Tiêu đề Feb-March_2025_9709_42_MS - Crack A Level.pdf ✅

[Turn over Cambridge International Examinations Cambridge International Advanced Subsidiary and Advanced Level MATHEMATICS 9709/42 Paper 4 Mechanics February/March 2025 MARK SCHEME Maximum Mark : 50

9709/42 Qualification - Mark Scheme March/2025 Final draft Page 3 of 10 Maths-Specific Marking Principles 1. Unless a particular method has been specified in the question, full marks may be awarded for any correct method. However, if a calculation is required then no marks will be awarded for a scale drawing. 2. Unless specified in the question, answers may be given as fractions, decimals or in standard form. Ignore superfluous zeros, provided that the degree of accuracy is not affected. 3. Allow alternative conventions for notation if used consistently throughout the paper, e.g. commas being used as decimal points. 4. Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored (isw). 5. Where a candidate has misread a number in the question and used that value consistently throughout, provided that number does not alter the difficulty or the method required, award all marks earned and deduct just 1 mark for the misread. 6. Recovery within working is allowed, e.g. a notation error in the working where the following line of working makes the candidate’s intent clear.
9709/42 Qualification - Mark Scheme March/2025 Final draft Page 4 of 10 Question Answer Marks Guidance 1 For resolving and balancing forces in any direction *M1 Allow sin/cos mix; correct number of terms X cos θ = 40 X sin θ = 30 A1 For correct resolving in two directions. X = √402 + 302 or θ = tan−1 30 40 DM1 For attempt to solve for either X = 50, θ = 36.9 A1 36.869... Total Q1 4 Alternative for Q1 using triangle of forces X = √402 + 302 M1 For attempt to solve for X using Pythagoras X = 50 A1 tan θ = 3 4 M1 For attempt to solve for θ θ = 36.9 A1 Vetter suggestion: alternative - consider M1 for attempting to draw a triangle of forces, A1 for drawing it correctly and then M1A1 as for main scheme. Also, SCB2 for correct answers with no working. 2(a) 0 = V 2 − 2 × 2 × 16 M1 For use of constant acceleration to get an equation in V only V = 8 A1 Total Q2(a) 2 2(b) Acceleration section : 42 = (4+V) 2 t [t = 7] Deceleration section: 0 = their 8 − 2t [t = 4] M1 For attempt to find time during acceleration or deceleration.