Mathematics 9709/45 May/June 2025: question paper and mark scheme

Mark scheme

1 PE = ±25 g × 12sin8[ = ±3000sin8 = ±417.519] B1 If not seen separately allow B1 for ± ( 25 g sin8 ± 50 ) × 12 . 1 B1 If the candidate mistakenly thinks that the distance is KE = ± × 25 × 2.42 [ = ±72] 12m vertically upwards then allow only this mark. 2 1 M1 Correct number of relevant terms with component WD = 25 g × 12sin 8 + × 25 × 2.42 + 50 × 12 included in PE term, dimensionally correct. Could 2 have 12F in place of WD . Allow sign errors; allow sin/cos mix. WD = 1090 J Allow to subsequently find F (ISW) but A0 if state ‘work done A1 1089.519303 Allow 1089 − 1090 J. = 90.793’ OE. Alternative method using Newton’s second law 2.42 = 02 + 2a × 12 [ a = 0.24] *M1 Or other suvat method using u = 0, v = 2.4 and s = 12 . F − 50 − 25 g sin 8 = 25 × their a *DM1 For use of Newton’s second law with their value for a. W Correct number of relevant terms with weight May have instead of F. component included. Allow sign errors; allow sin/cos 12 mix. May if correct see F − 50 − 34.793= 6 or F = 90.793 Work done = their F ×12 DM1 W OE e.g. F = Dep on both M marks. 12 Work done = 1090 J A1 Allow 1089 − 1090 J. 4

Mark scheme

2(a) 1.5 × 32 = (1.5 + 0.005 ) v M1 Conservation of momentum. Must have three non-zero terms. Allow sign errors. Must have correct masses with relevant velocities. Note: M1A0 if g included with the masses speed = 31.894 m s-1 Allow ‘31.894 = 31.9’ A1 ISW. Must be given to 3dp as specified in question. 2 2(b) (1.5 + 0.005) g − 25000 = (1.5 + 0.005 ) a [→ a = −16601.29568] *M1 Use of N2L with correct number of relevant terms; allow sign errors, but masses must be added, not subtracted. Mass must be (1.5 + 0.005 ) . 0 = ( their 31.894) + 2 × ( their −16601.29568) s DM1 Use of constant acceleration to get an equation in s 2 using their negative a and their speed (allow rounded to 3sf or better). Must use suvat correctly. s = 0.0306 m A1 0.030636983… Note: answer 0.0306184 from omitting weight terms in N2L giving a = 16611.29... This gets M0M0A0. Use of speed = 31.9 gives s = 0.030649 which gets full credit if correctly obtained. Alternative method using energy PE ‒ work done against friction = ± ( (1.5 + 0.005) g − 25000) s [=±24984.95s ] B1 M1 Allow sign errors including in PE but all masses must ( (1.5 + 0.005) g − 25000) s = − 12 × (1.5 + 0.005) × their 31.8942 be correct. Correct number of terms and dimensionally correct. Using their speed (allow rounded to 3sf or better). s = 0.0306 m A1 0.030636983… 3

Mark scheme

3 s1 = 0.5 × 30 × t1 s3 = 0.5 × 30 × t3 s2 = 30 × T *M1 Use of suvat for displacement s= for at least two of the 900 900 three or for s1+3 = 0.5 × 30 × (180 − T ) . Allow s1 = s3 = s = 30 × T 2a1 2a3 2 Must not use the same variable for all three. 15 (180 − T ) + 30T 4800 = 15t1 + 30t2 + 15t3 = DM1 Adding displacements in terms of times and equating to 4800. Allow if use 4.8 rather than 4800. 4800 = 15 (180 − T ) + 30T oe A1 For correct equation in any time variable. T = 140 A1 Condone the assumption that magnitude of acceleration and deceleration are equal (which should come to 1.5) and allow full credit if everything else correct. Allow X or any other variable for time Distance 4200m B1FT FT their T 30 × theirT Alternative method for question 3 Use of a trapezium (may be indicated by diagram) M1 If no correct subsequent working, then allow M1 for Or 0.5 × 30 × (180 + T ) trapezium drawn with height 30, base 180 and top of length T (do not need to indicate that area = 4800). 0.5 × 30 × (180 + T ) = 4800 B2 OE. T = 140 A1 Allow X or any other variable for time. Distance 4200m B1FT FT their T 30 × theirT . 5

Mark scheme

4(a) Resolving either direction. Each force must have a component *M1 With correct number of relevant terms. Allow consistent sin/cos mix but must have more than simply 15 8 sin α and / or cosα. Allow sign errors. Do not allow Note: sin α = , cos α = ‘forces to the left = forces to the right’ e.g. 17 17 May see X = ± ( 51cos α + 34cos α − 17sin α ) 8 8 15 51 × + 34 × = 17 × unless subsequently 17 17 17 and Y = ± ( 51sin α − 34sin α − 17cos α ) but no marks until values of sin α ‘corrected’. and cos α used. 15 Allow sin tan −1 etc. for M1A1A1 if correct. 8 A1 Allow use of 62° or 28° or better [leading to [ X = R cos β =] ± 51× 8 8 15 + 34 × − 17 × = ± ( 24 + 16 − 15 ) = ±25 17 17 17 ±24.8949… or better]. Note: consistent sin/cos mix leads to 67 which gets M1A0A0. A1 Allow use of 62° or 28° or better [leading to [Y = R sin β =] ± 51× 15 15 8 − 34 × − 17 × = ± ( 45 − 30 − 8 ) = ±7 17 17 17 ±7.0290… or better]. Note: consistent sin/cos mix leads to −7 which gets M1A0A0. R = 252 + 7 2 DM1 Attempt to solve for R Attempt to find R. From equations with the correct number of relevant terms. All forces resolved as appropriate but allow 25 7 Or R = or consistent sin/cos muddle. (cos their β ) (cos their β ) Allow use of their β provided correctly derived from equations with the correct number of relevant terms. 7 25 Or R = or (sin their β ) (sin their β ) 4(a) 7 25 DM1 Attempt to solve for β . β = tan −1 or β = tan −1 25 7 From equations with the correct number of relevant terms. All forces resolved as appropriate but allow 25 7 consistent sin/cos muddle. Allow use of their R Or β = cos −1 or cos −1 or provided correctly derived from equations with the their R their R correct number of relevant terms. 7 25 Or β = sin −1 or sin −1 their R their R R = 26 [.0] N or 674 A1 Both correct Allow resultant sketched with angle indicated and β = 15.6° above positive x − axis arrowed. If not specified in working, please check Allow 15.6° right of positive x − axis , anticlockwise 15.6 from positive x original diagram to see if direction specified there axis ‘74.4° clockwise from the y-axis’ ‘74.4° to right of the y-axis’ instead. Allow any explanation of direction that could Allow 15.6° North of East or ‘ 15.6° above the east’. Bearing of [0]74.4°, be drawn uniquely. N74.4°E Do not allow ‘East north 15.6°’ Alternative method resolving in the directions of the 34N and 17N forces Resolving either direction. M1 With correct number of relevant terms. Allow consistent sin/cos mix but must have more than simply sin α and / or cosα. Allow sign errors. Do not allow ‘forces to the left = forces to the right’ unless subsequently correct. ± (17 − 51cos33.855) [ = 25.3529] A1 ± ( 34 − 51cos56.144) [ = ±5.5882] A1 R = 25.35292 + 5.58822 M1 Attempt to solve for R. 5.5882 8 M1 OE. β = tan −1 + tan −1 Attempt to solve for β . −25.3529 15 4(a) R = 26 [.0] N β = 15.6° above positive x − axis or otherwise as above A1 6 4(b) 15 15 8 15 15 8 15 M1 With correct number of relevant terms. Allow sign P× − 34 × − 17 × = 0 or P × = 34 × + 17 × or P × = 38 or errors. Do not allow sin/cos mix. Must be an equation, 17 17 17 17 17 17 17 P − 34 8 NOT just an expression. = 17 15 May see P sin α − 34sin α −17cosα = 0 but no marks until values of sin α and cos α used P = 43.1 A1 646 Allow . 15 2

Mark scheme

5(a) 24000 B1 OE e.g. 24000 = D × 32 [ R] = D = [ = 750] 32 Attempt at N2L M1 Use of N2L with 3 relevant terms; allow sign errors. 28000 Allow R or their R. − R = 1500 × a 36 28000 24000 A1 − = 1500 × a or 777.777− 750 = 1500 × a 36 32 1 A1 a = 0.0185ms −2 or ms −2 54 4 5(b) 20000 *M1 Use of power = F × v OE, 20000 = ( 340 + 4v ) × v or = 340 + 4v v with F given in question but allow miscopy of F. Allow if candidate uses 20 instead of 20000. Attempt to solve a three-term quadratic to obtain at least one value of v . DM1 Correct quadratic is 4v2 + 340v − 20000 = 0 . Must be at least one correct real value for their 3 Three-term quadratic. If no solving seen, must have three-terms and one correct value for their three-term quadratic to be awarded this mark. Allow if candidate uses 20 instead of 20000. Speed = 40 m s–1 (only) A1 No solving needed for A1 if no wrong working. 3

Mark scheme

6(a) 5 B1 t = 2.5 or 2 25 B1 t = 12.5 or 2 2 6(b) [ a =] 2 (15 − 2t )( −2 )[ = 8t − 60] *M1 For attempt at differentiation. Must have expression of the form pt + q with p ≠ 0, p ≠ 4, q ≠ 0 . p and q need v not be simplified. Use of a = scores M0. t 8t − 60 = 0 AND (15 − 2t ) = 0 AND attempt to solve at least one of these DM1 No need for working to solve either equation. 2 Could have one of the two equations with their 7.5 equations. Allow 225 − 60t + 4t 2 = 0 Allow their a = 0. substituted rather than t. t = 7.5 (twice) A1 AG CWO Any error seen A0. Allow substitution of 7.5 for t instead into second equation after first one solved. i.e. (15 − 2 × 7.5) = 0 2 OR 8 × 7.5 − 60 = 0 OE. Candidates who solve a = v would only get the first M1 unless they show that 7.5 gives zero for either acceleration or velocity. 3 6(c) For attempt at integration *M1 The power of t must increase by 1 with a change of ( ) [ s =] 4t 2 − 60t + 225 dt coefficient in the same term. Use of s = vt scores M0. E.g. s = a (15 − 2t ) for a ≠ 1or a ≠ 0 3 4 A1 Correct integral, allow un-simplified. [ s =] t 3 − 30t 2 + 225t [ +c ][c = 0] If constant wrong can still get A1. 3 1 1 or [ s = ] (15 − 2t )3 × [ +c ][c = 562.5] 3 −2 4 4 7.5 DM1 Correct use of limits (their t from part (b) and zero) s = t 3 − 30t 2 + 225t = × 7.53 − 30 × 7.52 + 225 × 7.5 [ −0] and their integral (but allow one mistake in 3 0 3 calculation). 7.5 1 1 1 1 Could instead find c = 562.5 then substitute their t = or s = (15 − 2t ) × = 0 − (15 − 2 × 0 ) × 3 3 3 −2 0 3 −2 7.5. If they go on to add another displacement area to correctly obtained value allow M1A0. 1125 A1 Allow DM1A1 without working if correct answer after s = 562.5 or or 563 finding correct integral. 2 SCB1 for correct answer with no attempt at integration seen. If a candidate includes +c in their final answer they do not get the final A1 mark. 4

Mark scheme

7(a) T = mg or T − mg = 0 B1 12 B1 Allow cos23 or better but must have more than simply R = 6.5 g × [ = 60] cos α . Must be identified as reaction here or when 13 friction term is formed. 5 *M1 Resolving along the plane, either case to form an T − 6.5 g × + F = 0 or T − 25 + F = 0 equation. Must be component of weight and g must be 13 Can instead use system equation mg = 25 − F present. Allow any non-zero single term F. Allow any non-zero single term T in terms of mg . 5 T − 6.5 g × − F = 0 or T − 25 − F = 0 If either of the system equations seen, then first B1 can 13 be implied (even if friction or component of 6.5g Can instead use system equation mg = 25 + F missing). Allow sin/cos mix (need not be consistent) but must have more than simply sin α and / or cosα with α = 23 or better. A1 For both correct. Must have more than simply sin α and / or cosα but α = 23 or better. Use F = 0.4 R [ = 24] to get an equation in m only DM1 Where R is a component of weight. m = 0.1 or m = 4.9 A1 0.1 m 4.9 or m : 0.1 m 4.9 or [ 0.1, 4.9] A1 7 7(b) 5 B1 Can be implied from equation with appropriate signs. PEchange = ± 12 g × 0.6 − 6.5 g × 0.6 × = 13 ± ( 72 − 15 ) = ±57 5 Must have either or sin 23 not just sin α . 13 1 B1 Can be implied from equation with appropriate signs KEchange = ± × (12 + 6.5 ) v 2 2 12 B1 12 WD friction = ±0.4 × 65 × × 0.6 [ = ±24 × 0.6 = ±14.4] Must have either or cos23 not just cos α 13 13 5 12 M1 Attempt at work energy equation with correct number 12 g × 0.6 − 6.5 g × 0.6 × − 0.4 × 65 × × 0.6 = 0.5 × (12 + 6.5 ) v 2 of relevant terms; dimensionally correct. PE term must 13 13 consist of two parts, with components as required. 72 − 15 − 14.4 = 0.5 × (12 + 6.5 ) v 2 Allow sign errors. Do NOT allow sin/cos mix. 71 × 0.6 = 0.5 × (12 + 6.5 ) v 2 42.6 = 0.5 × (12 + 6.5 ) v 2 42.6 852 A1 2.1460… v= = 2.15 Allow v = 9.25 185 Alternative method finding tension then using energy . Use of Newton’s second law for A AND B *M1 Must have correct number of relevant terms. Allow For B: 12 g − T = 12a sign errors. Do not allow sin/cos mix. 5 12 Forces must have components (or not) as required. For A: T − 6.5 g × − 0.4 × 6.5 g × = 6.5a 5 13 13 Must have either or sin 23 not just sin α and 13 likewise with cos α . Must not use their value of T from part (a). 2736 A1 T= = 73.945 37 7(b) 2736 DB1 Energy from tension = × 0.6 37 1 2736 DM1 For either For B: × 12v 2 = 12 g × 0.6 − × 0.6 2 37 1 5 12 2736 For A: × 6.5v 2 + 6.5 g × × 0.6 + 0.4 × 6.5 g × × 0.6 = 0.6 × 2 13 13 37 42.6 852 A1 If no marks scored allow SCB1 for work done v= = 2.15 Allow v = 12 9.25 185 = ±0.4 × 65 × × 0.6 [ = ±24 × 0.6 = ±14.4] . 13 Alternative method using energy but treating the particles separately 1 *B1 Must not use their value of T from part(a). For B: × 12v 2 = (12 g − T ) × 0.6 2 Attempt to find an energy equation for particle A *M1 Must have correct number of terms. Allow sign errors. Do NOT allow sin/cos mix. 5 Must have either or sin 23 not just sin α and 13 likewise with cos α . Must not use their value of T from part (a). 1 5 12 A1 5 × 6.5v 2 = T − 6.5 g × − 0.4 × 6.5 g × × 0.6 Must have either or sin 23 not just sin α and 2 13 13 13 likewise with cos α . Must not use their value of T from part (a). For attempt to solve to get to ‘v =’ or ' v2 = ' (by eliminating T) DM1 7(b) 42.6 852 A1 If no solving seen allow M1A1 for correct answer or v= = 2.15 Allow v = M1 if correct for their simultaneous equations. 9.25 185 If no marks scored allow SCB1 for work done 12 = ±0.4 × 65 × × 0.6 [ = ±24 × 0.6 = ±14.4] . 13 Alternative method finding acceleration then forces then using energy. Use of Newton’s second law for system or for A AND B *M1 Must have correct number of relevant terms Allow 5 12 sign errors. Do not allow sin/cos mix. For system: 12 g − 6.5 g × − 0.4 × 6.5 g × = (12 + 6.5 ) a Forces must have components (or not) as required. 13 13 [120 − 25 − 24 = 18.5a] 5 Must have either or sin 23 not just sin α and 13 For B: 12 g − T = 12a [120 − T = 12a ] likewise with cos α . 5 12 Must not use their value of T from part (a). For A: T − 6.5 g × − 0.4 × 6.5 g × = 6.5a [T − 25 − 24 = 6.5a ] 13 13 142 A1 a= = 3.8378 37 142 1704 142 923 DB1 FB = 12 × = or FA = 6.5 × = AND any one of 37 37 37 37 923 1704 1704 923 [ EN A =] × 0.6 or [ EN B =] × 0.6 or [ ENTotal =] + × 0.6 37 37 37 37 1 923 1 1704 DM1 × 6.5v 2 = 2 × 0.6 or × 12v = × 0.6 2 37 2 37 1 1704 923 or × (12 + 6.5) v 2 = + × 0.6 2 37 37 7(b) 42.6 852 A1 If no marks scored allow SCB1 for work done v= = 2.15 Allow v = 12 9.25 185 = ±0.4 × 65 × × 0.6 [ = ±24 × 0.6 = ±14.4] 13 5