Cambridge A Level Mathematics 9709 — 2002 May/June 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) MAY/JUNE 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 10m s-?. (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 3 printed pages and 1 blank page. University of CAMBRIDGE Local Examinations Syndicate [Turn over © CIE 2002

04ms" One end of a light inextensible string is attached to a ring which is threaded on a fixed horizontal bar. The string is used to pull the ring along the bar at a constant speed of 0.4ms~!. The string makes a constant angle of 30° with the bar and the tension in the string is 5 N (see diagram). Find the work done by the tension in 10s. (3) A basket of mass 5 kg slides down a slope inclined at 12° to the horizontal. The coefficient of friction between the basket and the slope is 0.2. {i) Find the frictional force acting on the basket. 2] Gi) Determine whether the speed of the basket is increasing or decreasing. (3] By \ ‘ 10N 10N oA 5 A Two forces, each of magnitude 10N, act at a point O in the directions of OA and OB, as shown in the diagram. The angle between the forces is 6. The resultant of these two forces has magnitude 12 N. @ Find @. 3] Gi) Find the component of the resultant force in the direction of OA. 2) A box of mass 4.5kg is pulled at a constant speed of 2ms7! horizontal force of magnitude 15 N. along a rough horizontal floor by a (i) Find the coefficient of friction between the box and the floor. 3] The horizontal pulling force is now removed. Find ii) the deceleration of the box in the subsequent motion, 2) (iii) the distance travelled by the box from the instant the horizontal force is removed until the box comes to rest. 2) 8709/4202

3 i) A cyclist travels in a straight line from A to B with constant acceleration 0.06ms~*. His speed atA is 3ms7! and his speed at B is 6ms~'. Find (a) the time taken by the cyclist to travel from A to B, (2) (b) the distance AB. (2 Gi) A car leaves A at the same instant as the cyclist. The car starts from rest and travels in a straight line to B. The car reaches 8 at the same instant as the cyclist. At time fs after leaving A the speed of the car is km sl, where x is a constant. Find (a) the value of k, [4 (b) the speed of the car at B. {1 (i) A lorry P of mass 15000kg climbs a straight hill of length 800m at a steady speed. The hill is inclined at 2° to the horizontal. For P’s journey from the bottom of the hill to the top, find {a) the gain in gravitational potential energy, [2] (b) the work done by the driving force, which has magnitude 7000 N, es (c) the work done against the force resisting the motion. [2] (ii) A second lorry, Q@, also has mass 15 000 kg and climbs the same hill as P. The motion of Q is subject to a constant resisting force of magnitude 900 N, and Q's speed falls from 20m s” | at the bottom of the hill to 10ms*! at the top. Find the work donc by the driving force as Q climbs from the bottom of the hill to the top. [5] A B Particles A and B, of masses 0.15kg and 0.25kg respectively, are attached to the ends of a light inextensible string which passes over a smooth fixed pulley. The system is held at rest with the string taut and with A and 8 at the same horizontal level, as shown in the diagram. The system is then teleased. @ Find the downward acceleration of B. [4] After 2s B hits the floor and comes to rest without rebounding. The string becomes slack and A moves freely under gravity. {ii) Find the time that elapses until the string becomes taut again. [41 (iii) Sketch on a single diagram the velocity-time graphs for both particles, for the period from their release until the instant that B starts to move upwards. (3] S70gaMNNO?

CAMBRIDGE INTERNATIONAL EXAMINATIONS JUNE 2002 GCE Advanced Subsidiary Level MARK SCHEME MAXIMUM MARK : 50 SYLLABUS/COMPONENT :9709 /4 MATHEMATICS (Mechanics 1) eS University of CAMBRIDGE Local Examinations Syndicate

Page 1 Mark Scheme | Syliabus | Paper AS Level Examinations — June 2002 L.9709 | 4 For using WD = Fdcosa@ or P = Fycosa@ and WD = Pr [Mi WD = 5(0.4 x 10)cos30° Work done is 17.3 J (or 10V3 ) SR For candidates who calculate power (only) (max | out of 3) Power is 1.73 W. Notes: MI - their distance; cos or sin but not just 5x4 Radians M1 Al AO (max 2 out of 3); answer 3.085 does not score final A mark but may imply the previous Al (i) For using N= mg cos @ [5g cos 12° (= 48.9)] and F= yw N [0.2x 48.9] Frictional force is 9.78 N (9.59 from g = 9.8 and 9.60 from g = 9.81) Component of weight = 5g sin 12° (=10.4) {ft absence of g and/or sin/cos mix only) For comparing component of weight with frictional force or for finding the acceleration (0.123) using both the component of weight and the frictional force Ml Alternative: For comparing s with tan 12° or for comparing the ‘angle’ of friction with angle of inclination M1 0.2 < tan 12° or tan'0.2 < 12° Al Speed increasing (ft for arithmetic errors only) Notes: {i) ML accept absence of g and/or sin/cos mix (ii) B1 can be earned in (i) Illustration: ‘Sa = 1.04 ~ 9.78 } a <0 speed decreasing’ scores Al ft, whereas *Sa= 10.4 +9.78 D a>0 > speed increasing’ scores AO Radians: Can score both M marks as per scheme, and allow one A mark for both 8.44 and -26.8 (or -27 or -30) (max 3 out of 5)

Page 2 Mark Scheme Syllabus Paper AS Leval Examinations — June 2002 9709 4 {may be implied) or recognising that resultant acts along bisector or 12cos# =10+10cos@ and 12sin 8 = 10sin? or X= 10-10 cosa@ and ¥=10sin a Bl _| Complete method for a [@ = 2sin- = or 12? = 107 + 10? - 2x10’cosa@ } or resolving forces along the bisector [ 2x10cos— = 12) or squaring and adding and usingc?B +s°f =landc’6+s’@ =1 [144 = 100 + 200cos@+ 100 J M1 9= 106.3° or 1.85 rads For using component = 12c0s (12x 0.6] or 10-10 cos a Component is 7.2 N (ft only when B1 in part (i) is scored} SR for candidates whose diagram in (i) (actual or implied) has triangle with sides 10, 10, 12 and angle@ opposite the 12. (max 1 cut of 2) Component is + 7.2 N Bi Alternative: For candidates who draw a scale diagram. As for first mark in scheme above Bl Value of @ in the range 105° to 107° obtained Bl = 106.3° Bl 12 For drawing relevant perpendicular and measuring appropriate length M1 Component is 7.2 N Al Notes: Accept 7.19 or 7.20 or 7.21 (as well as 7.2) for final Al. The wrong diagram case (diagram may or not appear). Triangle has sides 10, 10, 12 with angle @ opposite the 12. (i) MO, 12? = 10? + 10° 2x10°cos@ MI AO (max | out of 3) (ii) Allow MI as per scheme if appropriate, otherwise use SR.

Page 3 Mark Scheme | Syllabus | Paper | AS Level Examinations — June 2002 [9709 | 4 | 4 (i) | N=4.5g, F=15 Bl For using ys = F/N M1 | Coefficient is 1/3 or 0.333 (0.340 from g = 9.8 or 9.81) Al 43 ii) For using Newton’s 2™ law {-15 = 4.5a ] M1 Deceleration is 10/3 ms“ (or 3.33) or a = -10/3 (or -3.33) Al | 2 iii : uty ai) For using v? =u? + 2as or v =u t+ atand s= t 0=4+ 2(-10/3)s MI Distance is 0.6 m Alft | 2 Notes: Allow inequality for M mark in (i) 4.$a= 15 Da=10/3 in (ii) scores M1] AO (unless a is said to be deceleration) v= 2,u=0 and a = 10/3 is OK for ML in (iii) even if a = +10/3 is found in (ii). Allow AI as well if 0.6m is found. Accept 0.601 from a = -3.33 for A mark in (iii) S() |(@)_ | Forusingv=u + at [6 =3 + (0.06) } MI 7] Time taken is 50s Al 2 (b) | For using 7 = u? + 2as (36 = 9 + 2(0.06)s] ors =ut+ “at? [s = 3(50) + % (0.06)2500] or s= 4 oe [s=%G +6)50] Ml Distance is 225m Al 2 For attempting to integrate A” Mi Al For finding & by substituting for s and ¢ in the expression for s obtained by integration or by using appropriate limits in the integration [ k50°/3 = 225) | k= 0.0054 or 27/5000 ft for 3 x (ans i(b))} / (ans i(a)y {b) | Speed is 13.5ms ft for (ans ii(a)) x (ans i(a))’ | BI ft SR (For candidates who use constant acceleration formulae in part {ii)} (max 1 out of 5) For k = 0.0036 and speed at 2 is 9ms" (in either order) Bl

Page 4 Mark Scheme Syllabus [Paper | AS Level Examinations ~ June 2002 9709 4 | (a | For using PE =mgh — [15 000x10(800sin2°)] Gain in PE is 4 190 000 J (4 187 900} (4 100 000 from g = 9.8 and 4 110 000 from g = 9.81) (b) | WD by driving force is 5 600 000 J For using WD = ans (b) ~ ans (a) or WD = (7000 - mgsin2°) x 800 WD against resistance is 1410000J (ft candidate’s ans (b) - ans (a) or (7000 — mgsin2°) x 800 providing the value found is +ve) | (1 500 000 from _g = 9.8 and 1 490 000 from g= 9.81} For using KE loss = 4 m(u’ - v’) [ % 15 000(400 — 100)] KE loss is 2 250 000 J WD against resistance is 900 x 800 (ii) May be implied by final answer For using WD as a linear combination of 3 terms reflecting the PE, the KE and the resistance [4.190 000 - 2 250 000 + 720 000 WD by driving force is 2 660 000 J (2 657 900) (2. 570 000 from g = 9.8 and 2 580 000 from g = 9.81) SR For candidates who assume, explicitly or implicitly, that the acceleration is constant. {max 3 out of 5} For using v= 7 + 2as (a = -0.1875) and DF = ma +900+ mgsin 2° ML For multiplying by 800 Mi WD by driving force is 2 660 000 J Al For incorrect use of multiple units (eg kJ) withold the A or B mark at the first occurrence, but do not penalise subsequently. Allow cos or (1 — cas) instead of sin for M mark in (i)(a), but g must be present Accept — 5 600 000 in (i)(b) and — 2 660 000 in (ii) Allow + the expressions for WD for M mark in (i)(c), but not for the A mark (including the ft) Answer 2 250 000 in (ii) is almost certainly worth 0 out of 5 (unless it is an answer for the loss in KE); see notes distributed at meeting.

Page 5 Mark Scheme Syllabus Paper AS Level Examinations — June 2002 9709 4 7 @ For applying Newton’s 2" law to A or B or for using (m, + m,)a = (m, -m,)g Mi! 0.15a =T—0.15g |_| 0.25a = 0.25g -T | | Alternative for the above 2 A marks: (0.15 + 0,25)a = (0.25-0.15)g_ A2 ‘Acceleration is 2.5ms™ (ft only for 0.25 following the absence of g) (2.45 from g = 9.8 or g = 9.81) Alft } 4 (ii) | v=5 ft for 2 x ans(i) (4.9 from g = 9.8 and 4.90(5) from g = 9.81) Bi ft | For using v = u + af to find time up or time down or total time up and down; acceleration must be +z Mt 5 f= 2x0 or-5 = 5 ~ 102 AIA Slack for Is Al |4 (iii) For 2 line segments representing motion with the string taut Bi For the line segment representing | Bi motion of A with the string slack For the line segment v = 0 Bl 3 representing B stationary with the string slack Notes: Allow absence of g for the M mark in (i) Allow —q instead of a for the first two A marks in (i) if, and only if, it applies to both equations. Third A mark is for 2.5 and if it follows a = -2.5 the answer must be properly justified. For answer 1s + 2s = 3s in (ii) allow final A mark (ISW for + 2s = 3s) Line segments must appear to be symmetric for first B mark in (iii) The graphs can have v positive downwards, but for 1* B mark the line segments must appear to be teflections of each other in the ¢ axis. Accept separate graphs for particles A and B, providing the direction of positive v is the same for both.