Equilibrium of a rigid body

4(a) Centre of mass M1 Volume from base of cylinder 1 kh Cone πr2kh h+ 3 4 h Cylinder πr2h 2 Combined πr2h   1 k +1   x 3  Take moments about base: A1 2 terms correct. πr2h   1 k +1   x = 1 πr2kh  h+ kh  + πr2h h A1 All terms correct. 3  3  4  2 ( ) h k2 + 4k +6 A1 AG. Shown convincingly. x = 4 ( 3+ k ) 4 4(b) r M1 tanθ= x 4 6h ( k +3 ) M1 4 = Equate to and simplify to quadratic. ( ) 3 h k2 +4k +6 3 2k2 −k −15=0 k=3 A1 CAO. No other solutions. 3 © UCLES 2021 Page 9 of 13

4(a) Centre of mass M1 Volume from base of cylinder 1 kh Cone πr2kh h+ 3 4 h Cylinder πr2h 2 Combined πr2h   1 k +1   x 3  Take moments about base: A1 2 terms correct. πr2h   1 k +1   x = 1 πr2kh  h+ kh  + πr2h h A1 All terms correct. 3  3  4  2 ( ) h k2 + 4k +6 A1 AG. Shown convincingly. x = 4 ( 3+ k ) 4 4(b) r M1 tanθ= x 4 6h ( k +3 ) M1 4 = Equate to and simplify to quadratic. ( ) 3 h k2 +4k +6 3 2k2 −k −15=0 k=3 A1 CAO. No other solutions. 3 © UCLES 2021 Page 9 of 13

1 Centre of mass B1 All distances correct. Area from DB ABD 24a2 – a ABCD can be split in other ways, for example ADC and ABC. BCD 48a2 2a Combined 72a2 x Taking moments about DB: M1 Moments equation with masses in correct ratio. 72a2x = 24a2 ×−a + 48a2 ×2a OR Taking moments about A: 72a2x = 24a2×2a+48a2 ×5a OR Taking moments about G: 24a 2( x+a )= 48a2 ×( 2a−x ) x =a A1 CWO Alternative method for question 1 6a−3a B1 One calculation. ADC: distance of centre of mass from BD = =a 3 6a−3a ABC: distance of centre of mass from BD = =a 3 Second calculation or statement about symmetry M1 x =a A1 3 © UCLES 2021 Page 6 of 16

4(a) M1 Attempt at moments with three terms. Area Centre of mass from AD 3 Square 9a2 a 2 3 CDF ah a 2 3 1 BEC ah 3a− h 2 3 Resulting AEFC 9a2 −3ah x Taking moments about AD: A1 Two terms correct. ( )  3  3  3  1  A1 All correct. 9a2 −3ah x =  9a2× a  −  ah×a  −  ah×  3a− h   2  2  2  3  27a2 −12ah+h2  9a−h A1 AEF x = 6 ( 3a−h )   = 6   y =x B1 By symmetry or equal to their x. 5 4(b) For equilibrium, x ⩽ 3a – h B1 Accept strict inequality. 27a2 – 12ah + h2 ⩽ 6(3a – h)2 27a2– 24ah + 5h2 ⩾ 0 M1 Homogeneous 3-term quadratic inequality. 9 A1 CAO. h ⩽ a 5 3 © UCLES 2021 Page 9 of 12

4(a) M1 Attempt at moments, 3 terms. Volume Centre of mass from AB 2 3 Hemisphere πa3 a 3 8 Cylinder πka( a )2 ka 2 2 2 2 a Remainder πa3 −πka   x 3 2 Taking moments about AB: A1 Any 2 terms correct. 2 a 2 2 3   a 2 ka A1 All correct.  πa3 −πka   ×x =  πa3× a  − πka   ×  3 2  3 8   2 2      ( ) 3a 2−k2 A1 Shown convincingly, AG. x = 2 ( 8−3k ) 4 4(b) x B1 tanθ= a ( ) 3 2−k2 7 = 2 ( 8−3k ) 18 27k2 −21k+2=0 M1 Rearrange to form quadratic. 2 1 A1 Both answers correct. k = and k = 3 9 3 © UCLES 2021 Page 9 of 12

4(a) M1 Attempt at moments with three terms. Area Centre of mass from AD 3 Square 9a2 a 2 3 CDF ah a 2 3 1 BEC ah 3a− h 2 3 Resulting AEFC 9a2 −3ah x Taking moments about AD: A1 Two terms correct. ( )  3  3  3  1  A1 All correct. 9a2 −3ah x =  9a2× a  −  ah×a  −  ah×  3a− h   2  2  2  3  27a2 −12ah+h2  9a−h A1 AEF x = 6 ( 3a−h )   = 6   y =x B1 By symmetry or equal to their x. 5 4(b) For equilibrium, x ⩽ 3a – h B1 Accept strict inequality. 27a2 – 12ah + h2 ⩽ 6(3a – h)2 27a2– 24ah + 5h2 ⩾ 0 M1 Homogeneous 3-term quadratic inequality. 9 A1 CAO. h ⩽ a 5 3 © UCLES 2021 Page 9 of 12

4(a) M1 Moments equation, condone missing ends of cylinder. Area Centre of mass from AB One expression on the RHS correct. 1 Cylinder 2πah2πa2 h 2 Shell 2π2a2 ha Moments about AB A1 A1 One correct expression on RHS correct scores A1. x  2πah2πa2 2π2a2 2π2a2 ha  2πah2πa2  1 h   2  2h10ax h2 ah8ah8a2 A1 h2 9ah8a2 x  2h5a 4 4(b) a B1 tan x h2 9ah8a2 3 M1 Form inequality and rearrange to quadratic, condone x   a equation. 2h5a 2 h2 6ah7a20 hah7a0 M1 Attempt to solve, condone equation. 7a ha A1 4 © UCLES 2022 Page 8 of 14

4(a) M1 Moments equation, condone missing ends of cylinder. Area Centre of mass from AB One expression on the RHS correct. 1 Cylinder 2πah2πa2 h 2 Shell 2π2a2 ha Moments about AB A1 A1 One correct expression on RHS correct scores A1. x  2πah2πa2 2π2a2 2π2a2 ha  2πah2πa2  1 h   2  2h10ax h2 ah8ah8a2 A1 h2 9ah8a2 x  2h5a 4 4(b) a B1 tan x h2 9ah8a2 3 M1 Form inequality and rearrange to quadratic, condone x   a equation. 2h5a 2 h2 6ah7a20 hah7a0 M1 Attempt to solve, condone equation. 7a ha A1 4 © UCLES 2022 Page 8 of 14

1 M1 Attempt at moments equation with all necessary terms. Area Distance from Oy Other options possible for RHS of moments equation, for Triangle OCD 6 2 example: (1) OAC: 30  6 and ABC: 12  9 Rectangle (2) OBC: 12  4 and OAB: 30 8 24 6 DEBC (3) Subtraction: 607.5611213 Triangle BAE 12 11 Trapezium 42 x OCBA where D is point with coordinates (3, 0) and E is a point with coordinates (9, 0). Parts that would give correct total area 42 B1 Moments about Oy A1 Correct equation. 42x 622461211 (=288) 288 A1 x  6.86 42 4 © UCLES 2022 Page 6 of 13

7(a) Frictional force =  normal reaction at D and E B1 F R, F N AB BC Moments about B, NaRaWa(sincos) B1 One moments equation about any point involving all relevant forces, resolved if necessary (AEF). Moments about centre, F aF aWa(cossin) AB BC Moments about D, F aNaWacossin BC Moments about E, RaF aWacossin AB Parallel to AB, N F WsinWsin B1 Two resolutions: all relevant terms, different frictional forces AB Perpendicular to AB, F RWcosWcos [Vertical: RcosF BC cosNsinF AB sinW W BC Horizontal: F sinF cosRsin Ncos] BC AB Alternative approach using two moments equations can earn the B1B1 1 M1 Combine appropriate equations. N R 1N 1R 2  1   1  M1 Collect terms to obtain ratio/fraction in terms of  only (CWO), N1 1 R1 1   2   2  any equivalent simplified form. 1 1  1 1  N  R   2 2  2 2  R :N 1  :1  A1 6 © UCLES 2022 Page 12 of 13 7(b) N R M1 Must include  terms. Divide resolution equations: tan N R 1 M1 FT their answer to part (a). N 3 Use R2N: tan 7 N 3 1 A1 tan 7 3 © UCLES 2022 Page 13 of 13

3(a) Let F and R be friction and normal reaction at A M1 Correct terms, allow sign errors and cos/sin mix. Take moments about A, for rod N3a=W2acos+kW4acos 4 A1 At least one intermediate line of working. 3N =(2+4k)W 5 8 N = W(1+2k) AG 15 2 3(b) Ncos+R=W +kW B1 Resolve (to include R) for rod. 6 B1 Both. →F=Nsin and F = R 7 28 21 M1 Find R or N . so R= W(1+2k) or R= W(1+k) 75 45 Eliminate to find k M1 Complete method. 1 A1 k = 3 5 © UCLES 2022 Page 7 of 12

2 Area Distance from AB ABC 27a2 2a 9 1 ABD ax x 2 3 9 Shape ADC 27a2 − ax x 2 Taking moments about AB M1 Moments equation with 3 terms.  9  9 1 x 27a2 − ax= 27a2 2a− ax  x A1 At least 2 terms correct.  2  2 3  54a3− 3 ax2  A1 All correct.   2 x =  9  27a2 − ax   2  For equilibrium, x x, B1 Use correct condition: allow strict inequality. Can be implied by correct final answer x 3a. 3  9  54a3− ax2 x27a2 − ax 2  2  54a2 −27ax+3x2 0 M1 Simplify and attempt to solve a quadratic inequality or equation. (x−3a)(x−6a) 0 (0 ) x 3a [only] A1 CAO © UCLES 2022 Page 7 of 15 2 Alternative method for question 2 Taking moments with B as origin. M1 1 1 A2 x = (0+x+6a)=2a+ x 3 3 1 B1 Allow strict inequality. For equilibrium, x x, so x 2a+ x 3 (0 ) x 3a M1 A1 6 © UCLES 2022 Page 8 of 15

3(a) Let F and R be friction and normal reaction at A M1 Correct terms, allow sign errors and cos/sin mix. Take moments about A, for rod N3a=W2acos+kW4acos 4 A1 At least one intermediate line of working. 3N =(2+4k)W 5 8 N = W(1+2k) AG 15 2 3(b) Ncos+R=W +kW B1 Resolve (to include R) for rod. 6 B1 Both. →F=Nsin and F = R 7 28 21 M1 Find R or N . so R= W(1+2k) or R= W(1+k) 75 45 Eliminate to find k M1 Complete method. 1 A1 k = 3 5 © UCLES 2022 Page 7 of 12

4(a) [Mass is proportional to volume] M1 A1 Distance of Distance of Volume centre of mass centre of mass from vertical axis from OC 2 3 Hemisphere 2a3 2a  2a 3 8 1 Cylinder a2d a d 2 Object 2 2a3 a2d x y 3   2 2a3 a2d  x  16 a32a a2da Moments equation, dimensionally correct, correct 3  3 number of terms. Allow sign errors. 32a2 3ad A1 AG. At least one line of intermediate working. Simplify to x  16a3d   2 2a3 πa2d  y  16 πa3   3 2a   πa2d 1 d M1 M nu o m m b e e n r t o s f e t q e u rm ati s o . n A , l d lo im w e s n i s g i n o n e a rr ll o y r s c . o rrect, correct 3  3  8  2 3  d2 8a2 A1 AEF y  216a3d 5 © UCLES 2023 Page 10 of 15 4(b) 2ax B1 sin 2a 1 32a2 3ad M1 Remove fractions 2a 2a 6 16a3d 5 16a3d32a3d 3 8 A1 d  a 3 3 © UCLES 2023 Page 11 of 15

4(a) [Mass is proportional to volume] M1 A1 Distance of Distance of Volume centre of mass centre of mass from vertical axis from OC 2 3 Hemisphere 2a3 2a  2a 3 8 1 Cylinder a2d a d 2 Object 2 2a3 a2d x y 3   2 2a3 a2d  x  16 a32a a2da Moments equation, dimensionally correct, correct 3  3 number of terms. Allow sign errors. 32a2 3ad A1 AG. At least one line of intermediate working. Simplify to x  16a3d   2 2a3 πa2d  y  16 πa3   3 2a   πa2d 1 d M1 M nu o m m b e e n r t o s f e t q e u rm ati s o . n A , l d lo im w e s n i s g i n o n e a rr ll o y r s c . o rrect, correct 3  3  8  2 3  d2 8a2 A1 AEF y  216a3d 5 © UCLES 2023 Page 10 of 15 4(b) 2ax B1 sin 2a 1 32a2 3ad M1 Remove fractions 2a 2a 6 16a3d 5 16a3d32a3d 3 8 A1 d  a 3 3 © UCLES 2023 Page 11 of 15

3(a) [Mass is proportional to area] B1 All correct for ABC and DEC. Area Centre of mass from AC ABC 1 2a .6a.8a (24a2) 2 DEC 1 1 x.5a x 2 3 ADEB 5 x 24a2  xa 2  5  1 5 M1 All moment terms present, dimensionally correct, Moments [about AC] x  24a2  xa  24a22a x ax allow sign error.  2  3 2 A1 All correct moments about AC. 288a2 5x2 A1 AEF x  348a5x 4 © UCLES 2023 Page 6 of 14 3(b) 288a2 5x2 B1 FT FT their expression for x from part (a). On the point of toppling about E: x x, x 348a5x Rearrange to 3-term quadratic: 10x2 144ax288a2 0 M1 Allow 3-term inequality. 25x12ax12a0, A1 Single correct answer, no inequality, CWO. 12 x a 5 3 © UCLES 2023 Page 7 of 14

3 Let N be normal reaction at B and F the frictional force acting downwards B1 Tcos=F+W →Tsin=N B1 Moments about B: Ta=Wsina+Wcosa M1A1 A moments equation with all relevant forces. 1 M1 F = N used 2  1  M1 Combine to obtain equation in . (cos+sin)  cos− sin  =1 oe Equation in trigonometric functions only.  2  1 3 M1 Solve trigonometric equation. cossin= (sin)2 2 2 sin(cos−3sin)=0 1 A1 tan= 3 8 © UCLES 2023 Page 8 of 13

3(a) Mass is proportional to area B1 At least two areas correct, at least one distance correct Centre of mass Area from AC 1 1 ABC 2aa a 2 3 1 a 2 2a BDE 2  a−   2 k 3k 2 1 a ADEC − 2  +a2 x 2 k  1 a 2  1  2a a 2 M1A1 Moments equation, dimensionally correct, correct number x− 2  +a2= aa2 −a−    of terms.   2 k   3  3k k a ( k2 +k−2 ) A1 a ( k3−3k+2 ) x = Allow unsimplified single fraction . 3k(k+1) 3k ( k2 −1 ) 4 © UCLES 2023 Page 7 of 15 3(b) x  5  B1 With their x from part (a). tan= =   a  18 So 18 ( k2 +k−2 ) =5 ( 3k2 +3k ) M1 Obtain a polynomial in k only, e.g. k3−13k+12=0, may be implied. k2 +k−12=0 (k+4)(k−3)=0, A1 CWO k=3 only 3 Question Answer Marks Guidance

3 Let N be normal reaction at B and F the frictional force acting downwards B1 Tcos=F+W →Tsin=N B1 Moments about B: Ta=Wsina+Wcosa M1A1 A moments equation with all relevant forces. 1 M1 F = N used 2  1  M1 Combine to obtain equation in . (cos+sin)  cos− sin  =1 oe Equation in trigonometric functions only.  2  1 3 M1 Solve trigonometric equation. cossin= (sin)2 2 2 sin(cos−3sin)=0 1 A1 tan= 3 8 © UCLES 2023 Page 8 of 13

4(a) Frictional force F and normal reaction R at point of contact of ring with plane. M1 Only allow cos/sin errors for T and W Resolve parallel to plane: F TcosWsin components, sign errors. Accept equations for vertical and horizontal (both needed). Moments about O: FaTasin B1 Combine and substitute for : M1 Expression for T in terms of W . T  1W A1 CAO 3 Alternative solution for question 4(a) Moments about point where ring touches plane: TasinTacosWasin M1 A1 Only allow cos/sin errors, sign errors. Must be dimensionally correct. Rearrange and substitute for : M1 Expression for T in terms of W . T  1W A1 CAO 3 4 4(b) Resolve perpendicular to plane: RTsinWcos M1 Only allow cos/sin errors for T and W components, sign errors. Use F R and combine to reach an equation in  only. M1 From part (a), F TcosWsin or F Tsin.  1 A1 7 3 © Cambridge University Press & Assessment 2024 Page 9 of 13

4(a) Frictional force F and normal reaction R at point of contact of ring with plane. M1 Only allow cos/sin errors for T and W Resolve parallel to plane: F TcosWsin components, sign errors. Accept equations for vertical and horizontal (both needed). Moments about O: FaTasin B1 Combine and substitute for : M1 Expression for T in terms of W . T  1W A1 CAO 3 Alternative solution for question 4(a) Moments about point where ring touches plane: TasinTacosWasin M1 A1 Only allow cos/sin errors, sign errors. Must be dimensionally correct. Rearrange and substitute for : M1 Expression for T in terms of W . T  1W A1 CAO 3 4 4(b) Resolve perpendicular to plane: RTsinWcos M1 Only allow cos/sin errors for T and W components, sign errors. Use F R and combine to reach an equation in  only. M1 From part (a), F TcosWsin or F Tsin.  1 A1 7 3 © Cambridge University Press & Assessment 2024 Page 9 of 13

5(a) M1 A1 Note that moments about OB is M0 (y 6a). OBC OAC ABC Moments equation about OC with all terms present, allow sign error, dimensionally correct. 1 Area 24a18a 9ax 216a2 9ax All correct for A1. 2 Centre of mass 8a 1x x from OC 3 Moments about OC  216a2 9ax  x 216a28a9ax 1 x 3 576a2 x2 x24a A1 Accept any equivalent form. x  or 72a3x 3 Alternative solution to question 5(a) Consider system as equivalent to particles at (0, 18a), (x, 0) and (24a, 0) B1 Then the x-coordinate of the centre of mass is at 1x24a M1 A1 3 3 © Cambridge University Press & Assessment 2024 Page 10 of 15 5(b) 18a6a 18a y y 18a M1 A1 Either way up (their value for x may be tan or or or x x 1 5ax 1 2a 1x substituted in). 2 x2 30ax144a2 0 M1 Obtain homogeneous (quadratic) equation Note that if simplified form of x is used, equation Or, with simplified form, 30ax24a will be linear. x6a A1 Single correct answer only. 4 Question Answer Marks Guidance

4(a) B1 Correct volumes and distances for large and small. Large Small Object M1 Moments equation with 3 terms, dimensionally correct. 2  2  4  Volume a2h   a  kh a2h1− k 3   9  A1 Correct, unsimplified. Centre of 1 1 mass from h kh x AB 2 2 Moments about AB: 2  4  1 2  1 a2h1− k y = a2h h−  a  kh kh  9  2 3  2 ( 9−4k2) h A1 y = 2(9−4k) 4 4(b) y ( 9−4k2) h 3 B1 FT FT their part (a) tan= : = a 2(9−4k)a 2 8 M1 Use h= a and simplify to quadratic in k: 32k2 −36k+9=0 3 3 3 A1 k = , 8 4 3 © Cambridge University Press & Assessment 2024 Page 8 of 12

4(a) In this question, allow equivalent marks for resolutions in different directions and B1 moments about other points. Apply the guidance given in the main scheme. Tcos+F =2W →Tsin=R B1 Moments about A: M1 All relevant terms included, dimensionally correct, Tcos3asin+Tsin3acos=W3asin+Wkasin forces must be resolved if appropriate. Allow sin/cos mix, allow sign errors. LHS: any equivalent expression, for example 3Tasin2, 3Tasin(180−2). OR All relevant terms included, dimensionally correct, Moments about C: forces must be resolved if appropriate. R3acos=F3asin+W(ka−3a)sin Allow sin/cos mix, allow sign errors. T M1 1 [6aTcos=(3+k)aW and (3cos+sin)=2W give]  and giveand give Use F = R and eliminate T and W to obtain an 3 3 1 expression in k and , dependent on a 12cos=(3+k)(cos+ sin) 3 dimensionally correct moments equation. k=5 A1 5 © Cambridge University Press & Assessment 2024 Page 11 of 19 4(b) A complete method to find F in terms of W M1 Any complete method to find F. For example, substitute into moments equation to 2 obtain T [T = 13 W, R=2W] . 3 2 A1 Correct. F = W 3 2 Question Answer Marks Guidance

4(a) B1 Correct volumes and distances for large and small. Large Small Object M1 Moments equation with 3 terms, dimensionally correct. 2  2  4  Volume a2h   a  kh a2h1− k 3   9  A1 Correct, unsimplified. Centre of 1 1 mass from h kh x AB 2 2 Moments about AB: 2  4  1 2  1 a2h1− k y = a2h h−  a  kh kh  9  2 3  2 ( 9−4k2) h A1 y = 2(9−4k) 4 4(b) y ( 9−4k2) h 3 B1 FT FT their part (a) tan= : = a 2(9−4k)a 2 8 M1 Use h= a and simplify to quadratic in k: 32k2 −36k+9=0 3 3 3 A1 k = , 8 4 3 © Cambridge University Press & Assessment 2024 Page 8 of 12

4(a) M1 Allow one error for cone or hemisphere. Volume CoM above base SOI, may be seen in moments equation. Cone 18πr3 3r 2 Hemisphere 16πr3 3r 3 4 Total object 18πr3 −16πr3 y 3 ( 18πr3−16πr3) y =18πr3(3r)−16πr3(3r) M1 A moments equation with 3 terms, allow sign 3 2 3 4 errors, must be dimensionally correct. A1 Fully correct equation, does not need to be simplified. y = 69r A1 38 4 4(b) When on point of toppling: M1 3r their y For attempt at tan using or . 3r tan= their y   = 3 2 8 3   their y 3r A1 FT FT their y . =58.8o A1 CAO 3 © Cambridge University Press & Assessment 2025 Page 12 of 17

4(a) M1 Allow one error for cone or hemisphere. Volume CoM above base SOI, may be seen in moments equation. Cone 18πr3 3r 2 Hemisphere 16πr3 3r 3 4 Total object 18πr3 −16πr3 y 3 ( 18πr3−16πr3) y =18πr3(3r)−16πr3(3r) M1 A moments equation with 3 terms, allow sign 3 2 3 4 errors, must be dimensionally correct. A1 Fully correct equation, does not need to be simplified. y = 69r A1 38 4 4(b) When on point of toppling: M1 3r their y For attempt at tan using or . 3r tan= their y   = 3 2 8 3   their y 3r A1 FT FT their y . =58.8o A1 CAO 3 © Cambridge University Press & Assessment 2025 Page 12 of 17

4 B1 Allow one error in first two columns. Areas 2a2 1 2 a(2a) 2a2 + 1 2 a(2a) Correct distances about any point. SOI, may be seen in moments equation. Distances from D −1a 2a x 2 3 −2a2(1a)+1 ( 2a2)(2a)= ( 3a2) x *M1 A moments equation with 3 terms, must be 2 2 3 dimensionally correct. (x =1a or x =8a) 9 9 m (1a ) =km(2a) DM1 Moments equation about toppling point D. 9 2 terms only, OE. Must include k. k = 1 A1 18 Alternative method for question 4 B1 Correct ratios including k. Ratio of masses 2a2 1 2 a(2a) k ( 2a2 + 1 2 a(2a)) Must simply to 2:1:3k. SOI, may be seen in moments equation. 2 (1a ) =1 (2a ) +3k(2a) *M1 A moments equation about D with 3 terms, must 2 3 be dimensionally correct. Must include k. DM1 Using correct distances. k = 1 A1 18 4 © Cambridge University Press & Assessment 2025 Page 12 of 17

5(a) Split into rectangle AGED and triangle GFE (DA parallel to EG). M1 All terms present. All terms must be dimensionally correct. Area CoM from AD CoM from AB Moments about AD: 18a2x =6ah (1h ) + ( 18a2 −6ah )(1h+2a ) Rectangle 6ah 1h 3a 2 3 2 Triangle 18a2 −6ah 1h+2a 2a 3 AFED 18a2 x y A1 Fully correct equation, does not need to be simplified. h2 −6ah+36a2 A1 AG, shown convincingly. x = 18a Moments about AB: 18a2y =6ah(3a)+ ( 18a2 −6ah )(2a) M1 All terms present. All terms must be dimensionally correct. y =2a+1h A1 Must expand brackets and collect like terms. 3 Alternative method for question 5(a) Split into triangle AED and triangle AEF. M1 All terms present. All terms must be dimensionally correct. Area CoM from AD CoM from AB Moments about AD: 18a2x =3ah (1h ) + ( 18a2 −3ah )(2a) AED 3ah 1h 4a 3 3 AEF 18a2 −3ah 2a 2a AFED 18a2 x y A1 Fully correct equation, does not need to be simplified. h2 −6ah+36a2 A1 AG, shown convincingly. x = 18a Moments about AB: 18a2y =3ah(4a)+ ( 18a2 −3ah )(2a) M1 All terms present. All terms must be dimensionally correct. © Cambridge University Press & Assessment 2025 Page 13 of 17 5(a) Alternative method for question 5(a) Subtract triangle EFH from rectangle AFHD M1 All terms present. All terms must be dimensionally correct. (H is on DC, FH parallel to AD). Area CoM from AD CoM from AB AFHD 36a2 −6ah 3a− 1h 3a Moments about AD: 2 18a2x = ( 36a2 −6ah )( 3a−1h ) − ( 18a2 −6ah )( 4a−1h ) EFH 18a2 −6ah 4a−1 3 h 4a 2 3 AFED 18a2 x y A1 Fully correct equation, does not need to be simplified. h2 −6ah+36a2 A1 AG, shown convincingly. x = 18a Moments about AB: M1 All terms present. All terms must be dimensionally correct. 18a2y = ( 36a2 −6ah )(3a)− ( 18a2 −6ah )(4a) y =2a+1h A1 Must expand brackets and collect like terms. 3 5 5(b) x B1 Correct expression for tan. tan= 6a− y h2 −6ah+36a2 M1 FT Equate their expression for tan (must involve x and y ) to 7 , 15 18a = 7 leading to 5h2−16ah+12a2 =0 . substitute the correct x and their y and simplify to a 3-term 6a−(2a+1h) 15 3 h2 −6ah+36a2    (h−2a)(5h−6a)=0  quadratic. May see 15  18a   =7 ( 6a−2a−1 3 h ) . © Cambridge University Press & Assessment 2025 Page 14 of 17

5(a) x =ka+1h M1 May be seen in a table. 3 x = 1(3ka+h) A1 AG, shown convincingly. 3 y =a+1(h−a)= 1(2a+h) B1 May be seen in a table. 3 3 Alternative method for question 5(a) Coordinates A(ka, 0), B(ka+h, h), C(ka, 2a) M1 Complete method to find x. x =1(ka+ka+h+ka)= 1(3ka+h) 3 3 A1 AG, shown convincingly. y =1(0+h+2a)= 1(h+2a) B1 3 3 3 © Cambridge University Press & Assessment 2025 Page 13 of 17 5(b) Taking moments about the x-axis: M1 Moments equation with 3 terms. x ( 2ka2 +ah ) = 2ka21ka+ah ( ka+1h ) Area From OD 2 3 OACD 2ka2 1ka 2 ABC ah ka+1h 3 Shape OABCD 2ka2 +ah x A1 Fully correct moments equation. May be seen in part 5(a). For equilibrium, x≤ka, so M1 Use correct condition with their x for OABCD. a2k2 +ahk+1h2 Allow strict inequality. 3 ≤ka M1 for simplifying and attempting to solve an inequality. 2ak+h May be implied by correct final answer. (0≤)h≤ 3ka A1 CAO 4 5(c) Point of toppling means h= 3ka=a. B1 ( ) x =1a 3+1 3 [h=a, meaning triangle ABC is isosceles.] B1 y =a (by symmetry) 2 © Cambridge University Press & Assessment 2025 Page 14 of 17

3(a) Vertical distance of COM of sector from OA: B1 SOI ( 2acos(1π)) sin(1π) 4a Allow unsimplified. 4 4 = 3(1π) 3π 4 ka2(1a)+ 1πa2  a− 4a = ( ka2 + 1πa2) y M1 3 eq -t u e a r t m io n d . i mensionally correct moments 2 4  3π 4   6ka+3πa−4a=3(4k+π)y  A1 FT Correct moments equation, allow unsimplified. FT their vertical distance of COM of sector from OA. 6k+3π−4 A1 AEF, but do not allow fractions within y = a 3(4k+π) numerator and/or denominator. 4 3(b) 4a M1 Dimensionally correct moments equation ka2(1 2 ka)= 1 4 πa2  3π   about OB using their horizontal distance of COM of sector from OB. k2 = 2 k = 2 =0.816 A1  3 3 Alternative method for question 3(b) ( 6k2 +3kπ+4 ) a M1 Use of x =ka with their x, which may be On the point of toppling, x =ka: =ka seen in part 3(a). 3(4k+π) 6k2 +4=12k2 k = 2 =0.816 A1   3 2 © Cambridge University Press & Assessment 2025 Page 11 of 17

5(a) x =ka+1h M1 May be seen in a table. 3 x = 1(3ka+h) A1 AG, shown convincingly. 3 y =a+1(h−a)= 1(2a+h) B1 May be seen in a table. 3 3 Alternative method for question 5(a) Coordinates A(ka, 0), B(ka+h, h), C(ka, 2a) M1 Complete method to find x. x =1(ka+ka+h+ka)= 1(3ka+h) 3 3 A1 AG, shown convincingly. y =1(0+h+2a)= 1(h+2a) B1 3 3 3 © Cambridge University Press & Assessment 2025 Page 13 of 17 5(b) Taking moments about the x-axis: M1 Moments equation with 3 terms. x ( 2ka2 +ah ) = 2ka21ka+ah ( ka+1h ) Area From OD 2 3 OACD 2ka2 1ka 2 ABC ah ka+1h 3 Shape OABCD 2ka2 +ah x A1 Fully correct moments equation. May be seen in part 5(a). For equilibrium, x≤ka, so M1 Use correct condition with their x for OABCD. a2k2 +ahk+1h2 Allow strict inequality. 3 ≤ka M1 for simplifying and attempting to solve an inequality. 2ak+h May be implied by correct final answer. (0≤)h≤ 3ka A1 CAO 4 5(c) Point of toppling means h= 3ka=a. B1 ( ) x =1a 3+1 3 [h=a, meaning triangle ABC is isosceles.] B1 y =a (by symmetry) 2 © Cambridge University Press & Assessment 2025 Page 14 of 17

3(a) Area From AD From AB B1 Areas and distances from AD correct. ABCD 4a2 a a Does not need to be presented in a table. AED a2 1a 2a 3 3 BCF ah 2a−1h 4a 3 3 Shape BFDE 3a2 −ah x y Taking moments about AD: M1 Dimensionally correct moments equation with at least ( 3a2 −ah ) x =4a2a−a21a−ah(2a−1h) three correct terms. 3 3 11a3−2a2h+1ah2 1a(h2 −6ah+11a2) h2 −6ah+11a2 A1 AG, shown convincingly. x = 3 3 = 3 = 3a2 −ah a(3a−h) 3(3a−h) Taking moments about AB: M1 Dimensionally correct moments equation with at least ( 3a2 −ah ) y =4a2a−a22a− ah4a three correct terms. 3 3 2a(5a−2h) A1 y = 3(3a−h) © Cambridge University Press & Assessment 2025 Page 11 of 18 3(a) Alternative method for question 3(a) Area From AD From AB B1 Areas and distances from AD correct. DEF a(2a−h) 1(3a−h) 4a Does not need to be presented in a table. 3 3 EBF a2 1(5a−h) 2a 3 3 Shape BFDE 3a2 −ah x y Taking moments about AD: M1 Dimensionally correct moments equation with at least ( 3a2 −ah ) x =a (2a−h)1(3a−h)+a21(5a−h) two correct terms. 3 3 1a(6a2 −5ah+h2 +5a2 −ah) 1a(h2 −6ah+11a2) h2 −6ah+11a2 A1 AG, shown convincingly. x = 3 = 3 = 3a2 −ah a(3a−h) 3(3a−h) Taking moments about AB: M1 Dimensionally correct moments equation with at least ( 3a2 −ah ) y =a(2a−h)4a+a22a two correct terms. 3 3 2a(5a−2h) A1 y = 3(3a−h) 5 3(b) x≥a meaning h2 −3ah+2a2≥0 M1 Use condition x≥a and simplify to three-term homogeneous quadratic inequality. [0]h≤a, h=2a. A1 Accept h=0. 2 © Cambridge University Press & Assessment 2025 Page 12 of 18

1(a) (1x)πr2x+(9r+x)18πr3 =x ( πr2x+18πr3) M1 Moments equation with three terms, dimensionally 2 8 correct so must be of the form r2x2 +(r+x)r3 =x ( r2x+r3) where ,,,, are all non-zero constants. For reference: Shape Distance Volume Cylinder © Cambridge University Press & Assessment 2026 Page 9 of 22 12 x π r 2 x Hemisphere 38  3 r + x 12  43 π ( 3 r ) 3 = 1 8 π r 3 Total x πr2(x+18r) B1 At least two terms correct. A1 All correct. 2x2 +81r2 +72xr A1 x = OE, for example, 4x+72r 12 π x 2 r 2 π + 2 r 8 14 x + π r 1 4 8 + π r 1 3 8 π r 3 x . All individual terms must be simplified. ISW after correct answer seen. 4 Trace ID: PC-76076FFBDF Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/3a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/31 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 1(b) x−x *M1 tan= For 3r © Cambridge University Press & Assessment 2026 Page 10 of 22 t a n 3  = x − r x or t a n 3  = x − r x or t a n 3  = x r − x or t a n 3  = x − r x only (allow equivalents with x replaced with r or vice versa). 2x2 +81x2 +72x2 DM1 −x For 4x+72x tan= 3x t a n 3  = x − x x or t a n 3  = x − x x only with their x correctly substituted – their xmust be dimensionally correct and have come from an expression of the form a x 2 + d b x r + 2 e + r c x r with a, b, c, d and e being non- zero. Must be in terms of either x or r only. 155x 79 If correct, may see −x= x in numerator. 76 76 For reference: t a n 7 9 2 2 8  =  . =19.1 A1 AWRT 19.1 (19.110752…) or 0.334 radians (0.3335455…). 79 Allow tan=− =19.1for full marks (but 228 A0 if given as a negative angle). 3 Trace ID: PC-76076FFBDF Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/3a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/31 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/

1(a) © Cambridge University Press & Assessment 2026 Page 9 of 22 ( 12 x )  π r 2 x + ( 98 r + x )  1 8 π r 3 = x  ( π r 2 x + 1 8 π r 3 ) M1 Moments equation with three terms, dimensionally correct so must be of the form r2x2 +(r+x)r3 =x ( r2x+r3) where ,,,, are all non-zero constants. For reference: Shape Distance Volume Cylinder 12 x π r 2 x Hemisphere 38  3 r + x 14π(3r)3 =18πr3 2 3 Total x π r 2 ( x + 1 8 r ) B1 At least two terms correct. A1 All correct. x = 2 x 2 + 4 8 x 1 r + 2 7 + 2 r 7 2 x r A1 OE, for example, 12 π x 2 r 2 π + 2 r 8 14 x + π r 1 4 8 + π r 1 3 8 π r 3 x . All individual terms must be simplified. ISW after correct answer seen. 4 Trace ID: PC-852B15CDF8 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/3a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/32 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 1(b) © Cambridge University Press & Assessment 2026 Page 10 of 22 t a n 3  = x − r x *M1 For t a n 3  = x − r x or t a n 3  = x − r x or t a n 3  = x r − x 3r or tan= only (allow equivalents with x x−x replaced with r or vice versa). t a n 2 2 4 8 1 2 73 2 7 2 2  = x + x + x + x x x − x DM1 For t a n 3  = x − x x or t a n 3  = x − x x only with their x correctly substituted – their xmust be dimensionally correct and have come from an expression of the form a x 2 + d b x r + 2 e + r c x r with a, b, c, d and e being non-zero. Must be in terms of either x or r only. If correct, may see 1 5 7 5 6 x − x = 7 7 9 6 x in numerator. For reference: t a n 7 9 2 2 8  =  . 1 9 .1  =  A1 AWRT 19.1 (19.110752…) or 0.334 radians (0.3335455…). Allow t a n 7 9 2 2 8 1 9 .1   = −  =  for full marks (but A0 if given as a negative angle). 3 Trace ID: PC-852B15CDF8 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/3a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/32 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/

4(a) © Cambridge University Press & Assessment 2026 Page 13 of 20 ( d 3 + 1 3 d 2 h ( k + 1 ) ) x = 13 k h d 2 ( 12 d + 14 k h ) M1 Dimensionally correct moments equation, which must be of the form ( d3+d2h(k+1)) x =khd2(d+kh) where , , ,  are all non-zero constants. For reference: Volume From AB Cube d 3 0 Vertical pyramid 13 h d 2 0 Horizontal pyramid 1khd2 3 12 d + 14 k h Object d 3 + 13 d 2 h ( k + 1 ) x B1 Either side of equation correct. A1 Fully correct moments equation. kh(2d +kh) A1 1khd2(1d +1kh) x = OE, for example, 3 2 4 or 4(hk+h+3d) d3+ 1d2h(k+1) 3 1 2 2 d d k + h + 4 k k h 2 h + 2 4 h . ISW 4 Trace ID: PC-202C151F9E Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/3a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/33 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 4(b) © Cambridge University Press & Assessment 2026 Page 14 of 20 x = 12 d B1FT SOI, equate their x to 12 d . k 2 h 2 − 2 d h − 6 d 2  = 0  M1 Rearrange to obtain a 3-term quadratic in h of the form 2 2 2  0     k h + d h + d = . Allow, for example, an equation of the form k 2 h 2 − 2 d h = 6 d 2 . Their x kh(d+kh) must be of the form , where (hk+h+d) , , , ,   are all non-zero constants h = d + d k 1 2 + 6 k 2  = d k 2 ( 1 + 1 + 6 k 2 )  A1 OE, for example, h = 2 d + 4 d 2 2 k + 2 2 4 k 2 d 2 . If seen, must reject d k 2 ( 1 − 1 + 6 k 2 ) . 3 4(c) 1 .5 d = d k 2 ( 1 + 1 + 6 k 2 ) M1 Dependent on M1 in 4(b). Equating their h to 1.5d or substituting h = 1.5d into k 2 h 2 − 2 d h − 6 d 2 = 0 . k=2 only A1 If seen, k=0 and k = − 2 must be rejected. 2 Trace ID: PC-202C151F9E Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/3a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/33 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/

2(a) EITHER: Taking moments about AD: © Cambridge University Press & Assessment 2026 Page 10 of 25 ( a 2 − 18 a 2 ( 1 + 12 ) ) x = a 2  12 a − 18 a 2  12 ( a − 16 a  12 ) − 18 a 2  56 a 13a2x = 1a3− 1 a2(11a ) − 5 a3  16 2 16 12 48  Taking moments about AB: ( a 2 − 18 a 2 ( 1 + 12 ) ) y = a 2  12 a − 18 a 2  12  16 a − 18 a 2  56 a   11 36 a 2 y = 12 a 3 − 19 6 a 3 − 54 8 a 3  (M1 Dimensionally correct moments equation about AD with all terms present. At least two correct terms. For reference: Area From AD From AB ABCD a2 12 a 12 a BFE 18 a 2  12 a − 16 a  12 16 a CFG 18 a 2 56 a 56 a Shape BCDEF a 2 − 18 a 2 ( 1 + 12 ) x y M1 Dimensionally correct moments equation about AB with all terms present. At least two correct terms. A1 One correct unsimplified equation. x = 51 2 a A1 AG WWW, A0 if any errors seen. Allow a correct equation followed by this answer with no intermediate working for this mark. y = 37 78 a A1) WWW but no working required. OE, answer must be exact. Trace ID: PC-051B44EFC7 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/3a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/34 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 2(a) OR: Alternative – rectangle and two triangles. Taking moments about AD: © Cambridge University Press & Assessment 2026 Page 11 of 25 ( 12 a 2 + 18 a 2 + 31 6 a 2 ) x = 12 a 2  14 a + 18 a 2  71 2 a + 31 6 a 2  34 a Taking moments about AB: ( 12 a 2 + 18 a 2 + 31 6 a 2 ) y = 12 a 2  12 a + 18 a 2  13 a + 31 6 a 2  12 a (M1 Dimensionally correct moments equation about AD with all terms present. At least two correct terms. M1 Dimensionally correct moments equation about AB with all terms present. At least two correct terms. A1 One correct unsimplified equation. x = 5 a A1 AG WWW, A0 if any errors seen. Allow a 12 correct equation followed by this answer with no intermediate working for this mark. y = 37 78 a A1) WWW but no working required. OE, answer must be exact. Available marks 5 2(b) 1a− 5 a M1 1a− 5 a a− y tan= 2 12 For tan= 2 12 or tan= OE a− y a− y 1a− 5 a 2 12 with their y substituted. Their y must be of the form ka, where 0  k  1 . Note that use of y = 37a leads to tan=13. 78 82 =9.0 A1 WWW, must come from a correct (but possibly inexact) value of y =(0.4743...)a. Allow AWRT 9 (9.0085037…) or 0.157 (0.157228…) in radians. 2 Trace ID: PC-051B44EFC7 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/3a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/34 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/