 2nd law to the car with a = 0, three relevant terms. May see term 1400g sin 5.7° . v = 11.25 m s-1 A1 AG From exact working only, may be implied if using 5.7°. 3 Question Answer Marks Guidance 2(b) DF – 1400g sin 2 – 600 = 1400a M1 Use of Newton’s second law for the car, 4 relevant terms. 22500 11.25 – 1400g sin 2 – 600 = 1400a A1 a = 0.651 m s-2 (3sf) A1 3

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3 P Q 60Å 30Å R A particle Q of mass 0.2 kg is held in equilibrium by two light inextensible strings PQ and QR. P is a fixed point on a vertical wall and R is a fixed point on a horizontal floor. The angles which strings PQ and QR make with the horizontal are 60Å and 30Å respectively (see diagram). Find the tensions in the two strings. [5] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 3 For attempting to resolve forces in either direction. M1 Correct number of relevant terms. TP cos 60 = TR cos 30 A1 TP sin 60 = TR sin 30 + 0.2g A1 Attempt to solve simultaneously for either tension. M1 From 2 equations, with correct number of relevant terms. TP = 3.46 N and TR = 2 N A1 Both correct. Allow TP = 2√3 N. Alternative method for question 3 0.2 sin60 sin150 sin150 = = P R T T g M1 Attempt one pair of Lami’s equations. Correct angles. One pair correct A1 Equations all correct A1 Solve for TP or TR M1 From equations of the correct form. TP = 3.46 N and TR = 2 N A1 Both correct. Allow TP = 2√3 N 5

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4 v (m s−1) 2 0 t (s) 0 1.5 6 7 13 15 20 21.5 −V An elevator moves vertically, supported by a cable. The diagram shows a velocity-time graph which models the motion of the elevator. The graph consists of 7 straight line segments. The elevator accelerates upwards from rest to a speed of 2 m s−1 over a period of 1.5 s and then travels at this speed for 4.5 s, before decelerating to rest over a period of 1 s. The elevator then remains at rest for 6 s, before accelerating to a speed of V m s−1 downwards over a period of 2 s. The elevator travels at this speed for a period of 5 s, before decelerating to rest over a period of 1.5 s. (a) Find the acceleration of the elevator during the first 1.5 s. 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(c) The combined weight of the elevator and passengers on its upward journey is 1500 kg. Assuming that there is no resistance to motion, find the tension in the elevator cable on its upward journey when the elevator is decelerating. 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Mark scheme: 4(a) Acceleration = 4 3 m s–2 1 4(b) ( ) ( ) 1 1 7 4.5 2 8.5 5 2 2 + × = + ×V M1 Equate expressions for the two areas (distances) leading to an equation in V. V = 1.7[0] (3sf) A1 Allow V = 46 27 . 2 4(c) Acceleration = −2 m s–2 B1 Or Deceleration = 2. T – 1500g = 1500× (−2) M1 Apply Newton’s second law to the lift, using an acceleration 4 ( 3 ≠ or their 4(a)). Correct dimensions and number of relevant terms. T = 12 000 N A1 3

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5 X N 30Å 5 kg A block of mass 5 kg is being pulled along a rough horizontal floor by a force of magnitude X N acting at 30Å above the horizontal (see diagram). The block starts from rest and travels 2 m in the first 5 s of its motion. (a) Find the acceleration of the block. 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(b) Given that the coefficient of friction between the block and the floor is 0.4, find X. 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The block is now placed on a part of the floor where the coefficient of friction between the block and the floor has a different value. The value of X is changed to 25, and the block is now in limiting equilibrium. (c) Find the value of the coefficient of friction between the block and this part of the floor. 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Mark scheme: 5(a) [2 = 1 25 2 × × a ] a = 0.16 m s–2 A1 Allow a = 4 25 . 2 5(b) R = 5g – X sin 30 B1 X cos 30 – F = 5a M1 Apply Newton’s 2nd law to the block, using their a. X cos 30 – 0.4(5g – X sin 30) = 5 × 0.16 M1 Use F = 0.4R to obtain an equation in X only, using their R which must involve 5g and a component of X only. X = 19.5 (3sf) A1 4 5(c) R = (5g – 25 sin 30) [R = 37.5] B1 F = 25 cos 30 25 3 2 F   =     B1 µ = F R = 0.577 (3sf) B1 Allow µ = 3 3 or µ = 1 3 . 3

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6 A particle moves in a straight line. It starts from rest from a fixed point O on the line. Its velocity at 3 time t s after leaving O is v m s−1, where v = t2 −8t 2 + 10t. (a) Find the displacement of the particle from O when t = 1. 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(b) Show that the minimum velocity of the particle is −125 m s−1. 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Mark scheme: 6(a) [ ] 3 2 2 8 10 d s t t t t   = − +        *M1 For attempting to integrate v. [ ] [ ] 5 3 2 2 1 16 5 3 5 = − + + s t t t C A1 Allow unsimplified. For correct use of correct limits. DM1 Use of limit at t = 0 may be implied. Displacement = 2.13 m (3sf) A1 Allow displacement = 32 15 . 4 Question Answer Marks Guidance 6(b) For attempting to differentiate v. *M1 [ ] 1 2 2 12 10 = − + a t t A1 Allow unsimplified. 1 2 0 2 12 10 0 =  − + = a t t DM1 Dependent on *M1. Set a = 0 and attempt to solve their 3 term equation in t or t or (= p t ) by treating it as a quadratic equation. 1 1 2 2 2 5 1 0    − − =          t t leading to t = 1 or t = 25 A1 Both correct. 1 2 d 2 6 d − = − a t t *DM1 Dependent on *M1. Determine the nature of the stationary point by: Either differentiating a and testing the sign of d d a t or by substituting values either side of their t value(s) and attempt to determine the nature of the stationary point(s). If using d d a t then must evaluate it at a t value for M1. Allow use with any t value from their ‘quadratic’. Use t = 25 in 1 2 d 2 6 25 d − = − × a t Evaluating d d a t correctly, hence a minimum. A1 Or by using a convincing argument to show that t = 25 gives a minimum value of v. If evaluated then d d a t must be 0.8. Minimum velocity = 3 2 2 25 8 25 10 25 −× + × = −125 m s-1 B1 AG This mark is awarded only if the previous 6 marks are awarded. 7

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7 0.5 kg P 0.8 N m kg Q 30Å 45Å Two particles P and Q of masses 0.5 kg and m kg respectively are attached to the ends of a light inextensible string. The string passes over a fixed smooth pulley which is attached to the top of two inclined planes. The particles are initially at rest with P on a smooth plane inclined at 30Å to the horizontal and Q on a plane inclined at 45Å to the horizontal. The string is taut and the particles can move on lines of greatest slope of the two planes. A force of magnitude 0.8 N is applied to P acting down the plane, causing P to move down the plane (see diagram). (a) It is given that m = 0.3, and that the plane on which Q rests is smooth. Find the tension in the string. 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(b) It is given instead that the plane on which Q rests is rough, and that after each particle has moved a distance of 1 m, their speed is 0.6 m s−1. The work done against friction in this part of the motion is 0.5 J. Use an energy method to find the value of m. 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Mark scheme: 7(a) Attempt Newton’s 2nd law for either P, Q or the system. M1 Correct number of relevant terms, dimensionally correct. For P: 0.8 + 0.5g sin 30 – T = 0.5a For Q: T – 0.3g sin 45 = 0.3a System: 0.8 + 0.5g sin 30 – 0.3g sin 45 = 0.8a A1 For any one correct equation. A1 For two correct equations. Attempt to solve for T. M1 Using two equations, each with the correct number of relevant terms. [a = 1.4733 may be seen]. T = 2.56 N (3sf) A1 Allow 99 75 2 80 + = T . 5 Question Answer Marks Guidance 7(b) KE and PE for m kg particle: 1 0.36 0.18 2 × = m m and sin45 5 2 = mg m B1 Any 2 correct PE or KE terms. KE and PE for 0.5 kg particle: 1 0.5 0.36 0.09 2 × × = and 0.5 sin30 2.5 = g B1 All 4 correct PE and KE terms. Apply the work-energy equation to the system as: PE loss + WD by 0.8 N = KE gain + 0.5 M1 Must include at least 5 relevant terms only and no extra terms. All terms dimensionally correct. 0.5g × 1× sin 30 – mg × 1× sin 45 + 0.8 × 1 = ½ × (0.5 + m) × 0.36 + 0.5 A1 May be seen as: 2.5 –5 2 0.8 0.09 0.18 0.5 + = + + m m m = 0.374 A1 Alternative method for question 7(b) KE and PE for m kg particle: 1 0.36 0.18 2 × = m m and sin45 5 2 = mg m B1 Correct KE and PE for m kg particle. 0.18 = a and 3.3 0.5(0.18) leading to 3.21 − = = T T B1 Evaluate the tension in the string using Newton’s second law applied to the 0.5 kg particle. For m kg particle: WD by T = KE gain + PE gain + 0.5 M1 At least 3 relevant terms including tension. All terms dimensionally correct. 1 3.21 1 0.36 sin 45 0.5 2 × = × + + m mg A1 m = 0.374 A1 Question Answer Marks Guidance 7(b) Alternative method for question 7(b) KE and PE for m kg particle: 1 0.36 0.18 and sin 45 5 2 2 × = = m m mg m KE and PE for 0.5 kg particle 1 0.5 0.36 0.09 2 × × = and 0.5 sin30 2.5 = g B1 Any 2 correct PE or KE terms. B1 All 4 correct PE and KE terms. Apply the work-energy equation to both particles as: 1 0.8 1 0.5 sin30 0.5 0.36 1 2 × + = × × + × g T and 1 1 0.36 sin45 0.5 2 × = × + + T m mg M1 Must include at least 5 relevant terms only and tension terms in both. [ ] 3.21 = T All terms dimensionally correct. 1 1 0.8 1 0.5 sin30 0.5 0.36 0.36 sin45 0.5 2 2 × + − × × = × + + g m mg A1 m = 0.374 A1 5

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