Equilibrium of a rigid body — Paper 3 · A Level Mathematics - Further 9231

4 B A E 7.5 cm C x cm F D A uniform square lamina ABCD has sides of length 10 cm. The point E is on BC with EC = 7.5cm , and the point F is on DC with CF = xcm . The triangle EFC is removed from ABCD (see diagram). The centre of mass of the resulting shape ABEFD is a distance x cm from CB and a distance y cm from CD. 400 - x 2 (a) Show that x = and find a corresponding expression for y. [4] 80 - 3x … … … … … … … … … … … … … … … … The shape ABEFD is in equilibrium in a vertical plane with the edge DF resting on a smooth horizontal surface. (b) Find the greatest possible value of x, giving your answer in the form a + b 2 , where a and b are constants to be determined. [3] … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 4(a) Area Centre of mass from BC Centre of mass from DC Square 100 5 5 Triangle ½ x. 15/2 1 3 x 5 2 Shape ABEFD 15 100 4 − x x y Take moments about BC: 15 100 4 σ   −     x . 15 1 500 . 4 3 σ σ = − x x x (M1 for all terms present) M1 2 400 80 3 − = − x x x AG A1 Take moments about DC: 15 100 4   −     x . y = 15 5 100 5 . 4 2 × − x M1 y = 800 15 160 6 − − x x A1 4 4(b) Use condition: x x  B1 2 2 80 400 0 − + x x  M1 20 10 2 = − x A1 3

This question in 9231/31 May/June 2020

4 B A E 7.5 cm C x cm F D A uniform square lamina ABCD has sides of length 10 cm. The point E is on BC with EC = 7.5cm , and the point F is on DC with CF = xcm . The triangle EFC is removed from ABCD (see diagram). The centre of mass of the resulting shape ABEFD is a distance x cm from CB and a distance y cm from CD. 400 - x 2 (a) Show that x = and find a corresponding expression for y. [4] 80 - 3x … … … … … … … … … … … … … … … … The shape ABEFD is in equilibrium in a vertical plane with the edge DF resting on a smooth horizontal surface. (b) Find the greatest possible value of x, giving your answer in the form a + b 2 , where a and b are constants to be determined. [3] … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 4(a) Area Centre of mass from BC Centre of mass from DC Square 100 5 5 Triangle ½ x. 15/2 1 3 x 5 2 Shape ABEFD 15 100 4 − x x y Take moments about BC: 15 100 4 σ   −     x . 15 1 500 . 4 3 σ σ = − x x x (M1 for all terms present) M1 2 400 80 3 − = − x x x AG A1 Take moments about DC: 15 100 4   −     x . y = 15 5 100 5 . 4 2 × − x M1 y = 800 15 160 6 − − x x A1 4 4(b) Use condition: x x  B1 2 2 80 400 0 − + x x  M1 20 10 2 = − x A1 3

This question in 9231/32 May/June 2020

4 4r 2r kr 3r A uniform solid circular cone, of vertical height 4r and radius 2r, is attached to a uniform solid cylinder, of height 3r and radius kr, where k is a constant less than 2. The base of the cone is joined to one of the circular faces of the cylinder so that the axes of symmetry of the two solids coincide (see diagram). The cone and the cylinder are made of the same material. (a) Show that the distance of the centre of mass of the combined solid from the vertex of the cone is ( 99 k 2 + 96) r 2 . [4] 18k + 32 … … … … … … … … … … … … … The point C is on the circumference of the base of the cone. When the combined solid is freely suspended from C and hanging in equilibrium, the diameter through C makes an angle a with the downward vertical, where tan a = 18 . (b) Given that the centre of mass of the combined solid is within the cylinder, find the value of k. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 4(a) Mass Centre of mass from vertex of cone Cylinder ( ) 2 π .3 kr rρ 11 2 r Cone ( ) 2 1 π 2 .4 3 r rρ 3r Combined ( ) ( ) 2 2 1 π .3 π 2 .4 3 kr r r r ρ   +     x B1 Take moments about vertex: ( ) ( ) 2 2 1 11 π .3 π 2 .4 3 2 r x kr r r r   × + = ×     ( ) 2 π .3 3 kr r r + × ( ) 2 1 π 2 .4 3 r r leading to M1A1 ( ) 2 2 99 96 18 32 + = + k r x k A1 4 Question Answer Marks 4(b) 4 tan 2 α − = x r r ( = 1/8) M1 17 4 = x r A1 Equate to answer in (a) and attempt to find k M1 2 4 45 80: 3 = = k k A1 4

This question in 9231/33 May/June 2020

7 4 3 ga a A B O One end of a light spring of natural length a and modulus of elasticity 4mg is attached to a fixed point O. The other end of the spring is attached to a particle A of mass km, where k is a constant. Initially the spring lies at rest on a smooth horizontal surface and has length a. A second particle B, of mass m, is moving towards A with speed 43 ga along the line of the spring from the opposite direction to O (see diagram). The particles A and B collide and coalesce. At a point C in the subsequent motion, the length of the spring is 34 a and the speed of the combined particle is half of its initial speed. (a) Find the value of k. [6] … … … … … … … … … … … … … … … … … … … At the point C the horizontal surface becomes rough, with coefficient of friction n between the combined particle and the surface. The deceleration of the combined particle at C is 209 g . (b) Find the value of n. [4] … … … … … … … … … … … … … … … … … … … … … … … … … … Additional Page If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown. … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 7(a) Collision: ( )1 + = k mv mu so 1 = + u v k B1 Loss in KE = ( ) 2 2 1 1 1 2 4   + −     m k v v B1 Gain in EPE = ½ .4mg/a. 2 4       a B1 So, ( ) ( ) 2 2 1 3 . . 1 2 4 8 1 + = + u mga m k k M1 Use value of u and solve M1 3 = k A1 6 Question Answer Marks 7(b) 9 4 . 20 + = g T F m M1 4 9 4 . 5 a mg mg F a + = M1 4 / 5 μ = R mg and compare with 4 = R mg B1 1 5 μ = A1 4

This question in 9231/33 May/June 2020

4 E D 2r C h A B 6r The diagram shows the cross-section ABCD of a uniform solid object which is formed by removing a cone with cross-section DCE from the top of a larger cone with cross-section ABE. The perpendicular distance between AB and DC is h, the diameter AB is 6r and the diameter DC is 2r. (a) Find an expression, in terms of h, for the distance of the centre of mass of the solid object from AB. [4] … … … … … … … … … … … … … … … … … The object is freely suspended from the point B and hangs in equilibrium. The angle between AB and the downward vertical through B is i. (b) Given that h = 134 r , find the value of tani. [2] … … … … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 4(a) Volume Centre of mass from AB Small cone 2 1 . 3 2 h r π 1 9 . 4 2 8 h h h   + =     Large cone ( ) 2 1 3 3 . 3 2 h r π 1 3 3 . 4 2 8 h h   =     Object ( ) 2 26 6 r h π x B1 For 9h/8 or 3h/8 (unsimplified) Take moments about AB 2 2 2 13 27 3 1 9 . . . 3 6 8 6 8 h h r h x r h r h π π π = − M1 A1 Moments equation: Allow use of relative masses 1, 26, 27 9 26 h x = A1 4 4(b) tan 3 x r θ = M1 (= 3 ) 26 h r Use 13 4 h r = 3 tan 8 θ = A1 2

This question in 9231/31 Oct/Nov 2020

3 An object consists of a uniform solid circular cone, of vertical height 4r and radius 3r, and a uniform solid cylinder, of height 4r and radius 3r. The circular base of the cone and one of the circular faces of the cylinder are joined together so that they coincide. The cone and the cylinder are made of the same material. (a) Find the distance of the centre of mass of the object from the end of the cylinder that is not attached to the cone. [4] … … … … … … … … … … … … … … … … … … … … … … … … (b) Show that the object can rest in equilibrium with the curved surface of the cone in contact with a horizontal surface. [3] … … … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 3(a) Volume Centre of mass from base Cone ( ) 2 1 3 . 4 3 r r π 4r r + Cylinder ( ) 2 3 . 4 r r π 2r Combined ( ) 2 4 3 . 4 3 r r π x B1 Distances correct Taking moments about base of cylinder: ( ) ( ) ( ) 2 2 2 4 1 . 3 .4 3 .4 .5 3 .4 .2 3 3 x r r r r r r r r π π π = + M1 A1 Moments equation 11 4 x r = A1 4 3(b) Condition: cos OG OA θ < (where O is vertex of cone and OA is slant height of cone) B1 Correct condition for equilibrium 5 4 4 5 4 5 r r r   + × <     M1 Expression in terms of r 21 < 25 True A1 Correct conclusion, with correct working 3

This question in 9231/32 Oct/Nov 2020

4 E D 2r C h A B 6r The diagram shows the cross-section ABCD of a uniform solid object which is formed by removing a cone with cross-section DCE from the top of a larger cone with cross-section ABE. The perpendicular distance between AB and DC is h, the diameter AB is 6r and the diameter DC is 2r. (a) Find an expression, in terms of h, for the distance of the centre of mass of the solid object from AB. [4] … … … … … … … … … … … … … … … … … The object is freely suspended from the point B and hangs in equilibrium. The angle between AB and the downward vertical through B is i. (b) Given that h = 134 r , find the value of tani. [2] … … … … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 4(a) Volume Centre of mass from AB Small cone 2 1 . 3 2 h r π 1 9 . 4 2 8 h h h   + =     Large cone ( ) 2 1 3 3 . 3 2 h r π 1 3 3 . 4 2 8 h h   =     Object ( ) 2 26 6 r h π x B1 For 9h/8 or 3h/8 (unsimplified) Take moments about AB 2 2 2 13 27 3 1 9 . . . 3 6 8 6 8 h h r h x r h r h π π π = − M1 A1 Moments equation: Allow use of relative masses 1, 26, 27 9 26 h x = A1 4 4(b) tan 3 x r θ = M1 (= 3 ) 26 h r Use 13 4 h r = 3 tan 8 θ = A1 2

This question in 9231/33 Oct/Nov 2020

4 kh h r A uniform solid circular cone has vertical height kh and radius r. A uniform solid cylinder has height h and radius r. The base of the cone is joined to one of the circular faces of the cylinder so that the axes of symmetry of the two solids coincide (see diagram, which shows a cross-section). The cone and the cylinder are made of the same material. (a) Show that the distance of the centre of mass of the combined solid from the base of the cylinder h ( k 2 + 4k + 6) is . [4] 4 ( 3 + k) … … … … … … … … … … … … … … … … The solid is placed on a plane that is inclined to the horizontal at an angle i. The base of the cylinder is in contact with the plane. The plane is sufficiently rough to prevent sliding. It is given that 3h = 2r and that the solid is on the point of toppling when tan i = 43 . (b) Find the value of k. [3] … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

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

This question in 9231/31 May/June 2021

4 kh h r A uniform solid circular cone has vertical height kh and radius r. A uniform solid cylinder has height h and radius r. The base of the cone is joined to one of the circular faces of the cylinder so that the axes of symmetry of the two solids coincide (see diagram, which shows a cross-section). The cone and the cylinder are made of the same material. (a) Show that the distance of the centre of mass of the combined solid from the base of the cylinder h ( k 2 + 4k + 6) is . [4] 4 ( 3 + k) … … … … … … … … … … … … … … … … The solid is placed on a plane that is inclined to the horizontal at an angle i. The base of the cylinder is in contact with the plane. The plane is sufficiently rough to prevent sliding. It is given that 3h = 2r and that the solid is on the point of toppling when tan i = 43 . (b) Find the value of k. [3] … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

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

This question in 9231/32 May/June 2021

1 D A C 16a 3a O 6a B A uniform lamina ABCD consists of two isosceles triangles ABD and BCD. The diagonals of ABCD meet at the point O. The length of AO is 3a, the length of OC is 6a and the length of BD is 16a (see diagram). Find the distance of the centre of mass of the lamina from DB. [3] … … … … … … … … … … … … … … …

3 marks

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

This question in 9231/33 May/June 2021

4 D 3a C h F 3a h A E B A uniform lamina AECF is formed by removing two identical triangles BCE and CDF from a square lamina ABCD. The square has side 3a and EB = DF = h (see diagram). (a) Find the distance of the centre of mass of the lamina AECF from AD and from AB, giving your answers in terms of a and h. [5] … … … … … … … … … … … … … … … … … … The lamina AECF is placed vertically on its edge AE on a horizontal plane. (b) Find, in terms of a, the set of values of h for which the lamina remains in equilibrium. [3] … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

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

This question in 9231/31 Oct/Nov 2021

4 A 1 2 a B ka a An object is formed by removing a solid cylinder, of height ka and radius 12 a , from a uniform solid hemisphere of radius a. The axes of symmetry of the hemisphere and the cylinder coincide and one circular face of the cylinder coincides with the plane face of the hemisphere. AB is a diameter of the circular face of the hemisphere (see diagram). 3a ( 2 - k 2 ) (a) Show that the distance of the centre of mass of the object from AB is . [4] 2 ( 8 - 3k) … … … … … … … … … … … … … … … … … When the object is freely suspended from the point A, the line AB makes an angle i with the downward vertical, where tan i = 187 . (b) Find the possible values of k. [3] … … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

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

This question in 9231/32 Oct/Nov 2021

4 D 3a C h F 3a h A E B A uniform lamina AECF is formed by removing two identical triangles BCE and CDF from a square lamina ABCD. The square has side 3a and EB = DF = h (see diagram). (a) Find the distance of the centre of mass of the lamina AECF from AD and from AB, giving your answers in terms of a and h. [5] … … … … … … … … … … … … … … … … … … The lamina AECF is placed vertically on its edge AE on a horizontal plane. (b) Find, in terms of a, the set of values of h for which the lamina remains in equilibrium. [3] … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

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

This question in 9231/33 Oct/Nov 2021

1 A i 7.5 N A particle of weight 10 N is attached to one end of a light elastic string. The other end of the string is attached to a fixed point A on a horizontal ceiling. A horizontal force of 7.5 N acts on the particle. In the equilibrium position, the string makes an angle i with the ceiling (see diagram). The string has natural length 0.8 m and modulus of elasticity 50 N. (a) Find the tension in the string. [2] … … … … (b) Find the vertical distance between the particle and the ceiling. [3] … … … … … … … … … … … … … …

5 marks

Mark scheme: 1(a) B1   1 2 2 2 7.5 10 T   = 12.5 N B1 2 1(b) Hooke’s law: 50 0.8 x T  , 0.2 x  B1   10 0.8 sin 1 12.5 x    M1 4 Vertical distance 0.8 5   A1 3

This question in 9231/31 May/June 2022

4 2a h a A B An object is composed of a hemispherical shell of radius 2a attached to a closed hollow circular cylinder of height h and base radius a. The hemispherical shell and the hollow cylinder are made of the same uniform material. The axes of symmetry of the shell and the cylinder coincide. AB is a diameter of the lower end of the cylinder (see diagram). (a) Find, in terms of a and h, an expression for the distance of the centre of mass of the object from AB. [4] … … … … … … … … … … … … … The object is placed on a rough plane which is inclined to the horizontal at an angle i, where tan i = 23 . The object is in equilibrium with AB in contact with the plane and lying along a line of greatest slope of the plane. (b) Find the set of possible values of h, in terms of a. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

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

This question in 9231/31 May/June 2022

1 A i 7.5 N A particle of weight 10 N is attached to one end of a light elastic string. The other end of the string is attached to a fixed point A on a horizontal ceiling. A horizontal force of 7.5 N acts on the particle. In the equilibrium position, the string makes an angle i with the ceiling (see diagram). The string has natural length 0.8 m and modulus of elasticity 50 N. (a) Find the tension in the string. [2] … … … … (b) Find the vertical distance between the particle and the ceiling. [3] … … … … … … … … … … … … … …

5 marks

Mark scheme: 1(a) B1   1 2 2 2 7.5 10 T   = 12.5 N B1 2 1(b) Hooke’s law: 50 0.8 x T  , 0.2 x  B1   10 0.8 sin 1 12.5 x    M1 4 Vertical distance 0.8 5   A1 3

This question in 9231/32 May/June 2022

4 2a h a A B An object is composed of a hemispherical shell of radius 2a attached to a closed hollow circular cylinder of height h and base radius a. The hemispherical shell and the hollow cylinder are made of the same uniform material. The axes of symmetry of the shell and the cylinder coincide. AB is a diameter of the lower end of the cylinder (see diagram). (a) Find, in terms of a and h, an expression for the distance of the centre of mass of the object from AB. [4] … … … … … … … … … … … … … The object is placed on a rough plane which is inclined to the horizontal at an angle i, where tan i = 23 . The object is in equilibrium with AB in contact with the plane and lying along a line of greatest slope of the plane. (b) Find the set of possible values of h, in terms of a. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

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

This question in 9231/32 May/June 2022

1 A uniform lamina OABC is a trapezium whose vertices can be represented by coordinates in the x-y plane. The coordinates of the vertices are O (0, 0), A (15, 0), B (9, 4) and C (3, 4). Find the x-coordinate of the centre of mass of the lamina. [4] … … … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

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

This question in 9231/33 May/June 2022

7 B D A i a E a C A uniform cylinder with a rough surface and of radius a is fixed with its axis horizontal. Two identical uniform rods AB and BC, each of weight W and length 2a, are rigidly joined at B with AB perpendicular to BC. The rods rest on the cylinder in a vertical plane perpendicular to the axis of the cylinder with AB at an angle i to the horizontal. D and E are the midpoints of AB and BC respectively and also the points of contact of the rods with the cylinder (see diagram). The rods are about to slip in a clockwise direction. The coefficient of friction between each rod and the cylinder is n. The normal reaction between AB and the cylinder is R and the normal reaction between BC and the cylinder is N. (a) Find the ratio R : N in terms of n. [6] … … … … … … … … … … … … … … … … … … … … … (b) Given that n = 13 , find the value of tani. [3] … … … … … … … … … … … … … … … … Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 7(a) , AB BC F R F N Moments about B, (sin cos Na Ra Wa      ) Moments about centre, (cos sin AB BC F a F a Wa      ) Moments about D,   cos sin BC F a Na Wa      Moments about E,   cos sin AB Ra F a Wa      B1 One moments equation about any point involving all relevant forces, resolved if necessary (AEF). Parallel to AB, sin sin AB N F W W      Perpendicular to AB, cos cos BC F R W W      B1 Two resolutions: all relevant terms, different frictional forces [Vertical: cos cos sin sin BC AB R F N F W W          Horizontal: sin cos sin cos BC AB F F R N        ] Alternative approach using two moments equations can earn the B1B1       1 1 1 2 N R N R        M1 Combine appropriate equations.     1 1 1 1 1 1 2 2 N R                    1 1 1 1 2 2 2 2 N R                  M1 Collect terms to obtain ratio/fraction in terms of only (CWO), any equivalent simplified form. : 1 :1 R N     A1 6 Question Answer Marks Guidance 7(b) Divide resolution equations: tan N R N R       M1 Must include  terms. Use 1 3 2 : tan 7 3 N R N N    M1 FT their answer to part (a). 1 tan 7  A1 3

This question in 9231/33 May/June 2022

3 B C A i A smooth cylinder is fixed to a rough horizontal surface with its axis of symmetry horizontal. A uniform rod AB, of length 4a and weight W, rests against the surface of the cylinder. The end A of the rod is in contact with the horizontal surface. The vertical plane containing the rod AB is perpendicular to the axis of the cylinder. The point of contact between the rod and the cylinder is C, where AC = 3a . The angle between the rod and the horizontal surface is i where tan i = 34 (see diagram). The coefficient of friction between the rod and the horizontal surface is 6.7 A particle of weight kW is attached to the rod at B. The rod is about to slip. The normal reaction between the rod and the cylinder is N. (a) Show that N = 8 W ( 1 + 2 k) . [2] 15 … … … … … … … … … … … … … … … (b) Find the value of k. [5] … … … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 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  2a cos+ kW  4a cos 4 A1 At least one intermediate line of working. 3 N = ( 2 + 4k ) W  5 8 N = W (1 + 2 k ) AG 15 2 3(b)  N cos+ R = W + kW B1 Resolve (to include R) for rod. 6 B1 Both. → F = N sin 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

This question in 9231/31 Oct/Nov 2022

2 A 9a B D C x 6a A uniform lamina is in the form of a triangle ABC in which angle B is a right angle, AB = 9a and BC = 6a . The point D is on BC such that BD = x (see diagram). The region ABD is removed from the lamina. The resulting shape ADC is placed with the edge DC on a horizontal surface and the plane ADC is vertical. Find the set of values of x, in terms of a, for which the shape is in equilibrium. [6] … … … … … … … … … … … … … … … …

6 marks

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

This question in 9231/32 Oct/Nov 2022

3 B C A i A smooth cylinder is fixed to a rough horizontal surface with its axis of symmetry horizontal. A uniform rod AB, of length 4a and weight W, rests against the surface of the cylinder. The end A of the rod is in contact with the horizontal surface. The vertical plane containing the rod AB is perpendicular to the axis of the cylinder. The point of contact between the rod and the cylinder is C, where AC = 3a . The angle between the rod and the horizontal surface is i where tan i = 34 (see diagram). The coefficient of friction between the rod and the horizontal surface is 6.7 A particle of weight kW is attached to the rod at B. The rod is about to slip. The normal reaction between the rod and the cylinder is N. (a) Show that N = 8 W ( 1 + 2 k) . [2] 15 … … … … … … … … … … … … … … … (b) Find the value of k. [5] … … … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 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  2a cos+ kW  4a cos 4 A1 At least one intermediate line of working. 3 N = ( 2 + 4k ) W  5 8 N = W (1 + 2 k ) AG 15 2 3(b)  N cos+ R = W + kW B1 Resolve (to include R) for rod. 6 B1 Both. → F = N sin 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

This question in 9231/33 Oct/Nov 2022

4 y a d 2a O C x An object is formed from a solid hemisphere, of radius 2a, and a solid cylinder, of radius a and height d. The hemisphere and the cylinder are made of the same material. The cylinder is attached to the plane face of the hemisphere. The line OC forms a diameter of the base of the cylinder, where C is the centre of the plane face of the hemisphere and O is common to both circumferences (see diagram). Relative to axes through O, parallel and perpendicular to OC as shown, the centre of mass of the object is ( x , y ). 32 a 2 + 3 ad (a) Show that x = and find an expression, in terms of a and d, for .y [5] 16a + 3 d … … … … … … … … … … … … … The object is placed on a rough plane which is inclined to the horizontal at an angle i where sin i = 16 . The object is in equilibrium with CO horizontal, where CO lies in a vertical plane through a line of greatest slope. (b) Find d in terms of a. [3] … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 4(a) [Mass is proportional to volume] Volume Distance of centre of mass from vertical axis Distance of centre of mass from OC Hemisphere   3 2 2 3 a  2a 3 2 8 a   Cylinder 2 a d  a 1 2 d Object   3 2 2 2 3 a a d    x y   3 2 3 2 2 16 2 2 3 3 a a d x a a a d a                M1 A1 Moments equation, dimensionally correct, correct number of terms. Allow sign errors. Simplify to x 2 32 3 16 3 a ad a d    A1 AG. At least one line of intermediate working.   3 2 3 2 2 16 3 1 2 π π 2 π 3 3 8 2 a a d y a a a d d                    M1 Moments equation, dimensionally correct, correct number of terms. Allow sign errors. y     2 2 3 8 2 16 3 d a a d    A1 AEF 5 Question Answer Marks Guidance 4(b) 2 sin 2 a x a    B1 2 1 32 3 2 2 6 16 3 a ad a a a d          5 16 3 32 3 3 a d a d    M1 Remove fractions 8 3 d a  A1 3

This question in 9231/31 May/June 2023

4 y a d 2a O C x An object is formed from a solid hemisphere, of radius 2a, and a solid cylinder, of radius a and height d. The hemisphere and the cylinder are made of the same material. The cylinder is attached to the plane face of the hemisphere. The line OC forms a diameter of the base of the cylinder, where C is the centre of the plane face of the hemisphere and O is common to both circumferences (see diagram). Relative to axes through O, parallel and perpendicular to OC as shown, the centre of mass of the object is ( x , y ). 32 a 2 + 3 ad (a) Show that x = and find an expression, in terms of a and d, for .y [5] 16a + 3 d … … … … … … … … … … … … … The object is placed on a rough plane which is inclined to the horizontal at an angle i where sin i = 16 . The object is in equilibrium with CO horizontal, where CO lies in a vertical plane through a line of greatest slope. (b) Find d in terms of a. [3] … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 4(a) [Mass is proportional to volume] Volume Distance of centre of mass from vertical axis Distance of centre of mass from OC Hemisphere   3 2 2 3 a  2a 3 2 8 a   Cylinder 2 a d  a 1 2 d Object   3 2 2 2 3 a a d    x y   3 2 3 2 2 16 2 2 3 3 a a d x a a a d a                M1 A1 Moments equation, dimensionally correct, correct number of terms. Allow sign errors. Simplify to x 2 32 3 16 3 a ad a d    A1 AG. At least one line of intermediate working.   3 2 3 2 2 16 3 1 2 π π 2 π 3 3 8 2 a a d y a a a d d                    M1 Moments equation, dimensionally correct, correct number of terms. Allow sign errors. y     2 2 3 8 2 16 3 d a a d    A1 AEF 5 Question Answer Marks Guidance 4(b) 2 sin 2 a x a    B1 2 1 32 3 2 2 6 16 3 a ad a a a d          5 16 3 32 3 3 a d a d    M1 Remove fractions 8 3 d a  A1 3

This question in 9231/32 May/June 2023

3 A 3a D 5a x C B E 6a A uniform lamina is in the form of a triangle ABC, with AC = 8a, BC = 6a and angle ACB = 90°. The point D on AC is such that AD = 3a. The point E on CB is such that CE = x (see diagram). The triangle CDE is removed from the lamina. (a) Find, in terms of a and x, the distance of the centre of mass of the remaining object ADEB from AC. [4] … … … … … … … … … … … … … … … … The object ADEB is on the point of toppling about the point E when the object is in the vertical plane with its edge EB on a smooth horizontal surface. (b) Find x in terms of a. [3] … … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 3(a) [Mass is proportional to area] Area Centre of mass from AC ABC 1.6 .8 2 a a ( 2 24 )  a 2a DEC 1 .5 2 x a 1 3 x ADEB 2 5 24 2  a xa x B1 All correct for ABC and DEC. Moments [about AC] 2 2 5 1 5 24 24 2 2 3 2            x a xa a a x ax M1 All moment terms present, dimensionally correct, allow sign error. A1 All correct moments about AC.   2 2 288 5 3 48 5    a x x a x A1 AEF 4 Question Answer Marks Guidance 3(b) On the point of toppling about E:   2 2 288 5 , 3 48 5     a x x x x a x B1 FT FT their expression for x from part (a). Rearrange to 3-term quadratic: 2 2 10 144 288 0    x ax a M1 Allow 3-term inequality.    2 5 12 12 0,    x a x a 12 5  x a A1 Single correct answer, no inequality, CWO. 3

This question in 9231/33 May/June 2023

3 P A i 2a E B D C A uniform square lamina of side 2a and weight W is suspended from a light inextensible string attached to the midpoint E of the side AB. The other end of the string is attached to a fixed point P on a rough vertical wall. The vertex B of the lamina is in contact with the wall. The string EP is perpendicular to the side AB and makes an angle i with the wall (see diagram). The string and the lamina are in a vertical plane perpendicular to the wall. The coefficient of friction between the wall and the lamina is 1.2 Given that the vertex B is about to slip up the wall, find the value of tani. [8] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

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

This question in 9231/31 Oct/Nov 2023

3 B D E A C 2a A uniform lamina is in the form of an isosceles triangle ABC in which AC = 2a and angle ABC = 90° . The point D on AB is such that the ratio DB : AB = 1: k. The point E on CB is such that DE is parallel to AC. The triangle DBE is removed from the lamina (see diagram). (a) Find, in terms of k, the distance of the centre of mass of the remaining lamina ADEC from the midpoint of AC. [4] … … … … … … … … … … … … … … … … … … … When the lamina ADEC is freely suspended from the vertex A, the edge AC makes an angle i with the downward vertical, where tan i = 185 . (b) Find the value of k. [3] … … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 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  2 a  a 2 3a 1  a  2 2  a  BDE 2  a −   2  k  3  k  1  a  2 2 ADEC −  2   + a x 2  k   1  a  2 2  1 2  2  a    a  2 M1A1 Moments equation, dimensionally correct, correct number x  −  2   + a  = a  a −  a −       of terms.   2  k  3  3  k    k    2 A1 3 a k + k − 2 a k − 3k + 2 ( ) ( ) x = Allow unsimplified single fraction . k 2 − 1 3k ( k + 1) 3k ( ) 4 3(b) x  5  B1 With their x from part (a). tan= =   a  18  So 18 k 2 + k − 2 = 5 3k 2 + 3k M1 Obtain a polynomial in k only, e.g. k 3 − 13k + 12 = 0 , ( ) ( ) 2 may be implied. k + k − 12 = 0 ( k + 4 )( k − 3 ) = 0, A1 CWO k = 3 only 3

This question in 9231/32 Oct/Nov 2023

7 5a A B 3a 4a P F A particle P of mass m is attached to one end of a light rod of length 3a. The other end of the rod is able to pivot smoothly about the fixed point A. The particle is also attached to one end of a light spring of natural length a and modulus of elasticity kmg. The other end of the spring is attached to a fixed point B. The points A and B are in a horizontal line, a distance 5a apart, and these two points and the rod are in a vertical plane. Initially, P is held in equilibrium by a vertical force F with the stretched length of the spring equal to 4a (see diagram). The particle is released from rest in this position and has a speed of 65 2ag when the rod becomes horizontal. (a) Find the value of k. [5] … … … … … … … … … … … … … … … (b) Find F in terms of m and g. [2] … … … … … … … … … … … … … (c) Find, in terms of m and g, the tension in the rod immediately before it is released. [2] … … … … … … … … … … … … … Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 7(a) 2 B1 1  6  Gain in KE = m 2ag and Gain in GPE = mg  3a sin   2  5  1 B1 kmg 2 2 2 Loss in EPE = ( 3a ) − a ( ) a 1 M1 KE, GPE and at least one EPE present, allow sign errors, 2 kmg 2 2 dimensionally correct. 1  6 2 ( 3a ) − a Energy equation: m  2 ag + mg  3a sin= ( ) 2  5  a A1 All correct. 36 12 A1 mg + mg = 4kmg 25 5 24 k = 25 5 7(b) 3a 72 M1 Hooke’s law. In lower position, tension T in spring = kmg  = 3kmg = mg a 25 Perpendicular to rod, T = ( F + mg )cos 72 A1 So, ( F + mg ) cos= mg , 25 19 F = mg 5 Alternative method for question 7(b) 3a 72 M1 In lower position, tension T in spring = kmg  = 3kmg = mg Hooke’s law. a 25 Eliminate T. T  cos= Tsin T  sin+ Tcos= F + mg 19 A1 F = mg 5 2 7(c) Let tension in rod = T ' M1 Parallel to rod, T  = ( F + mg )sin 96 A1 T  = mg 25 Alternative method for question 7(c) T  cos= Tsin M1 At least one equation seen with their T and/or F. T  sin+ Tcos= F + mg 96 A1 T  = mg 25 2

This question in 9231/32 Oct/Nov 2023

3 P A i 2a E B D C A uniform square lamina of side 2a and weight W is suspended from a light inextensible string attached to the midpoint E of the side AB. The other end of the string is attached to a fixed point P on a rough vertical wall. The vertex B of the lamina is in contact with the wall. The string EP is perpendicular to the side AB and makes an angle i with the wall (see diagram). The string and the lamina are in a vertical plane perpendicular to the wall. The coefficient of friction between the wall and the lamina is 1.2 Given that the vertex B is about to slip up the wall, find the value of tani. [8] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

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

This question in 9231/33 Oct/Nov 2023

4 P Q a O a A ring of weight W, with radius a and centre O, is at rest on a rough surface that is inclined to the horizontal at an angle a where tan a = 12 . The plane of the ring is perpendicular to the inclined surface and parallel to a line of greatest slope of the surface. The point P on the circumference of the ring is such that OP is parallel to the surface. A light inextensible string is attached to P and to the point Q, which is on the surface, such that PQ is horizontal (see diagram). The points O, P and Q are in the same vertical plane. The system is in limiting equilibrium and the coefficient of friction between the ring and the surface is n. (a) Find, in terms of W, the tension in the string PQ. [4] … … … … … … … … … … … … … … … … … (b) Find the value of n. [3] … … … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 4(a) Frictional force F and normal reaction R at point of contact of ring with plane. Resolve parallel to plane: cos sin     F T W M1 Only allow cos/sin errors for T and W components, sign errors. Accept equations for vertical and horizontal (both needed). Moments about O: sin  Fa Ta B1 Combine and substitute for : M1 Expression for T in terms of W . 1 3 T W  A1 CAO Alternative solution for question 4(a) Moments about point where ring touches plane: sin cos sin      Ta Ta Wa 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 . 1 3 T W  A1 CAO 4 4(b) Resolve perpendicular to plane: sin cos     R T W 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), cos sin F T W    or sin  F T . 1 7   A1 3

This question in 9231/31 May/June 2024

4 P Q a O a A ring of weight W, with radius a and centre O, is at rest on a rough surface that is inclined to the horizontal at an angle a where tan a = 12 . The plane of the ring is perpendicular to the inclined surface and parallel to a line of greatest slope of the surface. The point P on the circumference of the ring is such that OP is parallel to the surface. A light inextensible string is attached to P and to the point Q, which is on the surface, such that PQ is horizontal (see diagram). The points O, P and Q are in the same vertical plane. The system is in limiting equilibrium and the coefficient of friction between the ring and the surface is n. (a) Find, in terms of W, the tension in the string PQ. [4] … … … … … … … … … … … … … … … … … (b) Find the value of n. [3] … … … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 4(a) Frictional force F and normal reaction R at point of contact of ring with plane. Resolve parallel to plane: cos sin     F T W M1 Only allow cos/sin errors for T and W components, sign errors. Accept equations for vertical and horizontal (both needed). Moments about O: sin  Fa Ta B1 Combine and substitute for : M1 Expression for T in terms of W . 1 3 T W  A1 CAO Alternative solution for question 4(a) Moments about point where ring touches plane: sin cos sin      Ta Ta Wa 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 . 1 3 T W  A1 CAO 4 4(b) Resolve perpendicular to plane: sin cos     R T W 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), cos sin F T W    or sin  F T . 1 7   A1 3

This question in 9231/32 May/June 2024

4 A 2 a B 3 kh h a An object is formed by removing a cylinder of radius 2 a and height kh ( k 1 1 ) from a uniform solid 3 cylinder of radius a and height h. The vertical axes of symmetry of the two cylinders coincide. The upper faces of the two cylinders are in the same plane as each other. The points A and B are the opposite ends of a diameter of the upper face of the object (see diagram). (a) Find, in terms of h and k, the distance of the centre of mass of the object from AB. [4] … … … … … … … … … … … … … … … When the object is suspended from A, the angle between AB and the vertical is i, where tan i = 3 . 2 (b) Given that h = 8 a , find the possible values of k. [3] 3 … … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

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

This question in 9231/31 Oct/Nov 2024

4 B D 3a E C ka i 3a A The end A of a uniform rod AB of length 6a and weight W is in contact with a rough vertical wall. One end of a light inextensible string of length 3a is attached to the midpoint C of the rod. The other end of the string is attached to a point D on the wall, vertically above A. The rod is in equilibrium when the angle between the rod and the wall is i, where tan i = 3 . A particle of weight W is attached to the point 2 E on the rod, where the distance AE is equal to ka (3 1 k 1 6) (see diagram). The rod and the string are in a vertical plane perpendicular to the wall. The coefficient of friction between the rod and the wall is 1. The rod is about to slip down the wall. 3 (a) Find the value of k. [5] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … (b) Find, in terms of W, the magnitude of the frictional force between the rod and the wall. [2] … … … … … … … … …

7 marks

Mark scheme: 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.  T cos+ F = 2W → T sin= R B1 Moments about A: M1 All relevant terms included, dimensionally correct, T cos 3a sin+ T sin 3a cos= W  3a sin+ W  ka sin forces must be resolved if appropriate. Allow sin/cos mix, allow sign errors. LHS: any equivalent expression, for example 3Ta sin 2, 3Tasin (180 − 2) . OR All relevant terms included, dimensionally correct, Moments about C: forces must be resolved if appropriate. R  3a cos = F  3a sin+ W  ( ka − 3a ) sin Allow sin/cos mix, allow sign errors. T M1 1 [6 aT cos= ( 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 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

This question in 9231/32 Oct/Nov 2024

4 A 2 a B 3 kh h a An object is formed by removing a cylinder of radius 2 a and height kh ( k 1 1 ) from a uniform solid 3 cylinder of radius a and height h. The vertical axes of symmetry of the two cylinders coincide. The upper faces of the two cylinders are in the same plane as each other. The points A and B are the opposite ends of a diameter of the upper face of the object (see diagram). (a) Find, in terms of h and k, the distance of the centre of mass of the object from AB. [4] … … … … … … … … … … … … … … … When the object is suspended from A, the angle between AB and the vertical is i, where tan i = 3 . 2 (b) Given that h = 8 a , find the possible values of k. [3] 3 … … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

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

This question in 9231/33 Oct/Nov 2024

2 A particle P of mass m is attached to one end of a light elastic string of natural length a and modulus of elasticity mg. The other end of the string is attached to a fixed point O on a rough plane inclined at an angle of 30° to the horizontal. The particle P is held at rest at point O before being released. The frictional force acting on P as it slides down the plane is 11 mg . 30 (a) Find, in terms of a, the distance that P moves down the plane before coming to rest. [5] … … … … … … … … … … … … … … … … … … … … … … … … … (b) It is given that P remains at rest in this new position. Find, in terms of m and g, the magnitude of the frictional force in this position. [3] … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 2(a) mg 2 11 M1 Energy equation with all three terms. ( x − a ) = mgx sin30 − 30 mgx Dimensionally correct. 2 a B1 Two terms correct (need not be in equation). A1 Correct energy equation. 15 x 2 − 34ax + 15a 2 = 0 M1 Simplify to three-term homogeneous quadratic equation and attempt to solve. (3 x − 5a )(5 x − 3a ) = 0 x = 53 a A1 CAO, must reject x = 53 a if seen. 5 2(b) T = 23 mg B1 F = T − mg sin30   = 23 mg − 12 mg  M1 Equation for F. Allow sin/cos mix. Allow sign errors. 1 6 mg A1 3

This question in 9231/33 May/June 2025

4 B C a E 2a F 2a A D a An object consists of a uniform lamina with a particle attached. The uniform lamina ABCEFD of mass m is formed from a rectangle ABCD and an isosceles triangle CEF, where F is the midpoint of CD. The rectangle has sides AB = 2a and AD = a . The triangle CEF has base a and height 2a. The particle of mass km is attached to the lamina at E. The object rests in a vertical plane with its edge AD on horizontal ground (see diagram). Given that the object is on the point of toppling in its vertical plane about the vertex D, find the value of k. [4] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: 4 B1 Allow one error in first two columns. Areas 2 a ) 2 a 2a 2 12 a ( 2 + 12 a ( 2 a ) Correct distances about any point. SOI, may be seen in moments equation. Distances from D − 12 a 23 a x 1 1 2 *M1 A moments equation with 3 terms, must be 2 a 2 3a 2 x −2 a 2 ( 2 a ) + 2 ( )( 3 a ) = ( ) 1 dimensionally correct. ( x =  9 a or x = 89 a ) 1 DM1 Moments equation about toppling point D. m ( 9 a ) = km ( 2a ) 2 terms only, OE. Must include k. k = 181 A1 Alternative method for question 4 Ratio of masses 2a 2 12 a ( 2 a ) k ( 2 a 2 + 12 a ( 2 a ) ) B1 CorrectMust simplyratiostoincluding2:1:3k .k. SOI, may be seen in moments equation. 1 2 2 ( 2 a ) = 1( 3 a ) + 3k ( 2a ) *M1 A moments equation about D with 3 terms, must be dimensionally correct. Must include k. DM1 Using correct distances. k = 181 A1 4

This question in 9231/33 May/June 2025

5 D h E C 6a A F h B ABCD is a uniform square lamina of side 6a. Points E and F are on DC and AB respectively and are such that DE = FB = h . The quadrilateral BCEF is removed from the square lamina (see diagram). (a) Show that the distance of the centre of mass of the resulting lamina AFED from AD is h 2 - 6 ah + 36a 2 and find a corresponding expression for the distance of the centre of mass from AB. 18 a [5] … … … … … … … … … … … … … … … … … When the lamina AFED is suspended from the point D, the edge DA makes an angle i with the downward vertical, where tan i = 7 . 15 (b) Find, in terms of a, the two possible values of h. [3] … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 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 1 1 1 18a 2 − 6 ah 2 h 3a Moments about AD: 18a 2 x = 6 ah ( 3 h + 2 a ) Rectangle 6ah 2 h ) + ( )( Triangle 18a 2 − 6ah 13 h + 2 a 2a AFED 18a 2 x y A1 Fully correct equation, does not need to be simplified. h 2 − 6 ah + 36 a 2 A1 AG, shown convincingly. x = 18a 2 2 M1 All terms present. All terms must be dimensionally correct. 18a − 6 ah 2 a ) Moments about AB: 18a y = 6 ah ( 3a ) + ( )( y = 2 a + 13 h A1 Must expand brackets and collect like terms. 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 1 18a 2 − 3ah 2 a ) AED 3ah 13 h 4a Moments about AD: 18a 2 x = 3ah ( 3 h ) + ( )( AEF 18a 2 − 3ah 2a 2a AFED 18a 2 x y A1 Fully correct equation, does not need to be simplified. h 2 − 6 ah + 36 a 2 A1 AG, shown convincingly. x = 18a 2 2 M1 All terms present. All terms must be dimensionally correct. 18a − 3ah 2 a ) Moments about AB: 18a y = 3ah ( 4 a ) + ( )( y = 2 a + 13 h A1 Must expand brackets and collect like terms. 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 36a 2 − 6ah 3a − 12 h 3a Moments about AD: 18a 2 − 6ah 4a − 13 h 4a 18a 2 x = 36 a 2 − 6 ah 18 a 2 − 6 ah 4 a − 13 h ) EFH ( )( 3a − 12 h ) − ( )( AFED 18a 2 x y A1 Fully correct equation, does not need to be simplified. h 2 − 6 ah + 36 a 2 A1 AG, shown convincingly. x = 18a Moments about AB: M1 All terms present. All terms must be dimensionally correct. 18a 2 y = 36 a 2 − 6 ah 18a 2 − 6 ah 4 a ) ( )( 3a ) − ( )( y = 2 a + 13 h A1 Must expand brackets and collect like terms. 5 5(b) x B1 Correct expression for tan. tan = 6 a − y h 2 − 6 ah + 36 a 2 M1 FT Equate their expression for tan (must involve x and y ) to 157 , 7 18a = leading to 5h2 − 16ah + 12a 2 = 0 . substitute the correct x and their y and simplify to a 3-term 1 6 a − ( 2 a + 3 h ) 15 2 2  h − 6 ah + 36 a  1  = 7 ( 6 a − 2 a − 3 h ) .  18 a ( h − 2a )( 5h − 6a ) = 0  quadratic. May see 15  

This question in 9231/34 May/June 2025

5 y D C B h 2a 45° O ka A x A uniform lamina OABCD consists of a rectangle OACD and a triangle ABC. The length of OA is ka, the length of OD is 2a, the height of triangle ABC is h and angle CAB is 45° (see diagram). Relative to axes through O, parallel and perpendicular to OA as shown, the centre of mass of triangle ABC is ( x, y ) . (a) Show that x is 1 ( 3ka + h) , and find an expression for y. [3] 3 … … … … … … … … … … … … … … … … The lamina OABCD is placed vertically on its edge OA on a horizontal plane. (b) Find, in terms of a and k, the set of values of h for which the lamina is in equilibrium. [4] … … … … … … … … … … … … … … … … … 3 It is now given that k = and that the lamina is on the point of toppling. 3 (c) Find, in terms of a, the coordinates of the centre of mass of the triangle ABC. [2] … … … … … … …

9 marks

Mark scheme: 5(a) x = ka + 13 h M1 May be seen in a table. x = 13 ( 3ka + h ) A1 AG, shown convincingly. 1 y = a + 13 ( h − a ) = 3 ( 2 a + h ) B1 May be seen in a table. Alternative method for question 5(a) Coordinates A ( ka , 0 ) , B ( ka + h , h ) , C ( ka , 2 a ) M1 Complete method to find x . 1 x = 13 ( ka + ka + h + ka ) = 3 ( 3ka + h ) A1 AG, shown convincingly. 1 y = 13 ( 0 + h + 2 a ) = 3 ( h + 2 a ) B1 3 5(b) Taking moments about the x-axis: M1 Moments equation with 3 terms. 1 x 2ka 2 + ah = 2ka 2  12 ka + ah ( ka + 3 h ) Area From OD ( ) OACD 2ka 2 12 ka ABC ah ka + 13 h Shape OABCD 2 x 2ka + ah 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. a 2 k 2 + ahk + 13 h 2 Allow strict inequality. ≤ ka M1 for simplifying and attempting to solve an inequality. 2 ak + h May be implied by correct final answer. ( 0 ≤) h ≤ 3 ka A1 CAO 4 5(c) B1 Point of toppling means h = 3 ka  = a  . 3 + 1 x = 13 a ( ) [ h = a , meaning triangle ABC is isosceles.] B1 y = a (by symmetry) 2

This question in 9231/31 Oct/Nov 2025

3 A uniform lamina OABCD is in the form of a rectangle, OBCD, joined along the edge OB to a quarter circle OAB. The length of DO is ka and the length of OB is a. The lamina rests in a vertical plane with its edge CB on a horizontal surface (see diagram). ka D O A a C B (a) Find, in terms of k, a and r, an expression for the distance of the centre of mass above the horizontal surface. [4] [You may use without proof the result for the centre of mass of a circular sector in the list of formulae (MF19).] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … The lamina is on the point of toppling about B. (b) Find the value of k. [2] … … … … … … … … …

6 marks

Mark scheme: 3(a) Vertical distance of COM of sector from OA: B1 SOI ( 2 a cos ( 14 π ) ) sin ( 14 π ) 4 a Allow unsimplified. = 1 3 ( 4 π ) 3π 2 1 1 2  4 a  2 1 2 M1 3-term dimensionally correct moments y ka ( 2 a ) + 4 π a  a −  = ( ka + 4 π a ) equation.  3π   6ka + 3π a − 4a = 3 ( 4k + π ) y  A1 FT Correct moments equation, allow unsimplified. FT their vertical distance of COM of sector from OA. 6 k + 3π − 4 A1 AEF, but do not allow fractions within y = a 3 ( 4 k + π ) numerator and/or denominator. 4 3(b) 2 1 1 2  4 a  M1 Dimensionally correct moments equation ka ( 2 ka ) = 4 π a   about OB using their horizontal distance of  3π  COM of sector from OB.   k 2 = 23  k = 23 = 0.816 A1 Alternative method for question 3(b) 2 M1 Use of x = ka with their x , which may be 6 k + 3kπ + 4 a ( ) On the point of toppling, x = ka : = ka seen in part 3(a). 3 ( 4 k + π )  6k 2 + 4 = 12k 2  k = 23 = 0.816 A1   2

This question in 9231/32 Oct/Nov 2025

5 y D C B h 2a 45° O ka A x A uniform lamina OABCD consists of a rectangle OACD and a triangle ABC. The length of OA is ka, the length of OD is 2a, the height of triangle ABC is h and angle CAB is 45° (see diagram). Relative to axes through O, parallel and perpendicular to OA as shown, the centre of mass of triangle ABC is ( x, y ) . (a) Show that x is 1 ( 3ka + h) , and find an expression for y. [3] 3 … … … … … … … … … … … … … … … … The lamina OABCD is placed vertically on its edge OA on a horizontal plane. (b) Find, in terms of a and k, the set of values of h for which the lamina is in equilibrium. [4] … … … … … … … … … … … … … … … … … 3 It is now given that k = and that the lamina is on the point of toppling. 3 (c) Find, in terms of a, the coordinates of the centre of mass of the triangle ABC. [2] … … … … … … …

9 marks

Mark scheme: 5(a) x = ka + 13 h M1 May be seen in a table. x = 13 ( 3ka + h ) A1 AG, shown convincingly. 1 y = a + 13 ( h − a ) = 3 ( 2 a + h ) B1 May be seen in a table. Alternative method for question 5(a) Coordinates A ( ka , 0 ) , B ( ka + h , h ) , C ( ka , 2 a ) M1 Complete method to find x . 1 x = 13 ( ka + ka + h + ka ) = 3 ( 3ka + h ) A1 AG, shown convincingly. 1 y = 13 ( 0 + h + 2 a ) = 3 ( h + 2 a ) B1 3 5(b) Taking moments about the x-axis: M1 Moments equation with 3 terms. 1 x 2ka 2 + ah = 2ka 2  12 ka + ah ( ka + 3 h ) Area From OD ( ) OACD 2ka 2 12 ka ABC ah ka + 13 h Shape OABCD 2 x 2ka + ah 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. a 2 k 2 + ahk + 13 h 2 Allow strict inequality. ≤ ka M1 for simplifying and attempting to solve an inequality. 2 ak + h May be implied by correct final answer. ( 0 ≤) h ≤ 3 ka A1 CAO 4 5(c) B1 Point of toppling means h = 3 ka  = a  . 3 + 1 x = 13 a ( ) [ h = a , meaning triangle ABC is isosceles.] B1 y = a (by symmetry) 2

This question in 9231/33 Oct/Nov 2025