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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_419709_w19_ms_419709/41/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 crane is lifting a load of 1250 kg vertically at a constant speed V m s−1 . Given that the power of the crane is a constant 20 kW, find the value of V . [2]
Mark scheme
1 20 000 = V × 1250g M1 Use of P = Fv with F = mg V = 1.6 A1 2
Question
The total mass of a cyclist and her bicycle is 75 kg. The cyclist ascends a straight hill of length 0.7 km inclined at 1.5Å to the horizontal. Her speed at the bottom of the hill is 10 m s−1 and at the top it is 5 m s−1 . There is a resistance to motion, and the work done against this resistance as the cyclist ascends the hill is 2000 J. The cyclist exerts a constant force of magnitude F N in the direction of motion. Find F. [5]
Mark scheme
2 Initial KE = 12 × 75 × 102 B1 Either correct Final KE = 12 × 75 × 52 PE gained = 75 g × 700 sin1.5 [=13 743] B1 WD by F = F × 700 B1 For WD by F = F × d WD by F + Initial KE = Final KE + PE gain + 2000 M1 Use of work-energy equation. 5 dimensionally correct terms. F = 18.5 A1 5
Question
A block of mass 3 kg is at rest on a rough plane inclined at 60Å to the horizontal. A force of magnitude 15 N acting up a line of greatest slope of the plane is just sufficient to prevent the block from sliding down the plane. (i) Find the coefficient of friction between the block and the plane. [5] The force of magnitude 15 N is now replaced by a force of magnitude X N acting up the line of greatest slope. (ii) Find the greatest value of X for which the block does not move. [2]
Mark scheme
3(i) R = 3 g cos 60 B1 Use F = µR M1 [3gsin 60 – µ3gcos60 – 15 = 0] M1 Resolve forces parallel to the plane, 3 terms A1 Correct equation µ = 0.732 A1 Allow µ = 3 − 1 5 3(ii) [Maximum force = 3gsin60 + F M1 = 3 sin60 + µ3gcos60] X = 37(.0) ( A1 Allow X = 15 2 3 − 1 ) 2
Question
36 N 5 kg 4 kg B A ! Two blocks A and B of masses 4 kg and 5 kg respectively are joined by a light inextensible string. The 7 . The string is blocks rest on a smooth plane inclined at an angle ! to the horizontal, where tan ! = 24 parallel to a line of greatest slope of the plane with B above A. A force of magnitude 36 N acts on B, parallel to a line of greatest slope of the plane (see diagram). (i) Find the acceleration of the blocks and the tension in the string. [5] (ii) At a particular instant, the speed of the blocks is 1 m s−1 . Find the time, after this instant, that it takes for the blocks to travel 0.65 m. [2]
Mark scheme
4(i) Apply Newton’s second law to either or to the system M1 Block A: T – 4g × 257 = 4a A1 Any two correct. Allow α = 16.3 used. Block B: 36 – T – 5g × 257 = 5a System: 36 – 5g × 257 – 4g × 257 = 9a Either solving the system for a or solving a pair of simultaneous M1 equations for either a or T a = 1.2 ms–2 A1 T = 16 N A1 5 4(ii) 1 2 M1 Use constant acceleration equation(s) with u = 1 and solve a 3 0.65 = 1× t + 2 × 1.2 t term quadratic equation to find t t = 0.5 s A1 Alternative method for question 4(ii) 1 M1 Use relevant constant acceleration equations with u = 1 in a v2 = 12 + 2 × 1.2 × 0.65 [v = 1.6] and 0.65 = (1 + v ) × t complete method to find t 2 t = 0.5 s A1 2
Question
FN 4.5 N P 1Å 20Å A B 60Å 7.5 N A small ring P is threaded on a fixed smooth horizontal rod AB. Three horizontal forces of magnitudes 4.5 N, 7.5 N and F N act on P (see diagram). (i) Given that these three forces are in equilibrium, find the values of F and 1. [6] (ii) It is given instead that the values of F and 1 are 9.5 and 30 respectively, and the acceleration of the ring is 1.5 m s−2 . Find the mass of the ring. [2]
Mark scheme
5(i) Resolve forces either horizontally or vertically M1 7.5cos60 + 4.5cos20 = Fcosθ [= 7.97861] A1 7.5sin60 – 4.5sin20 = Fsinθ [= 4.95609] A1 F= ( 7.98 + 4.96 ) 2 2 M1 Use Pythagoras or use the value found for θ to find F θ = tan–1( 4.96 7.98 ) M1 Use trigonometry or the value found for F to find θ F = 9.39 and θ = 31.8 A1 Alternative method for question 5(i) F 4.5 7.5 M1 Attempt to use Lami = = sin 80 sin (120 + θ ) sin (160 − θ ) A1 One correct pair of terms A1 A second correct pair of terms [4.5sin(160 – θ) = 7.5sin(120 + θ)] M1 Attempt to solve for θ Use the θ value found by valid trigonometry to find F M1 F = 9.39 and θ = 31.8 A1 5(i) Alternative method for question 5(i) Forces 4.5, 7.5, F opposite angles 60 – θ, θ + 20, 100 M1 Illustrate a triangle of forces [F2 = 4.52 + 7.52 – 2 × 4.5 × 7.5 × cos100] M1 For application of cosine rule to find F A1 Correct equation 9.39 4.5 7.5 M1 One application of the sine rule to find θ = = sin100 sin ( 60 − θ ) sin (θ + 20 ) A1 Correct equation F = 9.39 and θ = 31.8 A1 6 5(ii) 9.5cos30 – 7.5cos60 – 4.5cos20 = m × 1.5 M1 Apply Newton’s second law to the ring along AB (4 terms) m = 0.166 kg A1 2
Question
A particle of mass 0.4 kg is released from rest at a height of 1.8 m above the surface of the water in a tank. There is no instantaneous change of speed when the particle enters the water. The water exerts an upward force of 5.6 N on the particle when it is in the water. (i) Find the velocity of the particle at the instant when it reaches the surface of the water. [2] (ii) Find the time that it takes from the instant when the particle enters the water until it comes to instantaneous rest in the water. You may assume that the tank is deep enough so that the particle does not reach the bottom of the tank. [4] (iii) Sketch a velocity-time graph for the motion of the particle from the instant at which it is released until it comes to instantaneous rest in the water. [3]
Mark scheme
6(i) 1 M1 KE gain = PE lost 0.4 g × 1.8 = × 0.4 × v 2 2 v = 6 ms–1 A1 Alternative method for question 6(i) v2 = 02 + 2 × g × 1.8 M1 Use constant acceleration equation(s) with a = g to find v v = 6 ms–1 A1 2 6(ii) 0.4g – 5.6 = 0.4a M1 Use Newton’s second law for the particle in the vertical (3 terms) a = –4 ms–2 A1 0 = 6 – 4t M1 Use of constant acceleration equation(s) such as v = u + at to find t t = 1.5 s A1 4 6(iii) Straight line starting at (0,0) with positive gradient B1 Second straight line starting at end of the first line with negative B1 gradient and ending with v = 0 All correct, start at (0, 0) with max velocity v = 6 at t = 0.6 B1FT FT on their v from (i) and/or their t from (ii) i.e. (0.6, 6) and finishing at (2.1, 0) 3
Question
A particle moves in a straight line, starting from rest at a point O, and comes to instantaneous rest at a point P. The velocity of the particle at time t s after leaving O is v m s−1 , where v = 0.6t2 − 0.12t3 . (i) Show that the distance OP is 6.25 m. [5] On another occasion, the particle also moves in the same straight line. On this occasion, the displacement of the particle at time t s after leaving O is s m, where s = kt3 + ct5 . It is given that the particle passes point P with velocity 1.25 m s−1 at time t = 5. (ii) Find the values of the constants k and c. [5] (iii) Find the acceleration of the particle at time t = 5. [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) 0.6t2 – 0.12t3 = 0 M1 For attempting to solve v = 0 (t = 0 or) t = 5 A1 ∫v dt = 0.2t3 – 0.03t4 *M1 For integrating the velocity OP = [0.2 × 53 – 0.03 × 54] – [0] DM1 Use limits to find OP Distance = 6.25 m A1 AG 5 7(ii) k × 53 + c × 55 = 6.25 B1 Using s = 6.25 at t = 5 to set up equation in k and c v = 3kt2 + 5ct4 *M1 For differentiating s to find v 1.25 = 3k × 52 + 5c × 54 DM1 For using the given value of v = 1.25 in the expression for v 125k + 3125c = 6.25 M1 For attempting to solve a pair of simultaneous equations in k and 75k + 3125c = 1.25 c and finding a value of either k or c k = 0.1, c = –0.002 A1 5 7(iii) a = 0.6t – 0.04t3 M1 For differentiating their expression for v At t = 5, a = –2 Acceleration = –2 ms–2 A1 2
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Mathematics 9709/41 for October/November 2019 is out of 50 marks, with 1 hour 15 minutes to answer it.
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