Cambridge A Level Mathematics 9709 — 2018 Feb/March Paper 4 · Variant 2

1 Two particles A and B, of masses 0.8 kg and 0.2 kg respectively, are connected by a light inextensible string that passes over a fixed smooth pulley. The particles hang vertically. The system is released from rest. Show that the acceleration of A has magnitude 6 m s−2 and find the tension in the string. [4] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1 [T – 2 = 0.2a 8 – T = 0.8a] M1 Attempt Newton’s 2nd law for System is 0.8g – 0.2g = (0.2 + 0.8)a either particle or use a formula for and T = 2(0.2)(0.8)g/(0.8 + 0.2) the system for a and/or T A1 Two correct equations Attempt to solve for a or T M1 a = 6 T = 3.2 A1 Both correct NB a = 6 AG 4

More questions on Newton’s laws of motion

2 2P N 1Å 10 N 60Å P N The three coplanar forces shown in the diagram are in equilibrium. Find the values of 1 and P. [4] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 EITHER: (M1 Resolve vertically (2 terms) 2P sin θ = P sin 60 θ = 25.7 A1 2P cos θ + P cos 60 = 10 M1 Resolve horizontally (3 terms) P = 4.34 A1) OR1: (M1 Attempt Lami’s theorem using one  2 P P 10  pair of terms  = =   sin120 sin ( 180 − θ) sin ( 60 + θ)  θ = 25.7 A1 Solve for θ Use a second Lami equation M1 P = 4.34 A1) OR2: (M1 Use sine or cosine rule with triangle of forces using forces P, 2P and 10 and with angles 60, θ and 120 – θ between θ = 25.7 A1 Use a second relationship from the triangle of M1 forces P = 4.34 A1) 4

More questions on Forces and equilibrium

3 A 7.2 m B A girl, of mass 40 kg, slides down a slide in a water park. The girl starts at the point A and slides to the point B which is 7.2 metres vertically below the level of A, as shown in the diagram. (i) Given that the slide is smooth and that the girl starts from rest at A, find the speed of the girl at B. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) It is given instead that the slide is rough. On one occasion the girl starts from rest at A and reaches B with a speed of 10 m s−1. On another occasion the girl is pushed from A with an initial speed V m s−1 and reaches B with speed 11 m s−1. Given that the work done against friction is the same on both occasions, find V. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 3(i) 1 2 M1 Use of KE gain = PE loss × 40 × v = 40 × g × 7.2 2 v = 12 m s–1 A1 2 3(ii) Work done against friction(WDF) M1 May be calculated as 1 2 1 2 1 2 WDF = 40 × g × 7.2 − × 40 × 10 [ = 880 ] × 40 × 12 − × 40 × 10 2 2 2 1 2 1 2 M1 For 4-term work-energy equation × 40 × V + 40 × g × 7.2 = × 40 × 11 + 880 with numerical attempt at work 2 2 done or or using the fact that WDF is the same 1 2 1 2 1 2 × 40 × V = × 40 × 11 − × 40 × 10 in both cases, extra initial KE = 2 2 2 difference in final KEs V = 21 = 4.58 A1 3

More questions on Energy, work and power

4 A particle of mass 12 kg is on a rough plane inclined at an angle of 25Å to the horizontal. A force of magnitude P N acts on the particle. This force is horizontal and the particle is on the point of moving up a line of greatest slope of the plane. The coefficient of friction between the particle and the plane is 0.8. Find the value of P. [6] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 4 [R = 12g cos 25 + P sin 25 M1 Attempt resolving of forces in any P cos 25 = F + 12g sin 25] one direction, parallel to, or perpendicular to plane [P = F cos 25 + R sin 25 or R cos 25 = F sin 25 + 12g] horizontally, vertically A1 Any one correct equation A1 Any second correct equation F = 0.8R M1 Use of F = µR Complete method to find P from 2 M1 equations(3 terms each) P = 242 A1 6

More questions on Forces and equilibrium

5 A small rocket is fired vertically upwards, starting from rest at ground level, and moves with constant acceleration. The rocket reaches a height of 200 m after 10 s. (i) Show that the speed of the rocket after 10 s is 40 m s−1 and find the acceleration of the rocket during the first 10 s. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) After 10 s, the rocket’s fuel stops burning and there is no upward force acting on the rocket. Find the maximum height above ground level reached by the rocket. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (iii) Find the total time from the instant the rocket is fired until it returns to the ground. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 5(i) 200 = ½ × (0 + v) × 10 M1 Use of suvat v = 40 m s–1 A1 AG 200 = ½ × a × 102 M1 Second use of suvat a = 4 m s–2 A1 4 5(ii) 0 = 402 – 2 × g × s M1 Use of suvat with a = g s = 80 so height above ground = 280 m A1 2 5(iii) EITHER: (M1 Use of suvat to find extra time to 0 = 40 – gt1 highest point t1 = 4 A1 280 = ½gt2 2 M1 Use of suvat to find time from highest point to ground t2 = √56 = 7.48... so total time = 21.5 s A1) OR: (M1 Use of s = ut + ½at2 −200 = 40t3 − ½gt3 2 with 200, 40 and g used 5t32 – 40t3 – 200 = 0 o.e. A1 Correct quadratic for time under [t3 2 – 8t3 – 40 = 0] gravity [t3 = 4 ± √56 = 4 ± 7.48] M1 Solution of relevant 3-term quadratic t3 = 11.48 so total time is 21.5 s A1) 4

More questions on Kinematics of motion in a straight line

6 A car of mass 1200 kg has a greatest possible constant speed of 60 m s−1 along a straight level road. When the car is travelling at a speed of v m s−1 there is a resistive force of magnitude 35v N. (i) Find the greatest possible power of the car. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) The car travels along a straight level road. Show that, at an instant when its speed is 30 m s−1, the greatest possible acceleration of the car is 2.625 m s−2. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (iii) The car travels at a constant speed up a hill inclined at an angle of sin−1 7 to the horizontal. 48 Find the greatest possible speed of the car. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 6(i) Driving force = 35 × 60 M1 Power = 35 × 602 = 126000 W A1 2 6(ii) 126000 B1FT Driving force is DF = 30 DF − 35 × 30 = 1200 a M1 For 3-term Newton’s 2nd law equation, dimensionally correct 3150 21 A1 AG a = = = 2.625 m s–2 1200 8 3 6(iii) 126000 M1 P DF = For F = v v 126000 7 M1 For 3-term force equation, or = 35v + 1200 g × equivalent v 48 A1 For correct (unsimplified) equation 35v 2 + 1750v − 126000 = 0 M1 For simplifying and solving of a 3- 2 term quadratic attempted or v + 50v − 3600 = 0 v = 40 ms-1 A1 v = −90 rejected or ignored 5

More questions on Forces and equilibrium

7 A particle P moves in a straight line. The velocity v m s−1 at time t s is given by v = 4 + 0.2t for 0 ≤t ≤10, 800 v = −2 + for 10 ≤t ≤20. t2 (i) Find the acceleration of P during the first 10 s. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Find the acceleration of P when t = 20. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (iii) Sketch the velocity-time graph for 0 ≤t ≤20. [3] (iv) Find the total distance travelled by P in the interval 0 ≤t ≤20. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 7(i) 0.2 (m s–2) B1 1 7(ii) a = −1600t −3 M1 For attempted differentiation of 800 −+2 t 2 Acceleration at t = 20 is –0.2 (m s–2) A1 2 7(iii) Straight line joining B1 t = 0, v = 4 to t = 10, v = 6 Curve with correct concavity joining end of B1 line to t = 20, v = 0 Correct labelling on axes provided the curves B1 pass through (0,4), (10,6), (20,0) 3 7(iv) Trapezium area = 50 B1 or from integration of 4 + 0.2t ∫−+2 800t − 2 d t = − 2t − 800t − 1 M1 Integration attempted ( ) A1 Correct indefinite integral −1 20 M1 Correct use of the limits  −2t − 800t   10 t = 10 and t = 20 = −40 − 40 + 20 + 80 Distance is 50 + 20 = 70 m A1 Correct total 5

More questions on Kinematics of motion in a straight line