Cambridge A Level Mathematics 9709 — 2021 May/June Paper 4 · Variant 3

1 Particles P of mass 0.4 kg and Q of mass 0.5 kg are free to move on a smooth horizontal plane. P and Q are moving directly towards each other with speeds 2.5 m s−1 and 1.5 m s−1 respectively. After P and Q collide, the speed of Q is twice the speed of P. Find the two possible values of the speed of P after the collision. [4] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 0.4 × 2.5 – 0.5 × 1.5 M1 Attempt momentum before impact. 0.4 × 2.5 – 0.5 × 1.5 = 0.4v + 0.5 × 2v M1 Use of conservation of momentum, either case. 0.4 × 2.5 – 0.5 × 1.5 = 0.4v + 0.5 × 2v or 0.4 × 2.5 – 0.5 × 1.5 = –0.4v + 0.5 × 2v A1 One correct equation Speed is 0.179 m s–1 or 0.417 m s–1 A1 Both values 4

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2 A cyclist is travelling along a straight horizontal road. She is working at a constant rate of 150 W. At an instant when her speed is 4 m s−1, her acceleration is 0.25 m s−2. The resistance to motion is 20 N. (a) Find the total mass of the cyclist and her bicycle. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ The cyclist comes to a straight hill inclined at an angle 1 above the horizontal. She ascends the hill at constant speed 3 m s−1. She continues to work at the same rate as before and the resistance force is unchanged. (b) Find the value of 1. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2(a) Forward force exerted by cyclist = 150 4 N [= 37.5 N] B1 OE. P = Fv used correctly. 150 20 0.25 4 m − = × M1 Use of Newton’s second law m = 70 kg A1 3 2(b) 150/3 – 20 – 70gsin θ = 0 M1 For resolving up the plane θ = 2.5° to 1d.p. A1 FT From 2.456…. FT θ = sin–1 3 m       from (a) 2

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3 20 N F N 1Å 60Å "Å !Å 30 N 40 N Four coplanar forces act at a point. The magnitudes of the forces are 20 N, 30 N, 40 N and F N. The directions of the forces are as shown in the diagram, where sin !Å = 0.28 and sin "Å = 0.6. Given that the forces are in equilibrium, find F and 1. [6] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 3 F sin θ + 20sin 60 – 30sin α – 40 sin β = 0 M1 For resolving in either direction Vertical: F sin θ + 20sin 60 – 30 × 0.28 – 40 × 0.6 = 0 [F sinθ = 15.07949…] A1 Horizontal: F cosθ + 40 × 0.8 – 30 × 0.96 – 20cos 60 = 0 [F cosθ = 6.8] A1 θ = 1 15.0794 tan 6.8 − … M1 For method for finding θ F = 2 2 15.07949 6.8 … + M1 For method for finding F θ = 65.7, F = 16.5 A1 6

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4 A particle is projected vertically upwards with speed u m s−1 from a point on horizontal ground. After 2 seconds, the height of the particle above the ground is 24 m. (a) Show that u = 22. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) The height of the particle above the ground is more than h m for a period of 3.6 s. Find h. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(a) 24 = u × 2 − 1 2 g × 22 u = 22 A1 AG 2 Question Answer Marks Guidance 4(b) At maximum height 0 = 222 − 2gs M1 Use of v2 = u2 + 2as to find maximum height. Maximum height s = 24.2 m A1 Height down = 0.5g × 1.82 (=16.2) M1 Find distance travelled down in 1.8 s. h = 8 A1 Alternative method for Question 4(b) 0 = 22 – 10t M1 Use of v = u – gt with u = 22 and v = 0 to find time to reach maximum height t = 2.2 A1 h = 22 × (2.2 – 1.8) – 1 2 g × (2.2 – 1.8)2 M1 Use of s = ut + 1 2at2 to find value of h h = 8 A1 Alternative method for Question 4(b) 22t – 2 1 2 gt = 22 × (t + 3.6) – 1 2 g × (t + 3.6)2 M1 Use of s = ut + 1 2at2 for times t and t + 3.6 to find time taken to reach height h. t = 0.4 (or t + 3.6 = 4) A1 h = 22 × 0.4 – 1 2 g × 0.42 M1 Use s = ut + 1 2at2 to find value of h. h = 8 A1 4

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5 A car of mass 1400 kg is towing a trailer of mass 500 kg down a straight hill inclined at an angle of 5Å to the horizontal. The car and trailer are connected by a light rigid tow-bar. At the top of the hill the speed of the car and trailer is 20 m s−1 and at the bottom of the hill their speed is 30 m s−1. (a) It is given that as the car and trailer descend the hill, the engine of the car does 150 000 J of work, and there are no resistance forces. Find the length of the hill. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) It is given instead that there is a resistance force of 100 N on the trailer, the length of the hill is 200 m, and the acceleration of the car and trailer is constant. Find the tension in the tow-bar between the car and trailer. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 5(a) B1 May be implied by energy equation. Loss of PE = 1900 × g × s sin 5 [= 1655.95s J] B1 May be implied by energy equation. 1900 × g × s sin 5 + 150 000 = ½ × 1900×302 – ½ × 1900 × 202 M1 For attempt at work/energy equation A1 Correct s = [Length of hill =] 196 m A1 5 5(b) 302 = 202 + 2a × 200 M1 Use of v2 = u2 + 2as a = 1.25 m s–2 A1 T – 100 + 500g sin 5 = 500a M1 For applying Newton’s second law to the trailer. T = 289 N A1 4

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6 A particle moves in a straight line and passes through the point A at time t = 0. The velocity of the particle at time t s after leaving A is v m s−1, where v = 2t2 −5t + 3. (a) Find the times at which the particle is instantaneously at rest. Hence or otherwise find the minimum velocity of the particle. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Sketch the velocity-time graph for the first 3 seconds of motion. [3] (c) Find the distance travelled between the two times when the particle is instantaneously at rest. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 6(a) (2t – 3)(t – 1) = 0 leading to t = ….. M1 Attempt to solve v = 0 t = 1 or t = 1.5 A1 Minimum velocity when t = 1.25 leading to v = ….. or d d v t = 4t – 5 = 0 t = 1.25 leading to v = ….. or 2 5 25 2 3 4 16 v t     = − − +           leading to v = ….. M1 Uses roots or dv/dt=0 to find t for vmin and attempts substitution to obtain vmin. Alternatively completes square. Minimum velocity is −0.125 m s–1 A1 Allow 1 8 v = − 4 6(b) Quadratic curve (two roots and v(3) > v(0)) B1 Goes through (1.25, –0.125), (0, 3), (1, 0), (1.5, 0), (3,6) B1 3 of the 5 key points shown on axes or as coordinates All five points shown on a totally correct graph B1 3 6(c) s = 3 2 2 5 3 3 2 − + t t t M1 For use of s = ∫v dt ( ) ( ) ( ) 3 2 2 5 1.5 1.5 3 1.5 3 2   − +     – ( ) ( ) ( ) 3 2 2 5 1 1 3 1 3 2   − +     M1 Correct use of limits (their 1 and 1.5) Distance = 0.0417 m A1 A0 for –0.0417 3

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7 4 N P 0.3 kg 1 A particle P of mass 0.3 kg rests on a rough plane inclined at an angle 1 to the horizontal, where sin 1 = 25.7 A horizontal force of magnitude 4 N, acting in the vertical plane containing a line of greatest slope of the plane, is applied to P (see diagram). The particle is on the point of sliding up the plane. (a) Show that the coefficient of friction between the particle and the plane is 4.3 [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ The force acting horizontally is replaced by a force of magnitude 4 N acting up the plane parallel to a line of greatest slope. (b) Find the acceleration of P. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Starting with P at rest, the force of 4 N parallel to the plane acts for 3 seconds and is then removed. Find the total distance travelled until P comes to instantaneous rest. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 7(a) R = 0.3g cos θ + 4 sin θ = 24 7 3 4 25 25 × + × [=4] F = 4 cos θ – 0.3g sin θ = 24 7 4 3 25 25 × − × [=3] M1 Resolving forces perpendicular to the plane or parallel to the plane. Allow use of θ = 16.3° A1 Two correct equations 3 = µ × 4 M1 For use of F = µR µ = 3 4 A1 AG Must be from correct and exact working, not using 16.3 4 7(b) F = µ× 0.3g cos θ = 3 4 × 3 × 24 25 54 2.16 25   = =     B1 4 − 3 4 × 0.3g × 24 25 – 0.3g × 7 25 = 0.3a M1 Use of Newton’s second law a = 1 0 3 m s–2 A1 3 Question Answer Marks Guidance 7(c) s1 = 1 2 × 10 3 × 32 = 15 and v = 10 3 × 3 = 10 B1 FT Distance s1 in 3s and v after 3s; FT a from (b) –0.3g × sin θ – µ × 0.3g cos θ = 0.3a leading to a = –10 0 = 102 + 2 × (-10) × s2 M1 Apply Newton’s 2nd law after 4 N removed, find a and use v2 = u2 + 2as to find extra distance s2 [s2 = 5 leading to total distance = s1 + s2 = 15 + 5 =] 20 m A1 Alternative method for Question 7(c) Work done = 2 10 4 0.5 3 3 × × × [= 60 J] B1 FT WD = Fs and s = ½ at2 for 4 N force; FT a from (b) 60 = µ × 0.3g cos θ × d + 0.3g × d sin θ M1 WD by 4 N force = WD against F + PE gain d = 20 m A1 3

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