Cambridge A Level Mathematics 9709 — 2002 Oct/Nov Paper 4 · Variant 1

CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Subsidiary Level General Certificate of Education Advanced Level MATHEMATICS 9709/4 PAPER 4 Mechanics 1 (M1) OCTOBER/NOVEMBER SESSION 2002 1 hour 15 minutes Additional materials: Answer paper Graph paper List of Formulae (MF9) TIME 1 hour 15 minutes INSTRUCTIONS TO CANDIDATES Write your name, Centre number and candidate number in the spaces provided on the answer paper/answer booklet. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. Where a numerical value for the acceleration due to gravity is needed, use 10 m s
2. INFORMATION FOR CANDIDATES The number of marks is given in brackets [ ] at the end of each question or part question. The total number of marks for this paper is 50. Questions carrying smaller numbers of marks are printed earlier in the paper, and questions carrying larger numbers of marks later in the paper. The use of an electronic calculator is expected, where appropriate. You are reminded of the need for clear presentation in your answers. This question paper consists of 4 printed pages. CIE 2002 [Turn over

2 1 A car of mass 1000 kg travels along a horizontal straight road with its engine working at a constant rate of 20 kW. The resistance to motion of the car is 600 N. Find the acceleration of the car at an instant when its speed is 25 m s
1. [3] 2 A man runs in a straight line. He passes through a fixed point A with constant velocity 7 m s
1 at time t
0. At time t s his velocity is v m s
1. The diagram shows the graph of v against t for the period 0  t  40. (i) Show that the man runs more than 154 m in the first 24 s. [2] (ii) Given that the man runs 20 m in the interval 20  t  24, find how far he is from A when t
40. [2] 3 A light inextensible string has its ends attached to two fixed points A and B, with A vertically above B. A smooth ring R, of mass 0.8 kg, is threaded on the string and is pulled by a horizontal force of magnitude X newtons. The sections AR and BR of the string make angles of 50 Æ and 20 Æ respectively with the horizontal, as shown in the diagram. The ring rests in equilibrium with the string taut. Find (i) the tension in the string, [3] (ii) the value of X. [3] 9709/4/O/N/02

3 4 Two particles A and B are projected vertically upwards from horizontal ground at the same instant. The speeds of projection of A and B are 5 m s
1 and 8 m s
1 respectively. Find (i) the difference in the heights of A and B when A is at its maximum height, [4] (ii) the height of A above the ground when B is 0.9 m above A. [4] 5 A force, whose direction is upwards parallel to a line of greatest slope of a plane inclined at 35 Æ to the horizontal, acts on a box of mass 15 kg which is at rest on the plane. The normal component of the contact force on the box has magnitude R newtons (see Fig. 1). (i) Show that R
123, correct to 3 significant figures. [1] When the force parallel to the plane acting on the box has magnitude X newtons the box is about to move down the plane, and when this force has magnitude 5X newtons the box is about to move up the plane (see Fig. 2). (ii) Find the value of X and the coefficient of friction between the box and the plane. [7] [Questions 6 and 7 are printed overleaf.] 9709/4/O/N/02 [Turn over

4 6 (i) A particle P of mass 1.2 kg is released from rest at the top of a slope and starts to move. The slope has length 4 m and is inclined at 25 Æ to the horizontal. The coefficient of friction between P and the slope is 1 4. Find (a) the frictional component of the contact force on P, [2] (b) the acceleration of P, [2] (c) the speed with which P reaches the bottom of the slope. [2] (ii) After reaching the bottom of the slope, P moves freely under gravity and subsequently hits a horizontal floor which is 3 m below the bottom of the slope. (a) Find the loss in gravitational potential energy of P during its motion from the bottom of the slope until it hits the floor. [1] (b) Find the speed with which P hits the floor. [3] 7 A particle P starts to move from a point O and travels in a straight line. At time t s after P starts to move its velocity is v m s
1, where v
0.12t
0.0006t2. (i) Verify that P comes to instantaneous rest when t  200, and find the acceleration with which it starts to return towards O. [3] (ii) Find the maximum speed of P for 0  t  200. [3] (iii) Find the displacement of P from O when t  200. [3] (iv) Find the value of t when P reaches O again. [2] 9709/4/O/N/02

CAMBRIDGE INTERNATIONAL EXAMINATIONS NOVEMBER 2002 GCE Advanced Subsidiary Level UNIVERSITY of CAMBRIDGE Local Examinations Syndicate

Page 1 Mark Scheme Syllabus | Paper AS Level Examinations — November 2002 9709 4 1 Driving force = 20 000/25 Bl For using Newton’s 2 Taw (3 terms needed) Mi [20 000/25 — 600 = 10004] | “Acceleration is 0.2ms7 Al iB Notes: 20000 = 600 = 1000a scores BO M1; sae 2 = 1000a scores B1 MO 20000 = 25(1000a + 600) scoresB1 M1 20 000/25 — 600 = 1000ga scores B1 MO 2 (i) | For 20x7 or 140 and % 4x7 or 14 Bl Valid argument that s; + sz > 154 (AG) Bl [ 2 Alternatively: Approx distance is 20x7 + 4x7 k (where %<k <1) M1 Whose value (shown) is (clearly) > 154 Al (ii) | For using area property with correct signs [140 + 20 — % 10x8] | M1 Distance is 120m Al |2 Note: 140 + 20 + 20= 20 scores MO in (ii)

Page 2 Mark Scheme Syllabus | Paper AS Level Examinations — November 2002 9709 4 3 @ For resolving forces on R vertically G3 terms needed) M1 i Tsin50° = Tsin20° + 0.8g¢ Al Tension is 18.9 N (18.5 from g= 9.81 or g = 9.8) Al |3 (ii) | For resolving forces on R horizontally M1 4 X= Tcos50°+ Teos20° Al X=29.9 (8tan75°) (29.3 from g = 9.81 or g = 9.8) » | Alft | 3 Alternatively (by scale drawing): lr Correct quadrilateral drawn to scale Mil O 8 T T 18.4<T <19.4 A2 29.45 X $30.4 A2 T=18.9 and X= 29.9 Al Notes: F, = Tsin50 ~ Tsin20 — 0.8g scores MO in (i) and F, = X — Tcos50 — Tcos20 scores MO in (ii). Note that sin/cos mix can score M1 AO AO M1 AO AO at best (this error leads to negative values for T and X). None of the four A marks can be scored unless and until T, = T; is stated or implied, where T, and T, are the tensions in the two parts of the string. Many candidates try to use Lami’s theorem. In order to score any marks the candidate needs to reduce the system to one of 3 forces. Two examples of how this might be done, and how it should be marked, are shown below. [The general idea is that M1 is given for a complete method for X, Al for a correct equation in X (only) and Al for X = 29.9, and similarly for T.] For example reducing the system to 3 forces of magnitudes 2Tcos35, X and 8, attempting to find the angles 105 and 165 and applying Lami MI X/sin105 = 8/sin165 Al X= 29.9 Al Applying Lami to find T MI 2Tcos35/sin90 = 8/sin165 Al T=18.9 Al Reducing the system to 3 forces of magnitudes T, T and ¥ 8? +X? and attempting to find the angles 70, 145 and 145 and applying Lami Mi V8? +X? sin15=8 Al X=29.9 Al Applying Lami to find T Mil Tisin 145 =V8* + 29.97 /sin 70 Al T=18.9 Al

Page 3 Mark Scheme Syllabus Paper AS Level Examinations — November 2002 9709 4 4 @ For using v = u - gt, with v = 0, to find ¢ [5-10 = 0] Mi | -Time to maximum height of A is 5/g Al | For using h = ut—% gf and evaluating hg (0.5) - h4-(0.5) Mi Difference in heights is 1.5m (1.53 from g = 9.81 or g = 9.8) Al |4 SR For difference in maximum heights (max 1 out of 4) 1.95m__(1.99 from g = 9.81 or 9.8) Bl (ii) | For attempting to solve hg - hy =0.9 fort [8¢—5t=0.9] Mi t=03 Al [ For using h = ut — 4 gf” with the value of f found Mi [h=5 x 0.3—% 10x 0.09] Height of A is 1.05 m (1.06 from g = 9.81 or g = 9.8) Al |4 Notes: Using a = +g in v =u + at scores MO at the first stage in (i) and using a=+gins=ut+ 4% at” scores MO at the second stage in (i) (notwithstanding the resultant ‘correct’ answer). Allow error in sign of the terms 4 gt” in expressions for hg and ha for both the first M1 and the first Al in (ii). Using a= +g in s = ut + % at” scores MO at the second stage in (ii). Note that 5? = 8” — 2g(s + 0.9) leads entirely fortuitously to the ‘correct’ answer 1.05 in (ii) (but this doesn’t apply when g is taken as 9.8). This.solution scores 0 out of 4 in (ii).

Page 4 + Mark Scheme Syllabus | Paper AS Level Examinations — November 2002 9709 [| 4 ((2/3)tan 35°] () | R=15x 10x cos35°= 123 (AG) [= 1 (ii) _| For resolving forces along the plane (either case) Mi 150sin35° = X + F and 150sin35° = 5X- F Al For eliminating F or ¥ M1 | |X =28.7 (ft from wrong F or wrong positive uz ) Alft (28.1 from g = 9.81 or g = 9.8) F or R= 10gsin 35° or equivalent (may be implied) (57.36) | Al. ForusingF=uR [57.36 = 122.9 or 100 sin 35°= 150 cos35°] | M1 Coefficient of friction is 0.467 (ft for positive value from wrong X) Alft SR for the case where a candidate does not use F explicitly and uses F < wR (and not F = wR) implicitly (max 4 out of 7) For resolving forces along the plane (either case) Mi 150sin35° —X < wRand 5X -150sin35° < uR Al For eliminating X (it is not possible to eliminate puR ) Mi BR > 100sin35° or equivalent "Al Notes: Do not allow answers from g = 9.81 or g= 9.8 in (i). Accept any answer which rounds to 123 in (i). Accept sin instead of cos for first M1 in (ii). F, = 150sin35°-X- F and F,=5X—F - 150sin35° scores M0 in (ii). 150sin35° - X- F= 15a and 5X-F - 150sin35°= 15a D> 300sin35° - 6X = 0 D X= 28.7 scores MI M1 in (ii), but none of the three A marks unless and until a is set equal to zero. If F is taken in the wrong direction the candidate can score M1 AQ M1 AI (not fortuitous) AO M1 AO in Gi). Allow 4 = 0.466 (however the inaccuracy arises) - this is because it would be harsh to regard 57.36/123, which equals 0.466; as p.a., since 123 is a printed answer. The value of g should not affect the value of £2 , but allow 0.457 or 0.458 from a mix (56.27/123 or 56.21/123) because 123 is a printed answer.

Page 5 I Mark Scheme Syllabus | Paper [ AS Level Examinations —- November 2002 9709 4 6 @@) | For using F= wmgcos a [0.25x1.2gcos25°] | M1 | Frictional component is 2.72 N (2.719) (2.67 from g = 9.81 or 2.66 from g = 9.8) Al SR for the candidate who uses F' < wR instead of F = wR (max 1 out of 2) F $2.72 ‘Bl (b) | For using Newton’s 2"? law (3 terms needed) [1.2gsin25° — 2.719 = 1.2a] | Mil | Acceleration is 1.96 ms (1.92 from g=9.81 or 3. y Alft ft for positive value of a from incorrect F). (©) | Forusing 7 =2as [7 = 2x1.96x4] MI Speed is 3.96ms" (3.92 from g = 9.81 or 9.8) Alft (ft for 8.00 (accept 8.0 or 8) following a sin/cos mix) (ii)(a) | PE Loss is 36J (35.3 from g = 9.81 or 9.8) Bl (b) | For using PE loss = KE gain from bottom of slope, or va loss = KE gain WD against friction from top of slope, or . M1 = Wert + Vioriz” and Wert” = (-3.96sin35°)? + 2gx3 16 % 1.207 - 3: 96") or 1.2g(4sin25° + 3) = % 1.2v’ + 2.719x4 or Alft v=[(-3. 96sin35°)" +2gx3] + (3.96c0s35° y . Speed is 8.70 ms! (8.62 from g= 9.81 or 8.61 from g= 9.8) Al SR (max J 1 out of 3) v =3.967 +2gx3 Bl ft “Notes: Allow sin 25 instead of cos 25 for M1 in (i)(a). Allow cos 25 instead of sin 25 for M1 in (i)(b). 1,2gsin25° — 2.719 = 1.2ga scores MO in (i)(b). Accept + 36 in (ii)(a). Allow M1 for % 1.2v’ = 36 in (ii)(b). Accept 8.7 (for 8.70) in (ii)(b).

Page 6 Mark Scheme | Syllabus | Paper | AS Level Examinations — November 2002 9709 | 7 (i) (200) = 0.12x200 — 0.0006x40 000 = 0 Bl For using a = dv/dt and evaluating a(200) or a(200+ €) forsuitably | M1 small ¢ [a = 0.12 — 0.0012x200] _ Acceleration is 0.12 ms” (accept a=-0.12) (must be from ¢=0) | Al 13 (ii) | For attempting to solve dv/dr = 0 or using t = % 200 (may be implied) | M1 t= 100 (ft incorrect 2-term dv/d¢ in (i)) Alft | Maximum speed is 6ms™ Al i (iii) | For integrating v Mi s=0.06"—0.0002F (+C) Al [. Displacement is 800m Al [3 (iv) | For attempting to solve s = 0 + Ml £=300 Al 2 Notes; The M mark in (ii) is not dependent on the M mark in (i), the dv/dr used may be what the candidate thinks is dv/dr. 800 + C is not acceptable for seond A1 in (iii). T =0 or 300 is not acceptable for A1 in (iv).