Cambridge A Level Mathematics 9709 — 2019 May/June Paper 4 · Variant 2

1 40 N 17 N 55Å 20Å 1Å 32 N P N Coplanar forces of magnitudes 40 N, 32 N, P N and 17 N act at a point in the directions shown in the diagram. The system is in equilibrium. Find the values of P and 1. [6] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 [P cos θ = 32 cos 20 – 17 sin 55] [P sin θ = 40 + 17 cos 55 – 32 sin 20] M1 Resolve forces horizontally or vertically 3 terms horizontally, 4 terms vertically A1 One correct A1 Both correct [P sin θ = 38.8062 P cos θ = 16.1446] ( ) ( ) 2 2 17cos55 32sin 20 40 32cos20 17cos35 = − + + − P M1 Either use Pythagoras to find P or use their value of θ to find P 1 4 ) t ( an (17cos55 32sin20 0 32cos20 17cos35) θ −  =    − +  − M1 Either use trigonometry to find θ or use their value of P to find θ [tan θ = 2.4037] P = 42(.0) and θ = 67.4 A1 6

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2 A car moves in a straight line with initial speed u m s−1 and constant acceleration a m s−2. The car takes 5 s to travel the first 80 m and it takes 8 s to travel the first 160 m. Find a and u. [6] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 Possible equations include: t = 0 to t = 5 → 80 = 5u + 12.5a t = 0 to t = 8 → 160 = 8u + 32a t = 5 to t = 8 → 80 = 3(u + 5a) + 4.5a i.e. 80 = 3u + 19.5a M1 Use the equation s = ut + ½at 2 to set up one equation in u and a or using speeds as u (at t = 0), u + 5a (at t = 5), u + 8a (at t = 8) and then apply s = ½ × (u + v) × t 80 = 5u + ½ × a × 52 → 5u + 12.5a = 80 A1 One correct equation in a and u 160 = 8u + 0.5a × 82 → 8u + 32a = 160 A1 Second correct equation in a and u M1 Attempt to solve a pair of valid simultaneous equations for a or u a = 8 3 A1 Allow a = 2.67 u = 28 3 A1 Allow u = 9.33 6

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3 A particle of mass 13 kg is on a rough plane inclined at an angle of 1 to the horizontal, where tan 1 = 12.5 The coefficient of friction between the particle and the plane is 0.3. A force of magnitude T N, acting parallel to a line of greatest slope, moves the particle a distance of 2.5 m up the plane at a constant speed. Find the work done by this force. [5] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 3 B1 Resolve perpendicular to the plane F = 0.3 × 13g cos 22.6 [F = 36] M1 Using F = µR T = F + 13g sin 22.6 = F + 13g × (5/13), [T = 86] M1 Apply Newton’s second law parallel to the plane with a = 0 WD = T × 2.5 [= 86 × 2.5] M1 WD = T × d WD = 215 J A1 Alternative method for question 3 R = 13g cos 22.6 = 13g × (12/13), [R = 120] B1 Resolve perpendicular to the plane F = 0.3 × 13g cos 22.6 [F = 36] M1 Using F = µR PE gain = 13 × g × 2.5 × (5/13) [= 125] M1 Attempt PE gain. Allow sin 22.6 for 5/13 [WD by T = 13 × g × 2.5 × (5/13) + F × 2.5] M1 Using WD by T = PE gain + WD against F WD by T = 215 J A1 5

More questions on Energy, work and power

4 A constant resistance to motion of magnitude 350 N acts on a car of mass 1250 kg. The engine of the car exerts a constant driving force of 1200 N. The car travels along a road inclined at an angle of 1 to the horizontal, where sin 1 = 0.05. Find the speed of the car when it has moved 100 m from rest in each of the following cases. • The car is moving up the hill. • The car is moving down the hill. 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Mark scheme: 4 M1 Apply Newton’s second law for motion up the hill [a = 225/1250 = 0.18] A1 Correct Newton’s law for motion up the hill [1200 – 350 + 1250 × 10 × 0.05 = 1250a] M1 Apply Newton’s second law for motion down the hill [a = 1475/1250 = 1.18] A1 Correct Newton’s law for motion down the hill Up the hill: v 2 = 0 + 2 × 0.18 × 100 Down the hill: v 2 = 0 + 2 × 1.18 × 100 M1 Use their a in the constant acceleration equations either to find v going up or going down the hill Up the hill: v = 6 ms–1 A1 Down the hill: v = 15.4 ms–1 A1 Allow v = 2√59 Alternative method for question 4 [1200 × 100 = 350 × 100 + 1250g × 100 × 0.05 + ½ × 1250 × v 2] M1 Attempt the work-energy equation for motion up the hill A1 Correct work-energy equation for motion up the hill [1200 × 100 + 1250g × 100 × 0.05 = 350 × 100 + ½ × 1250 × v 2] M1 Attempt work-energy equation for motion down the hill A1 Correct work-energy equation for motion down the hill M1 Attempt to solve either energy equation to find either v going up the hill or v going down the hill Up the hill: v = 6 ms–1 A1 Down the hill: v = 15.4 ms–1 A1 Allow v = 2√59 7

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5 A B 0.4 kg 0.2 kg 0.5 m Two particles A and B, of masses 0.4 kg and 0.2 kg respectively, are connected by a light inextensible string which passes over a fixed smooth pulley. Both A and B are 0.5 m above the ground. The particles hang vertically (see diagram). The particles are released from rest. In the subsequent motion B does not reach the pulley and A remains at rest after reaching the ground. (i) For the motion before A reaches the ground, show that the magnitude of the acceleration of each particle is 10 m s−2 and find the tension in the string. 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(ii) Find the maximum height of B above the ground. 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Mark scheme: 5(i) A: 4 – T = 0.4a B: T – 2 = 0.2a System: 4 – 2 = (0.4 + 0.2)a M1 Apply Newton’ second law to particle A (3 terms) or to particle B (3 terms) or to the system (4 terms implied) A1 Two correct equations M1 Either solve the system equation for a or solve two simultaneous equations for a or T or verify the given value of a by finding the same T value in both equations a = 10 3 , T = 8 3 A1 Both correct AG 4 5(ii) M1 Apply v 2 = u 2 +2as to particle A or particle B with a = 10/3 v2 = 0 + 2 × 10/3 × 0.5 A1 [v = 1.83 but not needed specifically] 0 = 10/3 – 2 × 10 × s [s = 1 6 ] M1 Apply v 2 = u 2 + 2as to particle B to find s, the distance travelled by B after A has hit the ground Maximum height = 7 6 = 1.17 m A1 Maximum height = 1/2 + 1/2 + 1/6 = 7/6 = 1.17 4

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7 Particles P and Q leave a fixed point A at the same time and travel in the same straight line. The velocity of P after t seconds is 6t t −3 m s−1 and the velocity of Q after t seconds is 10 −2t m s−1. (i) Sketch, on the same axes, velocity-time graphs for P and Q for 0 ≤t ≤5. [3] (ii) Verify that P and Q meet after 5 seconds. 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(iii) Find the greatest distance between P and Q for 0 ≤t ≤5. 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Mark scheme: 7(i) Straight line, reaching positive v-axis and positive t-axis (negative gradient) B1 Quadratic (U shape, through (0,0) and cutting t-axis at t < 5) B1 Fully correct graphs with correct labelling with t = 3, t = 5, v = 10, v = 60 seen B1 3 7(ii) ( ) 2 10 2 d 10 = ∫ − = − s t t t t (+ c) or use area of a triangle ½ × 10 × 5 [= 25] B1 Use either integration to find s for Q or use a correct formula to find the area under the relevant triangle M1 Use integration to find the displacement for P ( ) ( ) 2 3 2 6 18 d 2 9 = ∫ − = − + s t t t t t c A1 Correct integration for P (unsimplified) ( ) 5 3 2 0 2 9 25   = − =   s P t t or solve 2 3 2 10 2 9 − = − t t t t B1 Either evaluation of s(P) at t = 5 and show that at t = 5, s(P) = s(Q) = 25 or show that t = 5 is a solution of the cubic by solving or verify t = 5 is a solution of the cubic by substitution. 4 Question Answer Marks Guidance 7(iii) Distance PQ = |sP – sQ| = ±(2t 3 – 8t 2 – 10t) M1 Find the distance between P and Q Allow either sign sP and sQ must have been found by integration Maximum s if 6t 2 – 16t – 10 = 0 M1 Differentiate to obtain an equation in t and attempt to solve t = 3.19 A1 Maximum Distance PQ = (–)48.4 m A1 Alternative method for question 7(iii) 6t 2 – 18t = 10 – 2t M1 State that greatest distance between P and Q occurs when vP = vQ 6t 2 – 16t – 10 = 0 M1 Rearrange and attempt to solve for t t = 3.19 A1 Maximum Distance PQ = (–)48.4 m A1 4

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