Mathematics 9709/43 October/November 2025: question paper and mark scheme

Mark scheme

1(a) 1.6 ms −2 B1 8 oe e.g. etc. 5 Condone −1.6 . 1 1(b) 1 1 M1 Correct expression for the area of either one of the right- [3 ×] ×10 × 8 or [6 ×] × 5 × 8 angled triangles with base length 5 or one of the isosceles 2 2 triangles of base length 10. Sight of 40 or 20 as part of a total distance calculation can score this mark. 120 m A1 2

Mark scheme

2(a) 105000 B1 OE e.g. F × 35 = 105000 . Driving force F = [=3000] 35 F − 35 × c = 400 × 0.5 M1 Use of N2L; allow sign errors but must be 3 terms. Allow F or any non-zero value for the driving force (but not 105000). If assuming a = 0 then e.g. 105000 = c × 352 ( c = 85.714...) scores B1 M0. 105000 A1 AG – a correct equation for c e.g. 3000 − 35 × c = 400 × 0.5 − 35 × c = 400 × 0.5 c = 80 35 followed by c = 80 scores A1. ( 2800 = 35c c = 80 ) 3 2(b) 105000 3 B1 3 − 80 × v − 400 g × [= 0] B1 for correct weight component: ±400 g × or v 20 20 ( ) 2 80v + 600v − 105000[ = 0] 3 ±400 g × sin sin −1 . 20 [ = ±600] . OR 80v 2 + 600v = 105000 M1 OE. M1 for applying N2L with the correct number of dimensionally correct terms (with a = 0) leading to a three term quadratic equation/expression in v. Allow sign errors and sin/cos mix on the weight component. Must be using c = 80 for this mark. Speed = 32.7 m s −1 ONLY A1 −15 + 5 849 CWO. Allow exact or AWRT 32.7. 4 (For reference: 32.6720057…). If negative value (−40.172...) stated, then must be rejected for this mark. 3

Mark scheme

3(a) R + 25sin36 = 4 g [ R = 25.30536...] *M1 Attempt to resolve vertically to form an equation for the normal contact force for the block; correct number of dimensionally correct terms, condone sign errors, and sin/cos mix on the 25 N force term. Use of R = 4g scores no marks in this part. 2 DM1 Use of Work done = 0.4R ×120 ONLY so M0 if Work done against friction = 120 × × ( 40 − 25sin 36 ) 5 subtracting/adding the Work done by the pulling force to this value e.g. a final answer of 1212 (or 1210) coming from 120 × ( 25cos36 − 0.4 × (40 − 25sin 36) ) scores M1 M0 A0. 1210 or 1215 J A1 CWO 1210 or final AWRT 1215 or 1214 (usually coming from using R = 25.3). For reference: 1214.6576… 3 3(b) 1 1 B1 SOI, by KE terms on opposite sides of an equation with the KE change = ± × 4 × vB 2 − × 4 × 7 2 2 2 same sign so must imply a change in KE (e.g. 2v 2 on one side and 98 on the other). Work Done by pulling force = ±25cos36 ×120 [ = ±2427.050...] B1 1 1 M1 Attempt at work-energy equation using their work done × 4 × vB 2 − × 4 × 72 = 25cos36 × 120 − '1214.65...' against friction from 3(a) with four dimensionally correct 2 2 terms but allow sign errors and sin/cos mix on the pulling 2vB − 98 = 2427.050... − '1214.65...' 2 force term. Dependent on both M1 marks in 3(a) unless the total work done is recovered in this part e.g. a candidate might, in 3(a), find the total work done by the block (which if correct is 1212.39…) rather than the required work done against friction. Therefore, an answer of 1212.39… 3(a) would score M1 M0 A0 but then seeing 2v 2 − 98 = 1212.39... (or even 1210 but only if coming from 1212.39…) followed by 25.6 would score full marks in this part. M0 if using 25 ×120 for the pulling force. vB = 25.6 m s −1 A1 AWRT 25.6. (For reference: 25.59680…). Special case for using N2L Maximum 2 marks 25cos36 − 0.4 ( 40 − 25sin 36 ) = 4 a a = 2.53 B1 This is for getting a = 2.53 . (For reference: a = 2.525819…). v 2 = 72 + 2 × 2.5258... × 120 v = 25.6 m s −1 B1 AWRT 25.6. (For reference: 25.59680…). 4

Mark scheme

4(a) Attempt to differentiate s = 0.01t 3 − 0.3t 2 − 2.07t *M1 Decrease power by 1 and change of coefficient on the same v term; a = is M0. t [v =] 0.03t 2 − 0.6t − 2.07 A1 Allow un-simplified. [a =] 0.06t − 0.6 A1 Allow un-simplified. Must follow from the correct 3 term quadratic in t for v so A0 if coming from v = 0.03t 2 − 0.6t . OR Or the correct completing the square form for the correct 2 [v] = 0.03(t − 10) − 5.07 expression for v. 0.06t − 0.6 = 0 ( t = 10 ) DM1 Setting their a = 0 where a is a two-term linear expression (of the form k1t + k2 where k1 , k2 ≠ 0 ) in t. Or setting their t – 10 (from completing the square on a three term quadratic in t) equal to 0. s = −40.7 A1 CWO (so first four marks must have been awarded). Must be negative so do not allow s = −40.7 s = 40.7 so no ISW. Saying that s = −40.7 or 40.7 is A0. v = −5.07 speed is 5.07 m s −1 A1 CWO (so first four marks must have been awarded or first three marks if completing the square). Must be positive – allow A1 if negative sign is dropped without justification but saying speed is either −5.07 or 5.07 is A0. 6 4(b) 0.03t 2 − 0.6t − 2.07 = 0 *M1 Set their 3 term quadratic expression in t for v equal to zero – dependent on first M1 mark in 4(a). This mark can be implied by a correct value of t = 23 appearing in this part. t = 23 A1 CWO Ignore mention of t = – 3. Working might appear in 4(a), but this value must be stated in 4(b) to earn this mark. a = 0.06 × 23 − 0.6 DM1 Their positive t substituted into their two-term linear expression for a – this mark can be implied by a correct answer. a = 0.78 m s −2 ONLY A1 39 OE e.g. , etc. if −0.78 found then must be explicitly 50 rejected. 4

Mark scheme

5(a) Attempt at N2L for either A or B or the system *M1 Correct number of dimensionally correct terms, allow sign errors. Condone using a = ±10 (or another incorrect value of a) in either the equation for A or B for this mark only. A: T − 2 g = 2a A1 For any two correct equations in a and T (or implied by B: 6 g − T = 6a correct answers) – note that T must be positive in one System: 6 g − 2 g = (2 + 6)a equation and negative in the other. Note that the three equations: 2 g − T = 2a, T − 6 g = 6a and 2 g − 6 g = 8a score M1 A1 and lead to T = 3g and a = −0.5g . Solving for T DM1 From two N2L equations (both containing a) with the correct number of dimensionally correct terms OE. a = 0.5g or 5 m s −2 and T = 3g or 30 N A1 a must be positive – allow those who say a = −0.5g a = 0.5g (or a = −5 a = 5 ) without justification (but must be from correct working). 4 5(b) 1 M1 1 −3.2 = 1.2t + × (−5) × t 2 Using s = ut + at 2 with s = ±3.2, u = ± 1.2 and their a 2 2 1 from 5(a) (not ±g but allow ± 10 if come from wrong OR 3.2 = −1.2t + × 5 × t 2 2 working in 5(a)) – dependent on either a system equation with the correct number of dimensionally correct terms from 5(a) or from equations for A and B that both contain the correct number of dimensionally correct terms – or other complete method to find t. 1.4(0) s A1 6 + 2 209 AWRT 1.4 or . 25 (For reference: 1.39654…). 2 5(c) Distance A travels from initial position to first time at rest: M1 Attempt at finding the distance of A below its starting 0 = 1.22 + 2 × ( −5) × s1 ( s1 = 0.144 ) position when it first comes to instantaneous rest. Using 1.2 and their 5 from 5(a) – same dependency as in OR 5(b) for the value of a from 5(a). 0 = (−1.2) 2 + 2 × (5) × s1 ( s1 = −0.144 ) OR If using s = 1.2 × t1 + 0.5at12 then their t1 must have come 1 s1 = 1.2 × 0.24 + × (−5) × 0.242 ( s1 = 0.144 ) from 0 = 1.2 + a × t1 with their value of a from 5(a). 2 1 Energy for A: 0 − × 2 × 1.22 = 2 ×10 × s1 − 30 × s1 . 2 1 Energy for B: 0 − × 6 ×1.22 = −6 ×10 × s1 + 30 × s1 . 2 Energy for system: . 1 1 0 − × 2 × 1.22 + × 6 × 1.22 = 2 × 10 × s1 − 6 × 10 × s1 . 2 2 If using energy – must have the correct number of dimensionally correct terms but allow sign errors. If considering either A or B then their value of T (= 30) is dependent on both M1 marks in 5(a). 2 2 w2 = 1.22 + 2 × 5 × 3.2 or w = 0 + 2 × 5 × (3.2 + 0.144) *M1 Correct method to find the speed or speed 2 of A/B after travelling 3.2 metres (oe) – using 1.2 and their a from 5(a) – ( w = 33.44, w = 5.7827...) 2 same dependency as in part (b) for the value of a from 5(a). If using v = u + at then t must come from considering t2 − t1 OR w = 1.2 + 5(1.1565... − 0.24) (see reference on previous page). Distance A travels from a point 6.4 metres above the ground to rest: DM1 Attempt to find the distance travelled by A after B hits the 0 = 33.44 + 2 × (−10) × s2 ( s2 = 1.672 ) ground, using their w, a = −g and v = 0. Difference in distances is s1 + 3.2 + s2 = 5.016 m A1 AWRT 5.02 or 5.01 CWO. 4

Mark scheme

6(a) 1 M1 Attempt at conservation of momentum with correct number m × u + 2 × (−3u ) = m × − u + 2 × v of terms. Allow sign errors ONLY so no misreads in this 3 part. However, if using g with either (or both) masses, then allow this mark but do not allow subsequent A marks. 2 A1 A correct (un-simplified) expression for the velocity of Q v = mu − 3u 3 (2m − 9)u 4 after the collision e.g. , mu − 3u, 3 6 1 mu − 6u + mu 3 , etc. 2 2 A1 AG CWO – accept a correct expression for v put > 0 v > 0 mu − 3u > 0 m > 4.5 followed by m > 4.5 . 3 2 Note that mu − 3u = 0 m = 4.5 m > 4.5 is A0. 3 Using trial and improvement to show that v is negative with value(s) of m that are less than 4.5 is A0 unless they also show that v = 0 when m = 4.5. Similarly, if attempting to show that v is positive with value(s) of m that are greater than 4.5 they must also show that v = 0 when m = 4.5 for A1. This A1 mark is dependent on the previous A1 mark. 3 6(b) [Q’s speed is reduced by 75% by its impact with the wall ] speed after *B1FT Follow through their v of the form k1mu + k2u (if simplified) 12 1 impact is mu − 3u = mu − u 3 for k1 , k2 ≠ 0 from 6(a) OE e.g. ( 2m − 9 ) u or possibly 4 3 6 4 12 implied by later working e.g. − u 1 ( 9 − 2m ) u . 3 12 12 Allow ± e.g. − mu − 3u . 4 3 Look out for this expression embedded in later (possibly incorrect) working. 1 1 3 DM1 1 1 No further collisions therefore u mu − u For u v with v of the form k1mu + k2u for k1 , k2 ≠ 0 3 6 4 3 4 from 6(a) – allow > or = for this mark 4 2 Note that u = mu − 3u (which leads to m = 6.5) scores 3 3 B1 M1 ONLY. Solve inequality to obtain m 6.5 DM1 Solve to obtain an inequality of the form ‘m <’ or ‘m ’. Dependent on both previous marks - M0 if an equation for m found rather than an inequality for a range of values of m. ( 4.5 < ) m 6.5 m = 6 A1 CWO – condone m = 6 from m < 6.5. 4

Mark scheme

7(a) For resolving in any direction to form an equation *M1 Correct number of relevant dimensionally correct terms; allow sin/cos mix; allow sign errors. Forces that need resolving should be resolved, forces that do not need resolving should not be resolved. Perpendicular to plane: R = 6 g cosθ + 12sin θ A1 Or 12 + F cosθ = R sinθ (resolving horizontally). Parallel to plane: F + 12cosθ = 6 g sin θ or 6 g sin θ − 12cosθ = μ (...) A1 Or R cosθ + F sin θ = 6 g (resolving vertically). For use of F = μ R to get an equation in μ and θ only DM1 From dimensionally correct equations with the correct number of relevant terms (forces that need resolving should be resolved, forces that do not need resolving should not). 5tan θ − 1 A1 AG – CWO must give some indication of how the equation 6 g sin θ − 12cos θ = μ ( 6 g cos θ + 12sin θ ) leading to μ = in sin/cos leads to the given answer e.g. 5 + tan θ With some evidence of intermediate step e.g. 6 g sin θ − 12cosθ μ= then ‘divide through by cos’ gives 5sin θ 6 g cosθ + 12sin θ 12cosθ − 1 cos θ OR 12cosθ ( 5tan θ − 1) OR ‘divide by cosine’ OR 5tan θ − 1 μ= scores A1 (so therefore mentioning ‘dividing sin θ 12cosθ (5 + tan θ ) 5 + tan θ 12cosθ 5 + cosθ by cos’). sin θ use/mention of tan θ = OR sinθ = tanθ cosθ , etc. 6 g sin θ − 12cosθ 5 tan θ − 1 cosθ Seeing μ = = only is A0. 6 g cosθ + 12sin θ 5 + tan θ 5 7(b) 1 M1 Set the numerator of μ either =, > or to 0 and obtain μ > 0 tan θ > 5 tan θ k where is =, > or and k ≠ 0 - ignore any consideration of an upper bound (e.g. saying that μ < 1 ). 11.3 < θ ( < 90 ) A1 AWRT 11.3 (11.30993…). Must be strict inequality. Allow θ > 0.197(39...) - if anything other than < 90 then A0. 2