Cambridge A Level Mathematics 9709 — 2017 Oct/Nov Paper 4 · Variant 2

1 A particle of mass 0.2 kg is resting in equilibrium on a rough plane inclined at 20Å to the horizontal. (i) Show that the friction force acting on the particle is 0.684 N, correct to 3 significant figures. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ The coefficient of friction between the particle and the plane is 0.6. A force of magnitude 0.9 N is applied to the particle down a line of greatest slope of the plane. The particle accelerates down the plane. (ii) Find this acceleration. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1(i) F = 0.2g sin 20 = 0.684 N B1 AG 1 1(ii) R = 0.2g cos 20 B1 F = µR [= 0.6 × 0.2g cos 20] M1 Using F = µR F = 1.1276… [0.9 + 0.2g sin 20 – F = 0.2a] M1 Use of Newton’s 2nd law along the plane (4 relevant terms) a = 2.28 ms-2 A1 4

More questions on Kinematics of motion in a straight line

2 45Å 1Å 120 N T N A block of mass 15 kg hangs in equilibrium below a horizontal ceiling attached to two strings as shown in the diagram. One of the strings is inclined at 45Å to the horizontal and the tension in this string is 120 N. The other string is inclined at 1Å to the horizontal and the tension in this string is T N. Find the values of T and 1. [6] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 EITHER: (M1 Attempt to resolve (either direction with correct number of terms and dimensionally correct) T sin θ + 120 sin 45 = 15g A1 Resolving vertically T cos θ = 120 cos 45 A1 Resolving horizontally (15 g –120sin 45 ) M1 For using division to find θ or for using [tan θ = Pythagoras to find T (120cos45 ) or T = 65.15 2 + 84.85 2 ] θ = 37.5 A1 T = 107 A1) OR1: (A1 One correct equation 120 T 15 g = = sin ( 90 + θ) sin135 sin (135 − θ) A1 A second correct equation M1 Attempt to solve for θ or T θ = 37.5 A1 T = 107 A1 M1) Attempt to use triangle of forces

More questions on Forces and equilibrium

3 A car travels along a straight road with constant acceleration. It passes through points A, B and C. The car passes point A with velocity 14 m s−1. The two sections AB and BC are of equal length. The times taken to travel along AB and BC are 5 s and 3 s respectively. (i) Write down an expression for the distance AB in terms of the acceleration of the car. Write down a similar expression for the distance AC. Hence show that the acceleration of the car is 4 m s−2. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Find the speed of the car as it passes point C. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 3(i) sAB = 14 × 5 + ½a × 52 B1 or sAB = ½(14 + 14 + 5a) × 5 OE sAC = 14 × 8 + ½a × 82 B1 or sAC = ½(14 + 14 + 8a) × 8 OE [112 + 32a = 2(70 + 12.5a)] M1 Using AC = 2AB and solving for a or for substituting a = 4 and finding AB and AC a = 4 m s–2 A1 AG, If substituting a = 4 must show AB = 120 and AC = 240 OE 4 3(ii) [v = 14 + 4 × 8] M1 Use of v = u + at or any complete method to find v Velocity = 46 m s-1 A1 2

More questions on Kinematics of motion in a straight line

4 A particle P is projected vertically upwards from horizontal ground with speed 12 m s−1. (i) Find the time taken for P to return to the ground. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ The time in seconds after P is projected is denoted by t. When t = 1, a second particle Q is projected vertically upwards with speed 10 m s−1 from a point which is 5 m above the ground. Particles P and Q move in different vertical lines. (ii) Find the set of values of t for which the two particles are moving in the same direction. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(i) [12t – ½gt2 = 0] M1 Using s = ut + ½at2 or equivalent such as or finding time T to highest point and [0 = 12 – gT] with t = 2T used doubling. t = 2.4 s A1 2 4(ii) Critical point at t = 1.2 B1 Seen in 4(ii) Critical point at t = 2 B1 Seen in 4(ii) Both moving in same direction B1 1 < t < 1.2 Both moving in same direction B1 2 < t < 2.4 4

More questions on Kinematics of motion in a straight line

5 A cyclist is riding up a straight hill inclined at an angle ! to the horizontal, where sin ! = 0.04. The total mass of the bicycle and rider is 80 kg. The cyclist is riding at a constant speed of 4 m s−1. There is a force resisting the motion. The work done by the cyclist against this resistance force over a distance of 25 m is 600 J. (i) Find the power output of the cyclist. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ The cyclist reaches the top of the hill, where the road becomes horizontal, with speed 4 m s−1. The cyclist continues to work at the same rate on the horizontal part of the road. (ii) Find the speed of the cyclist 10 seconds after reaching the top of the hill, given that the work done by the cyclist during this period against the resistance force is 1200 J. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 5(i) EITHER: (B1 600 Resistance force = = 24 N 25 Weight component = 80 g (0.04) B1 For correct unsimplified numerical form = 32 N of the weight component [Power = 56 × 4] M1 For use of P = Fv where F is from two relevant force terms Power = 224 W A1) 4 OR: (B1 For a correct unsimplified numerical PE gain = 80g × 25 (0.04) expression for PE = 800 25 B1 Time taken = = 6.25 4 [WD by cyclist = P × 6.25 = 800 + 600] M1 For using WD = P × t where WD is from two relevant terms Power = 224 W A1) 4 5(ii) Work done by cyclist B1 FT For stating WD = power × time = 224 × 10 ( = 2240J) FT on P value found in 5(i) Initial KE = ½ × 80 × 42 [= 640 J] B1 [½ × 80v2 = 640 + P×10 –1200] M1 For using Work/Energy equation Speed = 6.48 m s–1 A1 Allow speed = √42 4

More questions on Energy, work and power

6 P ! Q Two particles P and Q, each of mass m kg, are attached to the ends of a light inextensible string. The string passes over a fixed smooth pulley which is attached to the edge of a rough plane. The plane is inclined at an angle ! to the horizontal, where tan ! = 24.7 Particle P rests on the plane and particle Q hangs vertically, as shown in the diagram. The string between P and the pulley is parallel to a line of greatest slope of the plane. The system is in limiting equilibrium. (i) Show that the coefficient of friction between P and the plane is 43. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ A force of magnitude 10 N is applied to P, acting up a line of greatest slope of the plane, and P accelerates at 2.5 m s−2. (ii) Find the value of m. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 6(i) R = mg cos α (R = 9.6m) B1 Allow use of α = 16.3º throughout [T = mg M1 For resolving forces on P and Q and F = mg sin α + T ] eliminating T or for considering the equilibrium of the system F = mg sin α + mg A1 (F = 12.8m) M1 For use of F = µR 4 A1 AG so must be from exact working Coefficient of friction = 1⅓ = 3 5 6(ii) EITHER: (*M1 For applying Newton’s 2nd law to P equation is P (5 terms) or Q (3 terms) 10 – mg sin α – F – T = 2.5 m Q equation is T – mg = 2.5m *M1 For applying Newton’s 2nd law to the other particle and eliminate T 10 – mg sin α – µmg cos α A1 If evaluated then this is – mg = 2m (2.5) 10 – 2.8m – 12.8m – 10m = 5m DM1 For solving this equation for m as far as m = Dependent on one or other of the previous M marks having been scored m = 0.327 A1) 50 Allow m = 153 OR: (*M1 For applying Newton’s 2nd law to the [10 – mg sin α –F – mg = m(2.5 + 2.5)] system. Allow with 5 terms *M1 System equation with all 6 terms 10 – mg sin α – µmg cos α A1 – mg = 2m (2.5) DM1 For solving this equation for m as far as m = Dependent on one or other of the previous M marks having been scored m = 0.327 A1) 50 Allow m = 153 5

More questions on Forces and equilibrium

7 A particle starts from rest and moves in a straight line. The velocity of the particle at time t s after the start is v m s−1, where v = −0.01t3 + 0.22t2 −0.4t. (i) Find the two positive values of t for which the particle is instantaneously at rest. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Find the time at which the acceleration of the particle is greatest. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (iii) Find the distance travelled by the particle while its velocity is positive. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 7(i) –0.01t(t2 – 22t + 40) = 0 M1 Attempting to solve v = 0 for t for a –0.01t(t – 20)(t – 2) = 0 solvable quadratic using factors or quadratic formula and obtaining two non- zero solutions t = 2 or t = 20 A1 2 7(ii) a = – 0.03t2 + 0.44t – 0.4 M1 For differentiation a is greatest (maximum) when M1 For differentiation or finding values of 0.44 – 0.06t = 0 t = t1 and t = t2 where a = 0 and using t = ½(t1 + t2) or completing the square or other method to find maximum value Max acceleration when t = 7.33 A1 22 Allow t = 3 3 7(iii) ∫− 0.01t 3 + 0.22t 2 − 0.4t d t *M1 For using integration. ( ) 0.01 4 0.22 3 2 A1 Correct Integration s ( t ) = − t + t − 0.2t Allow + C included 4 3 s ( 20 ) − s ( 2 ) DM1 Limits 2 and 20 used correctly Dependent on previous M1 having been scored Distance = 107 m A1 2673 Distance = = 106.92 25 4

More questions on Differentiation