ADVANCED SUBSIDIARY GCE UNIT 2821 PHYSICS A Forces and Motion FRIDAY 8 JUNE 2007 Morning Additional materials: Electronic calculator Ruler (cm/mm) Protractor Time: 1 hour INSTRUCTIONS TO CANDIDATES Write your name, Centre Number and Candidate number in the boxes above. Answer all the questions. Use blue or black ink. Pencil may be used for graphs and diagrams only. Read each question carefully and make sure you know what you have to do before starting your answer. Do not write in the bar code. Do not write outside the box bordering each page. For Examiner s Use WRITE YOUR ANSWER TO EACH QUESTION IN THE SPACE PROVIDED. ANSWERS WRITTEN ELSEWHERE WILL NOT BE MARKED. Qu. Max. Mark INFORMATION FOR CANDIDATES The number of marks for each question is given in brackets [ ] at the end of each question or part question. The total number of marks for this paper is 60. You will be awarded marks for the quality of written communication where this is indicated in the question. You may use an electronic calculator. You are advised to show all the steps in any calculation. 1 11 2 8 3 9 4 8 5 12 6 12 Total 60 This document consists of 16 printed pages. SPA (DR/DR) T15139/7 [M/100/3700] OCR is an exempt Charity [Turn over
Data 2 speed of light in free space, c = 3.00 10 8 ms 1 permeability of free space, 0 = 4 10 7 Hm 1 permittivity of free space, 0 = 8.85 10 12 Fm 1 elementary charge, the Planck constant, unified atomic mass constant, rest mass of electron, rest mass of proton, e = 1.60 10 19 C h = 6.63 10 34 Js u = 1.66 10 27 kg m e = 9.11 10 31 kg m p = 1.67 10 27 kg molar gas constant, R = 8.31 J K 1 mol 1 the Avogadro constant, N A = 6.02 10 23 mol 1 gravitational constant, G = 6.67 10 11 Nm 2 kg 2 acceleration of free fall, g = 9.81 m s 2
Formulae 3 uniformly accelerated motion, s = ut + 1 2 at 2 refractive index, n = v 2 = u 2 + 2as 1 sinc capacitors in series, 1 C = 1 C 1 + 1 C 2 +... capacitors in parallel, C = C 1 + C 2 +... capacitor discharge, x = x 0 e t/cr pressure of an ideal gas, p = 1 Nm 3 V <c 2 > radioactive decay, x = x 0 e λt t = 0.693 λ critical density of matter in the Universe, ρ 0 = 3H 0 2 8 G relativity factor, = (1 v 2 c 2 ) current, I = nave nuclear radius, r = r 0 A 1/3 sound intensity level, = 10 lg ( I I 0 ) [Turn over
4 Answer all the questions. 1 Fig. 1.1 shows a vacuum cleaner of weight W being pushed with a force P. The force P acts at 30 to the horizontal. dust collector P = 24.0N 30 W = 65.0N Fig. 1.1 floor The weight W is 65.0 N and the magnitude of force P is 24.0 N. (a) (i) Calculate 1 the horizontal component of the force P 2 the vertical component of the force P. horizontal component =...N vertical component =...N [3]
(ii) Show that the total downward vertical force is 77.0 N. 5 [1] (iii) Hence determine the magnitude of the resultant of the forces W and P. resultant force =...N [2] (iv) The vacuum cleaner is not switched on and is pushed in such a way that it travels at a constant velocity to the left. There are other forces acting on the vacuum cleaner. State and explain the magnitude of the resultant of these other forces.............[2] (b) (i) The total area of the vacuum cleaner in contact with the floor is 4.2 x 10 3 m 2. Calculate the pressure exerted on the floor by the total downward vertical force. pressure =...Pa [2] (ii) State and explain what happens to this pressure if the handle is lifted so that its angle with the horizontal direction is more than 30. The force P and the total area in contact with the floor remain constant..........[1] [Total: 11] [Turn over
6 2 (a) Explain the term centre of gravity of an object....[2] (b) Fig. 2.1 shows a lawn mower which is carried by two people. A diagram has been removed due to third party copyright restrictions Details: A diagram of a lawn mower Fig. 2.1 (i) The two people apply forces A and B at each end of the lawn mower. The weight of the lawn mower is 350 N. 1 Explain why the weight of the lawn mower does not act in the middle of the lawn mower, that is 55 cm from each end.......[1]
2 Use the principle of moments to show that the force B is 64 N. 7 [2] 3 Determine the force A. A =...N [1] (ii) State and explain what happens to the forces A and B if the person that applies force B moves his hands along the handle towards the middle of the lawn mower.............[2] [Total: 8] [Turn over
3 Fig. 3.1 shows the path of a tennis ball after passing over the net. tennis ball 8 net 1.20m 11.9 m line Fig. 3.1 As the ball passes over the net it is travelling horizontally at a height of 1.20 m. The ball strikes the ground on a line 11.9 m from the net. (a) Assume that there is no air resistance acting on the ball. (i) Show that the time taken for the ball to reach the line after passing over the net is 0.495s. [3]
(ii) At the instant the ball strikes the line calculate 9 1 the horizontal component of its velocity 2 the vertical component of its velocity. horizontal component =...m s 1 [2] vertical component =...m s 1 [2] (b) The mass of the tennis ball is 6.00 x 10 2 kg. Calculate the loss in gravitational potential energy of the ball from the time it passes over the net until it hits the line. loss in potential energy =...J [2] [Total: 9] [Turn over
4 (a) Define the Young modulus. 10...[1] (b) Fig. 4.1 shows a violin. A diagram has been removed due to third party copyright restrictions Details: A diagram of a violin Fig. 4.1 Two of the wires used on the violin, labelled A and G are made of steel. The two wires are both 500 mm long between the pegs and support. The 500 mm length of wire labelled G has a mass of 2.0 x 10 3 kg. The density of steel is 7.8 x 10 3 kg m 3. (i) Show that the cross-sectional area of wire G is 5.1 x 10 7 m 2. [2]
11 (ii) The wires are put under tension by turning the wooden pegs shown in Fig. 4.1. The Young modulus of steel is 2.0 x 10 11 Pa. Calculate the tension required in wire G to produce an extension of 4.0 x 10 4 m. tension =...N [3] (iii) Wire A has a diameter that is half that of wire G. Determine the tension required for wire A to produce an extension of 16 x 10 4 m. tension =...N [1] (iv) State the law that has been assumed in the calculations in (ii) and (iii)....[1] [Total: 8] [Turn over
5 Fig. 5.1 shows a vehicle that is used for carrying people off-road. 12 An image has been removed due to third party copyright restrictions Details: An image of a 4x4 vehicle Fig. 5.1 An off-road vehicle is designed so that it can be driven on rough and uneven ground. The mass of the vehicle and occupants is 3000 kg. (a) Explain how the tyres are designed to reduce the pressure of the vehicle on the surface over which it is travelling....[1] (b) The braking distance of the vehicle when travelling on a normal road at 26 ms 1 is 52 m. Calculate (i) the kinetic energy of the vehicle and occupants before braking occurs kinetic energy =...J [3] (ii) the average deceleration of the vehicle when braking deceleration =... unit... [2]
(iii) 13 the average braking force acting during the deceleration. braking force =...N [2] (c) (i) State and explain how two different road conditions affect the braking distance of a car. 1....... 2.......[2] (ii) The braking distance for a small car is shorter than for the off-road vehicle described above when they are tested travelling on the same road surface at the same speed. Discuss one difference between the small car and the off-road vehicle that could explain this..........[2] [Total: 12] [Turn over
6 In this question, two marks are available for the quality of written communication. 14 A skydiver jumps from an aircraft that is flying horizontally at a height of 5000 m with a constant speed. (a) Describe and explain the motion of the skydiver as she descends towards the ground from the moment she jumps until she opens her parachute. In your description of her motion use the terms speed, acceleration and force....[6]
15 (b) In some competitions the skydiver attempts to travel at the highest possible speed over a vertical distance of 1000 m. Discuss and explain the different methods that could be used by the skydiver to achieve the highest possible speed....[4] Quality of Written Communication [2] [Total: 12] END OF QUESTION PAPER
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2821 Mark Scheme June 2007 Mark Scheme Unit Code Session Year Version Page 1 of 6 2821 JUNE 2007 FINAL Abbreviations, annotations and conventions used in the Mark Scheme / = alternative and acceptable answers for the same marking point ; = separates marking points NOT = answers which are not worthy of credit ( ) = words which are not essential to gain credit = (underlining) key words which must be used to gain credit ecf = error carried forward AW = alternative wording ora = or reverse argument Question 1 Expected Answers Marks 1 (a) (i) 1 2 (ii) (iii) (iv) (b) (i) (ii) Horizontal component = 24cos30 = 21 (20.8) (N) vertical component = 24sin30 = 12 (12.0) (N) vertical force = 65 + 12 = 77 horizontal force = 20.8 (note ecf for 20.8 component) resultant = [(77) 2 + (20.8) 2 ] 1/2 = 80 (79. 8) (N) (or by vector triangle need correct labels and arrows for mark) 80 (79.8)(N) / equal to (iii) allow ecf the resultant force needs to be zero or forces need to balance above value to give no acceleration or constant velocity P = F / A = 77 / 4.2 x 10-3 = 18000 (18333) (Pa) more / increases downward / vertical component (of P) will be greater (for larger angles) M1 A0 Total: 11 3
2821 Mark Scheme June 2007 Mark Scheme Unit Code Session Year Version Page 2 of 6 2821 JUNE 2007 FINAL Abbreviations, annotations and conventions used in the Mark Scheme / = alternative and acceptable answers for the same marking point ; = separates marking points NOT = answers which are not worthy of credit ( ) = words which are not essential to gain credit = (underlining) key words which must be used to gain credit ecf = error carried forward AW = alternative wording ora = or reverse argument Question Expected Answers Marks 2 (a) The(single) point where the weight of an object (may B2 be taken to) acts Weak answers score one only e.g. where the weight acts (b) (i) 1 2 3 The (distribution of the) mass of the lawn mower is not uniform One correct moment about A stated B x 110 or 350 x 20 B = (350 x 20) / 110 (moments equated) B = 63.6 (N) A = 350 63.6 = 286(.4) (N) A0 (b)(ii) A goes down and B goes up Turning effect of B is less / B needs greater force to produce the same moment / if distance goes down force needs to go up (to maintain the same turning effect) Total: 8 4
2821 Mark Scheme June 2007 Mark Scheme Unit Code Session Year Version Page 6 of 6 2821 JUNE 2007 FINAL Abbreviations, / = alternative and acceptable answers for the same marking point annotations and ; = separates marking points NOT = answers which are not worthy of credit conventions ( ) = words which are not essential to gain credit used in the = (underlining) key words which must be used to gain credit Mark Scheme ecf = error carried forward AW = alternative wording ora = or reverse argument Question 6 Expected Answers Marks 6 (a) Vertical motion: Initial speed (down) is zero Initial acceleration (down) is g / 9.8 m s -2 Initial force (down) is weight / no drag force Speed (down) increases Acceleration decreases Drag force increases (with speed) / force is weight air resistance Reaches terminal velocity / speed is constant Acceleration is zero Drag force equals weight Maximum of 5 marks for vertical motion Horizontal motion: Initial velocity is the same as the aircraft Air resistance will reduce this horizontal velocity Maximum of six marking points required B6 (b) Reduce air resistance Fall head first / fall vertically / reduce area in direction of fall Fall with arms to the side / in line with body / streamline body Wear tight fitting clothes / smooth surface Good physics mark QWC Maximum of four marking points required SPAG ( greater than two errors) TECHNICAL LANGUAGE B4 Total: 12 5
2821 Mark Scheme June 2007 Mark Scheme Unit Code Session Year Version Page 3 of 6 2821 JUNE 2007 FINAL Abbreviations, annotations and conventions used in the Mark Scheme / = alternative and acceptable answers for the same marking point ; = separates marking points NOT = answers which are not worthy of credit ( ) = words which are not essential to gain credit = (underlining) key words which must be used to gain credit ecf = error carried forward AW = alternative wording ora = or reverse argument Question Expected Answers Marks 3 (a) (i) s = ut + ½ at 2 (ii) 1 2 1.20 = 0 + ½ x 9.81 x t 2 or u = 0 and s = ½ x a x t 2 t 2 = 2.4 / 9.81 / 0.2446 t = 0.494(6) s horizontal component = distance / time = 11.9 / 0.495 = 24.0(6) (m s -1 ) vertical component = u + at = 0 + 9.81 x 0.495 = 4.86 (m s -1 ) A0 (b) (loss of / change in) potential energy = mgh = 6 x 10-2 x 9.81 x 1.2 = 0. 71 (0.706)(J) 2 sf needed Total: 9 6
2821 Mark Scheme June 2007 Mark Scheme Unit Code Session Year Version Page 4 of 6 2821 JUNE 2007 FINAL Abbreviations, annotations and conventions used in the Mark Scheme / = alternative and acceptable answers for the same marking point ; = separates marking points NOT = answers which are not worthy of credit ( ) = words which are not essential to gain credit = (underlining) key words which must be used to gain credit ecf = error carried forward AW = alternative wording ora = or reverse argument Question Expected Answers Marks 4 (a) Young modulus = stress / strain (b) (i) (ii) (iii) (iv) Density = mass / volume Area x length = mass / density Area = (2.0 x 10-3 ) / (7800 x 0.5) or 2.56 x 10-7 / 0.5 = 5.1(3) x 10-7 m 2 E = (F x l) / (A x e) / stress = F / A (1.6 x 10 8 and strain = e / l (8 x 10-4 ) F = (E x A x e) / l = (2 x 10 11 x 5.1 x 10-7 x 4.0 x 10-4 ) / 0.5 =82 (N) (81.6) Diameter for D is half G hence area is ¼ of G Extension is 4x greater Tension required is the same = 82 (N) The extension is proportional to the force / Hooke s law (OWTE) A0 Total: 8 7
2821 Mark Scheme June 2007 Mark Scheme Unit Code Session Year Version Page 5 of 6 2821 JUNE 2007 FINAL Abbreviations, annotations and conventions used in the Mark Scheme / = alternative and acceptable answers for the same marking point ; = separates marking points NOT = answers which are not worthy of credit ( ) = words which are not essential to gain credit = (underlining) key words which must be used to gain credit ecf = error carried forward AW = alternative wording ora = or reverse argument Question Expected Answers Marks 5 (a) Large surface area (of tyres) (in contact with road) large area in contact with the road (b)(i) (ii) K = ½ mv 2 = 0.5 x 3000 x (26) 2 = 1014000 (J) 1.01 x 10 6 (J) v 2 = u 2 + 2as a = - 26 2 / (2 x 52) deceleration = 6.5 unit: m s -2 (iii) using F = ma or work done = F x d = 3000 x 6.5 F = 1.01 x 10 6 / 52 (c)(i) (ii) = 19500 (N) = 19400 (N) smooth / rough surface less / more friction longer / shorter distance ice surface less friction greater distance dry / wet more / less friction longer / shorter distance up slope / down slope refer to weight component and effect smaller mass same force gives larger deceleration or less braking force (from off-road) gives less deceleration B2 max Total: 12 8