Mathematics 9709/43 October/November 2023: question paper and mark scheme

Mark scheme

1 Use of suvat to find expressions for s M1 𝑠 = 3𝑢 − 5 × 9 A1 𝑠 = 4𝑢 − 5 × 16 𝑢 = 35, 𝑠 = 60 A1 Alternative method for Question 1: Use of suvat to find expressions for u at max height M1 0 = 𝑢 − 10 × 3.5 A1 𝑢 = 35, 𝑠 = 60 A1 3

Mark scheme

2(a) Attempt at conservation of momentum M1 [1.2𝑣 = (1.2 + 0.004) × 40] 602 A1 oe 𝑣 = 15 2 2(b) 02 = ( 40) + 2 × 0.04 × a 2 [a = −20000] M1 Use of a ‘suvat’ method to get an equation in a. Allow sign errors. Allow ±20000 . 0 + 40 or 0.04 = t gets t = 0.002, so 0 = 40 + 0.002a [ a = −20000] Do not allow 4 in place of 0.04. 2 602 Allow use of 40.1 or for velocity in place of 40. 15 Attempt to use Newton’s Second Law vertically. M1 Must have the correct number of relevant terms. Allow R + (1.2 + 0.004) g = (1.2 + 0.004) × a − sign errors, but terms including masses must be effectively added. Do not allow any mass other than (1.2 [ − R + 12.04 = 1.204a ] + 0.004). 602301 A1 WWW. R = 24 100 N [24 092.04 = ] 25 Note: use of wrong sign for g leads to answers 24 067.96 which gets max M1M1A0. Note: Missing weight term gets 24 080 which gets Max M1M0A0. 3 2(b) Alternative method for Question 2(b) using energy [Change in PE =] 1.204 g × 0.04 [ = 0.4816] B1 602 Allow use of 40.1 or for velocity in place of 40. 1 15 × 1.204 × ( 40 ) [ = 963.2] 2 or [change in KE =] B0 for kinetic energy, if extra kinetic energy terms 2 present. 1 M1 Attempt at work energy equation. Must have correct 1.204 g × 0.04 + ×1.204 × ( 40 ) = 0.04R 2 2 number of relevant terms. dimensionally correct; allow sign errors. Do not allow 4 in place of 0.04. 602 Allow use of 40.1 or for velocity in place of 40. 15 602301 A1 WWW R = 24 100 N [24 092.04 = ] 25 Note: use of wrong sign for g leads to answers 24 067.96 which gets max B1M1A0. Note: Missing potential energy term gets 24 080, which gets maximum of B1M0A0.

Mark scheme

3(a) Correct force diagram with 3 forces in the correct directions. B1 No labels required on the 3 forces and ignore wrong labels. Arrows needed. Allow either or both components of weight if fully labelled. Allow sin/cos mix. If forces are not connected to the block, then the line of action of each force must go through the block. 1 3(b) R = 8g cos30 = 40 3 = 69.282 B1 Resolving perpendicular to the plane. Resolving parallel to the plane and attempt to apply Newton’s second law. M1* 3 terms. Allow sign errors, sin/cos mix. Allow g [8 g sin 30 − F = 8 × 2.4 F = 20.8] missing, otherwise dimensionally correct. Use of F = μ R to get an equation in μ only. DM1 Allow g missing in either or both of F and R. Allow sign errors, consistent sin/cos mix. 8 g sin 30 − 8 g μ cos30 = 8 × 2.4 40 − 40 3μ = 19.2 R must be a single component of a force. Allow the 3 masses to be cancelled. 20.8 20.8 A1 μ = 0.3[ 0] May first see or 13 3 104 3 69.282 Allow exact value or oe. 40 3 75 600 4 3(c) 6 10 B1 3.79473… (3.8 without a more accurate value seen gets [ v 2 = 2 × 2.4 × 3 greatest speed =] 3.79 ms–1 = B0 and should be annotated SF). 5 1

Mark scheme

4(a) P M1 P P = 480 × 24 or, e.g. − 480 = 0 For − F = 0 or P = Fv oe. 24 v P = 11.52 [kW] A1 Allow 11.5 M1A0 for 11 520 or 11 500. 2 4(b) 1 B1 For either correct. KEbefore = × 1600 × 242 [ = 460800] 2 1 Do not allow ×1600 × ( 32 − 24 ) . 2 1 2 KE after = × 1600 × 322 [ = 819200] 2 [ PEloss ] = 1600 g × 280 × 0.09 [ = 1600 g × 25.2 = 403200] B1 Allow 1600 g × 280 × sin 5.16° or 1600 g × 280 × sin5.2° but not simply 1600 g × 280 × sin θ (unless implied by correct final answer). Total WD = 12000 × 10 [ = 120000] B1 WD oe, e.g. 12 000 = . 10 4(b) Work done against resistance = or 280F = or WD = or W = oe M1 Attempt at work energy equation with 5 relevant terms (4 relevant terms plus work done against resistance); 1 1 12000 ×10 + 1600 g × 280 × 0.09 − ×1600 × 322 + ×1200 × 242 dimensionally correct. Allow sign errors. 2 2 M0 for use of constant acceleration. [ = 120000 + 403200 − 819 200 + 460800 ] 1 Do not allow ×1600 × ( 32 − 24 ) . 2 2 WD = 164 800 [J] A1 Or 164.8 kJ CAO but condone 165 kJ or 165 000 [J] Not from use of constant acceleration or Newton’s second law. ISW attempt to find force after correct WD found. 5

Mark scheme

5(a) Attempt to resolve in one direction and form equation. M1 Must have correct number of relevant terms (forces must have components as required). Allow sin/cos mix. Allow sign errors. If only one equation shown and it involves 32, it must be 32, not P. T sinθ = 32 and T cosθ = 80 A1 For both horizontal and vertical, or both parallel and or perpendicular. 0 = 80sinθ − 32cosθ and T = 80cosθ + 32sinθ Attempt to solve for T or θ M1 Must get to T or θ ; e.g. T = 322 + 802 or 32 −1 80 θ = tan −1 . Condone, e.g. θ = tan . 80 32 Must come from equations with correct number of relevant terms. T = 86.2 [N 86.1626….] or 16 29 or 7424 and θ = 21.8 [21.801…] A1 For both. 4 5(a) Alternative method using triangle of forces T 2 = 802 + 322 [−2 × 80 × 32cos90] or T sinθ = 32 or T cosθ = 80 M1 For any of the five; allow sign errors. 80tan θ = 32 or T = 80cosθ + 32sin θ oe A1 For any two equations. Attempt to solve for T or θ M1 Must get to T or θ ; e.g. T = 322 + 802 or 32 θ = tan −1 . 80 T = 86.2 [N 86.1626….] or 16 29 or 7424 and θ = 21.8 [21.801…] A1 Alternative Triangle of forces method using sine rule T 32 80 M1 For any two. = = sin 90 sin (θ ) sin ( 90 − θ ) A1 For all three. Attempt to solve for T or θ M1 32 e.g. θ = tan −1 . 80 T = 86.2 [N 86.1626….] or 16 29 or 7424 and θ = 21.8 [21.801…] A1 For both. 5(b) Attempt to resolve in one direction and form equation M1 Must have correct number of relevant terms (forces must have components as required). Allow sin/cos mix. Allow sign errors. Must use 120, not T. 120sinθ = P and 120cosθ = 80 A1 For both horizontal and vertical, or both parallel and or 0 = 80sinθ − P cosθ and 120 = 80cosθ + P sinθ perpendicular. Attempt to solve for P or θ M1 Must get to P or θ ; e.g. P = 1202 − 802 or 80 θ = cos −1 . 120 Must come from equations with correct number of relevant terms. P = 89.4 [89.4427] or 40 5 or 8000 θ = 48.2 [48.1896] A1 For both; allow P = 89.5 (from 120sin 48.2 ). 4 Alternative method using triangle of forces 1202 = P2 + 802 [−2 × 80 × P cos90] or 120sinθ = P or 120cosθ = 80 M1 For any of the five; allow sign errors. or 80 tan θ = P or 120 = 80cosθ + P sin θ oe A1 For any two equations. Attempt to solve for P or θ M1 Must get to P or θ ; e.g. P = 1202 − 802 or 80 θ = cos −1 , oe. 120 P = 89.4 [89.4427] or 40 5 or 8000 θ = 48.2 [ 48.1896] A1 For both; allow P = 89.5 (from 120sin 48.2) . 5(b) Alternative Triangle of forces method using sine rule 120 P 80 M1 For any two. = = sin 90 sin (θ ) sin ( 90 − θ ) A1 For all three. Attempt to solve for P or θ M1 Must get to P or θ ; 80 −1 80 e.g. θ = 90 − sin −1 or θ = cos . 120 120 P = 89.4 [89.4427] or 40 5 or 8000 θ = 48.2 [48.1896] A1 For both; allow P = 89.5 (from 120sin 48.2) .

Mark scheme

6(a) Attempt to integrate a M1* The power of t must increase by 1 with a change of coefficient in the t 2 term. Do not penalise missing c. Use of v = at scores M0. 6t 2 A1 Condone an integral sign in front of correct answer. [v =] 36t − 3t 2 [ +c ] or [v =]36t − [ +c ] 2 0 = 36t − 3t 2 − 33 DM1 Use t = 2 and v = 27 to find c . Must get to c = and set 3 term quadratic equal to zero. 27 = 36 × 2 − 3 × 22 + c c = −33 Solve 0 = 36t − 3t 2 − 33 to get t = 1 and t = 11 A1 Allow t = 1 or t = 11 ; t = 1, t = 11 oe. 4 6(b) Attempt to integrate an expression of the form at + bt 2 [ +c] with non-zero a M1* The power of t must increase by 1 with a change of coefficient in the same term. Use of s = vt scores M0. and b. 36t 2 3t 3 If correct [ s =] − − 33t [ +c '] or [ s =]18t 2 − t 3 − 33t [+c '] 2 3 Attempt to evaluate their 18t 2 − t 3 − 33t for t = 0 to t = 1 or DM1 Attempt using their limits (at least one strictly between 0 and 12) correctly. t = 1 to t = 11 or t = 11 to t = 12 0 to 1: –16 – 0 = –16 1 to 11: 484 – (–16) = 500 11 to 12: 468 – 484 = –16 For all three DM1 Allow 11 to 12 implied by symmetry instead of found separately. Distance = [16 + 500 + 16 ] = 532 m A1 4

Mark scheme

7(a) Resolving for both particles or for the system to form equation(s) M1* Must have correct number of terms. Allow sign errors. Allow sin/cos mix. Allow g missing. M0 if acceleration included unless subsequently equated to zero. Masses must be appropriate for their equation(s). Forces must have components (or not) as required. Either T − F − 2.4 g sin 30 = 0 AND 3.3g − T = 0 A1 Both correct or system correct. May get F = 21. Can be with a wrong non-zero F. Or 3.3g − F − 2.4 g sin30 = 0 R = 2.4 g cos30 = 12 3 = 20.7846 B1 Use of F = μ R to get an equation in μ only DM1 Must be from F dimensionally correct and single term R which is equal to a component the 2.4 kg weight. Allow [3.3g − 2.4 g μ cos30 − 2.4 g sin 30 = 0] consistent sin/cos mix but must be different components of weight. F and R must be numerical expressions. μ = 1.01 [sight of 1.01036… or 1.0104] A1 AG perhaps from one of 3.3g − 24sin30 33 − 12 21 7 3 21 μ= = = = = 2.4 g cos30 12 3 12 3 12 20.7846 21 = 20.8 Do not allow unless evidence of 30 substituted for θ . E.g.: sight of 1.01036… or 1.0104. 5 7(b) Using Newton’s second law for both particles or the system M1* Must have correct number of terms. Allow sign errors. Allow sin/cos mix. Allow g missing. Masses must be appropriate for their equation(s). Forces must have components (or not) as required. Either 3.3g − T = 3.3a and T − F − 2.4 g sin 20 = 2.4a A1 Both correct or system equation correct. [T − 22.778 − 8.208 = 2.4a ] or [T − 30.986 = 2.4a ] Can be with a wrong non-zero F. or 3.3 g − F − 2.4 g sin 20 = ( 2.4 + 3.3 ) a [ 2.013367 = 5.7a ] F = 1.01 × 2.4 g cos 20 [ = 22.778] B1 For correct expression for F. Attempt to solve for a a = 0.353 [0.353222…] DM1 Using their F Must get to ‘a =’. If sin/cos mix must be consistent. v 2 = 2 × 0.353 ×1 [= 0.706444…] or [ v = 0.841] A1FT FT their value of a ≠ ± g to get an expression for v 2 or 1 v. Or 1 = 0 + × 0.353t 2 t = 2.3795 v = 0.353 × 2.38 2 Can be implied by awrt 0.84 for v or awrt 0.71 for v 2 . This mark does not depend on previous A or B mark, but both Ms must have been awarded. Using Newton’s second law on A after B reaches the ground M1* Must have correct number of terms. Allow sign errors. −F − 2.4 g sin 20 = 2.4a Allow sin/cos mix. Allow g missing. [ −1.01× 2.4 g cos 20 − 2.4 g sin 20 = 2.4a ] [ a = −12.911] [ −22.78814− 8.20848 = 2.4a ] Use of suvat to find s DM1 Using their a ≠ ± g . 0 = their 0.8412 + 2 × their − 12.911× s s = 0.027358. Must get to ‘s =’. May find and use t = 0.0651. 7(b) Total distance = 1.03 m A1 8 Alternative method using energy for first 5 marks 1 B1 [ KE gained =] × ( 2.4 + 3.3 ) v2 = 2.85v2 2 [ PE lost =] 3.3g × 1 − 2.4 g × 1sin 20 [ = 24.791] = B1 Allow omission of 1 in either or both terms. [Friction =] 1.01× 2.4 g cos 20 [ = 22.778] B1 For correct expression for F. 1 M1 For attempt at energy equation. Allow sign errors, allow × ( 2.4 + 3.3 ) v2 = 3.3g ×1 − 2.4 g ×1sin 20 − 1.01× 2.4 g cos 20 ×1 2 sin/cos mix but must have sin/cos where needed. Correct Or 2.85v 2 = 24.791− 22.778 number of terms, dimensionally correct. Allow omission of 1 in any or all the three relevant terms. Must have cos 20 and sin 20. To get a correct expression for v 2 A1 Can be implied by awrt 0.84 for v or awrt 0.71 for v 2 if v 2 = 0.706444 or v = 0.841 expression not seen. 7(b) Alternative method using energy for final 3 marks 1 M1 Using their v 2 . KE = × 2.4 × 0.8412 2 1 M1 For attempt at 3 term energy equation and solved to get 1.01× 2.4 g cos 20 × s + 2.4 g sin 20 × s = × 2.4 × 0.8412 2 to ‘s =’. Allow sign errors, allow consistent sin/cos mix [ s = 0.027358..] but must have sin/cos where needed. Correct number of terms, dimensionally correct. Total distance = 1.03 m A1