General Certificate of Education Advanced Subsidiary Examination January 2011 Mathematics MM1B Unit Mechanics 1B Wednesday 19 January 2011 1.30 pm to 3.00 pm For this paper you must have: the blue AQA booklet of formulae and statistical tables. You may use a graphics calculator. Time allowed 1 hour 30 minutes Instructions Use black ink or black ball-point pen. Pencil should only be used for drawing. Fill in the boxes at the top of this page. Answer all questions. Write the question part reference (eg (a), (b)(i) etc) in the left-hand margin. You must answer the questions in the spaces provided. Do not write outside the box around each page. Show all necessary working; otherwise marks for method may be lost. Do all rough work in this book. Cross through any work that you do not want to be marked. The final answer to questions requiring the use of calculators should be given to three significant figures, unless stated otherwise. Take g ¼ 9.8 m s 2, unless stated otherwise. Condensed Information The marks for questions are shown in brackets. The maximum mark for this paper is 75. Unit Mechanics 1B has a written paper only. Advice Unless stated otherwise, you may quote formulae, without proof, from the booklet. P35479/Jan11/MM1B 6/6/6/ MM1B
2 1 A trolley, of mass 5 kg, is moving in a straight line on a smooth horizontal surface. It has a velocity of 6 m s 1 when it collides with a stationary trolley, of mass m kg. Immediately after the collision, the trolleys move together with velocity 2.4 m s 1. Find m. (3 marks) 2 The graph shows how the velocity of a train varies as it moves along a straight railway line. Velocity (m s 1 ) 7 4 0 0 10 20 30 40 Time (seconds) (a) Find the total distance travelled by the train. (4 marks) (b) Find the average speed of the train. (2 marks) (c) Find the acceleration of the train during the first 10 seconds of its motion. (2 marks) (d) The mass of the train is 200 tonnes. Find the magnitude of the resultant force acting on the train during the first 10 seconds of its motion. (2 marks) P35479/Jan11/MM1B
3 3 A car, of mass 1200 kg, tows a caravan, of mass 1000 kg, along a straight horizontal road. The caravan is attached to the car by a horizontal tow bar, as shown in the diagram. Assume that a constant resistance force of magnitude 200 newtons acts on the car and a constant resistance force of magnitude 300 newtons acts on the caravan. A constant driving force of magnitude P newtons acts on the car in the direction of motion. The car and caravan accelerate at 0.8 m s 2. (a) (i) Find P. (3 marks) (ii) (b) (i) Find the magnitude of the force in the tow bar that connects the car to the caravan. (3 marks) Find the time that it takes for the speed of the car and caravan to increase from 7ms 1 to 15 m s 1. (3 marks) (ii) Find the distance that they travel in this time. (3 marks) (c) Explain why the assumption that the resistance forces are constant is unrealistic. (1 mark) 4 A canoe is paddled across a river which has a width of 20 metres. The canoe moves from the point X on one bank of the river to the point Y on the other bank, so that its path is a straight line at an angle a to the banks. The velocity of the canoe relative to the water is 4 m s 1 perpendicular to the banks. The water flows at 2ms 1 parallel to the banks. Y 2ms 1 4ms 1 X a 20 m Model the canoe as a particle. Turn over s P35479/Jan11/MM1B
4 (a) Find the magnitude of the resultant velocity of the canoe. (2 marks) (b) Find the angle a. (2 marks) (c) Find the time that it takes for the canoe to travel from X to Y. (2 marks) 5 A particle moves with constant acceleration ð 0:4i þ 0:2jÞ ms 2. Initially, it has velocity ð4i þ 0:5jÞ ms 1. The unit vectors i and j are directed east and north respectively. (a) Find an expression for the velocity of the particle at time t seconds. (2 marks) (b) (i) Find the velocity of the particle when t ¼ 22:5. (2 marks) (ii) State the direction in which the particle is travelling at this time. (1 mark) (c) Find the time when the speed of the particle is 5 m s 1. (6 marks) 6 Two particles, A and B, are connected by a light inextensible string which passes over a smooth peg. Particle A has mass 2 kg and particle B has mass 4 kg. Particle A hangs freely with the string vertical. Particle B is at rest in equilibrium on a rough horizontal surface with the string at an angle of 30 to the vertical. The particles, peg and string are shown in the diagram. A (2 kg) 30 B (4 kg) (a) By considering particle A, find the tension in the string. (2 marks) (b) Draw a diagram to show the forces acting on particle B. (2 marks) (c) (d) Show that the magnitude of the normal reaction force acting on particle B is 22.2 newtons, correct to three significant figures. (3 marks) Find the least possible value of the coefficient of friction between particle B and the surface. (4 marks) P35479/Jan11/MM1B
5 7 An arrow is fired from a point at a height of 1.5 metres above horizontal ground. It has an initial velocity of 12 m s 1 at an angle of 30 above the horizontal. The arrow hits a target at a height of 1 metre above horizontal ground. The path of the arrow is shown in the diagram. 12 m s 1 30 1.5 metres 1 metre Model the arrow as a particle. (a) (b) Show that the time taken for the arrow to travel to the target is 1.30 seconds, correct to three significant figures. (5 marks) Find the horizontal distance between the point where the arrow is fired and the target. (2 marks) (c) Find the speed of the arrow when it hits the target. (4 marks) (d) (e) Find the angle between the velocity of the arrow and the horizontal when the arrow hits the target. (2 marks) State one assumption that you have made about the forces acting on the arrow. (1 mark) 8 A van, of mass 2000 kg, is towed up a slope inclined at 5 to the horizontal. The tow rope is at an angle of 12 to the slope. The motion of the van is opposed by a resistance force of magnitude 500 newtons. The van is accelerating up the slope at 0.6 m s 2. 12 5 Model the van as a particle. (a) Draw a diagram to show the forces acting on the van. (2 marks) (b) Show that the tension in the tow rope is 3480 newtons, correct to three significant figures. (5 marks) Copyright ª 2011 AQA and its licensors. All rights reserved. P35479/Jan11/MM1B
Key to mark scheme abbreviations M mark is for method m or dm mark is dependent on one or more M marks and is for method A mark is dependent on M or m marks and is for accuracy B mark is independent of M or m marks and is for method and accuracy E mark is for explanation or ft or F follow through from previous incorrect result CAO correct answer only CSO correct solution only AWFW anything which falls within AWRT anything which rounds to ACF any correct form AG answer given SC special case OE or equivalent A2,1 2 or 1 (or 0) accuracy marks x EE deduct x marks for each error NMS no method shown PI possibly implied SCA substantially correct approach c candidate sf significant figure(s) dp decimal place(s) No Method Shown Where the question specifically requires a particular method to be used, we must usually see evidence of use of this method for any marks to be awarded. Where the answer can be reasonably obtained without showing working and it is very unlikely that the correct answer can be obtained by using an incorrect method, we must award full marks. However, the obvious penalty to candidates showing no working is that incorrect answers, however close, earn no marks. Where a question asks the candidate to state or write down a result, no method need be shown for full marks. Where the permitted calculator has functions which reasonably allow the solution of the question directly, the correct answer without working earns full marks, unless it is given to less than the degree of accuracy accepted in the mark scheme, when it gains no marks. Otherwise we require evidence of a correct method for any marks to be awarded. 3
MM1B 1 5 6 = ( m + 5) 2.4 M1A1 M1: Equation for conservation of 30 = 2.4m + 12 momentum with correct number of terms. 30 12 m = = 7.5 A1 3 A1: Correct equation. 2.4 A1: Correct mass CAO Total 3 2(a) 1 1 1 s = 10 4+ 10 4 + (4+ 7) 10+ 7 10 2 2 2 ( = 20+ 40+ 55+ 35) = 150 m M1M1A1 A1 4 1 1 1 s = (10+ 20) 4 + (4+ 7) 10+ 7 10 2 2 2 (M1M1A1) ( = 60+ 55+ 35) = 150 m (A1) 1 1 1 s = 10 4+ 10 4+ 10 4+ 10 3+ 7 10 2 2 2 (M1M1A1) ( = 20+ 40+ 40+ 15+ 35) = 150 m (A1) Consistent use of weight instead of mass penalise final A1 mark. M1: Any one term correct. M1: A second term correct. A1: Correct expression for total distance. A1: Total distance correct. (b) 150 1 Average Speed = = 3.75 ms 40 (c) 4 2 a = = 0.4 ms 10 M1 A1F 2 M1 A1 2 M1: Their total distance divided by 40. A1F: Correct average speed based on their distance from part (a). Must be correct to three or more significant figures. M1: Any division involving the numbers 10 and 4. A1: Correct acceleration. CAO Note on use of constant acceleration equations: award M1 for correct equation with correct values and A1 for correct final answer. (d) F = 200000 0.4 = 80000 N M1A1F 2 M1: Multiplication of, 2 10 n, for any integer n, by candidate s acceleration from part (c). A1F: Correct force based on their answer to part (c) multiplied by 200000. Total 10 Note: use of a = 2.5 gives 500000 N Accept 80kN 4
MM1B (cont) 3(a)(i) P 500 = 2200 0.8 M1A1 M1: Equation of motion for car and P = 1760 + 500 caravan as a single body. Must see 2200 (or 1200+1000) multiplied by 0.8, and = 2260 A1 3 500 (or 200+300). Allow sign errors. A1: Correct equation. A1: Correct value for P. (Award full marks for: (P =) 1760 + 500 = 2260 (If finding the tension first) or similar to obtain correct final answer.) P 1100 200 = 1200 0.8 P = 960 + 1100 + 200 = 2260 (M1A1) (A1) M1: Equation of motion for car with their value for the tension. Must see 1200 multiplied by 0.8, 200 and their tension. Allow sign errors. A1: Correct equation. A1: Correct value for P. (Award full marks for: (P =) 960 + 200 + 1100 = 2260 or similar to obtain correct final answer.) (a)(ii) T 300 = 1000 0.8 T = 300 + 800 = 1100 2260 200 T = 1200 0.8 T = 2260 200 960 = 1100 N M1A1 A1 (M1A1) (A1) 3 M1: Equation of motion for caravan. Must see 300 and 1000 multiplied by 0.8. Allow sign errors. A1: Correct equation. A1: Correct tension. CAO M1: Equation of motion for car. Must see 2260 (or candidate s P), 200 and 1200 multiplied by 0.8. Allow sign errors. A1: Correct equation. A1: Correct tension. CAO If candidates find tension first it must be stated in part (a)(ii) to gain any marks. The working does not have to be repeated if seen in part (a)(i). 5
MM1B (cont) 3(b)(i) 15 = 7 + 0.8t M1A1 M1: Use of a constant acceleration 15 7 equation to find t, with 7, 15 and 0.8. t = = 10 seconds A1: Correct equation. 0.8 A1 3 A1: Correct time. CAO (b)(ii) 2 2 15 = 7 + 2 0.8s 2 2 15 7 s = = 110 m 1.6 M1A1 A1 3 M1: Use of a constant acceleration equation to find s, with 7, 15 and 0.8. A1: Correct equation A1: Correct distance. CAO 1 s = (7 + 15) 10 = 110 m 2 1 s = + = 2 2 7 10 0.8 10 110 m (M1A1F) (A1F) (M1A1F) (A1F) M1: Use of a constant acceleration equation to find s, with 7, 15 and candidate s time. A1F: Correct equation. A1F: Correct distance. M1: Use of a constant acceleration equation to find s, with 7, 0.8 and candidate s time. A1F: Correct equation. A1F: Correct distance. If candidates find distance first it must be stated in part (b)(ii) to gain any marks. The working does not have to be repeated if seen in part (b)(i). (c) Resistance forces vary with speed (or velocity) Speed (or velocity) changes (or increases) It accelerates B1 1 B1: Correct explanation. Must not mention friction in main argument Total 13 6
MM1B (cont) 4(a) 2 2 ( V = ) 2 + 4 = 20 M1A1 2 M1: Equation or expression to find V based on Pythagoras. Must be +. = 2 5 A1: Correct velocity. Accept 20, 2 5, 1 = 4.47 ms 4.47 or more accurate answer from 4.472135 (b) 4 tanα = 2 M1 M1: Trigonometric equation to find angle. α = 63.4 Can be any of those as shown. For tan, A1F 2 fraction can be inverted. For sin, 2 can be used instead of 4. For cos, 4 can be used instead of 2. Can use their V from part 4 4 (a). sin α = or (M1) 2 5 4.47 A1F: Correct angle. Accept 63 or AWRT α = 63.4 (A1F) 63.4 or 63.5. 2 2 cos α = or 2 5 4.47 α = 63.4 (M1) (A1F) (c) 20 t = = 5 seconds 4 500 t = = 5 seconds 20 M1 A1 2 Total 6 M1: Division of distance by speed (for 10 20 500 22.4 example, or or or ) 2 4 20 4.47 Do not award M1 if distance and speed don t correspond (eg 10 4 or 20 20 or 2 4.47 ) A1: Correct time CAO. Accept 5.00 or 5.0 7
MM1B (cont) 5(a) v= (4i+ 0.5 j) + ( 0.4i+ 0.2 j )t M1A1 2 M1: Use of constant acceleration equation to find v with u 0i+ 0j A1: Correct v. (Could be done as a column vector.) (b)(i) v= (4i+ 0.5 j) + ( 0.4i+ 0.2 j) 22.5 = 5i+ 5j M1 A1 2 M1: Substitution of 22.5 into their expression for the velocity, even if no marks awarded in part (a). A1: Correct velocity CAO (Could be done using column vectors.) (b)(ii) North-west B1 1 B1: Correct statement of direction. Accept 315. Must follow from correct answer to (b)(i). (c) ( ) v= (4 0.4 t) i+ (0.5 + 0.2 t) j B1 2 2 2 5 = (4 0.4 t) + (0.5+ 0.2 t) 2 0.2t 3t 8.75 = 0 2 t 15t 43.75 = 0 t = 17.5 or t = 2.5 t = 17.5 M1A1 A1 dm1 A1 6 Total 11 B1: Grouping i and j components at some point in the solution. (Could be done as column vectors.) Allow 5 = (4 0.4 t) i+ (0.5 + 0.2 t) j M1: Seeing both components of their velocity squared and added A1: Correct equation. (Condone including i and j.) For example: 2 2 5 = (4 0.4 t) i + (0.5 + 0.2 t) j scores B1M1A0 2 2 2 5 = (4 0.4 t) i + (0.5+ 0.2 t) j scores B1M1A1 A1: Any correct simplified quadratic equation, with exactly three terms. dm1: Solving the quadratic equation. (Allow one substitution error in correctly quoted formula) Candidates with an incorrect quadratic equation must show method to get dm1. A1:Correct positive solution stated. 8
MM1B (cont) 6(a) T = 2 9.8 = 19.6 N M1A1 2 M1: Equating tension and weight. A1: Correct tension CAO Accept 2g Accept 19.62 from g = 9.81 (b) T R or N B1 B1 2 B1: R, F (not μr) and mg correct B1: T correct, must be in roughly correct direction. F If more than four forces shown, do not award more than one mark. mg or W or 4g or 9.8m or 39.2 or 39.24 Note all forces must be shown as arrows and have labels. Note some candidates may draw the force diagram in the section with the question. Components can be ignored if shown in a different notation eg dashed arrows. (c) Tcos30 + R= 4 9.8 ( R = )39.2 19.6cos30 = 39.2 16.9741... = 22.2259... = 22.2 N (to 3sf) AG (d) Tcos60 = F F = 19.6cos60 = 9.8 ( ) 19.6cos 60 μ 39.2 19.6cos30 19.6cos 60 μ 39.2 19.6cos30 μ 0.441 M1 A1 A1 3 M1 A1 dm1 A1 4 Total 11 M1: Resolving vertically to form a three term equation. (May be implied) A1 Correct expression for R or equation for R. Must see 19.6cos30 or equivalent (eg 2gsin60) A1: Correct force. Must see intermediate working, for example third or fourth line of working in solution opposite. Example: 19.6sin 30 R = 4 9.8 scores M1A0A0. Use of g = 9.81 still gives 22.2 N as the final answer. M1: Resolving horizontally A1: Correct expression for friction dm1: Use of F = μr or F μr (do not allow F μr ) A1: Final answer of μ = 0.441 or μ 0.441 from correct working Use of g = 9.81 still gives 0.441 as the final answer. 9
MM1B (cont) 7(a) 2 12sin30 t 4.9t = 0.5 M1A1A1 M1: Three term equation for vertical 2 motion, with ± g, ± 0.5 (or ±1 and 4.9t 12sin30 t 0.5 = 0 ±1.5) and 12sin30 t or 12cos30 t. t = 1.30281... or 0.078323... dm1 A1: Correct terms. (one must be t = 1.30 seconds (to 3sf) AG A1 5 equivalent to ±0.5) A1: Correct signs. dm1: Solving the quadratic to find t. Must see use of quadratic equation formula or can be implied by seeing 1.303 or 1.302 or similar. A1: Correct time from correct working. Must see more than 3 significant figures in candidate s working before the final answer or two correct solutions to the quadratic (eg 1.3 and 0.08). Accept 1.3 M1:Adding time up to time down time up = 0.6122 having used a quadratic. time down = 0.6122+0.0783=0.6905 A1: 0.6122 total time = 0.6122+0.6905 = 1.30 (to 3sf) (M1A1 dm1: Finding time down with a dm1a1a1) quadratic A1: 0.6905 A1: Correct answer Accept 1.3 M1:Forms an equation to find t 6.767 = 12sin 30 gt (M1A1A1) having found v first 12sin30 + 6.767 A1: Correct terms t = = 1.30281 = 1.30 (to 3sf) (dm1a1) A1: Correct signs g dm1: Solving for t A1: Correct time from correct working. Must see more than 3 significant figures in candidate s working before the final answer. Accept 1.3 (b) 12cos30 1.303 = 13.5 m M1A1 2 M1: Finding horizontal displacement using 1.30 (or better) and 12cos30. Do not allow 12sin 30. A1: Correct distance. AWRT 13.5. 10
MM1B (cont) 7(c) = 12sin30 9.8 1.3028 ( = 6.767) v y v = (12cos30 ) + ( 6.767) = 12.4 ms 2 2 1 M1A1 dm1a1 4 M1: Finding vertical component of velocity or velocity squared at impact. Must include 12sin 30 or 12cos30 and ±g A1: Correct expression for vertical component. May have 1.3 or 1.30 instead of 1.3028. (Accept +6.767 or similar) dm1: Finding speed from two components. May use 6.74. A1: Correct speed. Allow 12.3 or AWRT 12.4. Note using g = 9.81 still gives 12.4. (d) 6.767 tanθ = 12cos30 θ = 33.1 6.767 sinθ = 12.4 θ = 33.1 10.4 cosθ = 12.4 θ = 33.1 M1 A1F 2 M1: Trigonometric equation to find angle. Can only be those shown opposite or described below. For tan, fraction can be inverted. For sin, 10.4 can be used instead of 6.767. For cos, 6.767 can be used instead of 10.4. Can use their values from part (c) (eg 6.74 or 6.77). A1F: Correct angle. Accept AWRT 33. Follow though vertical component or final speed from part (c). (e) The weight is the only force acting. No air resistance. B1 1 B1: Appropriate assumption. Total 14 11
MM1B (cont) 8(a) B1: R, 500 and mg correct B1: Tension in roughly correct R or N T direction. 500 If more than four forces shown, do not award more than one mark. mg or W or 2000g or 19600 or 19620 or 9.8m B1 B1 2 Note all forces must be shown as arrows and have labels. Note some candidates may draw the force diagram in the section with the question. Components can be ignored if shown in a different notation eg dashed arrows. (b) 2000 0.6 = T cos12 500 2000 9.8sin 5 1200 + 500 + 19600sin5 T = cos12 3408.25 = cos12 ( = 3484.4) = 3480 ( to 3sf) AG M1A1A1 dm1 A1 5 Total 7 TOTAL 75 M1: Resolving parallel to the slope to obtain a four term equation of motion. The weight and tension terms must be resolved. A1: Correct terms. A1: Correct signs. dm1: Solving for T. A1: Correct tension. AWRT 3480. Allow AWRT 3490 from use of g = 9.81. 12
Scaled mark unit grade boundaries - January 2011 exams A-level Max. Scaled Mark Grade Boundaries and A Conversion Points Code Title Scaled Mark A A B C D E MD01 GCE MATHEMATICS UNIT D01 75-61 55 49 43 37 MFP1 GCE MATHEMATICS UNIT FP1 75-63 56 49 43 37 MM1A GCE MATHEMATICS UNIT M1A 100 no candidates were entered for this unit MM1B GCE MATHEMATICS UNIT M1B 75-61 53 46 39 32 MPC1 GCE MATHEMATICS UNIT PC1 75-56 49 42 36 30 MS1A GCE MATHEMATICS UNIT S1A 100-84 74 64 54 44 MS/SS1A/W GCE MATHEMATICS UNIT S1A - WRITTEN 75 64 34 MS/SS1A/C GCE MATHEMATICS UNIT S1A - COURSEWK 25 20 10 MS1B GCE MATHEMATICS UNIT S1B 75-61 53 46 39 32 MD02 GCE MATHEMATICS UNIT D02 75 69 63 56 50 44 38 MFP2 GCE MATHEMATICS UNIT FP2 75 67 60 51 42 34 26 MM2B GCE MATHEMATICS UNIT M2B 75 63 55 47 40 33 26 MPC2 GCE MATHEMATICS UNIT PC2 75-61 54 47 40 33 MS2B GCE MATHEMATICS UNIT S2B 75 66 59 52 45 38 31 XMCA2 GCE MATHEMATICS UNIT XMCA2 125 105 93 81 70 59 48 MFP3 GCE MATHEMATICS UNIT FP3 75 66 59 52 45 38 31 MPC3 GCE MATHEMATICS UNIT PC3 75 66 59 52 45 38 31 MFP4 GCE MATHEMATICS UNIT FP4 75 63 55 47 40 33 26 MPC4 GCE MATHEMATICS UNIT PC4 75 68 61 54 47 41 35 MEST1 GCE MEDIA STUDIES UNIT 1 80-56 49 42 35 28 MEST2 GCE MEDIA STUDIES UNIT 2 80-63 54 45 36 28 MEST3 GCE MEDIA STUDIES UNIT 3 80 69 58 47 37 27 17 MEST4 GCE MEDIA STUDIES UNIT 4 80 72 65 53 42 31 20 PHIL1 GCE PHILOSOPHY UNIT 1 90-54 48 42 37 32 PHIL2 GCE PHILOSOPHY UNIT 2 90-62 56 50 44 38