A-level Mathematics. MM03 Mark scheme June Version 1.0: Final

Transcription

1 -leel Mathematics MM0 Mark scheme 660 June 0 Version.0: Final

2 Mark schemes are prepared by the Lead ssessment Writer and considered, together with the releant questions, by a panel of subject teachers. This mark scheme includes any amendments made at the standardisation eents 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 coers the students responses to questions and that eery associate understands and applies it in the same correct way. s preparation for standardisation each associate analyses a number of students scripts. lternatie answers not already coered by the mark scheme are discussed and legislated for. If, after the standardisation process, associates encounter unusual answers which hae not been raised they are required to refer these to the Lead ssessment Writer. It must be stressed that a mark scheme is a working document, in many cases further deeloped and expanded on the basis of students reactions to a particular paper. ssumptions about future mark schemes on the basis of one year s document should be aoided; 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 aailable from aqa.org.uk Copyright 0 Q and its licensors. ll rights resered. Q retains the copyright on all its publications. Howeer, registered schools/colleges for Q are permitted to copy material from this booklet for their own internal use, with the following important exception: Q cannot gie permission to schools/colleges to photocopy any material that is acknowledged to a third party een for internal use within the centre.

3 Key to mark scheme abbreiations M mark is for method m or dm 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 E mark is for explanation or ft or F follow through from preious incorrect result CO correct answer only CSO correct solution only WFW anything which falls within WRT anything which rounds to CF any correct form G answer gien SC special case OE or equialent, or (or 0) accuracy marks x EE deduct x marks for each error NMS no method shown PI possibly implied SC substantially correct approach c candidate sf significant figure(s) dp decimal place(s) No Method Shown Where the question specifically requires a particular method to be used, we must usually see eidence of use of this method for any marks to be awarded. Where the answer can be reasonably obtained without showing working and it is ery unlikely that the correct answer can be obtained by using an incorrect method, we must award full marks. Howeer, the obious penalty to candidates showing no working is that incorrect answers, howeer 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 gien to less than the degree of accuracy accepted in the mark scheme, when it gains no marks. Otherwise we require eidence of a correct method for any marks to be awarded.

4 MRK SCHEME LEVEL Mathematics MM0 June Question Solution Marks Total Comments - [ F ] MLT : Correct dimensions of F : Substituting the - - ( ) ( ) - MLT LT L ( ML ) γ dimensions of the quantities into the gien equation to obtain RHS correctly. γ α+ β-γ -α M L T : Collecting indices on RHS. Could be implied by later work. γ α + β γ α 6 : γ : Two correct equations for α and β. α, β Total 6 : Correct alues for α and β. Condone use of units instead of dimensions. of 7

5 MRK SCHEME LEVEL Mathematics MM0 June Question Solution Marks Total Comments (a) x u cosαt (b)(i) (ii) t x u cosα y u sinα t gt x y u sinα u cosα ( 9.8) x u cosα.9x y x tαnα G u cos α.9s s s tan cos ( + tan ) cos s.9 s 7.9 : Correct expression for horizontal displacement. : Correct expression for t. : Correct expression for ertical displacement. llow sign errors. : Elimination of t from equation for ertical displacement. : Correct result from correct working. Penalise use of g 9.8. : Substituting ±s for x and y. x cos.0 y sin cos or y ( sin ) ( )( ) y.9 tan.9 cos 7 or 7 Total : Making s the subject of their equation. : WRT 7.9 Condone use of g 9.8 which gies 7.8. : Correct expression or alue for horizontal component of elocity. : Correct expression or alue for ertical component of elocity, with their answer to (b)(i). : Correct expression or alue. : Use of tan with their elocity components. : Correct angle to nearest degree. CO. of 7

6 MRK SCHEME LEVEL Mathematics MM0 June (b)(ii) lternatie:.9x y x tαnα u cos α dy (.9) x tαnα dx u cos α tan cos.8 The angle tan 7 ( )( ) (.8) - - or 7 : Correct deriatie. : Substituting alues. : Correct alue of the deriatie : Use of tan to find the angle. : Correct angle to nearest degree. CO. 6 of 7

7 MRK SCHEME LEVEL Mathematics MM0 June Question Solution Marks Total Comments (a) I. kg. kg m s m s : Momentum Impulse triangle with right angle. Can be implied by a correct equation. (b).. + I I N s : Use of Pythagoras to obtain a correct equation. OE for example I + 0. : Correct impulse. fter the impact: m s m s e : Sight of perpendicular component as e. Could be implied by a correct equation. : Correct elocity diagram, PI by a correct equation ( ) ( e ) + e or 0.7 Total 7 : Use of Pythagoras to obtain a correct equation. : Correct coefficient of restitution. 7 of 7

8 MRK SCHEME LEVEL Mathematics MM0 June (a) lternatie: Ij 0. cosαi + sinαj ( ) 0.( i).cosα. 0 cosα 0.6 sinα 0.8 I 0.( 0.8) I : Correct ector equation. : Correct alue for sina. : Correct impulse. (b) lternatie: cos β e e sin β 0. or 0.7 :Correct equation for motion parallel t : Value for cosβ or β. :Correct expression for : Correct impulse. e or correct eq 8 of 7

9 MRK SCHEME LEVEL Mathematics MM0 June Question Solution Marks Total Comments (a) (i) mu m + m OE : Equation with three momentum terms. u + : Correct equation. u - OE : Newton s Law of (ii) Restitution. (llow sign u errors.) : Correct equation. (b) u G 9 6 : Correct speed of, 0 from correct working. u u 9 u 9 The speed of u 9 m is u 9 e 0 6m m u m + 6m 9 0 u e u + 9 OE OE : Correct speed of. Do not accept negatie speed : Equation with three momentum terms. : Correct equation : Newton s Law of Restitution. (llow sign errors.) : Correct equation (c) 0 u + 6 ue ue u 9 ue u 6 OE F 8 : Soling equations to find the speed of after the second collision. F: Correct speed of after the second collision. FT their equations 9 of 7

10 MRK SCHEME LEVEL Mathematics MM0 June Q Solution Marks Total Comments second collision ue u > u ue > u 6 e > or 0.6 Equal radii Velocities are parallel to the line of centre F : For the inequality > F: Correct alue of k. FT their >. The alue of k must be less than and greater than 0 to score F : Comment about equal radii or same size. : Comment about the line of centres. Total 6 0 of 7

11 MRK SCHEME LEVEL Mathematics MM0 June (b) lternatie: u 9 m e 0 6m m u m + 6m 9 0 u e u 9 0 u + 6 ue u ue 9 u ue 6 OE F : Equation with three momentum terms. : Correct equation. : Newton s Law of Restitution. (llow sign errors.) : Correct equation. : Soling equations to find the elocity of after the second collision. F: Correct elocity of after the second collision. FT their equations. second collision u ue < u 6 9 : For the inequality < 9 ue > u 6 e > or 0.6 F F: Correct alue of k. The alue of k must be less than and greater than 0 to score F of 7

12 MRK SCHEME LEVEL Mathematics MM0 June Question Solution Marks Total Comments cos a or 0.6 and cosβ ( cosα ) + (.6cos β ) + (.) + ( ) + or : Correct alues for cos α and cos β. : Four term momentum equation along the line of centres. : Correct equation. May be in terms of α and β. 7 ( cosα.6 cos β ) (. ) ms : Newton s Law of Restitution. (llow sign errors.) : Correct equation. : Correct elocity of. WRT.67 7 ms : Correct elocity of. WRT.7 V V + + ( sinα ) - (.).6 ms : Finding speed of with their. May be in terms of α and β. : Correct speed. WRT.6 V V (.6sin β ) - (.). ms : Finding speed of with their. May be in terms of α and β. : Correct speed. WRT. Total of 7

13 MRK SCHEME LEVEL Mathematics MM0 June 6 (a)(i) (ii) α sinα sin 0 0 α.8 β. F S β or -0 0 sin β sin 0 0 N : For one elocity triangle, could be implied by later working. : For the other elocity triangle drawn together or separately, could be implied by the correct nd angle : Correct use of sine rule to find α or β. : Either angle correct. (b) earings : 6 07 ngle for shorter time : F S sin0. F S sin 0 km h.8 : Two correct bearings. ccept 7. : Selecting the smaller of their two angles from part (a). : Using the sine rule to find the speed of the frigate relatie to the ship, with their angle. : Correct speed. t h or 7.08 min F : Using distance oer speed. F: 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 mark. of 7

14 MRK SCHEME LEVEL Mathematics MM0 June F F S -0 0 N : Correct right angled elocity triangle. Could be implied by later working. F 0sin 0 OE F - kmh : Use of trigonometry to find speed. : Correct speed. CO. Total of 7

15 MRK SCHEME LEVEL Mathematics MM0 June (a)(ii) lternatie: ngle for shorter time :.8 t ( 0cos0 + cos.8 ) 8 t 8 + cos.8 0cos0 : Selecting the smaller of their two angles from part (a). : For 0 cos0 ± cos 6 : Correct expression. : Using distance oer speed. t 0.8 h or lternatie: ngle for shorter time : d sin 0 d.0.0 t t 0.8 h 8 sin0. km or 7.08 min 7.08 min.8 F F F: 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 do not award final mark. : Selecting the smaller of their two angles from part (a). : Using the sine rule to find the distance traelled by the frigate with their angle. : Correct distance : Using distance oer speed. : 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 do not award final mark. of 7

16 MRK SCHEME LEVEL Mathematics MM0 June Question Solution Marks Total Comments 7 (a) y u sin( α ϑ) t g cosϑt 0 u sin( α ϑ) t g cosϑt u sin( α ϑ) t (b) g cosϑ u sin α gt 0 : Expression for perpendicular height of particle aboe the plane. ccept wrong angles for but not sin and cos in wrong places. : Correct expression with y 0. : Soling for non-zero t. : Correct t. u sinα t g : Velocity equation to find time to. ( α ϑ) u sinα u sin g g cosϑ : Correct time. ( α ϑ) sin α cosϑ sin sinα cosϑ sinα cosϑ cosα sinϑ : Forming an equation using their time from part (a) and this time. sinα cosϑ cosα sin sinα sinϑ cosα cosϑ tαn α tαnϑ ϑ : 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. : Seeing required expression deried with k. Total 9 TOTL 7 6 of 7

17 MRK SCHEME LEVEL Mathematics MM0 June (b) lternatie: Taking x and y axes parallel and perpendicular to the plane respectiely and y using tan θ or equialent, x u cos ( α θ) ( α θ) u sin g sin θ tan θ g cosθ gu sin ( ) ( α θ) u sin α θ + cosθ g cosθ : Correct terms, allow sign errors. : ll correct cos ( α θ) tan θ sin( α θ)( tan θ + ) ( cos α cosθ + sin α sin θ) tan θ ( sin α cosθ sin θ cos α)( tan θ + ) : Use of identities to eliminate compound expressions. tan α tan θ + tan α tan θ tan θ 0 ( + tan θ) tan θ( tan θ) tan α + : Rearranging to the required form. tan α tan θ : Seeing required expression deried with k. 7 of 7