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Mathematics 9709/42 May/June 2020: question paper and mark scheme

Cambridge AS & A Level Mathematics (9709), Paper 4 Mechanics, May/June 2020. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.

9709_s20_qp_429709_s20_ms_429709/42/M/J/20

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Paper
Paper 4 Mechanics
Total marks
50
Time
1 hour 15 minutes
Code on the paper
9709/42/M/J/20
9709_s20_ms_42.pdf

9709/42 mark scheme, question by question

Each question as it's printed, and the mark scheme's points for it, as text. Diagrams, tables and graphs are in the PDFs above.

Question 16 marks

Question

A tram starts from rest and moves with uniform acceleration for 20 s. The tram then travels at a constant speed, V m s−1 , for 170 s before being brought to rest with a uniform deceleration of magnitude twice that of the acceleration. The total distance travelled by the tram is 2.775 km. (a) Sketch a velocity-time graph for the motion, stating the total time for which the tram is moving. [2] (b) Find V . [2] (c) Find the magnitude of the acceleration. [2]

Mark scheme

1(a) Trapezium, deceleration steeper than acceleration B1 Time from 0 to 200 B1 2 1(b) 0.5(170 + 200)v = 2775 M1 v = 15 A1 2 1(c) a = 15 ÷ 20 M1 a = 0.75 A1 2

Question 26 marks

Question

4P N 3P N 30Å 30Å 1Å PN 20 N Coplanar forces of magnitudes 20 N, P N, 3P N and 4P N act at a point in the directions shown in the diagram. The system is in equilibrium. Find P and 1. [6]

Mark scheme

2 Resolving forces in either direction M1 20 cos θ = 4 P cos 30 A1 4 P + 2 P sin 30 = 20sin θ A1 3 M1 cos θ = P 10 P sin θ = 4 3 2 1 2 P + P =1 100 16 P = 3.29 A1 θ = 55.3 A1 6

Question 38 marks

Question

TN 60Å 2.5 kg 20Å A particle of mass 2.5 kg is held in equilibrium on a rough plane inclined at 20Å to the horizontal by a force of magnitude T N making an angle of 60Å with a line of greatest slope of the plane (see diagram). The coefficient of friction between the particle and the plane is 0.3. Find the greatest and least possible values of T . [8]

Mark scheme

3 T sin 60 + R = 25cos 20 B1 Attempt at resolving in any direction M1 T cos 60 = F + 25 sin 20 A1 T cos 60 + F = 25 sin 20 A1 Use of F = μ R M1 T cos 60 = 25sin 20 ± 0.3(25cos 20 − T sin 60) M1 25sin 20 ± 0.3 × 25cos 20 T= cos 60 ± 0.3sin 60 T = 6.26 A1 T = 20.5 A1 8

Question 410 marks

Question

Small smooth spheres A and B, of equal radii and of masses 4 kg and 2 kg respectively, lie on a smooth horizontal plane. Initially B is at rest and A is moving towards B with speed 10 m s−1 . After the spheres collide A continues to move in the same direction but with half the speed of B. (a) Find the speed of B after the collision. [2] A third small smooth sphere C, of mass 1 kg and with the same radius as A and B, is at rest on the plane. B now collides directly with C. After this collision B continues to move in the same direction but with one third the speed of C. (b) Show that there is another collision between A and B. [3] (c) A and B coalesce during this collision. Find the total loss of kinetic energy in the system due to the three collisions. [5]

Mark scheme

4(a) 4 × 10 [ +0] = 4 × 0.5v + 2v M1 v A = 5 and vB = 10 A1 2 4(b) Conservation of momentum B, C M1 2 × 10 [+0] = 2 × v + 3v v=4 A1 v A > vB , hence another collision A1 3 4(c) Conservation of momentum A, B M1 14 A1 4 × their 5 + 2 × their 4 = 4v + 2v v= (ms −1 ) 3 1 M1 KE initial = × 4 × 102 2 1 14 1 A1 KE final = × 6 × their ( )2 + × 1× their 122 2 3 2 412 188 A1 Loss of KE = 200 − = 3 3 5

Question 510 marks

Question

A car of mass 1250 kg is moving on a straight road. (a) On a horizontal section of the road, the car has a constant speed of 32 m s−1 and there is a constant force of 750 N resisting the motion. (i) Calculate, in kW, the power developed by the engine of the car. [2] (ii) Given that this power is suddenly decreased by 8 kW, find the instantaneous deceleration of the car. [3] (b) On a section of the road inclined at sin−1 0.096 to the horizontal, the resistance to the motion of the car is 1000 + 8v N when the speed of the car is v m s−1 . The car travels up this section of the road at constant speed with the engine working at 60 kW. Find this constant speed. [5]

Mark scheme

5(a)(i) DF = 750 B1 Power = their (750) × 32 B1 FT = 24kW 2 5(a)(ii) 16000 = DF × 32 M1 DF = 500 500 − 750 = 1250 × a M1 a = [ −]0.2 A1 3 5(b) DF = 1000 + 8v + 1250 × 10 × 0.096 M1 2200 + 8v A1 60000 = (2200 + 8v)v M1 8v 2 + 2200v − 60000 = 0 A1 v = 25 A1 5

Question 610 marks

Question

A particle P moves in a straight line. The velocity v m s−1 at time t s is given by v = 2t + 1 for 0 ≤ t ≤ 5, v = 36 − t2 for 5 ≤ t ≤ 7, v = 2t − 27 for 7 ≤ t ≤ 13.5. (a) Sketch the velocity-time graph for 0 ≤ t ≤ 13.5. [3] (b) Find the acceleration at the instant when t = 6. [2] (c) Find the total distance travelled by P in the interval 0 ≤ t ≤ 13.5. [5] Additional Page If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.

Mark scheme

6(a) Correct for 0 ≤ t ≤ 5 B1 Correct for 5 ≤ t ≤ 7 B1 Correct for 7 ≤ t ≤ 13.5 B1 3 6(b) a = −2t by differentiating M1 a = −12 A1 2 6(c) 5 6 7 13.5 M1 s = (2t + 1) d t + (36 − t 2 ) d t + (36 − t 2 ) d t + 0 5 6 7 (2t − 27) d t 5 6 7 13.5 A1 s = (2t + 1) d t + (36 − t 2 ) d t + (36 − t 2 ) d t + 0 5 6 7 (2t − 27) d t t3 M1 s = [t 2 + t ] + [36t − ] + t 2 − 27t 3 All correct A1 s = 84.25 A1 5

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Where can I find the 9709/42 May/June 2020 mark scheme?

On this page, free: the mark scheme for AS & A Level Mathematics 9709/42, May/June 2020 (9709_s20_ms_42), and the question paper (9709_s20_qp_42), to read or download, with each question's marking points written out underneath.

How many marks is 9709/42 out of?

Mathematics 9709/42 for May/June 2020 is out of 50 marks, with 1 hour 15 minutes to answer it.

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