Cambridge A Level Mathematics 9709 — 2007 Oct/Nov Paper 4 · Variant 1

1 A car of mass 900 kg travels along a horizontal straight road with its engine working at a constant rate of P kW. The resistance to motion of the car is 550 N. Given that the acceleration of the car is 0.2 m s−2 at an instant when its speed is 30 m s−1, find the value of P. [4]

Mark scheme: 1 M1 For using Newton’s second law (3 terms) DF – 550 = 900x0.2 A1 [P = 730x30 ÷ 1000] M1 For using P = (DF)v P = 21.9 A1 4 2

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2 A particle is projected vertically upwards from a point O with initial speed 12.5 m s−1. At the same instant another particle is released from rest at a point 10 m vertically above O. Find the height above O at which the particles meet. [5]

Mark scheme: 2 M1 For applying s = ut + ½ at2 or (u + at)2 = u2 +2as with a = ± g (either particle) s1 = 12.5t – ½ gt2, s2 = ± ½ gt2 or A1 (12.5 – gt)2 = 12.52 – 2gs1 and (gt)2 = 2gs2 [12.5t – ½ gt2 + ½ gt2 = 10] M1 For using s1 + s2 = 10 t = 0.8s or A1 2s1 = 25 2 − 0 . 2 s 1 − ( 20 − 2 s 1 ) (or better) Height is 6.8m A1ft 5 ft for 12.5t – 5t2 or 10 – 5t2 with candidate’s t (requires both M marks) 2 2 2

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3 A particle is in equilibrium on a smooth horizontal table when acted on by the three horizontal forces shown in the diagram. (i) Find the values of F and θ. [4] (ii) The force of magnitude 7 N is now removed. State the magnitude and direction of the resultant of the remaining two forces. [2]

Mark scheme: 3 (i) [7 = Fcosθ and 4 = Fsinθ M1 For stating F2 = 72 + 42 directly or for F2 = 72 + 42 (or tanθ = 4/7)] resolving in the i and j directions and eliminating θ or F F = 8.06 A1 Allow 8.07 from 4÷ sin29.7o [7 = 8.06cosθ or 4 = 8.06sinθ ] M1 For stating tanθ = 4/7 directly or for (or 7 = Fcos29.7o or 4 = Fsin29.7o) substituting for F or forθ into 7 = Fcosθ or 4 = Fsinθ θ = 29.7 A1 4 Allow 29.8 from sin-1(4÷ 8.06) SR for candidates who mix sine and cosine (max 3/4) Fsinθ = 7, Fcosθ = 4 F2 = 72 + 42 M1 For tanθ = 7/4 M1 For F = 7 and θ = 60.3o A1 (ii) Magnitude 7 N B1 Direction opposite to that of the force B1 2 Any equivalent form of magnitude 7 N 2

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4 The diagram shows the vertical cross-section of a surface. A and B are two points on the cross-section, and A is 5 m higher than B. A particle of mass 0.35 kg passes through A with speed 7 m s−1, moving on the surface towards B. (i) Assuming that there is no resistance to motion, find the speed with which the particle reaches B. [3] (ii) Assuming instead that there is a resistance to motion, and that the particle reaches B with speed 11 m s−1, find the work done against this resistance as the particle moves from A to B. [3]

Mark scheme: 4 (i) M1 For using KE = ½ mv2 [½ mv2- ½ m72 = mgx5] M1 For equation from KE gain = PE loss (3 terms) Speed is 12.2ms-1 A1 3 SR for candidates who treat AB as straight and vertical (max 1mark out of 3) v2 = 72 +2g5 v = 12.2 B1 (ii) M1 For using WD = PE loss – KE gain or WD = KE at B in (i) – actual KE at B WD = 0.35x10x5 – ½ 0.35(112 – 72) or A1ft ft wrong v in part (i) or for 12.2 scored by WD = ½ 0.35(12.22 – 112) B1in (i) Work done is 4.9 J A1 3 This mark is not available if v = 12.2 is used, having been scored by B1 in part (i) SR for candidates who treat AB as straight and vertical, and resistance as constant (max 1mark out of 3) a = 7.2 ms-2, R = 0.98 N, WD = 4.9 J B1 SR for candidates who write ‘Resistance =’ instead of ‘WD =’ (max 2/3) 0.35x10x5 – ½ 0.35(112 – 72) or ½ 0.35(12.22 – 112) seen B1 Answer 4.9J (NB J seen) B1 GCE A/AS LEVEL – October/November 2007 9709 04

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5 A ring of mass 4 kg is threaded on a fixed rough vertical rod. A light string is attached to the ring, and is pulled with a force of magnitude T N acting at an angle of 60◦to the downward vertical (see diagram). The ring is in equilibrium. (i) The normal and frictional components of the contact force exerted on the ring by the rod are R N and F N respectively. Find R and F in terms of T. [4] (ii) The coefficient of friction between the rod and the ring is 0.7. Find the value of T for which the ring is about to slip. [3]

Mark scheme: 5 (i) M1 For resolving horizontally (normal force must have a horizontal component) R = Tsin60o A1 [F = W + Tcos60o ] M1 For resolving vertically (allow if normal force is not horizontal but equation must contain F , W and T) F = 40 + Tcos 60o A1ft 4 ft – allow F = 40 + Tsin 60o following R = Tcos60o (ii) M1 For using F = µ R 40 + 0.5T = 0.7x0.866T A1ft Any correct form ft unsimplified with candidate’s F(T) (with 2 terms) and R(T) T = 377 A1 3 2 2

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6 (i) A man walks in a straight line from A to B with constant acceleration 0.004 m s−2. His speed at A is 1.8 m s−1 and his speed at B is 2.2 m s−1. Find the time taken for the man to walk from A to B, and find the distance AB. [3] (ii) A woman cyclist leaves A at the same instant as the man. She starts from rest and travels in a straight line to B, reaching B at the same instant as the man. At time t s after leaving A the cyclist’s speed is k(200t −t2) m s−1, where k is a constant. Find (a) the value of k, [4] (b) the cyclist’s speed at B. [1] (iii) Sketch, using the same axes, the velocity-time graphs for the man’s motion and the woman’s motion from A to B. [3]

Mark scheme: 6 (i) [2.2 = 1.8 + 0.004t] M1 For using v = u + at (or v2 = u2 + 2as) Time taken is 100s A1 (or Distance is 200 m) Distance is 200 m A1ft 3 ft s = 2t or 1.8t + 0.002t2 (or t = s/2) (or Time taken is 100s) (ii) (a) M1 For integrating v(t) to find s(t) s = k(100t2 – t3/3) (+C) A1 [k(100x1002 – 1003/3) = 200] DM1 For using s(0) = 0 (may be implied) and s(100) = 200 k = 0.0003 A1 4 (b) Speed is 3 ms-1 B1ft 1 ft candidate’s t and/or k. (iii) M1 For straight line segment, v(t) +ve and increasing throughout (including at t = 0) M1 For parabolic segment through origin, with +ve slope Parabolic segment has decreasing A1 3 Depends on both M marks slope; sketches correct relative to each other (line crosses curve once) GCE A/AS LEVEL – October/November 2007 9709 04

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7 A rough inclined plane of length 65 cm is fixed with one end at a height of 16 cm above the other end. Particles P and Q, of masses 0.13 kg and 0.11 kg respectively, are attached to the ends of a light inextensible string which passes over a small smooth pulley at the top of the plane. Particle P is held at rest on the plane and particle Q hangs vertically below the pulley (see diagram). The system is released from rest and P starts to move up the plane. (i) Draw a diagram showing the forces acting on P during its motion up the plane. [1] (ii) Show that T −F > 0.32, where T N is the tension in the string and F N is the magnitude of the frictional force on P. [4] The coefficient of friction between P and the plane is 0.6. (iii) Find the acceleration of P. [6]

Mark scheme: 7 (i) R B1 1 The components F and R may be represented by a single contact force, which must be T shown at an acute angle to the downward slope. F W (ii) M1 For finding the resultant upward force (RUF) (3 terms required) T – F – 0.13g (16/65) A1 [T – F – 0.13g (16/65) > 0] M1 For use of RUF > 0 (since P starts to move upwards). T – F > 0.32 A1 4 AG (iii) R = 0.13g(63/65) or B1ft ft 0.13g cos 75.7….. 0.13g cos14.25… (= 1.26) F = 0.6 x 1.26 (= 0.756) M1 For using F = µ R M1 For applying Newton’s second law to P (4 terms required) or to Q (3 terms required) or for using WQ - WPsinα - F = (mP + mQ)a T – F – 0.32 = 0.13a and A1ft ft1.26 instead of 0.32 following a consistent 0.11g – T = 0.11a sin/cos mix throughout (i) and (ii) or 0.11g – F – 0.32 = (0.13 + 0.11)a M1 For substituting for F and solving for a. Acceleration is 0.1 ms-2 A1 6

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