Mark Scheme (Results) Summer 08 Pearson Edexcel GCE In Mechanics M (6678/0)
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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. When examiners are in doubt regarding the application of the mark scheme to a candidate s response, the team leader must be consulted. Crossed out work should be marked UNLESS the candidate has replaced it with an alternative response.
General Instructions for Marking PEARSON EDEXCEL GCE MATHEMATICS. The total number of marks for the paper is 7.. 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.
3. 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.. 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 3 SF. Use of g = 9.8 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 R 00 N a v ms - F θ Motion down the plane: 70g F 70gsin 00 Dimensionally correct. Condone sign errors and sin/cos confusion. F 40 00 9000 Use of P Fv : F v B Award in (b) if not seen in (a) 9000 3 70g 00 v 49 9000 v 70 (4) R 4. ms - F b 00 N θ F ma : F 70g sin 00 70a 70g 9000 3 70g 00 70a 4. 49 Dimensionally correct. Condone sign errors and sin/cos confusion. Unsimplified equation with at most one error Correct unsimplified equation a 0.47 (0.467) (m s - ) or 3 sf only not 7 (4) [8]
Q Scheme Marks Notes a Using column vectors or i and j: Must be subtracting. I cos 7cos3 4 0. 0. Need both components I sin 7sin 3 0 Could consider components separately 0.347 0.803 Correct unsimplified equation. Accept +/- I 0.347 0.803 D Use Pythagoras to find magnitude Dependent on the previous I 0.87 0.87 or better Alternative using vector triangle (4) a alt 3.4 θ I Allow with velocities rather than impulse/momentum 0.8 Cosine rule: I.4 0.8.4 0.8 cos3 D I 0.87 (N s) (0.87 or better) (4) Dependent on the previous b 0.803 tan 0.347 0.347 or cos 0.87 ft Trig ratio of a relevant angle (using velocities or impulse/momentum) Correct expression for correct Ft on values from (a) Do not ISW 66.6 (67) Or better from correct work. (3) b alt Sine rule: sin sin 3.4 I ft 66.6 (67) Or better (3) [7]
Q Scheme Marks Notes 3a a Mass ratios : 4 a : : a 4 B Or equivalent 4a Distances relative to BD: a,, d 3 B Or equivalent. Condone sign errors Moments about BD (or a parallel axis) Dimensionally correct. All terms required. Condone sign errors. Accept in a vector equation. a 4a 4a a 4 a d 3 Correct unsimplified equation 4 a 4 4 d (Distance from AE = 8 6 a 6 38 ) Obtain given answer from correct 0 0a working. d a Condone -ve becoming positive with no 3 8 38 explanation at the end () a D d θ θ 3b. E Use trig to find relevant angle Using the given value of d a 38 tan d 0 0.98 Correct expression for required angle 9... 9 NB:The question asks for the nearest degree (3) [8]
Q Scheme Marks Notes T θ 4 R A M(A): at mga cos T mg cos M(B): mga cos Fr asin R acos Resolve : θ Fr a mg a Fr T sin mg cos sin : R T cos mg B First equation Need all terms. Condone sign errors and sin/cos confusion Second equation Need all terms. Condone sign errors and sin/cos confusion Third equation Need all terms. Condone sign errors and sin/cos confusion Use Fr R : R Tsin B Condone correct inequality Form equation in and : Eliminate T and R Dependent on first 3 M marks R mg mg coscos D and R mg cos sin Solve for mg cossin Dependent on previous M D mg mg coscos cossin cos [0] Obtain given answer from correct working Must explain if inequality becomes equality
Alt Moments (about B): mga cos Fr asin R acos Resolving (parallel to rod): M Fr cos Rsin mg sin A Use of Fr R : mg cos Rsin R cos B Rcos Rsin mg sin Form equation in and : mg sin Rcos Rsin mg cos Rcos Rsin D sin cos sin cos cos sin Solve for : cos sin sin cos cos sin D Correct unsimplified - each error sin cos sin cos cos sin cos NB for alternatives using moments and resolving: e.g. Resolve : Fr T sin M(centre): at acosr asinfr Obtain given answer from correct working First equation Sufficient equations to solve MA
Alt θ 3 concurrent forces atanα T Res a B A α θ a mg tan tan tan tan tan a tan tan a tan tan tan tan tan tan tan sin cos cos sin F R B tan D sincos sin cos cossin D cos (0) Obtain given answer from correct working
Q Scheme Marks Notes u u a A(3m) B(m) CLM: 3m u mu 3mv mw 4u 3v w v w Impact law: w v u u u 3 Solve for simultaneous equations for w or v: 3w 3v 3u, w3v 4u D Dependent on both previous M marks Must see working - Given Answer 7 w 7u, w u v u Or equivalent. Must be positive (7) b Speed of B after collision with wall: 7 7 B Accept +/- u u 0 Total time for either particle B Equate the time travelled for each particle: x y x y Correct unsimplified 7 7 u u u 0 x 0y x y, 7u 7u u u 0 x 0 y 3 x 3 y D Dependent on previous y x, Or equivalent. y x 0.4x or better (6)
Alt Alt Alt 3 Speed of B after collision with wall: 7 7 B Accept +/- u u 0 Time of travel for B: x y x 0y 7 7 u u 7u 0 B Distance moved by A: Correct method for distance x 0y x 4y u 7u 7 Correct unsimplified x 4y y x, x 4y 7y 7x Dependent on previous. D 7 Form equation in x and y Or equivalent. y x 0.4x or better (6) Speed of B after collision with wall: 7 7 B Accept +/- u u 0 x x distance moved by A x u x B Distance apart when B hits the wall 7u 7 Gap closing at 7 u u u 0 0 0x Time to collision: x u Use of 9 x x for is M0 7 0 77u 7 7 7 0x Dependent on previous D Distance moved by B: y u x Or equivalent. 0 77u 0.4x or better (6) Speed of B after collision with wall: 7 7 B Accept +/- u u 0 x x distance moved by A x u x B Distance apart when B hits the wall 7u 7 Ratio of speeds 4:7 7 x Distance moved by B: y x 7 D (6) [3] Dependent on previous Use of 9 x x for is M0 7 7 Or equivalent. 0.4x or better
Alt 4 Speed of B after collision with wall: 7 7 B Accept +/- u u 0 x x distance moved by A x u x B Distance apart when B hits the wall 7u 7 Equate times for each particle to cover the residual distance. x 0 y y, x Use of 9 x x for is M0 y y 7 7 u 7 7u 7 7 Dependent on previous D Distance moved by B: y x Or equivalent. 0.4x or better (6)
Q Scheme Marks Notes 6a Differentiate v: 4 6t 8 t a i j Anywhere in (a) Use of F ma and substitute t 3 : F 0. 4 63 i 8 3 j 7i j Use of Pythagoras theorem: D D F 49 0 7.07... 7. or better For v, i component= j component: 4t 3t 40 8t t Solve for t: 4t t 40 0, t 3t0 0 t t 0, t D a 4 30 i 8 0 j 6i j (ms - ) Only (9) Dependent on the first Dependent on the first NB Could use Pythagoras and then use F ma. st st step. nd - nd step With no incorrect equations in t seen Dependent on the previous M, Must see method if solving an incorrect quadratic Only - could be implied by later rejection of - 6b Integrate v: 3 3 r t t p i 40t 4t t qj 3 A - ee 43 3 r i j, r 93 j AB 3 r r D 49 AB i 49 j i j 3 3 49.7 or better () [4] 3 80, 3 3 Use limits in a definite integral or to evaluate a constant of integration Dependent on the previous
Q Scheme Marks Notes 7a Horizontal distance: x ucost B Vertical distance: y u sint gt x g x y usin ucos ucos D gx gx xtan x u cos u ut ut gt Condone sign errors and sin/cos confusion Substitute for t and Dependent on previous g x u x () Obtain given answer from exact working 7b g x36, y 0 : 0 36, Use given equation or a complete method u using suvat to find u. g 36 u, 4 u (m s - ) Accept 4g () ( u cos 9.39 (m s 7c Min speed ucos - ) ) Consistent with their B in (a) Minimum KE: D 0.3 Dependent on previous 0.3u cos 3. (3) (J) (3) 7c dy u Max ht when 0, x 8 Or from 36 alt dx g (symmetry) Conservation of energy: mu mgh mv 0.3 0.3 g g 3. (J) (3)
Q Scheme Marks Notes 7d 7d alt 7d alt 7d alt3 Gradientof trajectory at B B Accept tan dy g Differentiate and equate : x dx u (their u) solve for x: g x u gx u D Dependent on previous x. g () Gradient of trajectory at B B Accept tan Use components of velocity: u sin gt u cos usin ucos 0 t g g t.40 Horizontal distance: ucost. (3) (m) D Dependent on previous () Can be implied by downward velocity Gradient of trajectory at B B u cos or tan vy, gt Use suvat to find t t.39 g Horizontal distance: ucost. (3) (m) D Dependent on previous () ucos ucos 0 u sin u sin gt B Scalar product = 0 u u gt cos sin sin 0 u sin. gt Horizontal distance: u u cos. t u cos. gsin g () [] Must have gt in second vector Solve for t
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