AS MATHEMATICS MECHANICS 1B

Transcription

1 AS MATHEMATICS MECHANICS 1B MB Final Mark Scheme 660 June 017 Version/Stage: v1.0

2 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 017 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.

3 MARK SCHEME AS MATHEMATICS MB JUNE 017 Annotations Annotation ^ B1 BOD Cross FT H Line H Wavy ISW MR Not Relevant Text box SEEN Tick Tick Per FIW Description Accuracy mark awarded one Omission Mark Independent mark one Benefit of the doubt Incorrect point Follow through Dynamic, Horizontal line that can be expanded Dynamic, Horizontal Wavy line that can be expanded Ignore subsequent work Method mark awarded one MR Not Relevant On Page Comment Indicates that the point has been Correct point Different Perspectives From incorrect work

4 MARK SCHEME AS MATHEMATICS MB JUNE 017 Key to mark scheme abbreviations M mark is for method m or dm mark is dependent on one or more M marks and is for method A mark is dependent on M or m marks and is for accuracy B 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 previous incorrect result CAO correct answer only CSO correct solution only AWFW anything which falls within AWRT anything which rounds to ACF any correct form AG answer given SC special case OE or equivalent A,1 or 1 (or 0) accuracy marks x EE deduct x marks for each error NMS no method shown PI possibly implied SCA 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 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.

5 MARK SCHEME AS MATHEMATICS MB JUNE (a) P cos0 = 0 : Seeing P cos 0 = 0 Psin 0 = 0 oe or P = 0 cos0 = = = 6. (to sf) : Correct equation. : Correct answer. Accept exact forms as shown. AWRT 6. P sin 0 = Q 0 Q = sin 0 cos0 0 0 = = =.1(to sf) F : Seeing P cos0 = Q or P sin 0 = Q oe where P is their answer to (a). F: Correct equation. Follow through their P from part (a). : Correct answer. Accept exact forms as shown. AWRT.1 0cos 60 = Q cos0 () : Resolving perpendicular to P. : Correct equation. 0cos 60 Q = cos0 0 0 = = =.1(to sf) () () () : Correct answer. Accept exact forms as shown. AWRT.1 Total 6

6 MARK SCHEME AS MATHEMATICS MB JUNE 017 (a) 0 = 100v : Seeing 0 or 0. : Seeing 100v v = m s : Correct speed. Final answer must be positive. Allow either of these solutions depending on the candidate s interpretation. Interpretation = 80v : Seeing momentum term of 0. : Seeing 80v. : Seeing ±±v or v = = 0.9 m s 80-1 : Final speed as w 0 = 0 w = 0. () () () : Seeing momentum term of 0. : Seeing 80 Velocity of the trolley after the second bag has been thrown relative to its velocity before, for example 80w. : Adding or subtracting 0. to/from their w. v = 0. + w = 0.9 () : Final speed as 0.9. Interpretation 0 = 80v 0( v) 80 = 100v v = 0.8 () () () () : : Seeing momentum term of 0. : Seeing 80v. : Seeing ±±v or 1.6 : Final speed as w + 0 ( w ) = 0 w = 0. () () () : Seeing 80w. : Seeing ±±w : Adding or subtracting 0. to/from their w. v = 0. + w = 0.8 () : Final speed as 0.8. NOTE: Deduct 1 mark from each part if g included consistently with the mass. Total 7 6

7 MARK SCHEME AS MATHEMATICS MB JUNE 017 (a) Time = 0 = 1 s B1 B1: Correct time. PI. BC = 1. 1 = 18 m 1. tanα = ( α = 0.96 ) 1. BC = 0 tanα = 0 = 18 m (B1) () () () : Their time multiplied by 1.. : Correct distance. B1: Correct expression for tanα or correct angle α. Allow tan β =, ( β = 9.0 ) 1. :Using 0tanα with their angle. : Obtaining 18 m. AWRT sinθ = = 0.6 θ = 6.87 Time = 0 cos6.87 = 18.8 s : Equation to find an angle. Accept sin θ = 0.6 oe or cos θ = 0. 6 oe. : Angle as 6.87 allow.1 AWRT 7 or : Correct expression for time 0 Time = or cos Time = sin.1 Allow angles as AWRT 7 or. : Correct time. AWRT Accept 18.7 V = = () () : Use of Pythagoras with, 1. and a minus sign. : Correct resultant velocity. Time = () : Correct expression for time. = 18.8 s () () : Correct time. AWRT Accept 18.7 Total 7 7

8 MARK SCHEME AS MATHEMATICS MB JUNE 017 (a) 0 = + a. : Use of v = u + as with v = 0. : Correct equation. a = =. m s - : Correct acceleration. Must be negative. 1. = ( + 0) t t = = + 1.6a a = 1.6 =. m s - () () () () : Use of a constant acceleration equation with their t and v = 0 to find a. : Correct equation. : Correct acceleration. Must be negative. R = 9.8 = 19.6 N B1 1 B1: Correct final answer. Allow 19.6 from g =9.81. Accept g. (c). = 19.6µ µ = 19.6 = 0. : Use of F = µ R with their R. : Use of µ R = ma with ±their answer to (a). : Correct final answer AWRT 0.. (d) Less because the friction force would be less (due to the presence of air resistance). B1 B1 B1: Less. B1: Must say that the friction force would be less. Do not award second B1 mark if they say that coefficient of friction is greater or unchanged. Total 9 8

9 MARK SCHEME AS MATHEMATICS MB JUNE 017 (a) 1 : Use of a constant acceleration = 0 + a 1 equation, with u = 0, to find a. : Correct equation. a = = 0.6 m s 7 - : Correct acceleration. Accept - m s. 8 (i) T 80 = F : Equation of motion for trailer with T and 80. Must have mass 00 their acceleration. Allow sign errors. F: Correct equation with their acceleration from (a). T = 0 : Correct T. (ii) If P is found first, then the equation below applies: P T 00 = T = 0 P = P = 180 () (F) () F : Equation of motion for car with T, 00 and their P. Must have mass 1600 their acceleration. F: Correct equation with their acceleration from (a). : Correct T. : Equation of motion for car and trailer combined with P, 00 and 80. Must have mass 000 their acceleration. Allow sign errors. F: Correct equation with their acceleration from (a). : Correct P. P T 00 = () (F) : Equation of motion for car with their T, 00 and P. Must have mass 1600 their acceleration. F: Correct equation with their acceleration from (a). P = 180 () : Correct P. Total 9 9

10 MARK SCHEME AS MATHEMATICS MB JUNE (a) B1: All forces shown with arrows and R labels. Accept: 80 or T for the tension. 80 F R or N for reaction force. B1 1 0g, mg, W or 196 for weight. 0g F or µr for friction. (c) R + T sin 0 = mg R + 80sin 0 = 196 R = 16 N T cos0 µ R = ma 80cos = 0a : Correct terms used to form an equation. Allow sign errors Allow cos0. : Correct fully substituted equation. : Correct reaction force. AWRT 16. : Correct terms used to form an equation of motion. Allow sign errors. Allow sin0. : Correct, fully substituted, equation. a = 0. m s - : Correct acceleration. AWRT 0.. Allow 0. or 0.0 from use of g = (d) (i) 80cos = 0a a = m s : Seeing - inserted into their equation of motion from part (c) or subtracting 0. from their answer to part (c). : Correct acceleration. Allow or from use of g = Accept AWRT (d) (ii) The air resistance force will (not be constant but) vary with speed or velocity. B1 1 B1: Statement about air resistance varying with the speed. Stating that the box is accelerating implies that the speed is changing. However, stating that the resistance varies with/as the acceleration is not acceptable. Total 10 10

11 MARK SCHEME AS MATHEMATICS MB JUNE (a) v = ( + 0. ) i + (1. 1.6) j : Correct expression for the velocity =.i + 0.8j when t =. : Correct simplified velocity. v =. =.7 m s v = ( + 0.t) i + (1.t 1.6) j d d: Finding magnitude of the velocity. : Correct magnitude. AWRT.7. : Correct expression for velocity in any form, eg ( i 1.6j) + (0.i + 1.j) t : Correct velocity as the sum of two components. PI. 10 = ( + 0.t) 0 = 1.0t + (1.t 1.6) 1.8t 81. t = 8 or t = t = 8 1 r = i = 0i +.6j 0i +.6j Average Velocity = 8 = i +.j j d d : Quadratic equation to find time when speed is 10, including both their velocity components. : Correct expanded quadratic, that is equal to zero. (Allow any multiple of the equation.) d: One or two times obtained from their quadratic. : Concludes that t = 8. d: Finding the position vector at their time. : Correct position vector. AWRT 0.0i +.6j : Correct average velocity. v = ( ) i + ( ) j = 6i + 8j (d) () (i 1.6j) + (6i + 8j) Average Velocity = = i +.j () 9 Total 1 d: Finding the velocity at their time. : Correct velocity vector. : Correct average velocity. AWRT.0i +.j 11

12 MARK SCHEME AS MATHEMATICS MB JUNE (a) sinα =,cosα = B1 B1: Correct values for sinα and/or cosα. PI. Or finding α = AWRT 7. 8 = t +.9t.9t +.t 8 = 0 t = 1.06 or t = 1. : Seeing three term equation with: sinαt or cosαt, 8 and ±.9t. : Correct terms with possible sign errors. : Correct equation. : Correct time obtained. x = 1.06 x =.8 m d 7 d: cosα their time. : Correct distance. Accept final answers between.7 and.9. v v x y = =. = = 16.6 v = = 1.1m s -1 B1 d B1: Correct horizontal component seen. : Equation to find vertical component, with correct terms but possible sign errors. : Correct component or the square of the correct component. d: Finding the magnitude of the velocity. : Correct speed. Accept final answers between 1.1 and 1. from correct method. (c) v v x y = =. = = 1.7 v = = 1.1m s -1 (B1) () () (d) () () As the particle starts at rest it will not move if µ tanα therefore for B1B1 movement µ <. Total 1 B1: Correct horizontal component seen. : Equation to find vertical component, with their time and sinα but possible sign errors. : Correct component or the square of the correct component. d: Finding the magnitude of the velocity. : Correct speed. Accept final answers between 1.1 and 1. from correct method. Allow 1. from the use of g = B1: Conclusion that µ <. B1: Correct justification. 1