Cambridge A Level Mathematics 9709 — 2019 Oct/Nov Paper 4 · Variant 2

1 A particle moves in a straight line. The displacement of the particle at time t s is s m, where s = t3 −6t2 + 4t. Find the velocity of the particle at the instant when its acceleration is zero. [4] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 *M1 Attempt at differentiation of s to find v (a =) 6t – 12 *M1 Attempt at differentiation of v to find a [When a = 0, t = 2] DM1 Solve to find t when a = 0 and find v at this time v = –8 ms–1 A1 Alternative method for question 1 (v =) 3t2 – 12t + 4 M1 Attempt at differentiation of s to find v (v =) 3(t – 2)2 – 8 or 12 2 2 6 − = = = b t a M1 For using the method of completing the square or using the value of‘ ' ' 2 b a − to find the t value of the minimum velocity M1 Use of the t value at minimum velocity to find v v = –8 ms–1 A1 4

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2 v (m s−1) 12 V U 0 t (s) 0 5 30 35 50 The diagram shows a velocity-time graph which models the motion of a tractor. The graph consists of four straight line segments. The tractor passes a point O at time t = 0 with speed U m s−1. The tractor accelerates to a speed of V m s−1 over a period of 5 s, and then travels at this speed for a further 25 s. The tractor then accelerates to a speed of 12 m s−1 over a period of 5 s. The tractor then decelerates to rest over a period of 15 s. (i) Given that the acceleration of the tractor between t = 30 and t = 35 is 0.8 m s−2, find the value of V. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Given also that the total distance covered by the tractor in the 50 seconds of motion is 375 m, find the value of U. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2(i) ( ) 12 0.8 35 30 − = − V or 12 = V + 0.8 × 5 an equation in V V = 8 A1 2 2(ii) ( ) 1 25 8 5 10 15 6 8 5 375 2   × + × + × + × + × =     U M1 Attempt to find total distance travelled by the tractor in 50s to set up an equation for U using EITHER areas OR suvat equations OR a combination of areas and suvat In either case total distance must be attempted A1FT Correct equation FT on their V from (i) U = 6 A1 3

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3 5 N P 0.3 kg 0.9 m 1.2 m A B A particle P of mass 0.3 kg is held in equilibrium above a horizontal plane by a force of magnitude 5 N, acting vertically upwards. The particle is attached to two strings PA and PB of lengths 0.9 m and 1.2 m respectively. The points A and B lie on the plane and angle APB = 90Å (see diagram). Find the tension in each of the strings. [5] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 3 4 3 0.3 5 5 5 × + × + = A B T T g M1 Resolving vertically 3 4 5 5 × = × A B T T M1 Resolving horizontally A1 Both correct M1 Solve for TA or TB TA = 1.6 N and TB = 1.2 N A1 Alternative method for question 3 5 3 sin90 sin126.9 sin143.1 −   = =     A B T T M1 Attempt one pair of Lami’s equations M1 Attempt a second pair of Lami equations A1 Equations all correct M1 Evaluate TA or TB TA = 1.6 N and TB = 1.2 N A1 Question Answer Mark Guidance 3 Alternative method for question 3 4 4 5cos36.9 3cos36.9 5 3 5 5 = − = × −× A T M1 Resolve along PA 3 3 5cos53.1 3cos53.1 5 3 5 5 = − = × −× TB M1 Resolve along PB A1 Both correct M1 Evaluate TA or TB TA = 1.6 N and TB = 1.2 N A1 Alternative method for question 3 Forces 2N, TA and TB with angles 36.9 and 53.1 M1 Attempt to illustrate a triangle of forces [TA = 2cos36.9, TB = 2cos53.1] M1 Use trigonometry in the triangle to find TA and TB A1 Both correct M1 Solve for TA or TB TA = 1.6 N and TB = 1.2 N A1 5

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4 A lorry of mass 25 000 kg travels along a straight horizontal road. There is a constant force of 3000 N resisting the motion. (i) Find the power required to maintain a constant speed of 30 m s−1. 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The lorry comes to a straight hill inclined at 2Å to the horizontal. The driver switches offthe engine of the lorry at the point A which is at the foot of the hill. Point B is further up the hill. The speeds of the lorry at A and B are 30 m s−1 and 25 m s−1 respectively. The resistance force is still 3000 N. (ii) Use an energy method to find the height of B above the level of A. 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Mark scheme: 4(i) M1 Use of P = Fv with F = resistance P = 90000 W = 90kW A1 2 4(ii) PE gained = 25000gh B1 Correct expression for PE Allow PE = 25 000 g d sin 2 Initial 2 1 KE 25000 30 2 = × × [= 11 250 000] Final 2 1 KE 25000 25 2 = × × [= 7 812 500] B1 For either correct [KE loss = 3 437 500] Initial KE = Final KE + 25000gh + 3000 sin2 h OR Initial KE = Final KE + 25000gdsin2 + 3000d M1 For a 4 term work-energy equation, correct dimensions A1 Correct work-energy equation involving h or d h = 10.2 m (10.2318…) A1 5

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5 Two particles A and B move in the same vertical line. Particle A is projected vertically upwards from the ground with speed 20 m s−1. One second later particle B is dropped from rest from a height of 40 m. (i) Find the height above the ground at which the two particles collide. 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(ii) Find the difference in the speeds of the two particles at the instant when the collision occurs. 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Mark scheme: 5(i) 2 1 20 10 2 A h t t = − × or ( ) 2 1 10 1 2 B h t = ± × − B1 OE ( ) ( ) 2 1 20 1 10 1 2 A h T T = + − × + or 2 1 10 2 B h T = ± × [Meet when ( ) 2 2 1 1 20 10 10 1 40 2 2 t t t − × + × − = ] *M1 Set up an equation using their hA, their hB and 40 10t – 35 = 0 DM1 Solve for t and attempt to find the height at collision. t = 3.5 so height at collision = 8.75 m A1 T = 2.5 and height at collision = 8.75 m Alternative method for question 5(i) 2 1 20 1 10 1 15, 20 10 1 10 2 A h v = × − × × = = − × = B1 Finding distance travelled by A and its speed after 1 second HA + HB = 25 2 2 1 1 10 10 10 25 2 2 T T T   − × × + × × =     *M1 T is the time beyond 1s until the particles reach same level HA and HB are distances travelled by A and B in T seconds. [10T = 25 → T = 2.5] DM1 Solve for T and attempt to find the height at collision t = 3.5 so height = 8.75 m A1 4 Question Answer Mark Guidance 5(ii) vA = 20 – gt = –15 or vA2 = 202 + 2(–g)(8.75) M1 Use of their t or their h ⩽ 20 from 5(i) in a constant acceleration formula which would lead to finding vA vB = – g(t – 1) = –25 or vB2 = 2(g)(40 – 8.75) M1 Use of their t ± 1 or their 40 – h from 5(i) in a constant acceleration formula which would lead to finding vB Difference = 10 ms–1 A1 CWO 3

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6 A block of mass 3 kg is initially at rest on a rough horizontal plane. A force of magnitude 6 N is applied to the block at an angle of 1 above the horizontal, where cos 1 = 2425. The force is applied for a period of 5 s, during which time the block moves a distance of 4.5 m. (i) Find the magnitude of the frictional force on the block. 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(ii) Show that the coefficient of friction between the block and the plane is 0.165, correct to 3 significant figures. 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(iii) When the block has moved a distance of 4.5 m, the force of magnitude 6 N is removed and the block then decelerates to rest. Find the total time for which the block is in motion. 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Mark scheme: 6(i) 2 1 4.5 0 5 2 a = + × × M1 For use of 2 1 2 s ut at = + to find a a = 0.36 A1 24 6 3 0.36 25 F × − = × M1 Resolving horizontally. Allow use of θ = 16.3 F = 4.68 N A1 4 6(ii) [ ] 7 3 6sin16.3 3 6 28.32 25 R g g = − = −× = B1 4.68 = µ × 28.32 M1 Use of F = µR µ = 0.165 (0.165254…) A1 AG. Allow µ 39 236 = 3 Question Answer Mark Guidance 6(iii) v = 5 × 0.36 [= 1.8] or ( )[ ] 2 0.36 4.5 1.8 v = × × = B1FT For velocity at t = 5 ft on their a from 6(i) 3a = –0.165 × 3g M1 Using Newton’s second law with new frictional force 0 = 1.8 – 0.165gt (t = 1.09) M1 Using constant acceleration equations which would lead to a positive value of t Total time = 5 + 1.09 = 6.09 s A1 4

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7 Q 0.2 kg P 0.3 kg 0.8 m 1 Two particles P and Q, of masses 0.3 kg and 0.2 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 edge of a smooth plane. The plane is inclined at an angle 1 to the horizontal, where sin 1 = 35. P lies on the plane and Q hangs vertically below the pulley at a height of 0.8 m above the floor (see diagram). The string between P and the pulley is parallel to a line of greatest slope of the plane. P is released from rest and Q moves vertically downwards. (i) Find the tension in the string and the magnitude of the acceleration of the particles. 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Q hits the floor and does not bounce. It is given that P does not reach the pulley in the subsequent motion. (ii) Find the time, from the instant at which P is released, for Q to reach the floor. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (iii) When Q hits the floor the string becomes slack. Find the time, from the instant at which P is released, for the string to become taut again. 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Mark scheme: 7(i) M1 Use of Newton’s second law for P or Q or the system For P: 3 0.3 0.3 sin36.9 0.3 5 T g T g a − × = − = For Q: 0.2g – T = 0.2a System: ( ) 3 0.2 0.3 0.2 0.3 5 g g a − × = + or 0.2g – 0.3g sin 36.9 = (0.2 + 0.3)a A1 Two correct equations Allow use of θ = 36.9 [0.2g – 0.18g = 0.5a] M1 For solving either the system for a or for solving a pair of simultaneous equations for a or T a = 0.4 ms–2 A1 T = 1.92 N A1 5 Question Answer Mark Guidance 7(ii) 2 1 0.8 0 0.4 2 t = + × × a M1 For use of the constant acceleration equations with their a from 7(i) and a ≠ ± g for a complete method to find t t = 2 s A1 2 7(iii) Speed when Q hits the floor = 2 × 0.4 (= 0.8) or ( )[ ] 2 0.4 0.8 0.8 v = × × = B1FT Using v = u + at with u = 0 Allow FT for their unsimplified v = at or v2 = 2as with a from (i), t from (ii) and s = 0.8 3 0.3 0.3 sin36.9 0.3 5 g g a − × = − = [a = –6] M1 Using Newton’s second law for P to find a ≠ ± g ( ) ( ) 2 1 0 0.8 6 0.2666... 2 t t t = + × − = or 0 = 0.8 – 6T (T = 0.13333 = 2 15 and t = 2T = 0.26666 = 4 15 ) M1 Use of the constant acceleration equation(s) to find the time taken for P to return to the position where the string first became slack. Total time = 2 + 0.266... = 2 + 4 15 = 2.27 = 34 15 s A1 4

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