Mark Scheme (Results) Summer 016 Pearson Edexcel GCE in Mechanics (6678_01)
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General Marking Guidance All candidates must receive the same treatment. Examiners must mark the first candidate in exactly the same way as they mark the last. Mark schemes should be applied positively. Candidates must be rewarded for what they have shown they can do rather than penalised for omissions. Examiners should mark according to the mark scheme not according to their perception of where the grade boundaries may lie. There is no ceiling on achievement. All marks on the mark scheme should be used appropriately. All the marks on the mark scheme are designed to be awarded. Examiners should always award full marks if deserved, i.e. if the answer matches the mark scheme. Examiners should also be prepared to award zero marks if the candidate s response is not worthy of credit according to the mark scheme. Where some judgement is required, mark schemes will provide the principles by which marks will be awarded and exemplification may be limited. Crossed out work should be marked UNLESS the candidate has replaced it with an alternative response.
General Instructions for Marking PEARSON EDEXCEL GCE MATHEMATICS 1. The total number of marks for the paper is 75.. The Edexcel Mathematics mark schemes use the following types of marks: M marks These are marks given for a correct method or an attempt at a correct method. In Mechanics they are usually awarded for the application of some mechanical principle to produce an equation. e.g. resolving in a particular direction, taking moments about a point, applying a suvat equation, applying the conservation of momentum principle etc. The following criteria are usually applied to the equation. To earn the M mark, the equation (i) should have the correct number of terms (ii) be dimensionally correct i.e. all the terms need to be dimensionally correct e.g. in a moments equation, every term must be a force x distance term or mass x distance, if we allow them to cancel g s. For a resolution, all terms that need to be resolved (multiplied by sin or cos) must be resolved to earn the M mark. M marks are sometimes dependent (DM) on previous M marks having been earned. e.g. when two simultaneous equations have been set up by, for example, resolving in two directions and there is then an M mark for solving the equations to find a particular quantity this M mark is often dependent on the two previous M marks having been earned. A marks These are dependent accuracy (or sometimes answer) marks and can only be awarded if the previous M mark has been earned. E.g. M0 is impossible. B marks These are independent accuracy marks where there is no method (e.g. often given for a comment or for a graph) A few of the A and B marks may be f.t. follow through marks.
. General Abbreviations These are some of the traditional marking abbreviations that will appear in the mark schemes. bod benefit of doubt ft follow through the symbol will be used for correct ft cao correct answer only cso - correct solution only. There must be no errors in this part of the question to obtain this mark isw ignore subsequent working awrt answers which round to SC: special case oe or equivalent (and appropriate) dep dependent indep independent dp decimal places sf significant figures The answer is printed on the paper The second mark is dependent on gaining the first mark 4. All A marks are correct answer only (cao.), unless shown, for example, as ft to indicate that previous wrong working is to be followed through. After a misread however, the subsequent A marks affected are treated as A ft, but manifestly absurd answers should never be awarded A marks. 5. For misreading which does not alter the character of a question or materially simplify it, deduct two from any A or B marks gained, in that part of the question affected. 6. If a candidate makes more than one attempt at any question: If all but one attempt is crossed out, mark the attempt which is NOT crossed out. If either all attempts are crossed out or none are crossed out, mark all the attempts and score the highest single attempt. 7. Ignore wrong working or incorrect statements following a correct answer.
General Principles for Mechanics Marking (But note that specific mark schemes may sometimes override these general principles) Rules for M marks: correct no. of terms; dimensionally correct; all terms that need resolving (i.e. multiplied by cos or sin) are resolved. Omission or extra g in a resolution is an accuracy error not method error. Omission of mass from a resolution is a method error. Omission of a length from a moments equation is a method error. Omission of units or incorrect units is not (usually) counted as an accuracy error. DM indicates a dependent method mark i.e. one that can only be awarded if a previous specified method mark has been awarded. Any numerical answer which comes from use of g = 9.8 should be given to or SF. Use of g = 9.81 should be penalised once per (complete) question. N.B. Over-accuracy or under-accuracy of correct answers should only be penalised once per complete question. However, premature approximation should be penalised every time it occurs. Marks must be entered in the same order as they appear on the mark scheme. In all cases, if the candidate clearly labels their working under a particular part of a question i.e. (a) or (b) or (c), then that working can only score marks for that part of the question. Accept column vectors in all cases. Misreads if a misread does not alter the character of a question or materially simplify it, deduct two from any A or B marks gained, bearing in mind that after a misread, the subsequent A marks affected are treated as A ft Mechanics Abbreviations M(A) Taking moments about A. NL NEL HL Newton s Second Law (Equation of Motion) Newton s Experimental Law (Newton s Law of Impact) Hooke s Law SHM Simple harmonic motion PCLM Principle of conservation of linear momentum RHS, LHS Right hand side, left hand side.
Q Scheme Marks Notes 1a t 0, v 11 r 11 t, v 4p q11, Accept 4p q r 4p q 8 Any equivalent unsimplified form with 11 used Differentiate to find acceleration OR use symmetry, t 4, v 11 a pt q 11 16 p 4q 11, 4pq 0 t, a 0 4p q 0 nd eqn in p & q and solve for p & q Dependent on both previous m marks q q 8, q 8, p 1a alt v t t 8 11 t, a 4t 8 4(ms - ) Min speed at t v pt qt r k t c v k t (8) Completed square form. Correct completed square form t 0, v 11 4k 11, Solve for k k v t t 8t 11 Differentiate to find acceleration Dependent on both previous m marks a4 t t, a 4(m s - ) (8) 1b Integrate: t dt t t C follow their coefficients found in (a) Accept in p, q, r or t 8t 11dt t 4t 11t C At most one error seen ft For their coefficients All correct ft For their coefficients provided 0 t t 9 Use of t, t as limits on a definite 0 6 or integral (or subtract distances to cancel C). t 4t 11t Dependent on having integrated. 16 Allow with p, q, r 18 6 16
Q Scheme Marks Notes (m) Accept exact equivalent or.7 or better (5) [1]
Q Scheme Marks Notes a Equation of motion up or down the road. Requires all terms. Condone sign errors and trig confusion. Must be dimensionally correct. F mg sin R F R 9 Correct equation up the road b sin G R 9 G mg R P P F or G 1.5 1.5 P P 9 R or R 9 1.5 1.5 P 9 4 P, R 1.5 1.5 Correct equation down the road P 4900 (500g), R 784 (80g) Must be using work-energy. KE lost = PE gained + WD against R 1 800 1.5 d 8009.8 their R d 0 ft (6) Use of P F at least once v Solve simultaneous equations for P or R, provided F G and P and P used correctly CSO. Both values correct. Accept sf, sf or an exact multiple of g Equation needs all terms and no extras. Condone sign errors. At most 1 error. Allow with R (with trig. substituted) 6500 9d Rd Correct equation in their R(with trig. substituted) 6500 d 5.1(m) CSO. Accept 5(m) 1176 (4) [10]
Q Scheme Marks Notes Since this question is about the magnitude of the impulse, condone subtraction in the "wrong. order" throughout. mv mu 0.6ci cj ci cj Impulse = change in momentum Marking the RHS only 0.6 cicj Magnitude 0.6 c 9c Correct use of Pythagoras' theorem on mv mu or vu Marking the RHS only. Dependent on the previous 0.6 10c 0.6 10c Accept 10c for change in velocity The next two marks are not available to a candidate who has equated a scalar to a vector. 10 0.6 10c Equate & solve for c Dependent on the previous 10 c Accept. or better (6) alt m m 0.6c c c c 0.6c c v u i j i j change in momentum Square of magnitude i j = 0.610c The next two marks are not available to a candidate who has equated a scalar to a vector. 40 0.6c 9c, Equate & solve for c 10 c (6) alt 10 cos c c 0.6 10 sin cc 1 0.6c 10 cos 0.6c 10 sin 0.6c tan 1 cos 10 10 cos 0.6c 10 c Impulse momentum equation Correct equation Compare coefficients and form equation for cos or sin correct
alt mu I Impulse momentum triangle Units used for the vectors must be dimensionally correct mv 10 10 Sides of magnitude 5 c, 5 c, or 5 c, 5 c, 10 5 5 u c c c c Use of scalar product c c 0 at 90...to show velocities perpendicular Use of Pythagoras' theorem in a right 0.6 5c 0.6 5c 10 angled triangle 18 40, c 5 (6)
Q Scheme Marks 4a Triangle sector mass 4.5 4 c of m from O 1.41... 16 1.56...40... : 16 101 4 4.5 4 4.5 d 6 101 d 6 4 4.5.951... Distance from DA.951... 0.80 (0.8) m (5) Mass ratios Distances Distances from AD are 16 and 0.80 1 Moments about an axis through O and parallel to DA. Terms must be dimensionally correct. Shapes combined correctly. Correct unsimplified equation (distance from O) Accept 101 6 4 4.5 4a alt Triangle sector 4 1.56.. mass 4.5 c of m from both axes OC,OB 1 16 1 16 16 4 4.51 4 4.5 x 101 x y 64 4.5 101 d 6 4 4.5 Mass ratios Distances Moments about an axis through O. Terms must be dimensionally correct. Condone sign error(s) Correct unsimplified equation Distance from O Distance from DA.951... 0.80 (0.8) m (5)
4b D θ (.1) 0.80 m 4balt G their 0.80.1 tan or tan.1 their 0.80 Use of tan to find a relevant angle: 1.4 or 68.6 Angle between DC and downward vertical Correct method for the required = 15 - their θ angle = 114 The Q asks for the angle to the nearest degree. (4) GD OD OG OD. OC cos 45 Complete method to find angle sin 45 sin GD.8 ODG DG OG 66.4 Correct method for the required angle Required angle 180 66.4 114 The Q asks for the angle to the nearest degree. (4) [9]
Q Scheme Marks Notes 5a M(A): d cos 5g 4P 5a alt Resolving horizontally: Psin F Resolving vertically: Pcos R 5g 5gd cos R5g 4 5gd cossin F 4 M(B): 5g cos 4 d F sin 4 Rcos 4 Resolve parallel to the rod: 5g sin Rsin F cos F cos R 5g sin 5g cos 4 d F sin 4 F cos 4cos 5g sin cos 4F sin sin 0g cos 0g cos 5gd cos (8) Terms must be dimensionally correct. Condone trig confusion Requires all terms. Condone trig confusion and sign errors Correct equation Substitute for P to find R or F Dependent on both previous M marks One force correct. Accept equivalent 0g 5gd 0g tan forms e.g. R 4 1 tan Both forces correct. Accept equivalent 5gd tan forms e.g. F 4sec Needs all three terms. Terms must be dimensionally correct. Condone trig confusion At most one error Requires all terms. Condone trig confusion and sign errors At most one error Correct equation Eliminate one variable to find F or R Dependent on both previous M marks 5gd cossin F 4 One force correct 5gd cos R5g 4 Both forces correct See next page for part (b)
5gd cossin 5b 4 5gd cos 5g 4 Use of F R 5g 4 4 1 5gd cos 5gd cos sin 4 dcos dcos sin 4169 10d 144d Use 5 tan and solve for d 1 169 d (=.6 m or better) 66 (4) 5balt 5balt 1 5 1 75gd 5 F 5gd Use tan 1 1 4 169 1 5gd 144 R 5g 4 169 Both unsimplified expressions 1 75gd 5169g 180gd Use of F Rand solve for d 169 150gd 180gd 845g, d 66 (=.6 m or better) (4) 1 5 R 5g P, F P 1 1 5 1 1 P 5g P 1 1 65 P g 4P 169 d 5g cos 66 Substitute trig in their equations from resolving. use F Rand solve for d [1]
Q Scheme Marks Notes 6a Horizontal motion: x t Vertical motion: g Correct use of suvat. y 4t t Condone sign error(s) x g x y 4 9,, 4 g 18 7 g 18 Use y x and form an equation in one variable solve for λ 4 or 4. (4.9) Not 0 g (6) 7 alta Horizontal motion: x t g Correct use of suvat. Vertical motion: y 4t t Condone sign error(s) 1 14 Use y x and form an equation in t 4t gt, t g one variable t Solve for λ 4. (4.9) (6) 6b At A: v (m s -1 ) 14 Complete method using suvat to find v 4 g g v with their t or λ 10 (m s -1 Accept +10 with direction confirmed by ) diagram their 10 Dependent on the first in (b) Speed 109 ( m s -1 ) (10.4) Allow for v =10 1 their 10 tan or 1 Use trig to find a relevant angle. tan their 10 Dependent on the first in (b) Direction 7.below the horizontal (1.8 radians) Accept direction i10j Do not accept a bearing (7) Alt 6b Loss in GPE : mg 4m 1 1 Terms must be dimensionally correct. Gain in KE : mv m 5 Condone sign error. 1 5 Solve for v: 4 v v 109 vcos Use trig. to find a relevant angle 7.below the horizontal (7) Accept correct angle marked correctly on a diagram. [1]
Q Scheme Marks Notes u 7a A (m) 4 B (m) e v w 7b 7b alt 7c CLM: 6mu mv mw Requires all terms. Must be dimensionally correct. Condone sign error(s) 6u v w This equation defines their directions 9 Impact: w v u Must be used with e on the correct side u 4 4 Penalise inconsistent directions here 9 Solve simultaneous equations for v or w 6u w u w Dependent on the previous M marks 1 One correct w u = vb 10 9 1 9 v w u u u, va = 4 10 4 0 0 u Both correct (7) 1 Speed of B after hitting wall 10 ue () e their w Impulse 7 1 1 mu m u ue 4 10 10 9 1 1 1 e, e 4 10 14 7 1 Impulse mu m u V 4 10, u V 0 () 1u u e, 10 0 1 e 14 () Speed of B after second impact = ft 1 1 u u 14 10 0 Same velocity (and A has a head start), so no collision. () [1] ew for their w. Must be trying to use the correct equation with m. () Use impulse to find V. Must be trying to use the correct equation with m. V e their w. Compare two relevant speeds. (ft on their V or their e x their w) From correct work only
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