Cambridge A Level Mathematics 9709 — 2021 Oct/Nov Paper 4 · Variant 1

1 A bus moves from rest with constant acceleration for 12 s. It then moves with constant speed for 30 s before decelerating uniformly to rest in a further 6 s. The total distance travelled is 585 m. (a) Find the constant speed of the bus. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the magnitude of the deceleration. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 1(a) + + = ( ) 0.5 30 48 585 + = V Complete method to set up an equation in V using constant acceleration equations or correct area formula in v-t graph. Speed of the bus = 15 ms–1 A1 Must be positive. 2 1(b) Magnitude of deceleration = 2.5 B1 FT OE. Do not allow a = –2.5. 1

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2 Two small smooth spheres A and B, of equal radii and of masses km kg and m kg respectively, where k > 1, are free to move on a smooth horizontal plane. A is moving towards B with speed 6 m s−1 and B is moving towards A with speed 2 m s−1. After the collision A and B coalesce and move with speed 4 m s−1. (a) Find k. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find, in terms of m, the loss of kinetic energy due to the collision. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2(a) Attempt at use of conservation of momentum M1 4 terms implied, i.e. m and km included before and after collision. Velocity after collision is the same for m and km. ( ) 6 2 4 × − × = + × km m km m A1 3 k = A1 3 2(b) KE initial = ( ) 2 2 1 1 6 2 2 2 × × + × × − km m KE after = ( ) 2 1 4 2 × + × km m M1 Attempt at any of the three possible KE terms, unsimplified. k need not be substituted here. Loss of KE = 24m J A1 FT KE loss = 56m – 32m FT on their ,k KE loss ( ) 10 6 = − k m, 0.6 > k . 2

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3 24 N ! 1 P N 20 N ! 36 N Coplanar forces of magnitudes 24 N, P N, 20 N and 36 N act at a point in the directions shown in the diagram. The system is in equilibrium. Given that sin ! = 35, find the values of P and 1. [6] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 3 Attempt at resolving in any direction M1 Correct number of terms. No substitution for α required. ( ) cos 36 24 cos36.9 θ = − P or ( ) cos 36 24 0.8 θ = − × P A1 ( ) sin 20 24 36 sin36.9 14.4 21.6 θ + = + = + P or sin 20 60 0.6 36 θ + = × = P A1 cos 9.6, sin 16 θ θ = = P P 2 2 16 9.6 = + P M1 Correct method for solving equations for P. OE 1 16 tan 9.6 θ −  =     M1 Correct method for solving equations for θ . OE e.g. 1 5 tan 3 θ −  =     18.7 P = [ ] 59 .0 θ = A1 Allow 16 34 5 P = . Allow 18.6 P = . 6

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4 A particle of mass 12 kg is stationary on a rough plane inclined at an angle of 25Å to the horizontal. A force of magnitude P N acting parallel to a line of greatest slope of the plane is used to prevent the particle sliding down the plane. The coefficient of friction between the particle and the plane is 0.35. (a) Draw a sketch showing the forces acting on the particle. [1] (b) Find the least possible value of P. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(a) Correct 4 force diagram 1 4(b) Attempt to resolve forces parallel to the plane M1 Three terms, allow sign errors. [ ] 12 sin25 50.7 + = = P F g A1 Must have correct direction of F here. 12 cos25 = R g [= 108.8] B1 [ ] 120cos25 0.35 38.1 = × = F 38.1 120sin 25 P + = M1 Attempt to solve for P using equations with the correct number of terms. R must be a single-term component of 12g. 12.6 = P A1 P = 12.64926… Allow 12.7 P = 5

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5 A car of mass 1600 kg travels at constant speed 20 m s−1 up a straight road inclined at an angle of sin−1 0.12 to the horizontal. (a) Find the change in potential energy of the car in 30 s. 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(b) Given that the total work done by the engine of the car in this time is 1960 kJ, find the constant force resisting the motion. 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(c) Calculate, in kW, the power developed by the engine of the car. 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(d) Given that this power is suddenly decreased by 15%, find the instantaneous deceleration of the car. 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Mark scheme: 5(a) × PE change 1 600 s 0.12 = × × × g [ ] PE change 1 600 20 30 0.12 = × × × × g M1 Attempt change in PE. May use angle = 6.9º. Allow sin/cos error only. Change in PE 1152000 J = A1 3 5(b) 1960 000 PE = + res WD their [ ] 1960 000 1152 000 = + res WD [ ] 808 000 J = res WD M1 Using work-energy, allow sign error. 600 = ÷ res R WD B1 Using 600 = × res WD R . Force resisting motion 1350 N = = R to 3sf A1 Allow R = 4040 3 N. Allow R negative. Alternative method for question 5(b) 1600 0.12 0 − − × = DF R g M1 R is the resisting force. 196000 9800 20 30 3 DF   = =   ×   B1 Force resisting motion = 4040 1350 N 3 = = R to 3sf A1 Allow R negative. 3 Question Answer Marks Guidance 5(c) 4040 1600 0.12 20 3 P g   = + × × ×     196 000 3   =     M1 For using = × P DF v . Allow use of their R. 65.3 = P kW A1 Alternative method for question 5(c) 1960 000 30 P = M1 For using Work done Time = ÷ P . P = 65.3 kW A1 Alternative method for question 5(c) 9800 20 3 = × P M1 For using = × P DF v . Allow use of their DF. P = 65.3 kW A1 2 Question Answer Marks Guidance 5(d) 196 000 0.85 20 3 × = × DF B1 FT 8330 3 P DF v DF   = × =     FT on their P. 1600 0.12 1600 − − × = DF R g a 8330 4040 1920 1600 3 3   − − =     a M1 Newton’s 2nd law, four terms, allow sin/cos error, their R and their DF. [ ] 2 0.306 ms a − = − A1 [ ] [ ] 490 49 1600 160 a = - = - Alternative method for question 5(d) 9800 20 = × DF B1 FT Using the reduction in power as the cause of the deceleration. 9800 0.15 = × = × their P DF v 1600 = DF d 9800 1600 20   =     d M1 [ ] 2 0.306 ms a − = − A1 [ ] [ ] 490 49 1600 160 a = - = - 3

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6 A particle P moves in a straight line starting from a point O and comes to rest 14 s later. At time t s after leaving O, the velocity v m s−1 of P is given by v = pt2 −qt 0 ≤t ≤6, v = 63 −4.5t 6 ≤t ≤14, where p and q are positive constants. The acceleration of P is zero when t = 2. (a) Given that there are no instantaneous changes in velocity, find p and q. 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(b) Sketch the velocity-time graph. [3] (c) Find the total distance travelled by P during the 14 s. 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Mark scheme: 6(a) = − *M1 Attempt to differentiate v. 36 6 36 − = p q 4 0 − = p q DM1 For attempting to set up 2 equations using a = 0 at t = 2 and matching the velocities at t = 6 and solve for p or q . 3, 12 = = p q A1 Both correct. 3 6(b) Correct quadratic from t = 0 to t = 6 or Correct straight line from 6 to 14 B1 No labelling necessary for this mark. Both quadratic and straight line correct B1 Must join and no labelling needed. All correct and key points shown B1 All correct, labelled at (4, 0), (6, 36) and (14, 0). 3 6(c) Attempt to integrate v *M1 Allow in terms of p and q. 3 2 6 = − s t t A1 FT FT on their p and q values. ( ) 4 6 3 2 3 2 4 0 quadratic 6 6     = − + −     s t t t t DM1 [ ] 32 32 = + Using limits correctly for t = 0 to t = 6. Allow in terms of p and q. ( ) 14 2 6 triangle 63 2.25 144   = − =   s t t or area of triangle 144 = B1 Total distance travelled in 14 s = 208 m A1 5

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7 A 2 kg 3 kg B 0.45 m 18Å Two particles A and B of masses 2 kg and 3 kg respectively are connected by a light inextensible string. Particle B is on a smooth fixed plane which is at an angle of 18Å to horizontal ground. The string passes over a fixed smooth pulley at the top of the plane. Particle A hangs vertically below the pulley and is 0.45 m above the ground (see diagram). The system is released from rest with the string taut. When A reaches the ground, the string breaks. Find the total distance travelled by B before coming to instantaneous rest. You may assume that B does not reach the pulley. 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Mark scheme: 7 Particle A: 2 2 − = g T a Particle B: 3 sin18 9.27 3 − = − = T g T a System: ( ) 2 3 sin18 2 9.27 2 3 − = − = + g g g a or the system. Correct number of terms. A1 A and B correct or system correct. 2.145898034 a = [ ] 5 10.72949017 = a M1 Attempt to find a using equations with correct number of terms. 2 2 0.45 v a = × × M1 Use of constant acceleration equations with their ≠± a g to find v2 when A reaches the ground. 2 2 2.145898034 0.45 1.931308 v = × × =  [ ] 1.389715162 = v A1 Allow unsimplified. 0, 3 sin18 3 = ± = T g a [ ] 3.0901699 = ± a M1 Attempt to find a for the motion of B when string becomes slack. Allow sin/cos error, no extra terms. [ ] 0 1.93 2 3.09 = −× ×s [ ] 0.312 = s M1 Use constant acceleration equations, using a new ≠± a g , to find the further distance, s, travelled by B before coming to rest. Total distance moved by B = 0.45 + 0.312 = 0.762 m A1 Alternative method for question 7 Attempt PE loss as A reaches the ground M1 Allow sin/cos error. PE loss = 2 0.45 3 0.45sin18 × − × g g [ ] 4.82827 = A1 Correct unsimplified. ( ) 2 1 2 0.45 3 0.45sin18 2 3 2 × − × = × + g g v *M1 Apply work-energy equation as PE loss = KE gain, allow sign error, sin/cos error, 4 terms implied. Question Answer Marks Guidance 7 Solve for v2 DM1 [ ] 2 1.931308 1.389715162 = … = v v A1 PE gain = 3 sin18 × g s M1 Attempt PE gain for B after string breaks, allow sign error, sin/cos mix, s = extra distance travelled by B along the plane. 1 3 sin18 3 1.931308 2 × = × × g s [ ] 0.312 = s M1 Work energy equation for B as PE gain = KE loss, 2 terms. Total distance moved by B = 0.45 + 0.312 = 0.762 m A1 8

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