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Cambridge AS & A Level Mathematics (9709), Paper 4, October/November 2019. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
9709_w19_qp_439709_w19_ms_439709/43/O/N/19
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
A crate of mass 500 kg is being pulled along rough horizontal ground by a horizontal rope attached to a winch. The winch produces a constant pulling force of 2500 N and the crate is moving at constant speed. Find the coefficient of friction between the crate and the ground. [3]
Mark scheme
1 F=µ×500g B1 Use of F=µR [2500=µ×500g] M1 Resolving horizontally µ=0.5 A1 3
Question
A train of mass 150 000 kg ascends a straight slope inclined at !Å to the horizontal with a constant driving force of 16 000 N. At a point A on the slope the speed of the train is 45 m s−1 . Point B on the slope is 500 m beyond A. At B the speed of the train is 42 m s−1 . There is a resistance force acting on the train and the train does 4 × 106 J of work against this resistance force between A and B. Find the value of !. [5]
Mark scheme
2 PE gain =150000g × 500sinα (=75000000gsinα) B1 Correct expression for PE gain 1 1 B1 Correct expression for KE loss ×150000 × 452 − ×150000 × 422 (=19575000) 2 2 M1 For 5 term work energy equation (or 4 terms if using loss in KE as 1 term) 150000g × 500sinα=19575000+16000×500 – 4×106 A1 α =1.8 A1 5
Question
50 N 20Å 60 N !Å 30Å RN 100 N Three coplanar forces of magnitudes 50 N, 60 N and 100 N act at a point. The resultant of the forces has magnitude R N. The directions of these forces are shown in the diagram. Find the values of R and !. [6]
Mark scheme
3 Resolving horizontally or vertically M1 50cos20 + 60 – 100sin30 (=56.984…) A1 100cos30 – 50sin20 (= 69.501…) A1 56.984... M1 Method to find either R or α R = (56.984...2 + 69.501...2 ) or α = tan −1 69.501... R=89.9 (89.876…) A1 α=39.3 (39.348…) A1 6
Question
A car travels along a straight road with constant acceleration. It passes through points P, Q, R and S. The times taken for the car to travel from P to Q, Q to R and R to S are each equal to 10 s. The distance QR is 1.5 times the distance PQ. At point Q the speed of the car is 20 m s−1 . (i) Show that the acceleration of the car is 0.8 m s−2 . [3] (ii) Find the distance QS and hence find the average speed of the car between Q and S. [3]
Mark scheme
4(i) sPQ = 20 × 10 − 0.5a × 10 2 or sQR = 20 × 10 + 0.5a × 102 M1 1 1 For use of s = vt − at 2 or s = ut + at 2 OE 2 2 suvat to find PQ or QR s = 200–50a and 1.5s = 200 + 50a A1 OE 1.5(200 – 50a) = 200 + 50a Æ 100 = 125a Æ a = 0.8 ms–2 B1 AG 3 4(ii) 1 M1 1 Distance QS = 20 × 20 + × 0.8 × 202 Using s = ut + at 2 2 2 Distance=560 m A1 560 B1 Average speed between Q and S = = 28ms −1 20 3
Question
A cyclist is travelling along a straight horizontal road. The total mass of the cyclist and his bicycle is 80 kg. His power output is a constant 240 W. His acceleration when he is travelling at 6 m s−1 is 0.3 m s−2 . (i) Show that the resistance to the cyclist’s motion is 16 N. [3] (ii) Find the steady speed that the cyclist can maintain if his power output and the resistance force are both unchanged. [2] (iii) The cyclist later ascends a straight hill inclined at 3Å to the horizontal. His power output and the resistance force are still both unchanged. Find his acceleration when he is travelling at 4 m s−1 . [3] 6 Particle P travels in a straight line from A to B. The velocity of P at time t s after leaving A is denoted by v m s−1 , where v = 0.04t3 + ct2 + kt. P takes 5 s to travel from A to B and it reaches B with speed 10 m s−1 . The distance AB is 25 m. (i) Find the values of the constants c and k. [6] (ii) Show that the acceleration of P is a minimum when t = 2.5. [3]
Mark scheme
5(i) 240 B1 Use of power = force × velocity Driving force = (= 40 N) 6 [40 – R = 80 × 0.3] M1 Use of Newton’s Second Law (3 terms) Resistance is 16 N A1 AG 3 5(ii) 240 M1 Use of P=Fv with DF=resistance v = 16 Steady speed is 15 ms–1 A1 2 5(iii) Use of Newton’s Second Law M1 (4 terms) 240 A1 − 16 − 80 g sin 3 = 80a 4 Acceleration is 0.0266 ms–2 A1 3 Q6(i) 10 = 0.04 × 53 + 52 c + 5k ( 5c + k = 1) B1 Use of t=5, v=10 *M1 For use of s = vdt s= 0.04 4 ct 3 kt 2 t + + + (C ) ∫ 4 3 2 53 52 DM1 Use of t = 0, s = 0 and t = 5, s = 25 25 = 0.01× 54 + c+ k 3 2 125 25 125 25 A1 6.25 + c + k = 25 c + k = 18.75 3 2 3 2 Solving for c or for k M1 c = – 0.3 and k = 2.5 A1 6 Q6(ii) a = 0.12t 2 − 0.6t + 2.5 M1 dv For use of a = dt a ' = 0.24t − 0.6 = 0 → t = ... or a = 0.12(t 2 − 5t + ...) = 0.12 (t − 2.5) 2 + ... M1 da Uses = 0 or completes the square for a dt Minimum when t = 2.5 A1 AG 3
Question
A B m kg km kg 0.81 m Two particles A and B have masses m kg and km kg respectively, where k > 1. The particles are attached to the ends of a light inextensible string. The string passes over a fixed smooth pulley and the particles hang vertically below it. Both particles are at a height of 0.81 m above horizontal ground (see diagram). The system is released from rest and particle B reaches the ground 0.9 s later. The particle A does not reach the pulley in its subsequent motion. (i) Find the value of k and show that the tension in the string before B reaches the ground is equal to 12m N. [7] At the instant when B reaches the ground, the string breaks. (ii) Show that the speed of A when it reaches the ground is 5.97 m s−1 , correct to 3 significant figures, and find the time taken, after the string breaks, for A to reach the ground. [4] (iii) Sketch a velocity-time graph for the motion of particle A from the instant when the system is released until A reaches the ground. [2] 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
7(i) 1 2 M1 1 0.81 = 0 + 2 × a × 0.9 For use of s = ut + at 2 2 a=2 A1 T – mg = ma or kmg – T = kma M1 Use of Newton’s Second Law for A or B or ( m − mA ) g use of a = B ( mB + mA ) T – mg = ma and kmg – T = kma or a = ( km − m ) g A1 ( km + m ) a= ( kg − g ) = 2 → k = ... M1 Solves to find k ( k + 1) k = 1.5 A1 T = 10m + 2m = 12m N B1 AG 7 7(ii) Velocity of A when string breaks = 2 × 0.9 (=1.8 ms–1 upwards) B1FT For use of v=u+at ft a from (i) v2=1.82+2g×1.62 → v=... M1 For use of suvat to find vA at ground Speed is 5.97 ms–1 A1 AG Time taken = (1.8 + 5.97 ) = 0.777 s (0.7769...) B1 g 4 7(iii) Straight line from (0, 0) to (0.9, 1.8) B1 Straight line from (0.9, 1.8) to approx. (1.7, –6) B1FT FT 0.9 + t from (ii) for 1.7 2
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Mathematics 9709/43 for October/November 2019 is out of 50 marks, with 1 hour 15 minutes to answer it.
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