AQA MM03 w MS JUN15

A-level Mathematics MM03 Mark scheme 6360 June 2015 Version 1.0: Final Mark schemes are prepared by the Lead Assessment Writer and considered, together with the relevant questions, by a panel of subject teachers This mark scheme includes any amendments made at the standardisation events which all associates participate in and is the scheme which was used by them in this examination The standardisation process ensures that the mark scheme covers the students’ responses to questions and that every associate understands and applies it in the same correct way As preparation for standardisation each associate analyses a number of students’ scripts Alternative answers not already covered by the mark scheme are discussed and legislated for If, after the standardisation process, associates encounter unusual answers which have not been raised they are required to refer these to the Lead Assessment Writer It must be stressed that a mark scheme is a working document, in many cases further developed and expanded on the basis of students’ reactions to a particular paper Assumptions about future mark schemes on the basis of one year’s document should be avoided; whilst the guiding principles of assessment remain constant, details will change, depending on the content of a particular examination paper Further copies of this mark scheme are available from aqa.org.uk Copyright © 2015 AQA and its licensors All rights reserved AQA retains the copyright on all its publications However, registered schools/colleges for AQA are permitted to copy material from this booklet for their own internal use, with the following important exception: AQA cannot give permission to schools/colleges to photocopy any material that is acknowledged to a third party even for internal use within the centre Key to mark scheme abbreviations M m or dM A B E or ft or F CAO CSO AWFW AWRT ACF AG SC OE A2,1 –x EE NMS PI SCA c sf dp mark is for method mark is dependent on one or more M marks and is for method mark is dependent on M or m marks and is for accuracy mark is independent of M or m marks and is for method and accuracy mark is for explanation follow through from previous incorrect result correct answer only correct solution only anything which falls within anything which rounds to any correct form answer given special case or equivalent or (or 0) accuracy marks deduct x marks for each error no method shown possibly implied substantially correct approach candidate significant figure(s) 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 MARK SCHEME A– LEVEL Mathematics – MM03 – June 15 Question [F ] = MLT ( Solution Marks -2 MLT -2 = LT -1 B1 ) (L ) (ML ) α β -3 γ m1 γ =1   α + β − 3γ = 1  − α = −2  A1 m1 Total A1: γ = m1: Two correct equations for α and β A1: Correct values for α and β Condone use of units instead of dimensions A1 β =1 Comments B1: Correct dimensions of F M1: Substituting the dimensions of the quantities into the given equation to obtain RHS correctly m1: Collecting indices on RHS Could be implied by later work M1 = M γ Lα + β -3γ T -α α=2 , Total of 17 MARK SCHEME A– LEVEL Mathematics – MM03 – June 15 Question (a) Solution Marks x = u cos αt t= Total M1 M1: Correct expression for horizontal displacement A1: Correct expression for t A1 x u cos α gt 2 x  x  y = u sin α × − (9.8)  u cos α  u cos α  Comments y = u sin αt − (b)(i) M1 M1: Correct expression for vertical displacement Allow sign errors m1 (ii) 4.9 x y = x tαn α − AG u cos α 4.9 s − s = s tan 55  − 21 cos 55  + tan 55  212 cos 55  s= 4.9 s = 71.9 ( A1 ) x = 21 cos 55  = 12.045 71.895 y = 21sin 55 − 9.8 × 21 cos 55  or y = 21sin 55 − 2(9.8)(− 71.895) M1 M1: Substituting ±s for x and y m1 m1: Making s the subject of their equation A1: AWRT 71.9 Condone use of g = 9.81 which gives 71.8 A1  ( ) B1 B1: Correct expression or value for horizontal component of velocity M1 M1: Correct expression or value for vertical component of velocity, with their answer to (b)(i) y = −41.292 − 41.292 21 cos 55  = −74  or 74  tan −1 A1 A1: Correct expression or value m1: Use of tan with their velocity components A1: Correct angle to nearest degree CAO m1 A1 Total m1: Elimination of t from equation for vertical displacement A1: Correct result from correct working Penalise use of g = 9.81 13 of 17 MARK SCHEME A– LEVEL Mathematics – MM03 – June 15 (b)(ii) Alternative: y = x tαn α − 4.9 x u cos α 2(4.9 )x dy = tαn α − dx u cos α = tan 55  − = −3.428 2(4.9 )(71.895) 212 × cos 55  The angle = tan -1 (- 3.428) = −74  or 74  B1 B1: Correct derivative M1 M1: Substituting values A1 A1: Correct value of the derivative m1: Use of tan to find the angle A1: Correct angle to nearest degree CAO m1 A1 of 17 MARK SCHEME A– LEVEL Mathematics – MM03 – June 15 Question Solution Marks Total Comments (a) 2.5 kg m s −1 I B1: Momentum – Impulse triangle with right angle Can be implied by a correct equation B1 1.5 kg m s −1 2.5 = 1.5 + I (b) M1 M1: Use of Pythagoras to obtain a correct equation OE for example  I  =3 +   0.5  I = Ns After the impact: A1 A1: Correct impulse B1 B1: Sight of perpendicular component as e Could be implied by a correct equation m s −1 B1: Correct velocity diagram, PI by a correct equation 4e ms (3 ) B1 −1 M1: Use of Pythagoras to obtain a correct equation A1: Correct coefficient of restitution M1 = (4e ) + 2 A1 e = or 0.75 Total 2 7 of 17 MARK SCHEME A– LEVEL Mathematics – MM03 – June 15 (a) Alternative: Ij = 0.5(5 cos αi + sin αj ) − 0.5(3i ) 2.5 cos α − 1.5 = 0  cos α = 0.6   sin α = 0.8  I = 0.5(5 × 0.8)  B1 B1 : Correct vector equation M1 M1 : Correct value for sina A1 A1 : Correct impulse I =2 (b) Alternative: = sin β cos β =   2  2  e= 0.5 e = or 0.75 B1 B1 B1 : Correct equation for motion parallel t B1 : Value for cosβ or β = 45° M1 A1 M1 : Correct expression for e or correct eq A1 : Correct impulse of 17 MARK SCHEME A– LEVEL Mathematics – MM03 – June 15 Question (a) (i) (ii) (b) Solution OE mu = mv1 + 2mv u = v1 + 2v 2 u = v - v1 3v = u v2 = u AG 10 v1 = u − u v1 = − u Marks M1 A1 M1 A1 OE A1 The speed of A is u u e 2m 6m v3 10 u = −2v3 + 6v 5  e u  = v3 + v 9  (c) Comments M1: Equation with three momentum terms A1: Correct equation M1: Newton’s Law of Restitution (Allow sign errors.) A1: Correct equation A1 A1: Correct speed of B, from correct working A1: Correct speed of A Do not accept negative speed M1 A1 v4 5  2m u  = −2mv3 + 6mv 9  Total M1: Equation with three momentum terms A1: Correct equation OE M1 A1 M1: Newton’s Law of Restitution (Allow sign errors.) A1: Correct equation OE 10  5 u = −2v3 + 6 ue − v3   9 10 10 8v3 = ue − u 5 OE v3 = ue − u 12 36 m1 A1F m1: Solving equations to find the speed of B after the second collision A1F: Correct speed of B after the second collision FT their equations of 17 MARK SCHEME A– LEVEL Mathematics – MM03 – June 15 Q Solution Marks Total Comments M1 second collision ⇒ 5 ue − u > u 12 36 9 ue > u 12 36 e > or 0.6 M1: For the inequality v3 > v1 A1F A1F: Correct value of k FT their v3 > v1 The value of k must be less than and greater than to score A1F B1 B1 B1: Comment about equal radii or same size B1: Comment about the line of centres Equal radii ⇒ Velocities are parallel to the line of centre Total 16 10 of 17 MARK SCHEME A– LEVEL Mathematics – MM03 – June 15 (b) Alternative: u e 2m 6m v3 v4 5  2m u  = 2mv3 + 6mv 9  10 u = 2v3 + 6v 5  e u  = v − v3 9  10  5 u = 2v3 + 6 ue + v3   9 10 10 8v3 = u − ue 5 v3 = u − ue 36 12 second collision ⇒ 5 u − ue < − u 36 12 9 ue > u 12 36 e> or 0.6 OE M1A1 M1: Equation with three momentum terms A1: Correct equation M1A1 M1: Newton’s Law of Restitution (Allow sign errors.) A1: Correct equation m1A1 F m1: Solving equations to find the velocity of B after the second collision A1F: Correct velocity of B after the second collision FT their equations M1 A1F M1: For the inequality v3 < v1 A1F: Correct value of k The value of k must be less than and greater than to score A1F 11 of 17 MARK SCHEME A– LEVEL Mathematics – MM03 – June 15 Question Solution cos a = Marks or 0.6 and cosβ = or 0.3846 13 2(4 cos α ) + 1(2.6 cos β ) = 2v A + 1v B Total B1 B1: Correct values for cos α and cos β M1: Four term momentum equation along the line of centres A1: Correct equation May be in terms of α and β M1A1 2(2.4 ) + 1(1) = 2v A + 1v B (4 cos α − 2.6 cos β ) = v B − v A M1 A1 M1: Newton’s Law of Restitution (Allow sign errors.) A1: Correct equation (2.4 −1) = v B − v A 5.8 = 2v A + v B  0.8 = v B − v A vA = ms −1 37 vB = ms −1 15 A1 5 V A =   + (4 sin α ) 3 5 V A =   + (3.2) = 3.61 ms -1 3 Comments 11 A1: Correct velocity of A AWRT 1.67 A1 A1: Correct velocity of B AWRT 2.47 m1 m1: Finding speed of A with their v A May be in terms of α and β A1: Correct speed AWRT 3.61 A1  37  VB =   + (2.6 sin β )  15  m1 m1: Finding speed of B with their v B May be in terms of α and β A1: Correct speed AWRT 3.44  37  VB =   + (2.4 ) = 3.44 ms -1  15  A1 Total 11 12 of 17 MARK SCHEME A– LEVEL Mathematics – MM03 – June 15 (a)(i) F vS N α B1 B1 B1: For one velocity triangle, could be implied by later working B1: For the other velocity triangle drawn together or separately, could be implied by the correct 2nd angle 35 35 β 30  -50 (ii) sin α sin 30 sin β sin 30  = = or 50 35 50 35  α = 45.58   β = 134.42   346   Bearings :  074   ngle for shorter time : 45.58  M1 A1 A1: Either angle correct A1 A1: Two correct bearings Accept 74° B1 vS 35 =  sin 104.42 sin 30  M1 v S = 67.79 km h −1 A1 F (b) F t= 67.79 = 0.118 h M1: Correct use of sine rule to find α or β B1: Selecting the smaller of their two angles from part (a) M1: Using the sine rule to find the speed of the frigate relative to the ship, with their angle A1: Correct speed m1 or 7.08 A1F m1: Using distance over speed A1F: Correct time FT their speed Full marks can be scored by using both angles and choosing the shorter time If both times calculated and none selected not award final A1 mark 13 of 17 MARK SCHEME A– LEVEL Mathematics – MM03 – June 15 F vS N B1 B1: Correct right angled velocity triangle Could be implied by later working vF 30  -50 vF = 50 sin 30 M1 OE M1: Use of trigonometry to find speed A1: Correct speed CAO A1 vF = 25 kmh -1 Total 13 14 of 17 MARK SCHEME A– LEVEL Mathematics – MM03 – June 15 (a)(ii) Alternative: ngle for shorter time : 45.58  t (50 cos 30  + 35 cos 45.58  ) = B1 B1: Selecting the smaller of their two angles from part (a) M1: For M1A1 50 cos 30  ± 35 cos 46    t =    50 cos 30 + 35 cos 45.58   t = 0.118 h or 7.08 Alternative: ngle for shorter time : 45.58  d =  sin 30 sin 104.42 d = 4.130 4.130   t =  35   t = 0.118 h km A1: Correct expression m1: Using distance over speed m1 A1F B1 B1: Selecting the smaller of their two angles from part (a) M1: Using the sine rule to find the distance travelled by the frigate with their angle A1: Correct distance m1: Using distance over speed M1 A1 m1 or 7.08 A1F A1F: Correct time FT their angle Full marks can be scored by using both angles and choosing the shorter time If both times calculated and none selected not award final A1 mark A1: Correct time FT their angle Full marks can be scored b using both angles and choosing the shorter time If both times calculated and none selected not award final A1 mark 15 of 17 MARK SCHEME A– LEVEL Mathematics – MM03 – June 15 Question (a) (b) Solution y = u sin (α − ϑ )t − g cos ϑt 2 = u sin (α − ϑ )t − g cos ϑt 2 2u sin (α − ϑ ) t= g cos ϑ Marks Total M1 A1 m1 A1 u sin α − gt = u sin α g Comments M1: Expression for perpendicular height of particle above the plane Accept wrong angles for M1 but not sin and cos in wrong places A1: Correct expression with y = m1: Solving for non-zero t A1: Correct t M1 M1: Velocity equation to find time to A u sin α 2u sin (α − ϑ ) = g g cos ϑ A1 A1: Correct time sin α cos ϑ = sin (α − ϑ ) sin α cos ϑ = sin α cos ϑ − cos α sin ϑ m1 m1: Forming an equation using their time from part (a) and this time M1 M1: Use of identity to eliminate compound expressions It is not enough to only expand sin (α − θ ) in the expression in part (a) without anything else t= sin α cos ϑ = cos α sin ϑ   sin α sin ϑ  =2  cos α cos ϑ tαn α = tαn ϑ A1 Total TOTAL A1: Seeing required expression derived with k = 75 16 of 17 MARK SCHEME A– LEVEL Mathematics – MM03 – June 15 (b) Alternative: Taking x and y axes parallel and perpendicular to the plane respectively and − y or equivalent, using tan θ = x   2u sin (α − θ )  u cos(α − θ ) − g sin θ  tan θ = g cos θ   g 2u sin (α − θ ) cos θ − u sin (α − θ ) + g cos θ ( M1 M1: Correct terms, allow sign errors A1: All correct A1 ) cos(α − θ ) tan θ = sin (α − θ ) tan θ + (cos α cos θ + sin α sin θ ) tan θ = (sin α cos θ − sin θ cos α )(2 tan θ + 1) M1 M1: Use of identities to eliminate compound expressions m1 m1: Rearranging to the required form A1 A1: Seeing required expression derived with k = tan α tan θ + tan α − tan θ − tan θ = ( ) ( tan α + tan θ = tan θ + tan θ tan α = tan θ ) 17 of 17 [...]... A1: Correct velocity of B AWRT 2.47 m1 m1: Finding speed of A with their v A May be in terms of α and β A1: Correct speed AWRT 3.61 A1 2  37  2 VB =   + (2.6 sin β )  15  m1 m1: Finding speed of B with their v B May be in terms of α and β A1: Correct speed AWRT 3.44 2  37  2 VB =   + (2.4 ) = 3.44 ms -1  15  A1 Total 11 12 of 17 MARK SCHEME A– LEVEL Mathematics – MM03 – June 15 6 (a)(i)... May be in terms of α and β M1A1 2(2.4 ) + 1(1) = 2v A + 1v B 4 (4 cos α − 2.6 cos β ) = v B − v A 7 M1 A1 M1: Newton’s Law of Restitution (Allow sign errors.) A1: Correct equation 4 (2.4 −1) = v B − v A 7 5.8 = 2v A + v B  0.8 = v B − v A 5 vA = ms −1 3 37 vB = ms −1 15 A1 2 5 2 V A =   + (4 sin α ) 3 2 5 2 V A =   + (3.2) = 3.61 ms -1 3 Comments 11 A1: Correct velocity of A AWRT 1.67 A1... Mathematics – MM03 – June 15 (b) Alternative: 5 u 9 0 e 2m 6m v3 v4 5  2m u  = 2mv3 + 6mv 4 9  10 u = 2v3 + 6v 4 9 5  e u  = v 4 − v3 9  10  5 u = 2v3 + 6 ue + v3  9  9 10 10 8v3 = u − ue 9 3 5 5 v3 = u − ue 36 12 second collision ⇒ 5 5 1 u − ue < − u 36 12 9 5 9 ue > u 12 36 3 e> or 0.6 5 OE M1A1 M1: Equation with three momentum terms A1: Correct equation M1A1 M1: Newton’s Law of Restitution... do not award final A1 mark 15 of 17 MARK SCHEME A– LEVEL Mathematics – MM03 – June 15 Question 7 (a) (b) Solution 1 y = u sin (α − ϑ )t − g cos ϑt 2 2 1 0 = u sin (α − ϑ )t − g cos ϑt 2 2 2u sin (α − ϑ ) t= g cos ϑ Marks Total M1 A1 4 m1 A1 u sin α − gt = 0 u sin α g Comments M1: Expression for perpendicular height of particle above the plane Accept wrong angles for M1 but not sin and cos in wrong places... Selecting the smaller of their two angles from part (a) M1: Using the sine rule to find the speed of the frigate relative to the ship, with their angle A1: Correct speed m1 or 7.08 min A1F m1: Using distance over speed A1F: Correct time FT their speed Full marks can be scored by using both angles and choosing the shorter time If both times calculated and none selected do not award final A1 mark 13 of 17... not award final A1 mark 13 of 17 MARK SCHEME A– LEVEL Mathematics – MM03 – June 15 3 F vS N B1 B1: Correct right angled velocity triangle Could be implied by later working vF 30  -50 vF = 50 sin 30 M1 OE M1: Use of trigonometry to find speed A1: Correct speed CAO A1 vF = 25 kmh -1 Total 13 14 of 17 MARK SCHEME A– LEVEL Mathematics – MM03 – June 15 (a)(ii) Alternative: ngle for shorter time : 45.58... velocity triangle, could be implied by later working B1: For the other velocity triangle drawn together or separately, could be implied by the correct 2nd angle 35 35 β 30  -50 (ii) sin α sin 30 sin β sin 30  = = or 50 35 50 35  α = 45.58   β = 134.42   346   Bearings :  074   ngle for shorter time : 45.58  M1 5 A1 A1: Either angle correct A1 A1: Two correct bearings Accept 74° B1 5 vS... compound expressions It is not enough to only expand sin (α − θ ) in the expression in part (a) without anything else t= sin α cos ϑ = 2 cos α sin ϑ   sin α sin ϑ  =2  cos α cos ϑ 5 tαn α = 2 tαn ϑ A1 Total TOTAL A1: Seeing required expression derived with k = 2 9 75 16 of 17 MARK SCHEME A– LEVEL Mathematics – MM03 – June 15 (b) Alternative: Taking x and y axes parallel and perpendicular to the plane... u sin (α − θ ) + g cos θ ( M1 M1: Correct terms, allow sign errors A1: All correct A1 ) cos(α − θ ) tan θ = sin (α − θ ) 2 tan 2 θ + 1 (cos α cos θ + sin α sin θ ) tan θ = (sin α cos θ − sin θ cos α )(2 tan 2 θ + 1) M1 M1: Use of identities to eliminate compound expressions m1 m1: Rearranging to the required form A1 A1: Seeing required expression derived with k = 2 tan α tan 2 θ + tan α − 2 tan 3 θ... of Restitution (Allow sign errors.) A1: Correct equation m1A1 F m1: Solving equations to find the velocity of B after the second collision A1F: Correct velocity of B after the second collision FT their equations M1 A1F M1: For the inequality v3 < v1 A1F: Correct value of k The value of k must be less than 1 and greater than 0 to score A1F 11 of 17 MARK SCHEME A– LEVEL Mathematics – MM03 – June 15 Question