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Mark Scheme (Results)January 2018Pearson EdexcelInternational Advanced Subsidiary LevelIn Mechanics M1 (WME01)Paper 01

Edexcel and BTEC QualificationsEdexcel and BTEC qualifications are awarded by Pearson, the UK’s largest awardingbody. We provide a wide range of qualifications including academic, vocational,occupational and specific programmes for employers. For further information visitour qualifications websites at www.edexcel.com or www.btec.co.uk. Alternatively,you can get in touch with us using the details on our contact us page atwww.edexcel.com/contactus.Pearson: helping people progress, everywherePearson aspires to be the world’s leading learning company. Our aim is to helpeveryone progress in their lives through education. We believe in every kind oflearning, for all kinds of people, wherever they are in the world. We’ve beeninvolved in education for over 150 years, and by working across 70 countries, in100 languages, we have built an international reputation for our commitment tohigh standards and raising achievement through innovation in education. Find outmore about how we can help you and your students at: www.pearson.com/ukJanuary 2018Publications Code WME01 01 1801 MSAll the material in this publication is copyright Pearson Education Ltd 2018

General Marking Guidance All candidates must receive the sametreatment. Examiners must mark the first candidatein exactly the same way as they mark the last. Mark schemes should be applied positively.Candidates must be rewarded for what they haveshown they can do rather than penalised foromissions. Examiners should mark according to the markscheme not according to their perception of where thegrade boundaries may lie. There is no ceiling on achievement. All markson the mark scheme should be used appropriately. All the marks on the mark scheme are designedto be awarded. Examiners should always award fullmarks if deserved, i.e. if the answer matches the markscheme. Examiners should also be prepared to awardzero marks if the candidate’s response is not worthy ofcredit according to the mark scheme. Where some judgement is required, markschemes will provide the principles by which markswill be awarded and exemplification may be limited. When examiners are in doubt regarding theapplication of the mark scheme to a candidate’sresponse, the team leader must be consulted. Crossed out work should be marked UNLESSthe candidate has replaced it with an alternativeresponse.

PEARSON EDEXCEL IAL MATHEMATICSGeneral Instructions for Marking1. The total number of marks for the paper is 752. The Edexcel Mathematics mark schemes use the following types of marks: M marks: Method marks are awarded for ‘knowing a method and attempting toapply it’, unless otherwise indicated. A marks: Accuracy marks can only be awarded if the relevant method (M)marks have been earned. B marks are unconditional accuracy marks (independent of M marks) Marks should not be subdivided.3. AbbreviationsThese are some of the traditional marking abbreviations that will appear in themark schemes. bod – benefit of doubt ft – follow through the symbol cao – correct answer only cso - correct solution only. There must be no errors in this part of thequestion to obtain this mark isw – ignore subsequent working awrt – answers which round to SC: special case o.e. – or equivalent (and appropriate) d or dep – dependent indep – independent dp decimal places sf significant figures The answer is printed on the paper or ag- answer given will be used for correct ftor d The second mark is dependent on gaining the first mark

4. All A marks are ‘correct answer only’ (cao.), unless shown, for example, as A1 ft toindicate 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 answersshould 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 theattempts 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 needresolving (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 previousspecified method mark has been awarded. Any numerical answer which comes from use of g 9.8 should be given to 2 or 3 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 onceper complete question. However, premature approximation should be penalised everytime 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 aquestion i.e. (a) or (b) or (c), then that working can only score marks for that part ofthe 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, thesubsequent A marks affected are treated as A ft Mechanics AbbreviationsM(A) Taking moments about A.N2LNewton’s Second Law (Equation of Motion)NELNewton’s Experimental Law (Newton’s Law of Impact)HLHooke’s LawSHM Simple harmonic motionPCLMPrinciple of conservation of linear momentumRHS, LHSRight hand side, left hand side.

January 2018Mechanics 1 – WME01Mark SchemeQuestionNumberSchemeMarksAB30 145 TBTACWN.B. If they assume that the tensions are the same, can scoremax:M0A0M1A0DM0A0A0.If they use the same angles, can score max: M1A0M1A0DM0A0A0Resolve parallel to AB: TA cos30 TB cos 45Resolve perpendicular to AB: W TA sin 30 TB sin 45Solve for TA or TB2TA W 0.73W (or better)1 3TB 61 3M1A1M1A1DM1A1W 0.90W (or better)A1(7)Alternative (triangle of forces):TA60 75 W45 TBTAW sin 45 sin 75TBWSine rule for TB: sin 60 sin 75Solve for TA or TB: TA 0.73W (or better )Sine rule for TA:TB 0.90W (or better)M1A1M1A1DM1A1A1(7)[7]

QuestionNumber1SchemeMarksNotes for question 1First M1 for resolving horizontally with usual rulesFirst A1 for a correct equationSecond M1 for resolving vertically with usual rulesSecond A1 for a correct equationThird DM1, dependent on both previous M marks, for solving for eitherTA or TBThird A1 for TA 0.73W or better or any correct surd answer but A0 forW, where k is a decimal. Allow ‘invisible brackets’kFourth A1 for TB 0.90W or better (0.9W is A0) or any correct surdWanswer but A0 for, where k is a decimal.kAlternative using sine rule or Lami’s TheoremTAWFirst M1A1 foroe (e.g. allow sin 105 or reciprocals) sin 45 sin 75TBWSecond M1 for(allow sin 30 and/or sin 105) sin 60 sin 75TBWSecond A1 for sin 60 sin 75Third DM1, dependent on either previous M mark, for solving for eitherTA or TBThird A1 for TA 0.73W or better or any correct surd answer but A0 forW, where k is a decimal.kFourth A1 for TB 0.90W or better or any correct surd answer but A0Wfor, where k is a decimal.k18

QuestionNumberSchemeMarksR20 N2.θPF40 NResolve horizontally: F 20cos Their F e.g. allow RResolve vertically:R 40 20sin 20cos 40 20sin Use of F R : 20cos 40 20sin cos 2 sin Given AnswerM1A1M1A1DM1A1[6]2Notes for question 2First M1 for resolving horizontally with usual rulesFirst A1 for a correct equationSecond M1 for resolving vertically with usual rulesSecond A1 for a correct equationThird DM1, dependent on both previous M marks, for use of F R togive inequality in only. (N.B. If they use F R in the horizontalresolution, this mark is not available)Third A1 for given answer19

QuestionNumberScheme2u3aMarksuA2mBkmuv2 u 2m 2u 2 Magnitude of impulse 5muImpulse on AM1A1A1(3)3b3b alt u CLM: 2m 2u km u 2m kmv 2 kmv 5mu kmu 0Use of v 0 : k 5Given AnswerAlternative: Impulse on B: 5mu km v u v Use of v 0 :5u u ORk5u u 0 k 5kk M1A1DM1A1(4)M1A15uu vOR if v 0, then k 5Given AnswerDM1A1(4)[7]3a3a altNotes for question 3M1 for using impulse change in momentum for A (M0 if clearlyadding momenta or if g is included or if not using 2m in both terms) butcondone sign errors. u u First A1 for 2m 2u or 2m 2u 2 2 Second A1 for 5mu (must be positive since magnitude) terms collected5uAlternative: Use CLM to find v u then usekImpulse on B: km ((5u/k - u) u) M1A1 for the complete method 5mu3bA1First M1 for CLM with correct no. of terms, all dimensionally correct.Condone consistent g’s or cancelled m’s and sign errors.First A1 for a correct equation (allow v in place of v)Second DM1 for use of v 0 or v 0 as appropriateSecond A1 for given answer correctly obtained.110

QuestionNumber3baltSchemeMarksFirst M1 for using their impulse on A change in momentum for B (M0if clearly adding momenta or if g is included or if not using km in bothterms) but condone sign errors.First A1 for a correct equation (allow v in place of v)Second DM1 for use of v 0 or v 0, as appropriate, but must be froma correct v or k, to deduce given answer.Second A1 for given answer correctly obtained.111

QuestionNumberSchemeMarksRF4a6g30 Perpendicular to plane: R 6 g cos30Parallel to plane: 6 g sin 30 F 6a N.B. Could be their F1F R seen. N.B. Could be their R4Solve for a : a 2.78 (2.8) (ms-2)B1M1A1B1M1A1(6)4bUse of suvat: v2 u 2 2as 2 2.78 10v 7.45417. 7.45 (7.5) (ms-1)M1A1(2)[8]4aNotes for question 4First B1 for R 6gcos 30 seenFirst M1 for resolving parallel to the plane with usual rulesFirst A1 for a correct equationN.B. F does not need to be substituted for this A mark1R seen4Second M1 for solving for aSecond A1 for 2.78 or 2.8Second B1 for F 4bN.B. could be their RM1 for a complete method for finding v, using their aA1 for 7.45 or 7.5112

QuestionNumberSchemeMarksSpeed5a20T4TTimeBasic shape20, 4T and T placed correctlyB1DB1(2)5bUse of v u at : constant speed 0.6 20 12 (ms-1)(Speed at end 12 0.3T )Using v-t graph:12TDistance: 705 4T 20 4T 12 12 0.3T 22 48T 120 12T 0.15T 2 60T 120 0.15T 2Form 3 term quadratic and solve for T: 3T 2 1200T 11700 0 T 2 400T 3900 0 T 10 T 390 0M1A1M1A2M1T 10 onlyA1(7)Alternative:Use of v u at : constant speed 0.6 20 12 (ms-1)M1A11Using s ut at 2 : 705 0.3 400 4T 12 12T 0.15T 2 2M1A222 0.15T 60T 585 0 T 400T 3900 0 T 10 T 390 0T 10 onlyM1A1(7)5cExtra time: (2 20) their TORTotal time: 20 5T 40 T(their T)12 0.3 theirT0.3B1M1A1 100 (s)(3)Alternative: Total time to decelerate to rest 12/0.3 40Total time A to C 20 4T 40 100B1M1A1[12]113

QuestionNumberSchemeMarksNotes for question 55aFirst B1 for basic shape. Allow if ‘extra triangle’ on end included,provided B clearly markedSecond DB1 : may use, 20, 20 4T, 20 5T5b5cFirst M1 for attempt to find constant speed (v u at or a gradient)20 x 0.6First A1 for 12Second (generous) M1 for clear attempt to use 705 total area under thegraph to give an equation in T only but must see ½ used somewhereN.B. M0 if just a trapezium oe is usedSecond A1 and Third A1: for any correct equation, -1 e.e.o.o.Third M1 for forming and attempt to solve a 3 term quadratic(need evidence of solving e.g. formula or factorising, if T values areincorrect) otherwise this M mark can be implied if they state that T 10with no working. (T 390 NOT needed)Fourth A1 for T 10.N.B. For total area, could see:Trapezium Rectangle Triangle12705 4T 20 4T T (12 0.3T ) 12 T 0.3T2Triangle Rectangle Trapezium705 12 .20.12 (4T 12) 12 T 12 12 0.3T Triangle Rectangle Rectangle Triangle705 12 .20.12 (4T 12) T (12 0.3T ) 12 T 0.3TTriangle Rectangle Trapezium (at top)1705 2 .20.12 5T (12 0.3T ) 12 0.3T (5T 4T )Rectangle – triangle– triangle705 12(20 5T ) 12 .20.12 12 T 0.3T1212B1 for either additional time is T or time to decelerate is0.30.3M1 for a correct method to find the total time, using their T1212 20 4T T or20 4T T0.30.3A1 for 100 cao114

QuestionNumber6aSchemeMarksResultant force 2i 3j 4i 5j 6i 2 j (N)M1Use of F ma : 6i 2 j 2a , a 3i jM1Magnitude:a 32 12 10 3.2 or better (ms-2)M1A1(4)6b(10i 2 j) ( ui uj) T (3i j)10 u 3T and 2 u TT 6(i)u 8(ii)M1DM1A1ftA1A1(5)[9]6a6bNotes for question 6First M1 for adding forces – must collect i’s and j’sSecond M1 for use of F ma or F maThird M1 for finding a magnitudeA1 for 10 3.2 or better First M1 for use of v u at with their a (M0 if clearly using F insteadof a)Second DM1, dependent on previous M, for equating cpts of i and jFirst A1ft for two correct equations following their aSecond A1 for T 6Third A1 for u 8115

QuestionNumberSchemeMarks2RRA7a1m CD4m1mBx8gN.B. If RC and RD reversed,can score max: M1A1(if vert res is used)M1A0DM1A0Consistent omission of g in both parts of this question can score all ofthe marks.Resolve vertically: 3R 8gM1A1M(C) : 8g x 1 4 2 R8 gx 8 g M1A164 g 88 g11, x 333Given AnswerDM1A1(6)N.B. (Allow RD instead of 2RC in either equation for M mark)M2 A2SC: M(G): R( x 1) 2R(5 x)11Given answerx 3DM1 A1(6)7bN.B. If they use a value for a reaction found in part (a) in their part (b),no marks for part (b) available.RDRFAF2m3mD1mB113g3m8gN.B. RD kRFResolve vert : RF kRF 11g(Allow RD instead of kRF for M mark))5M(F) : (kRF 3) (3g 2) 8 g 3(Allow RD instead of kRF for M mark)2k oe , 0.29 or better7M1A1M1A1DM1A1(6)[12]116

QuestionNumberSchemeMarksNotes for question 7First M1 for either resolving vertically or taking moments with usualrulesFirst A1 for a correct equationSecond M1 for taking moments with usual rulesSecond A1 for a correct equationN.B. Their moments equation(s) may not be in x, if they’ve clearlydefined a different distance and can score the A1 in each case.7aThird DM1, dependent on first two M marks, for solving for xThird A1 for “ x (or AG) 11/3”GIVEN ANSWER (Must be EXACT)M(A), (R 1) (2 R 5) 8 gxPossible equations: M(B), (R 5) (2 R 1) 8 g (6 x)M(D), (R 4) 8 g (5 x)N.B. (Allow RD instead of 2RC in all cases for M mark)First M1 for either resolving vertically or taking moments with usualrulesFirst A1 for a correct equationSecond M1 for taking moments with usual rulesSecond A1 for a correct equationThird DM1, dependent on first two M marks, for solving for kThird A1 for k 2/7, any equivalent fraction or 0.29 or better7bM(A), 2RF 5kRF 8 g 1137M(B), 4RF (1 kRF ) (8 g ) (3g 6)3Possible equations:4M(D), 3RF 8 g (3g 5)35411M(G ),RF kRF 3g 333N.B. (Allow RD instead of kRF in all cases for M mark)117

QuestionNumberSchemeMarksTTA3 kg8a5kg B3g5g40 Motion of A : T 3g sin 40 3aMotion of B :5g T 5aSolve for T30 (N) or 30.2 (N)M1A1M1A1DM1A1(6)8b1g 5g T 5 3sin 40 3.76 (ms-2)58Use of suvat : v u at 3.76 1.5 5.64 (ms-1) or 5.6 (ms-1)5g T 5a a M1DM1A1(3)8c1Distance in first 1.5 seconds: s a1.52 4.23 (m)2their (b)2s 4.23 (m)OR: v2 u 2 2as :2 aNew a g sin 40 (-ve sign not needed)B12their (b)(m)2 new aTotal distance: 6.76 (m) (6.8)Distance up plane : v2 u 2 2as , s M1A1DM1A1(5)[14]8a8b8cNotes for question 8First M1 for equation of motion for A, with usual rulesFirst A1 for a correct equationSecond M1 for equation of motion for B, with usual rulesSecond A1 for a correct equationN.B. Either of these can be replaced by the whole system equation:5g 3g sin 40 8aThird DM1, dependent on previous two M marks, for solving for TThird A1 for 30 or 30.2 (N)First M1 for finding a value for a (possibly incorrect) This mark couldbe earned in part (a) BUT MUST BE USED IN (b).Second DM1,dependent on previous M, for a complete method to findthe speed of B as it hits the groundA1 for 5.6 or 5.64 (m s-1)First M1 for a complete method to find distance fallen by BFirst A1 for 4.23 or better118

QuestionNumberSchemeMarksB1 for new a g sin 40 (- sign not needed) (seen or implied)Second DM1, dependent on having found a new a, for a completemethod to find extra distance moved by A up the plane BUT M0 if newa is g.Second A1 for 6.8 or 6.76 (m).Pearson Education Limited. Registered company number 872828with its registered office at 80 Strand, London, WC2R 0RL, United Kingdom119

Mar 08, 2018 · Pearson Edexcel International Advanced Subsidiary Level In Mechanics M1 (WME01) Paper 01 . Edexcel and BTEC Qualifications Edexcel and BTEC qualifications are awarded by Pearson, the UK’s largest awarding body. We provide a wid

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Examiners should mark according to the mark scheme not according to their perception of where the grade boundaries may lie. All marks on the mark scheme should be used appropriately. All 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.