Circular motion — Paper 3 · A Level Mathematics - Further 9231

2 R x i A B A light inextensible string of length a is threaded through a fixed smooth ring R. One end of the string is attached to a particle A of mass 3m. The other end of the string is attached to a particle B of mass m. The particle A hangs in equilibrium at a distance x vertically below the ring. The angle between AR and g BR is i (see diagram). The particle B moves in a horizontal circle with constant angular speed 2 . a Show that cos i = 13 and find x in terms of a. [5] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 2 For A: 3 = T mg For B: cosθ ↑ = T mg Equate: 3 cosθ = mg mg 1 cos 3 θ = A1 2 sinθ ω → = T mr with ( )sinθ = − r a x M1 Equate: ( ) 2 3 ω = − mg m a x A1 4 = a x A1 5

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7 A hollow cylinder of radius a is fixed with its axis horizontal. A particle P, of mass m, moves in part of a vertical circle of radius a and centre O on the smooth inner surface of the cylinder. The speed of P when it is at the lowest point A of its motion is 72 ga . The particle P loses contact with the surface of the cylinder when OP makes an angle i with the upward vertical through O. (a) Show that i = 60° . [5] … … … … … … … … … … … … … … … … … … … … … … (b) Show that in its subsequent motion P strikes the cylinder at the point A. [5] … … … … … … … … … … … … … … … … … … … … … … … … … … 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) ( ) 2 cosθ + = mv N mg a B1 ( ) 2 1 1 7 cos 2 2 2 θ − = − + ag mv m mg a a M1A1 Loses contact when N = 0, so combine and simplify M1 1 cos : 60 2 θ θ = = ° AG A1 5 Question Answer Marks 7(b) When P is vertically below O, its horizontal displacement is sin60 a , so time T = sin60 cos60 a v = 3 / a v 6 = a g M1 From (a), 2 1 2 = v ag A1 Vert: 2 3 1 . 2 2 = − v h T g T M1 3 3 3 2 2 − = −a a a A1 This corresponds to the point A A1 Alternative method for question 7(b) 2 4 3 = − x y x a M1A1 Coordinates of A: 1 3 3, 2 2 = = − x a y a B1 Substitute coordinates into 2 4 3 = − x y x a and show that these satisfy this equation M1A1 5

This question in 9231/31 May/June 2020

2 R x i A B A light inextensible string of length a is threaded through a fixed smooth ring R. One end of the string is attached to a particle A of mass 3m. The other end of the string is attached to a particle B of mass m. The particle A hangs in equilibrium at a distance x vertically below the ring. The angle between AR and g BR is i (see diagram). The particle B moves in a horizontal circle with constant angular speed 2 . a Show that cos i = 13 and find x in terms of a. [5] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 2 For A: 3 = T mg For B: cosθ ↑ = T mg Equate: 3 cosθ = mg mg 1 cos 3 θ = A1 2 sinθ ω → = T mr with ( )sinθ = − r a x M1 Equate: ( ) 2 3 ω = − mg m a x A1 4 = a x A1 5

This question in 9231/32 May/June 2020

7 A hollow cylinder of radius a is fixed with its axis horizontal. A particle P, of mass m, moves in part of a vertical circle of radius a and centre O on the smooth inner surface of the cylinder. The speed of P when it is at the lowest point A of its motion is 72 ga . The particle P loses contact with the surface of the cylinder when OP makes an angle i with the upward vertical through O. (a) Show that i = 60° . [5] … … … … … … … … … … … … … … … … … … … … … … (b) Show that in its subsequent motion P strikes the cylinder at the point A. [5] … … … … … … … … … … … … … … … … … … … … … … … … … … 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) ( ) 2 cosθ + = mv N mg a B1 ( ) 2 1 1 7 cos 2 2 2 θ − = − + ag mv m mg a a M1A1 Loses contact when N = 0, so combine and simplify M1 1 cos : 60 2 θ θ = = ° AG A1 5 Question Answer Marks 7(b) When P is vertically below O, its horizontal displacement is sin 60 a , so time T = sin60 cos60 a v = 3 / a v 6 = a g M1 From (a), 2 1 2 = v ag A1 Vert: 2 3 1 . 2 2 = − v h T g T M1 3 3 3 2 2 − = −a a a A1 This corresponds to the point A A1 Alternative method for question 7(b) 2 4 3 = − x y x a M1A1 Coordinates of A: 1 3 3, 2 2 = = − x a y a B1 Substitute coordinates into 2 4 3 = − x y x a and show that these satisfy this equation M1A1 5

This question in 9231/32 May/June 2020

1 A particle P of mass m is attached to one end of a light inextensible string of length a. The other end of the string is attached to a fixed point O on a smooth horizontal plane. The particle P moves in horizontal circles about O. The tension in the string is 4mg. Find, in terms of a and g, the time that P takes to make one complete revolution. [2] … … … … … … … … … … … … …

2 marks

Mark scheme: 1 2 4 ω = = T mg ma so 2 4 ω = g a B1 Time per revn = 2π ω = π a g B1 2

This question in 9231/33 May/June 2020

3 A particle Q of mass m is attached to a fixed point O by a light inextensible string of length a. The particle moves in complete vertical circles about O. The points A and B are on the path of Q with AB a diameter of the circle. OA makes an angle of 60° with the downward vertical through O and OB makes an angle of 60° with the upward vertical through O. The speed of Q when it is at A is 2 ag . Given that TA and TB are the tensions in the string at A and B respectively, find the ratio TA : TB . [6] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 3 4 cos60 . − = A ag T mg m a ( 9 2 = A T mg ) B1 2 cos60 . + = B v T mg m a B1 Energy: ( ) 2 1 4 .2 cos60 2 − = m ag v mg a 2 2 = v ag M1A1 3 2 = B mg T A1 Ratio is 3 : 1 A1 6

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2 O a P i 4 5 5ag A particle P is attached to one end of a light inextensible string of length a. The other end of the string is attached to a fixed point O. The particle P is held with the string taut and making an angle i with the downward vertical. The particle P is then projected with speed 45 5ag perpendicular to the string and just completes a vertical circle (see diagram). Find the value of cosi. [5] … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 2 At top, tension = 0, so 2 mv mg a = ( 2 ) v ag = B1 ( ) 2 2 1 1 1 cos 2 2 mv mu mga θ = − + M1 A1 Energy equation Substitute for u and v : ( ) 16 .5 2 1 cos 25 ag ag ag θ = − + M1 Eliminate 1 cos 10 θ = A1 5

This question in 9231/31 Oct/Nov 2020

3 One end of a light elastic string, of natural length a and modulus of elasticity 4mg, is attached to a fixed point O. The other end of the string is attached to a particle of mass m. The particle moves in g a horizontal circle with a constant angular speed with the string inclined at an angle i to the a downward vertical through O. The length of the string during this motion is ( k + 1) a . (a) Find the value of k. [4] … … … … … … … … … … … … (b) Find the value of cosi. [2] … … … … … … … … … … …

6 marks

Mark scheme: 3(a) 4 .ka T mg a = B1 Use Hooke’s law ( ) sin 1 sin . mrg g T m k a a a θ θ   = = +     M1 N2L horizontally. Must see T and k . ( )1 T mg k = + A1 Equate: 1 3 k = A1 4 3(b) cos T mg θ ↑ = M1 ( 4 ) 3 T mg = 3 cos 4 4 3 mg mg θ = = A1 2

This question in 9231/31 Oct/Nov 2020

1 1 6 ag A a a O A fixed smooth solid sphere has centre O and radius a. A particle of mass m is projected downwards a with the with speed 16 ag from the point A on the surface of the sphere, where OA makes an angle upward vertical through O (see diagram). The particle moves in part of a vertical circle on the surface of the sphere. It loses contact with the sphere at the point B, where OB makes an angle b with the upward vertical through O. Given that cos a = 23 , find the value of cosb. [5] … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 1 At B, 2 cos mv mg a β = : ( 2 cos ) v ag β = ( ) 2 2 1 1 cos cos 2 2 mv mu mga α β = + − M1A1 Energy equation with 4 terms and correct dimensions 2 3 ( 2 cos ) 2 ga v ag β = − Substitute for u, cosα and v: 2 cos 2 cos 6 3 ag ag ag β β   = + −     M1 Eliminate to find cosβ 1 cos 2 β = A1 5

This question in 9231/32 Oct/Nov 2020

4 A particle P of mass m is moving in a horizontal circle with angular speed ~ on the smooth inner surface of a hemispherical shell of radius r. The angle between the vertical and the normal reaction of the surface on P is i. g (a) Show that cos i = 2 . [3] ~ r … … … … … … … … … … … … … … … … … … … … … … … … … The plane of the circular motion is at a height x above the lowest point of the shell. When the angular speed is doubled, the plane of the motion is at a height 4x above the lowest point of the shell. (b) Find x in terms of r. [4] … … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 4(a) cos N mg θ ↑ = B1 2 sin sin N mr θ θω ← = B1 cos mg N θ = so 2 cos g r θ ω = B1 AG 3 4(b) cos r x r θ − = = 2 g r ω B1 Using trig of situation: must involve x In new situation: 2 4 4 g r x r r ω − = × M1 Using new situation with 4x and 2ω seen ( ) 4 4 r x r x − = − M1 Combining 1 5 x r = A1 4

This question in 9231/32 Oct/Nov 2020

2 O a P i 4 5 5ag A particle P is attached to one end of a light inextensible string of length a. The other end of the string is attached to a fixed point O. The particle P is held with the string taut and making an angle i with the downward vertical. The particle P is then projected with speed 45 5ag perpendicular to the string and just completes a vertical circle (see diagram). Find the value of cosi. [5] … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 2 At top, tension = 0, so 2 mv mg a = ( 2 ) v ag = B1 ( ) 2 2 1 1 1 cos 2 2 mv mu mga θ = − + M1 A1 Energy equation Substitute for u and v : ( ) 16 .5 2 1 cos 25 ag ag ag θ = − + M1 Eliminate 1 cos 10 θ = A1 5

This question in 9231/33 Oct/Nov 2020

3 One end of a light elastic string, of natural length a and modulus of elasticity 4mg, is attached to a fixed point O. The other end of the string is attached to a particle of mass m. The particle moves in g a horizontal circle with a constant angular speed with the string inclined at an angle i to the a downward vertical through O. The length of the string during this motion is ( k + 1) a . (a) Find the value of k. [4] … … … … … … … … … … … … (b) Find the value of cosi. [2] … … … … … … … … … … …

6 marks

Mark scheme: 3(a) 4 .ka T mg a = B1 Use Hooke’s law ( ) sin 1 sin . mrg g T m k a a a θ θ   = = +     M1 N2L horizontally. Must see T and k . ( )1 T mg k = + A1 Equate: 1 3 k = A1 4 3(b) cos T mg θ ↑ = M1 ( 4 ) 3 T mg = 3 cos 4 4 3 mg mg θ = = A1 2

This question in 9231/33 Oct/Nov 2020

2 A hollow hemispherical bowl of radius a has a smooth inner surface and is fixed with its axis vertical. A particle P of mass m moves in horizontal circles on the inner surface of the bowl, at a height x above the lowest point of the bowl. The speed of P is 83 ga . Find x in terms of a. [6] … … … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 2 cosθ ↑ = R 2 sinθ → = mv R r B1 sinθ = r a B1 ( ) ( ) 2 8cos 3 1 cos θ θ = − M1 Quadratic equation in cosθ . 1 cos 3 θ = A1 2 3 = x a A1 6

This question in 9231/31 May/June 2021

5 A particle P of mass m is attached to one end of a light inextensible string of length a. The other end of the string is attached to a fixed point O. The particle completes vertical circles with centre O. The points A and B are on the path of P, both on the same side of the vertical through O. OA makes an angle i with the downward vertical through O and OB makes an angle i with the upward vertical through O. The speed of P when it is at A is u and the speed of P when it is at B is ag. The tensions in the string at A and B are TA and TB respectively. It is given that TA = 7TB. Find the value of i and find an expression for u in terms of a and g. [8] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 5 2 cosθ − = A mu T mg a B1 cosθ + = B mag T mg a B1 TA = 7TB so 2 cos 7 cos θ θ   + = − +     mu mag mg mg a a ( ) 2 7 8cosθ = − u ag M1 Use given relationship and combine. Energy: ( ) 2 1 1 cos cos 2 2 θ θ − = + mu mag mg a a So ( ) 2 4cos 1 θ = + u ag M1 A1 Energy equation. Equate expressions for 2 u M1 1 cos , 60 2 θ θ = = ° A1 CAO 3 = u ga A1 CAO 8

This question in 9231/31 May/June 2021

2 A hollow hemispherical bowl of radius a has a smooth inner surface and is fixed with its axis vertical. A particle P of mass m moves in horizontal circles on the inner surface of the bowl, at a height x above the lowest point of the bowl. The speed of P is 83 ga . Find x in terms of a. [6] … … … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 2 cosθ ↑ = R 2 sinθ → = mv R r B1 sinθ = r a B1 ( ) ( ) 2 8cos 3 1 cos θ θ = − M1 Quadratic equation in cosθ . 1 cos 3 θ = A1 2 3 = x a A1 6

This question in 9231/32 May/June 2021

5 A particle P of mass m is attached to one end of a light inextensible string of length a. The other end of the string is attached to a fixed point O. The particle completes vertical circles with centre O. The points A and B are on the path of P, both on the same side of the vertical through O. OA makes an angle i with the downward vertical through O and OB makes an angle i with the upward vertical through O. The speed of P when it is at A is u and the speed of P when it is at B is ag. The tensions in the string at A and B are TA and TB respectively. It is given that TA = 7TB. Find the value of i and find an expression for u in terms of a and g. [8] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 5 2 cosθ − = A mu T mg a B1 cosθ + = B mag T mg a B1 TA = 7TB so 2 cos 7 cos θ θ   + = − +     mu mag mg mg a a ( ) 2 7 8cosθ = − u ag M1 Use given relationship and combine. Energy: ( ) 2 1 1 cos cos 2 2 θ θ − = + mu mag mg a a So ( ) 2 4cos 1 θ = + u ag M1 A1 Energy equation. Equate expressions for 2 u M1 1 cos , 60 2 θ θ = = ° A1 CAO 3 = u ga A1 CAO 8

This question in 9231/32 May/June 2021

3 R i B A Particles A and B, of masses 3m and m respectively, are connected by a light inextensible string of length a that passes through a fixed smooth ring R. Particle B hangs in equilibrium vertically below the ring. Particle A moves in horizontal circles on a smooth horizontal surface with speed 25 ga . The angle between AR and BR is i (see diagram). The normal reaction between A and the surface is 125 mg. (a) Find cosi. [3] … … … … … … … … … … … … … … … … … … … (b) Find, in terms of a, the distance of B below the ring. [3] … … … … … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 3(a) For B: = T mg For A: cos 3 θ + = R T mg M1 All 3 terms, allow sign errors, allow sin/cos mix. Use given R to obtain 3 cos 5 θ = A1 3 3(b) 2 3 sinθ = mv T r M1 May be seen in part (a), allow sin/cos mix. sinθ = r AR B1 Or equivalent. [Combine to give 3 4 = a AR , so] 1 4 = BR a A1 3 Question Answer Marks Guidance

This question in 9231/33 May/June 2021

4 1 2 u B O i a A u A particle of mass m is attached to one end of a light inextensible string of length a. The other end of the string is attached to a fixed point O. The particle is initially held with the string taut at the point A, where OA makes an angle i with the downward vertical through O. The particle is then projected with speed u perpendicular to OA and begins to move upwards in part of a vertical circle. The string goes slack when the particle is at the point B where angle AOB is a right angle. The speed of the particle when it is at B is 12 u (see diagram). Find the tension in the string at A, giving your answer in terms of m and g. [8] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 4 ( ) 2 2 1 1 cos sin 2 2 2 θ θ   − = +     u mu m mg a a *M1 Energy equation, with 2 KE terms and a two-part GPE term, allow cos/sin mix. ( ) 2 3 2 cos sin 4 u ag θ θ = + A1 At B, tension in string is zero, so 2 2 sin u m mg a θ       = ( 2 4 sinθ = u ag ) B1 N2L Eliminate 2 u DM1 tan 2 θ = OE A1 At A, 2 cosθ − = mu T mg a B1 N2L 9 5 5 = T mg ( = 4.02 ) mg M1 A1 Substitute to find T. Alternative method for question 4 ( ) 2 2 1 1 cos sin 2 2 2 θ θ   − = +     u mu m mg a a *M1 Energy equation, with 2 KE terms and a two-part GPE term, allow cos/sin mix. ( ) 2 3 2 cos sin 4 u ag θ θ = + A1 Question Answer Marks Guidance 4 At B, tension in string is zero, so 2 2 sin u m mg a θ       = ( 2 4 sinθ = u ag ) B1 N2L Eliminate θ : 2 8 5 5 = ag u DM1 A1 At A, 2 cosθ − = mu T mg a B1 N2L 9 5 5 = T mg ( = 4.02 ) mg M1 A1 Substitute to find T. 8 Question Answer Marks Guidance

This question in 9231/33 May/June 2021

1 One end of a light elastic string, of natural length a and modulus of elasticity 3mg, is attached to a fixed point O on a smooth horizontal plane. A particle P of mass m is attached to the other end of the string and moves in a horizontal circle with centre O. The speed of P is 43 ga . Find the extension of the string. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: 1 3 = mgx T a CAO. ( ) 4 3 = + mga T a x B1 2 2 9 9 4 0 + − = x ax a leading to ( )( ) 3 3 4 0 − + = x a x a M1 1 3 = x a A1 4 Must include logs. Condone missing modulus.

This question in 9231/31 Oct/Nov 2021

6 A particle P, of mass m, is attached to one end of a light inextensible string of length a. The other end of the string is attached to a fixed point O. The particle P moves in complete vertical circles about O with the string taut. The points A and B are on the path of P with AB a diameter of the circle. OA makes an angle i with the downward vertical through O and OB makes an angle i with the upward vertical through O. The speed of P when it is at A is 5ag . The ratio of the tension in the string when P is at A to the tension in the string when P is at B is 9 : 5. (a) Find the value of cosi. [6] … … … … … … … … … … … … … … … … … … … … … … … (b) Find, in terms of a and g, the greatest speed of P during its motion. [2] … … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 6(a) At A 5 cosθ − = × A ag T mg m a B1 N2L At B 2 cosθ + = × B v T mg m a B1 N2L 2 1 1 5 2cos 2 2 θ × − = × m ag mv mga M1 Energy equation with correct number of terms. 2 5 4 cosθ = − v ag ga A1 Accept multiplied by m and/or divided by a. Use ratio of tensions = 9 : 5 M1 Use ratio and simplify to an expression in cosθ . 2 cos 5 θ = A1 CAO 6 6(b) Greatest speed at lowest point ( ) 2 1 1 5 1 cos 2 2 θ − × + = × − m ag mV mga M1 Energy equation including lowest point, correct number of terms. 31 5 = ag V A1 FT Ft their cosθ from part (a). 2

This question in 9231/31 Oct/Nov 2021

3 R i 3m B m A Particles A and B, of masses m and 3m respectively, are connected by a light inextensible string of length a that passes through a fixed smooth ring R. Particle B hangs in equilibrium vertically below the ring. Particle A moves in horizontal circles with speed v. Particles A and B are at the same horizontal level. The angle between AR and BR is i (see diagram). (a) Show that cos i = 13 . [2] … … … … … … … … … … (b) Find an expression for v in terms of a and g. [4] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 3(a) = M1 Must see both of these separately. Combining, 1 cos 3 θ = A1 At least one step of working, AG. 2 3(b) (cos , θ − = a x x where x = AR) 3 1 or 4 4 AR a BR a = = or radius = 2 a B1 8 sin 3 θ   =       2 sinθ = mv T r M1 Combining to find an equation in v2, a and g only. DM1 2 2 = v ga , 2 = v ga A1 4

This question in 9231/32 Oct/Nov 2021

7 One end of a light inextensible string of length a is attached to a fixed point O. The other end of the string is attached to a particle P of mass m. The particle P is held vertically below O with the string taut and then projected horizontally. When the string makes an angle of 60° with the upward vertical, P becomes detached from the string. In its subsequent motion, P passes through the point A which is a distance a vertically above O. (a) The speed of P when it becomes detached from the string is V. Use the equation of the trajectory of a projectile to find V in terms of a and g. [4] … … … … … … … … … … … … … … … … … … … … … … … … (b) Find, in terms of m and g, the tension in the string immediately after P is initially projected horizontally. [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. … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 7(a) Coordinates of A: = = − B1 ( ) 2 2 2 3 3 2 3 1 2 2 2 .4 = − a g a a V M1 Substitute their (x, y) into correct trajectory equation. Rearrange to find 2 V . M1 2 3 3 , 2 2 = = V ag V ag A1 4 7(b) ( ) 2 2 1 1 1 cos60 2 2 − = + mu mV mga M1 Energy equation. 2 9 2 = u ag A1 u is the speed at P. 2 − = m T mg u a M1 N2L 11 2 = T mg A1 4

This question in 9231/32 Oct/Nov 2021

1 One end of a light elastic string, of natural length a and modulus of elasticity 3mg, is attached to a fixed point O on a smooth horizontal plane. A particle P of mass m is attached to the other end of the string and moves in a horizontal circle with centre O. The speed of P is 43 ga . Find the extension of the string. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: 1 3 = mgx T a CAO. ( ) 4 3 = + mga T a x B1 2 2 9 9 4 0 + − = x ax a leading to ( )( ) 3 3 4 0 − + = x a x a M1 1 3 = x a A1 4 Must include logs. Condone missing modulus.

This question in 9231/33 Oct/Nov 2021

6 A particle P, of mass m, is attached to one end of a light inextensible string of length a. The other end of the string is attached to a fixed point O. The particle P moves in complete vertical circles about O with the string taut. The points A and B are on the path of P with AB a diameter of the circle. OA makes an angle i with the downward vertical through O and OB makes an angle i with the upward vertical through O. The speed of P when it is at A is 5ag . The ratio of the tension in the string when P is at A to the tension in the string when P is at B is 9 : 5. (a) Find the value of cosi. [6] … … … … … … … … … … … … … … … … … … … … … … … (b) Find, in terms of a and g, the greatest speed of P during its motion. [2] … … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 6(a) At A 5 cosθ − = × A ag T mg m a B1 N2L At B 2 cosθ + = × B v T mg m a B1 N2L 2 1 1 5 2cos 2 2 θ × − = × m ag mv mga M1 Energy equation with correct number of terms. 2 5 4 cosθ = − v ag ga A1 Accept multiplied by m and/or divided by a. Use ratio of tensions = 9 : 5 M1 Use ratio and simplify to an expression in cosθ . 2 cos 5 θ = A1 CAO 6 6(b) Greatest speed at lowest point ( ) 2 1 1 5 1 cos 2 2 θ − × + = × − m ag mV mga M1 Energy equation including lowest point, correct number of terms. 31 5 = ag V A1 FT Ft their cosθ from part (a). 2

This question in 9231/33 Oct/Nov 2021

2 One end of a light inextensible string of length a is attached to a fixed point O. A particle of mass m is attached to the other end of the string. The particle is held at the point A with the string taut. The angle between OA and the downward vertical is equal to a, where cos a = 45 . The particle is projected from A, perpendicular to the string in an upwards direction, with a speed 3ga . It then moves along a circular path in a vertical plane. The string first goes slack when it makes an angle i with the upward vertical through O. Find the value of cosi. [5] … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 2   2 1 1 3 cos cos 2 2 mv m ga mga        M1 A1 Energy equation.   2 cos m T mg v a    B1 N2L 4 cos 3 2 cos 2 5 ag ga ga ga       M1 Combine to find cos. 7 cos 15  A1 5

This question in 9231/31 May/June 2022

5 C D i a i B 3a A A light inextensible string AB passes through two small holes C and D in a smooth horizontal table where AC = 3a and DB = a. A particle of mass m is attached at the end A and moves in a horizontal circle with angular velocity ~. A particle of mass 34 m is attached to the end B and moves in a horizontal circle with angular velocity k~. AC makes an angle i with the downward vertical and DB makes an angle i with the horizontal (see diagram). Find the value of k. [7] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 5 For A: 2 sin T mr    M1 N2L horizontal. 3 sin r a   B1 Correct expression for radius. 2 3 T m a   A1 Similarly, for B: 2 2 3 cos 4 T m r k     M1 N2L horizontal 2 2 3 4 T mak   A1 2 2 2 3 3 4 m a mak     M1 Equate expressions for T. 2 4, 2 k k   A1 Question Answer Marks Guidance 5 Alternative method for question 5 For A: 2 cos , sin T mg T mr      M1 N2L horizontal and vertical. 3 sin r a   B1 Correct expression for radius. 2 2 5 5 3 , 4 12 g T m a mg a        A1 Combine to obtain expression for 2 . Similarly, for B: 2 2 3 cos 4 T m r k     M1 N2L horizontal. 2 2 3 4 T mak   A1 2 5 3 5 4 4 12 g mg mak a   M1 Substitute for T and . 2 4, 2 k k   A1 7

This question in 9231/31 May/June 2022

2 One end of a light inextensible string of length a is attached to a fixed point O. A particle of mass m is attached to the other end of the string. The particle is held at the point A with the string taut. The angle between OA and the downward vertical is equal to a, where cos a = 45 . The particle is projected from A, perpendicular to the string in an upwards direction, with a speed 3ga . It then moves along a circular path in a vertical plane. The string first goes slack when it makes an angle i with the upward vertical through O. Find the value of cosi. [5] … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 2   2 1 1 3 cos cos 2 2 mv m ga mga        M1 A1 Energy equation.   2 cos m T mg v a    B1 N2L 4 cos 3 2 cos 2 5 ag ga ga ga       M1 Combine to find cos. 7 cos 15  A1 5

This question in 9231/32 May/June 2022

5 C D i a i B 3a A A light inextensible string AB passes through two small holes C and D in a smooth horizontal table where AC = 3a and DB = a. A particle of mass m is attached at the end A and moves in a horizontal circle with angular velocity ~. A particle of mass 34 m is attached to the end B and moves in a horizontal circle with angular velocity k~. AC makes an angle i with the downward vertical and DB makes an angle i with the horizontal (see diagram). Find the value of k. [7] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 5 For A: 2 sin T mr    M1 N2L horizontal. 3 sin r a   B1 Correct expression for radius. 2 3 T m a   A1 Similarly, for B: 2 2 3 cos 4 T m r k     M1 N2L horizontal 2 2 3 4 T mak   A1 2 2 2 3 3 4 m a mak     M1 Equate expressions for T. 2 4, 2 k k   A1 Question Answer Marks Guidance 5 Alternative method for question 5 For A: 2 cos , sin T mg T mr      M1 N2L horizontal and vertical. 3 sin r a   B1 Correct expression for radius. 2 2 5 5 3 , 4 12 g T m a mg a        A1 Combine to obtain expression for 2 . Similarly, for B: 2 2 3 cos 4 T m r k     M1 N2L horizontal. 2 2 3 4 T mak   A1 2 5 3 5 4 4 12 g mg mak a   M1 Substitute for T and . 2 4, 2 k k   A1 7

This question in 9231/32 May/June 2022

4 One end of a light inextensible string of length a is attached to a fixed point O. A particle of mass m is attached to the other end of the string and is held with the string taut at the point A. At A the string makes an angle i with the upward vertical through O. The particle is projected perpendicular to the string in a downward direction from A with a speed u. It moves along a circular path in the vertical plane. When the string makes an angle a with the downward vertical through O, the speed of the particle is 2u and the magnitude of the tension in the string is 10 times its magnitude at A. It is given that u = 23 ga . (a) Find, in terms of m and g, the magnitude of the tension in the string at A. [6] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … (b) Find the value of cosa. [2] … … … … … … … … … … … … …

8 marks

Mark scheme: 4(a)   2 2 1 1 cos cos 2 2 mv mu mga      M1 Energy equation with all necessary terms, GPE terms must be resolved, allow sin/cos mix, allow sign error.     2 2 1 1 2 cos cos 2 2 m u mu mga      A1 2u may be substituted later. Implied by   3 2 cos cos 2 3 ag ga      At A, 2 cos m T mg u a    B1 N2L Also, 2 10 cos 4 m T mg u a    B1 N2L and use of tension (10T). Use all three (two N2L and energy) equations to find T in terms of m and g only. M1 Might see   3 2 9 cos cos 3 m T mg ga a         cos cos 1       10cos cos 4     1 3 T mg  A1 6 4(b) Substitute back, 1 4 2 10 cos 3 3 m mg mg ga a      M1 Any appropriate method to obtain cos. 2 cos 3  A1 2

This question in 9231/33 May/June 2022

1 A particle of mass 2 kg is attached to one end of a light inextensible string of length 0.6 m. The other end of the string is attached to a fixed point on a smooth horizontal surface. The particle is moving in a circular path on the surface. The tension in the string is 20 N. Find how many revolutions the particle makes per minute. [3] … … … … … … … … … … … … … … … … … … … … … … … … …

3 marks

Mark scheme: Question Answer Marks Guidance 1 2  v 2 2 M1 Use F = ma: 20 = OR 20 = 2  0.6 0.6 2 2 50 A1 v = 6 OR = 3 60v 60 A1 FT 38.9848…. Number of revolutions per min = OR 0.6  2π 2π so 39(.0) revolutions 3

This question in 9231/31 Oct/Nov 2022

5 A particle P of mass m is attached to one end of a light inextensible string of length a. The other end of the string is attached to a fixed point O. The string is held taut with OP horizontal. The particle P is projected vertically downwards with speed 13 ag and starts to move in a vertical circle. P passes through the lowest point of the circle and reaches the point Q where OQ makes an angle i with the downward vertical. At Q the speed of P is kag and the tension in the string is 116 mg . (a) Find the value of k and the value of cosi. [4] … … … … … … … … … … … … … … … … … … … … … … … … At Q the particle P becomes detached from the string. (b) In the subsequent motion, find the greatest height reached by P above the level of the lowest point of the circle. [4] … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 5(a) 1 2 1 2 B1 Energy equation. mv − mu = mga cos 2 2  1  kag = ag + 2 ag cos    3  m 2 B1 N2L at B. T − mg cos= v a 11 m 11 So mg − mg cos= .kag , − cos= k 6 a 6 Solve simultaneously. M1 4 1 A1 Both. k = , cos= 3 2 4 5(b) Initial speed = kag sin B1 2 M1 2 2 Use v = u + 2 as : 0 = kag sin − 2 gs ( ) 1 A1 s = a 2 1 1 A1 FT Height above lowest point = s + a − a cos = a + a − a = a 2 2 4

This question in 9231/31 Oct/Nov 2022

1 A particle P of mass m is attached to one end of a light inextensible string of length a. The other end of the string is attached to a fixed point O. The string is held taut with OP making an angle a with the downward vertical, where cos a = 23 . The particle P is projected perpendicular to OP in an upwards direction with speed 3ag . It then starts to move along a circular path in a vertical plane. Find the cosine of the angle between the string and the upward vertical when the string first becomes slack. [4] … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: Question Answer Marks Guidance 1 m 2 2 B1 N2L When string goes slack, mg cos = v , v = ag cos  May include T, but B1 not awarded until T = 0. a 1 1 2 B1 Energy equation. m.3ag − mv = mg ( a cos+ a cos ) 2 2 2  2  M1 Combine. So u − ag cos  = 2 ag  cos +   3  2 4 A1 u − ag 3 5 cos  = = 3ag 9 4

This question in 9231/32 Oct/Nov 2022

6 R B h A A light inextensible string is threaded through a fixed smooth ring R which is at a height h above a smooth horizontal surface. One end of the string is attached to a particle A of mass m. The other end of the string is attached to a particle B of mass 67 m . The particle A moves in a horizontal circle on the surface. The particle B hangs in equilibrium below the ring and above the surface (see diagram). When A has constant angular speed ~, the angle between AR and BR is i and the normal reaction between A and the surface is N. a and the normal reaction When A has constant angular speed 32 ~, the angle between AR and BR is between A and the surface is 12 N . a . [5] (a) Show that cos i = 49 cos … … … … … … … … … … … … … … … … … … … (b) Find N in terms of m and g and find the value of cosa. [4] … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 6(a) 6 B1 May be implied. T = mg 7 T sin= mr2 = mh tan 2 B1 Allow r for radius. Radius of circle = h tan B1 2 6 g [So  = cos] 7 h 9 2 6 g M1 Second scenario, equivalent result . In second scenario,  = cos 4 7 h 6 g 4 6 g A1 Combine convincingly to obtain given result. Equate, cos=  cos giving 7 h 9 7 h 4 cos= cos AG 9 5 6(b) First scenario: N + T cos= mg 1 B1 Both. Second scenario, N + T cos= mg 2 6 12 M1 12cos− 6cos= 7 Equate: mg − mg cos= 2mg − mg cos 7 7 3 A1 cos= 4 5 A1 N = mg 7 4

This question in 9231/32 Oct/Nov 2022

1 A particle of mass 2 kg is attached to one end of a light inextensible string of length 0.6 m. The other end of the string is attached to a fixed point on a smooth horizontal surface. The particle is moving in a circular path on the surface. The tension in the string is 20 N. Find how many revolutions the particle makes per minute. [3] … … … … … … … … … … … … … … … … … … … … … … … … …

3 marks

Mark scheme: Question Answer Marks Guidance 1 2  v 2 2 M1 Use F = ma: 20 = OR 20 = 2  0.6 0.6 2 2 50 A1 v = 6 OR = 3 60v 60 A1 FT 38.9848…. Number of revolutions per min = OR 0.6  2π 2π so 39(.0) revolutions 3

This question in 9231/33 Oct/Nov 2022

5 A particle P of mass m is attached to one end of a light inextensible string of length a. The other end of the string is attached to a fixed point O. The string is held taut with OP horizontal. The particle P is projected vertically downwards with speed 13 ag and starts to move in a vertical circle. P passes through the lowest point of the circle and reaches the point Q where OQ makes an angle i with the downward vertical. At Q the speed of P is kag and the tension in the string is 116 mg . (a) Find the value of k and the value of cosi. [4] … … … … … … … … … … … … … … … … … … … … … … … … At Q the particle P becomes detached from the string. (b) In the subsequent motion, find the greatest height reached by P above the level of the lowest point of the circle. [4] … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 5(a) 1 2 1 2 B1 Energy equation. mv − mu = mga cos 2 2  1  kag = ag + 2 ag cos    3  m 2 B1 N2L at B. T − mg cos= v a 11 m 11 So mg − mg cos= .kag , − cos= k 6 a 6 Solve simultaneously. M1 4 1 A1 Both. k = , cos= 3 2 4 5(b) Initial speed = kag sin B1 2 M1 2 2 Use v = u + 2 as : 0 = kag sin − 2 gs ( ) 1 A1 s = a 2 1 1 A1 FT Height above lowest point = s + a − a cos = a + a − a = a 2 2 4

This question in 9231/33 Oct/Nov 2022

3 A particle P of mass m is attached to one end of a light inextensible string of length a. The other end of the string is attached to a fixed point O. The particle P is held at the point A, where OA makes an angle i with the downward vertical through O, and with the string taut. The particle P is projected perpendicular to OA in an upwards direction with speed u. It then starts to move along a circular path in a vertical plane. The string goes slack when P is at B, where angle AOB is 90° and the speed of P is 45 ag . (a) Find the value of sini. [2] … … … … … … … … … (b) Find, in terms of m and g, the tension in the string when P is at A. [5] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 3(a) At B, 4 sin 5 m ag mg a  until tension 0  used. Mass must be seen. No sign error. 4 sin 5  A1 2 3(b) At A, 2 cos mu T mg a    B1 Energy   2 1 1 4 cos sin 2 2 5 ag mu m mga       M1 A1 Energy equation with 4 terms, dimensionally correct. Mass must be present, allow sign errors. Must see 1 2 in the KE terms. Solve to find T M1 Complete method leading to an expression in mg for T . 21 5 T mg  A1 CWO 5

This question in 9231/31 May/June 2023

5 A light elastic string of natural length a and modulus of elasticity mmg has one end attached to a fixed point O on a smooth horizontal surface. When a particle of mass m is attached to the free end of the string, it moves with speed v in a horizontal circle with centre O and radius x. When, instead, a particle of mass 2m is attached to the free end of the string, this particle moves with speed 12 v in a horizontal circle with centre O and radius 34 x . (a) Find x in terms of a. [5] … … … … … … … … … … … … … … … … … … … … … … … … (b) Given that v = 12ag , find the value of m. [2] … … … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 5(a) Hooke’s law,   1 mg T x a a    or 2 3 4 mg T x a a          Also, 1T = 2 mv x and equate   2 mv mg x a x a    M1   2 gx x a v a    Dimensionally correct terms. Similarly: 2 3 1 2 4 2 3 4 x mg a m v a x                M1 2 3 3 2 4 gx v x a a          Must have 1 2 v and 3 4 x on the RHS. Their dimensionally correct 2. T Equate expressions for 2 v and solve for x in terms of a . M1 4 x a  A1 WWW 5 SC B3 for answer of 4a using  instead of mg  . 5(b)   2 a v xg x a   or 2 2 3 3 4 a v xg x a         and substitute 4 x a  , 12 v ag  M1 FT their expression for x. 1 A1 CAO 2

This question in 9231/31 May/June 2023

3 A particle P of mass m is attached to one end of a light inextensible string of length a. The other end of the string is attached to a fixed point O. The particle P is held at the point A, where OA makes an angle i with the downward vertical through O, and with the string taut. The particle P is projected perpendicular to OA in an upwards direction with speed u. It then starts to move along a circular path in a vertical plane. The string goes slack when P is at B, where angle AOB is 90° and the speed of P is 45 ag . (a) Find the value of sini. [2] … … … … … … … … … (b) Find, in terms of m and g, the tension in the string when P is at A. [5] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 3(a) At B, 4 sin 5 m ag mg a  until tension 0  used. Mass must be seen. No sign error. 4 sin 5  A1 2 3(b) At A, 2 cos mu T mg a    B1 Energy   2 1 1 4 cos sin 2 2 5 ag mu m mga       M1 A1 Energy equation with 4 terms, dimensionally correct. Mass must be present, allow sign errors. Must see 1 2 in the KE terms. Solve to find T M1 Complete method leading to an expression in mg for T . 21 5 T mg  A1 CWO 5

This question in 9231/32 May/June 2023

5 A light elastic string of natural length a and modulus of elasticity mmg has one end attached to a fixed point O on a smooth horizontal surface. When a particle of mass m is attached to the free end of the string, it moves with speed v in a horizontal circle with centre O and radius x. When, instead, a particle of mass 2m is attached to the free end of the string, this particle moves with speed 12 v in a horizontal circle with centre O and radius 34 x . (a) Find x in terms of a. [5] … … … … … … … … … … … … … … … … … … … … … … … … (b) Given that v = 12ag , find the value of m. [2] … … … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 5(a) Hooke’s law,   1 mg T x a a    or 2 3 4 mg T x a a          Also, 1T = 2 mv x and equate   2 mv mg x a x a    M1   2 gx x a v a    Dimensionally correct terms. Similarly: 2 3 1 2 4 2 3 4 x mg a m v a x                M1 2 3 3 2 4 gx v x a a          Must have 1 2 v and 3 4 x on the RHS. Their dimensionally correct 2. T Equate expressions for 2 v and solve for x in terms of a . M1 4 x a  A1 WWW 5 SC B3 for answer of 4a using  instead of mg  . 5(b)   2 a v xg x a   or 2 2 3 3 4 a v xg x a         and substitute 4 x a  , 12 v ag  M1 FT their expression for x. 1 A1 CAO 2

This question in 9231/32 May/June 2023

1 A particle P of mass m is attached to one end of a light inextensible string of length a. The other end of the string is attached to a fixed point O. The particle P is held at the point A, where OA makes an angle a with the downward vertical through O, and with the string taut. The particle P is projected perpendicular to OA in an upwards direction with speed 3ag . It then starts to move along a circular path in a vertical plane. The string goes slack when P is at B, where OB makes an angle i with the upward vertical. Given that cos a = 45 , find the value of cosi. [4] … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: 1   2 1 1 .3 cos cos 2 2      m ag mv mga M1 Energy equation, 4 terms, dimensionally correct, mass must be present, allow sign errors, allow sin in both terms on RHS 2 cosmv mg a B1 N2L, may include tension initially but not awarded until tension = 0 used 3 1 4 . cos cos 2 2 5            mag m ag mga 3 7 cos 2 10  M1 Dependent on tension = 0 and on an energy equation, eliminate 2v . 7 cos 15  A1 If no m in energy equation and no further errors, award SCB2 for correct final answer 4

This question in 9231/33 May/June 2023

5 One end of a light elastic string, of natural length 12a and modulus of elasticity kmg, is attached to a fixed point O. The other end of the string is attached to a particle of mass m. The particle moves with constant speed 32 3ag in a horizontal circle with centre at a distance 12a below O. The string is inclined at an angle i to the downward vertical through O. (a) Find, in terms of a, the extension of the string. [5] … … … … … … … … … … … … … … … … … … … … … … … … (b) Find the value of k. [3] … … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 5(a) cos   T mg 2 27 sin 4 mv ag T m r r            B1 12 tan  r a used M1 Divide: 27 tan , 4 12tan     so 3 tan 4  M1 Finds value for tan OE. Reduces to equation in  or x, no k. 9 ,  r a extension of string = 3a A1 Alternative method for question 5(a) Let L be stretched length of string. cos   T mg B1 Or 12a T mg L   2 27 sin 4 mv ag T m r r            B1 sin  r L used M1 Use 12 cos a L and   0.5 2 2 144 sin   L a L and eliminate T. M1 2 2 2 [ 144 81 ] 15 L a a L a    , extension of string = 3a A1 5 Question Answer Marks Guidance 5(b) Hooke’s law:   12 12  kmg L a T a B1 Eliminate T:   12 12 12   kmg L a mgL a a M1 5 12    L k L a A1 3

This question in 9231/33 May/June 2023

6 A particle P of mass m is attached to one end of a light inextensible rod of length 3a. An identical particle Q is attached to the other end of the rod. The rod is smoothly pivoted at a point O on the rod, where OQ = x . The system, of rod and particles, rotates about O in a vertical plane. At an instant when the rod is vertical, with P above Q, the particle P is moving horizontally with speed u. When the rod has turned through an angle of 60° from the vertical, the speed of P is 2 ag , and the tensions in the two parts of the rod, OP and OQ, have equal magnitudes. (a) Show that the speed of Q when the rod has turned through an angle of 60° from the vertical is 2x ag . [2] 3a - x … … … … … … … … … … … (b) Find x in terms of a. [5] … … … … … … … … … … … … … … … … … … … (c) Find u in terms of a and g. [4] … … … … … … … … … … … … … … … … … … Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. … … … … … … … … … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 6(a) Angular speeds of P and Q are equal, M1 vQ vP so = x 3a − x A1 Shown convincingly: angular speeds equal stated. 2 x ag vQ = AG 3a − x 2 6(b) m  4ag B1 For P: T + mg cos60= 3a − x mvQ2 B1 For Q: T − mg cos60= x m.4 ag mvQ2 M1 Eliminate T: − mg cos60+ = mg cos60+ 3a − x x m  4ag mx 4ag M1 Solve to find x . = 1 + 2 Obtain 3-term quadratic equation. 3a − x ( 3a − x ) 4a ( 3a − x ) = ( 3a − x ) 2 + 4ax , x 2 + 2 ax − 3a 2 = 0 ( x − a )( x + 3a ) = 0, x = a A1 5 6(c) Energy changes from initial position: B1 KEs correct. 1 2 Gain in KE of P: m 4 ag − u ( ) B1FT GPEs correct. 2 1   u 2 2  Loss in KE of Q: m   − vQ  2   2     Loss in GPE of P = mg ( 3a − x )(1 − cos60 ) ( = mga )  1  Gain in GPE of Q = mgx (1 − cos60 )  = mga   2   1 2 1  u  2 2  M1 Energy equation. m 4 ag − u − m  = − mgx (1 − cos60+) mg ( 3a − x )(1 − cos60 ) − vQ    ( )   2 2  2    5 2 A1 AEF Simplify: 4 ag − u + ag = ag 4 2 16 4 u = ag , u = 5ag 5 5 4

This question in 9231/31 Oct/Nov 2023

1 One end of a light inextensible string of length a is attached to a fixed point O. The other end of the string is attached to a particle of mass m. The string is taut and makes an angle i with the downward vertical through O, where cos i = 23 . The particle moves in a horizontal circle with speed v. Find v in terms of a and g. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: Question Answer Marks Guidance 1 ↑ T cosθ= mg B1 mv 2 B1 → T sinθ= a sinθ Eliminate T and substitute for 𝜃 M1 5 A1 v = ag 6 4

This question in 9231/32 Oct/Nov 2023

5 vB B i a O α A vA A bead of mass m moves on a smooth circular wire, with centre O and radius a, in a vertical plane. The a with the downward vertical bead has speed vA when it is at the point A where OA makes an angle through O, and cos a = 35 . Subsequently the bead has speed vB at the point B, where OB makes an angle i with the upward vertical through O. Angle AOB is a right angle (see diagram). The reaction of the wire on the bead at B is in the direction OB and has magnitude equal to 16 of the magnitude of the reaction when the bead is at A. (a) Find, in terms of m and g, the magnitude of the reaction at B. [6] … … … … … … … … … … … … … … … … … … … … … … … … (b) Given that v A = kag , find the value of k. [2] … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 5(a) mv A2 B1 At A: R A − mg cos= a mv B2 B1 At B: − RB + mg cos = a 1 2 1 2 M1 All terms present, allow sign errors, cos/sin mix. Energy: mv A − mvB = mga ( cos+ cos) 2 2 A1 AEF 1 M1 Any equivalent working, leading to a dimensionally Eliminate velocities, use RB = R A and substitute angle values 6 correct equation in RB and mg. 1 1 a ( RA − mg cos) − a ( − RB + mg cos) = mga ( cos+ cos) 2 2 3 4 7 6 RB − mg + RB − mg = 2mg  5 5 5 3 A1 RB = mg 5 6 5(b) 3 3 mv A2 M1 From first equation in (a), 6  mg − mg = 5 5 a v A2 = 3ag , so A1 CAO k = 3 2

This question in 9231/32 Oct/Nov 2023

6 A particle P of mass m is attached to one end of a light inextensible rod of length 3a. An identical particle Q is attached to the other end of the rod. The rod is smoothly pivoted at a point O on the rod, where OQ = x . The system, of rod and particles, rotates about O in a vertical plane. At an instant when the rod is vertical, with P above Q, the particle P is moving horizontally with speed u. When the rod has turned through an angle of 60° from the vertical, the speed of P is 2 ag , and the tensions in the two parts of the rod, OP and OQ, have equal magnitudes. (a) Show that the speed of Q when the rod has turned through an angle of 60° from the vertical is 2x ag . [2] 3a - x … … … … … … … … … … … (b) Find x in terms of a. [5] … … … … … … … … … … … … … … … … … … … (c) Find u in terms of a and g. [4] … … … … … … … … … … … … … … … … … … Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. … … … … … … … … … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 6(a) Angular speeds of P and Q are equal, M1 vQ vP so = x 3a − x A1 Shown convincingly: angular speeds equal stated. 2 x ag vQ = AG 3a − x 2 6(b) m  4ag B1 For P: T + mg cos60= 3a − x mvQ2 B1 For Q: T − mg cos60= x m.4 ag mvQ2 M1 Eliminate T: − mg cos60+ = mg cos60+ 3a − x x m  4ag mx 4ag M1 Solve to find x . = 1 + 2 Obtain 3-term quadratic equation. 3a − x ( 3a − x ) 4a ( 3a − x ) = ( 3a − x ) 2 + 4ax , x 2 + 2 ax − 3a 2 = 0 ( x − a )( x + 3a ) = 0, x = a A1 5 6(c) Energy changes from initial position: B1 KEs correct. 1 2 Gain in KE of P: m 4 ag − u ( ) B1FT GPEs correct. 2 1   u 2 2  Loss in KE of Q: m   − vQ  2   2     Loss in GPE of P = mg ( 3a − x )(1 − cos60 ) ( = mga )  1  Gain in GPE of Q = mgx (1 − cos60 )  = mga   2   1 2 1  u  2 2  M1 Energy equation. m 4 ag − u − m  = − mgx (1 − cos60+) mg ( 3a − x )(1 − cos60 ) − vQ    ( )   2 2  2    5 2 A1 AEF Simplify: 4 ag − u + ag = ag 4 2 16 4 u = ag , u = 5ag 5 5 4

This question in 9231/33 Oct/Nov 2023

5 Two particles A and B of masses m and km respectively are connected by a light inextensible string of length a. The particles are placed on a rough horizontal circular turntable with the string taut and lying along a radius of the turntable. Particle A is at a distance a from the centre of the turntable and particle B is at a distance 2a from the centre of the turntable. The coefficient of friction between each particle and the turntable is 1.5 2 g When the turntable is made to rotate with angular speed , the system is in limiting equilibrium. 5 a (a) Find the tension in the string, in terms of m and g. [4] … … … … … … … … … … … … … … … … … … … … … … … (b) Find the value of k. [3] … … … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 5(a) For A: 2     A F T m a M1 Only allow sign errors. 1 5 A F mg mg    B1 Accept with g replaced by 10. Combine: 1 4 5 25 T mg mg   M1 To reach an equation in T and mg only. Accept with g replaced by 10. 1 25 T mg  A1 CAO 4 5(b) For B: 2 2     B F T km a M1 Only allow sign errors. 1 5 B F kmg kmg    and combine to find k M1 To reach an equation in k only. 1 3 k  A1 3

This question in 9231/31 May/June 2024

7 A smooth sphere with centre O and of radius a is fixed to a horizontal plane. A particle P of mass m is projected horizontally from the highest point of the sphere with speed u, so that it begins to move along the surface of the sphere. The particle P loses contact with the sphere at the point Q on the sphere, where OQ makes an angle i with the upward vertical through O. u 2 + 2ag (a) Show that cos i = . [4] 3ag … … … … … … … … … … … … … … … … … … … … … … … … … It is given that cos i = 56 . (b) Find, in terms of a and g, an expression for the vertical component of the velocity of P just before it hits the horizontal plane to which the sphere is fixed. [3] … … … … … … … … … … … … … … … (c) Find an expression for the time taken by P to fall from Q to the plane. Give your answer in the a form k , stating the value of k correct to 3 significant figures. [2] g … … … … … … … … … 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) Energy: 1 1 2 2 2 2 1 cos mu mv mga     *M1 m must be present, dimensionally correct, no missing terms. Allow sin instead of cos . Allow sign errors. N2L: 2 cosmv mg a B1 No reaction when P loses contact. Eliminate 2 v DM1 2 2 cos 3 u ag ag    A1 AG 4 7(b) Vertical component of velocity of P when it leaves the sphere: sin  v 55 216          ag *B1 Must not come from u .     2 2 sin 2 1 cos       V v g a DM1 Use of 2 2 ' 2 .   v u as Allow sin for cos. Allow sign errors. 847 216  ag V A1 AEF 3 Question Answer Marks Guidance 7(c) 1 847 55 216 216           ag ag t g M1   1 847 55 1.48 6 6   a a g g A1 2

This question in 9231/31 May/June 2024

5 Two particles A and B of masses m and km respectively are connected by a light inextensible string of length a. The particles are placed on a rough horizontal circular turntable with the string taut and lying along a radius of the turntable. Particle A is at a distance a from the centre of the turntable and particle B is at a distance 2a from the centre of the turntable. The coefficient of friction between each particle and the turntable is 1.5 2 g When the turntable is made to rotate with angular speed , the system is in limiting equilibrium. 5 a (a) Find the tension in the string, in terms of m and g. [4] … … … … … … … … … … … … … … … … … … … … … … … (b) Find the value of k. [3] … … … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 5(a) For A: 2     A F T m a M1 Only allow sign errors. 1 5 A F mg mg    B1 Accept with g replaced by 10. Combine: 1 4 5 25 T mg mg   M1 To reach an equation in T and mg only. Accept with g replaced by 10. 1 25 T mg  A1 CAO 4 5(b) For B: 2 2     B F T km a M1 Only allow sign errors. 1 5 B F kmg kmg    and combine to find k M1 To reach an equation in k only. 1 3 k  A1 3

This question in 9231/32 May/June 2024

7 A smooth sphere with centre O and of radius a is fixed to a horizontal plane. A particle P of mass m is projected horizontally from the highest point of the sphere with speed u, so that it begins to move along the surface of the sphere. The particle P loses contact with the sphere at the point Q on the sphere, where OQ makes an angle i with the upward vertical through O. u 2 + 2ag (a) Show that cos i = . [4] 3ag … … … … … … … … … … … … … … … … … … … … … … … … … It is given that cos i = 56 . (b) Find, in terms of a and g, an expression for the vertical component of the velocity of P just before it hits the horizontal plane to which the sphere is fixed. [3] … … … … … … … … … … … … … … … (c) Find an expression for the time taken by P to fall from Q to the plane. Give your answer in the a form k , stating the value of k correct to 3 significant figures. [2] g … … … … … … … … … 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) Energy: 1 1 2 2 2 2 1 cos mu mv mga     *M1 m must be present, dimensionally correct, no missing terms. Allow sin instead of cos . Allow sign errors. N2L: 2 cosmv mg a B1 No reaction when P loses contact. Eliminate 2 v DM1 2 2 cos 3 u ag ag    A1 AG 4 7(b) Vertical component of velocity of P when it leaves the sphere: sin  v 55 216          ag *B1 Must not come from u .     2 2 sin 2 1 cos       V v g a DM1 Use of 2 2 ' 2 .   v u as Allow sin for cos. Allow sign errors. 847 216  ag V A1 AEF 3 Question Answer Marks Guidance 7(c) 1 847 55 216 216           ag ag t g M1   1 847 55 1.48 6 6   a a g g A1 2

This question in 9231/32 May/June 2024

2 A particle P of mass m is attached to one end of a light inextensible string of length a. The other end of the string is attached to a fixed point O. The particle P is held at the point A with the string taut. It is given that OA makes an angle i with the downward vertical through O, where tan i = 3 . The particle P 4 is projected perpendicular to OA in an upwards direction with speed 5ag , and it starts to move along a circular path in a vertical plane. When P is at the point B, where angle AOB is a right angle, the tension in the string is T. Find T in terms of m and g. [5] … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 2 mv 2 B1 At B, T + mg sin= a 1 2 1 2 M1A1 Energy A to B: mu − mv = mga ( cos+ sin) 2 2 Substitute for u and  to find T: M1  4 3  T = mg  5 −2 −3   5 5  8 A1 T = mg 5 5

This question in 9231/31 Oct/Nov 2024

6 O a 1 m P 0.8 m b Q 1.4 m A particle P of mass 0.05 kg is attached to one end of a light inextensible string of length 1 m. The other end of the string is attached to a fixed point O. A particle Q of mass 0.04 kg is attached to one end of a second light inextensible string. The other end of this string is attached to P. The particle P moves in a horizontal circle of radius 0.8 m with angular speed ~ rad s -1 . The particle Q moves in a horizontal circle of radius 1.4 m also with angular speed ~ rad s -1 . The centres of the circles are vertically below O, and O, P and Q are always in the same vertical plane. The strings OP and PQ remain at constant angles a and b respectively to the vertical (see diagram). (a) Find the tension in the string OP. [3] … … … … … … … … … … … … … … … … … (b) Find the value of ~. [3] … … … … … … … … … … … … … … (c) Find the value of b. [2] … … … … … … … … … … … …

8 marks

Mark scheme: 6(a) At P:  T1 cos= T2 cos + 0.05 g B1 At Q:  T2 cos = 0.04 g B1 OR: whole system: T1 cos= 0.09 g T1 = 0.15 g = 1.5 N B1 3 6(b) T1 sin − T2 sin  = 0.05  0.82 M1 Allow sin/cos mix T2 sin  = 0.04  1.42 M1 T1sin = 0.05  0.82 + 0.04  1.42 A1 2 5  = 12.5, = 2 2 3 6(c) T2 cos = 0.04 g and T2 sin  = 0.04  1.42 M1 From part (a) and part (b) 7 Divide: tan = 4 = 60.3 A1 2

This question in 9231/31 Oct/Nov 2024

1 A particle of mass 2 kg is attached to one end of a light elastic string of natural length 0.8 m and modulus of elasticity 100 N. The other end of the string is attached to a fixed point O on a smooth horizontal surface. The particle is moving in a horizontal circle about O with the string taut and with constant angular speed 5 radians per second. Find the extension of the string. [3] … … … … … … … … … … … … … … … … … … … … … … … … …

3 marks

Mark scheme: Question Answer Marks Guidance 1 100 B1 100 Hooke’s law: T = x T = ( r − 0.8 ) 0.8 0.8 N2L: T = 2  ( 0.8 + x )  52 B1 T = 2 r 52 Equate and solve: 50 ( 0.8 + x ) = 125 x , x = 0.533 B1 8 15 3

This question in 9231/32 Oct/Nov 2024

6 O θ a θ A 1 10ag 3 A particle P of mass m is attached to one end of a light inextensible string of length a. The other end of the string is attached to a fixed point O. The particle P is held with the string taut and the string makes an angle i with the downward vertical through O. The particle P is projected at right angles to the string with speed 1 10ag and begins to move downwards along a circular path. When the string is 3 vertical, it strikes a small smooth peg at the point A which is vertically below O. The circular path and the point A are in the same vertical plane. After the string strikes the peg, the particle P begins to move in a vertical circle with centre A. When the string makes an angle i with the upward vertical through A the string becomes slack (see diagram). The distance of A below O is 5 a . 9 (a) Find the value of cosi. [6] … … … … … … … … … … … … … … … … … … … … … … … (b) Find the ratio of the tensions in the string immediately before and immediately after it strikes the peg. [4] … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 6(a) For P to lowest point L: M1* 1 2 1 2 Dimensionally correct, all terms present, allow Energy: mv = mu + mga (1 − cos) sign errors, allow cos/sin error. 2 2 2 28 [ v = ag − 2ag cos] 9 From L to string goes slack: M1* 4 a 1 2 1 2 4 a Dimensionally correct, all terms present, with , Energy: mv = mw + mg (1 + cos) 9 2 2 9 allow sign errors, allow cos/sin error.  2 20 26  w = ag − ag cos    9 9  Both equations correct, allow unsimplified. A1 mw2 B1 4 a When string goes slack: mg cos= 9 Equate expressions for w 2 to find a value for cos DM1 2 A1 cos= 3 6 Alternative method for question 6(a) For P from start to string goes slack: M2 4 Dimensionally correct, all terms present, with a 9 1 2 1 2  4  Allow sign errors, allow cos/sin error Energy: mw = mu + mga (1 − cos) − (1 + cos)   2 2  9  5  4  RHS may appear with a − 1 + a cos   9  9  4 Allow M1 if is missing or if an attempt at both 9 heights, but all other conditions are met.  2 10  5 13  20 26  A1 Correct, allow unsimplified. w = ag + 2ag − cos = ag − ag cos      9  9 9  9 9  mw2 B1 4 a When string goes slack: mg cos= 9 Equate expressions for w 2 to find a value for cos DM1 2 A1 cos= 3 6 6(b) mv 2 M1 EITHER equation, dimensionally correct, allow Tension before: T1 − mg = sign error only a mv 2 Tension after: T2 − mg = 4 a 9  2 16  25 A1 EITHER tension correct v = ag , T1 = mg T2 = 5mg    9  9 Find the other tension (from a valid equation) and find ratio of tensions M1 mv 2 Equation must be of the form T − mg = r Ratio is 5 : 9 A1 5 Any equivalent ratio, allow 9 4

This question in 9231/32 Oct/Nov 2024

2 A particle P of mass m is attached to one end of a light inextensible string of length a. The other end of the string is attached to a fixed point O. The particle P is held at the point A with the string taut. It is given that OA makes an angle i with the downward vertical through O, where tan i = 3 . The particle P 4 is projected perpendicular to OA in an upwards direction with speed 5ag , and it starts to move along a circular path in a vertical plane. When P is at the point B, where angle AOB is a right angle, the tension in the string is T. Find T in terms of m and g. [5] … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 2 mv 2 B1 At B, T + mg sin= a 1 2 1 2 M1A1 Energy A to B: mu − mv = mga ( cos+ sin) 2 2 Substitute for u and  to find T: M1  4 3  T = mg  5 −2 −3   5 5  8 A1 T = mg 5 5

This question in 9231/33 Oct/Nov 2024

6 O a 1 m P 0.8 m b Q 1.4 m A particle P of mass 0.05 kg is attached to one end of a light inextensible string of length 1 m. The other end of the string is attached to a fixed point O. A particle Q of mass 0.04 kg is attached to one end of a second light inextensible string. The other end of this string is attached to P. The particle P moves in a horizontal circle of radius 0.8 m with angular speed ~ rad s -1 . The particle Q moves in a horizontal circle of radius 1.4 m also with angular speed ~ rad s -1 . The centres of the circles are vertically below O, and O, P and Q are always in the same vertical plane. The strings OP and PQ remain at constant angles a and b respectively to the vertical (see diagram). (a) Find the tension in the string OP. [3] … … … … … … … … … … … … … … … … … (b) Find the value of ~. [3] … … … … … … … … … … … … … … (c) Find the value of b. [2] … … … … … … … … … … … …

8 marks

Mark scheme: 6(a) At P:  T1 cos= T2 cos + 0.05 g B1 At Q:  T2 cos = 0.04 g B1 OR: whole system: T1 cos= 0.09 g T1 = 0.15 g = 1.5 N B1 3 6(b) T1 sin − T2 sin  = 0.05  0.82 M1 Allow sin/cos mix T2 sin  = 0.04  1.42 M1 T1sin = 0.05  0.82 + 0.04  1.42 A1 2 5  = 12.5, = 2 2 3 6(c) T2 cos = 0.04 g and T2 sin  = 0.04  1.42 M1 From part (a) and part (b) 7 Divide: tan = 4 = 60.3 A1 2

This question in 9231/33 Oct/Nov 2024

1 A particle P of mass m is attached to one end of a light inextensible string of length a. The other end of the string is attached to a fixed point O. The particle moves in a horizontal circle with constant angular speed ~ and with the string inclined at an angle of i to the downward vertical. Given that tan i = 4 , find ~ in terms of a and g. [3] 3 … … … … … … … … … … … … … … … … … … … … … … … … …

3 marks

Mark scheme: Question Answer Marks Guidance 1 T cos= mg B1 B1 for both equations, SOI. T sin= m2 r 3'5 mg' sin  = m2 a sin  M1 Combining and using r = a sin. Allow sin/cos mix in resolving. Must have a trigonometric ratio on both sides. 5 g A1 OE, CAO = 3a 3

This question in 9231/33 May/June 2025

5 B a A 3 rg O r 2 A hollow cylinder of radius r is fixed with its axis horizontal. Points A, B and O are in the same vertical plane perpendicular to the axis of the cylinder, with A and B on the smooth inner surface and O on the axis. OA and OB make angles 90° and a respectively with the upward vertical through O, with A and B on opposite sides of the vertical. A particle of mass m is projected vertically downwards from point A with speed 3 rg and moves in a vertical circle inside the cylinder (see diagram). The particle loses 2 contact with the cylinder at point B. (a) Find the value of a. [4] … … … … … … … … … … … … … … … … … (b) In the subsequent motion find, in terms of r, the greatest height above O reached by the particle. [4] … … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 5(a) 1  3 gr ) 2 mv 2 + mgr cos= 12 m ( 2  v 2 = 32 gr − 2 gr cos M1 Energy equation. All 3 terms. Allow sin/cos error. Allow sign errors. 2 B1 N2L at B. mv mgcos= r 3 mg cos= m ( 2 g − 2 g cos) M1 Combining to eliminate v. cos= 12 , = 60 A1 4 5(b) v 2 = 12 gr B1 FT FT their 60. 2 M1 Allow sign error. 2 0 = sin'60 ' ' v ' − 2 gh ( ) h = 163 r A1 FT FT their 60and their v2. Total height = 163 r + r cos60 = 1611 r A1 4

This question in 9231/33 May/June 2025

4 A h a O B A hollow cone with a smooth inner surface is fixed with its vertex O downwards. The semi-vertical angle of the cone is a, where tan a = 3 . A light inextensible string has a particle A of mass m attached 4 to one end and a particle B of mass m attached to the other end. The string passes through a small hole in the cone at O. Particle B hangs in equilibrium below O. Particle A is on the inner surface of the cone at a height h above the level of O and moves in horizontal circles with constant angular speed ~ (see diagram). Find ~ in terms of g and h. [6] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 4 For B: T = mg B1 Must be seen explicitly. = T cos+ mg *M1 Allow sin/cos mix, allow sign error. For A: (  ) R sin Allow this mark with mg instead of T. ( → ) R cos+ T sin  = mr2 *M1 Allow sin/cos mix, allow sign error. Allow this mark with mg instead of T. r = h tan B1 Seen anywhere. Combine equations to find an expression for 2 in terms of g and DM1 mg (1 + cos) May see R = 3mg or R = . h only. sin  g A1 If T = mg is not seen explicitly, maximum possible mark is = 2 h B0 M1 M1 B1 DM1 A0. 6

This question in 9231/34 May/June 2025

7 A particle P of mass m is attached to one end of a light inextensible string of length a. The other end of the string is attached to a fixed point O. The particle moves in complete vertical circles with centre O, with the string taut. When the string makes an angle i with the downward vertical through O the speed of P is 4ag . The ratio of the greatest and least tensions in the string during the motion is 11 : 1. Find the value of cosi. [8] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … 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. … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 7 Energy, highest to lowest: M1 Note that assuming the same speed at highest and lowest 1 2 mu 2 − 12 mv 2 = 2 amg  v 2 = u 2 − 4 ag  points could give the correct final answer but will not score full marks. Energy, start to lowest: One energy equation, all terms present, dimensionally correct, 1 allow sign error. u 2 = ag ( 6 − 2cos)  2 mu 2 − 12 m (4 ag ) = mg ( a − a cos)   M1 A second energy equation, all terms present, dimensionally Energy, start to highest, correct, allow sign error 1 v 2 = 2 ag (1 − cos)  2 m (4 ag ) − 12 mv 2 = mg ( a + a cos)   A1 One fully correct energy equation. mu 2 B1 Use of N2L at the lowest point, may see TL = mg ( 7 − 2cos) . At lowest point, TL − mg = a mv 2 B1 Use of N2L at the highest point, may see TH = mg (1 − 2cos) . At highest point, TH + mg = a Must use a different speed. TL = 11TH B1 mu 2 + mg a SOI, must be correct way around. For example, = 11 . mv 2 − mg a Combine to find equation involving cos only. M1 Must see an energy equation involving cos only. cos= 15 A1 8

This question in 9231/34 May/June 2025

2 A particle P of mass m is moving in a horizontal circle with angular speed ~ on the smooth inner 1 surface of a hemispherical shell of radius r. The angle between the upward vertical and the normal reaction of the surface on P is i , where tan i = 3 . 1 1 4 When the angular speed is increased to ~ , the angle between the upward vertical and the normal 2 reaction of the surface on P becomes i , where tan i = 4 . 2 2 3 ~ Find the ratio 1 . [4] ~ 2 … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: 2  N1 cos1 = mg B1 N2L horizontally and vertically for the first case. 4 g = r12 → N1 sin1 = m ( r sin1 ) 12 5  N 2 cos2 = mg B1 N2L horizontally and vertically for the second case. 3 g = r22 → N2 sin2 = m ( r sin2 )22 5 5 4 g r12 M1 Combine to find ratio. = 5 3 g r22 1 1 A1 2 3 =2 4

This question in 9231/31 Oct/Nov 2025

6 17ag P 60° a O A particle P of mass m is attached to one end of a light inextensible string of length a. The other end of the string is attached to a fixed point O. Initially P is held with the string taut and making an angle of 60° with the upward vertical through O. The particle P is projected perpendicular to the string in a downwards direction with speed 17ag . It then starts to move along a circular path in a vertical plane with centre O (see diagram). At the lowest point of its path, vertically below O, the particle P collides with a stationary particle Q. (a) Find, in terms of a and g, an expression for the speed of P immediately before the collision with Q. [2] … … … … … … … … … … … … … … … … … As a result of the collision, P rebounds and moves back along a circular path with centre O. The string becomes slack when P reaches the point on the circle vertically above O. (b) Find, in terms of a and g, an expression for the speed of P immediately after the collision with Q. [3] … … … … … … … … … … … The mass of particle Q is km and the collision between P and Q is perfectly elastic. (c) Find the value of k. [3] … … … … … … … … … … … …

8 marks

Mark scheme: 6(a) Let v be the speed of P before the collision. M1 Dimensionally correct energy equation with a GPE term and Energy conservation before the collision: two KE terms. 1 2 m (17ag ) + 2 3 mag = 12 mv 2 A1 v = 20ag 2 6(b) Let wP be speed of P after the collision and V be its speed at the top. B1 Dimensionally correct equation. 2 T = 0 must be seen or implied. mV N2L: = ( T ) + mg V = ag    a 1 2 1 2 M1 Dimensionally correct energy equation with a GPE term and Energy equation: 2 mwP = 2 mV + 2mga two KE terms. A1 wP = 5ag 3 6(c) Let wQ the speed of Q after the collision. B1 Momentum conserved (must see masses). mv = kmwQ − mwP wP + wQ = ev  = v  B1 NEL with consistent signs. OR OR Use conservation of kinetic energy. 1 2 mv 2 = 12 mwP2 + 12 kmwQ2 k = 3 B1 Must have e = 1 seen or implied. 3

This question in 9231/31 Oct/Nov 2025

1 A particle P of mass m is attached to two light inextensible strings each of length l. The end of one string is attached to a fixed point A and the end of the other string is attached to a fixed point B, with A vertically above B. Angle APB is a right angle. The particle P rotates in a horizontal circle at a constant angular speed ~ with both strings taut (see diagram). A l ~ P l B Find the tension in string AP in terms of m, g, l and ~. [4] … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: Question Answer Marks Guidance 1 TA cos45+ TB sin 45= m2 l sin 45 B1 Correct horizontal equation. Allow sin/cos mix. B0 B0 M0 A0 for using the same tension in each string. TA sin45− TB cos45= mg B1 Correct vertical equation. Allow sin/cos mix. 2TA = m2 l + 2 mg M1 Combine, isolating term in T A . 1 2 A1 AEF  l + 2 g TA = 2 m ( ) 4

This question in 9231/32 Oct/Nov 2025

6 u A B a θ O 3u β A fixed smooth sphere with radius a and centre O rests on horizontal ground. A particle is projected horizontally from the highest point, A, of the sphere with speed u. The particle begins to move in a vertical circle along the surface of the sphere. The particle loses contact with the sphere at the point B, where the angle AOB is i. After leaving the surface of the sphere, the particle moves freely under gravity before striking the horizontal ground with speed 3u at an angle b to the horizontal (see diagram). Find the value of b. [8] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 6 A to ground: 1 2 mu 2 + 2mga = 12 m ( 3u ) 2  u 2 = 12 ga  *M1 One energy equation, all terms present. *M1 Second energy equation, all terms present. A to B: 1 1 − cos)  2 mv 2 + mgacos= 12 mu 2 + mga  v 2 = u 2 + 2ag (  A1 At least one correct energy equation involving cos. B to ground: 1 v 2 = 9u 2 − 2ag (1 + cos)  2 mv 2 + mgacos= 12 m ( 3u ) 2 − mga   mv 2 2 B1 N2L at B. mgcos=  v = agcos   a gacos = 12 ga + 2 ag (1 − cos) DM1 Eliminate u and v. cos= 56 A1 Horizontally at ground: v cos= 3u cos . M1 A complete method to find . 5 5 6 ga ( 6 ) 5 cos = = 15 ( 54 ) 1 3 2 ga OR xv = v cos= 56 ga ( 56 ) 2 2 847 , yv = ( v sin) + 2 ga (1 + cos) = 216 ga 847 216 ga tan = 5 5 6 ga ( 6 ) = 69.0 A1 8

This question in 9231/32 Oct/Nov 2025

2 A particle P of mass m is moving in a horizontal circle with angular speed ~ on the smooth inner 1 surface of a hemispherical shell of radius r. The angle between the upward vertical and the normal reaction of the surface on P is i , where tan i = 3 . 1 1 4 When the angular speed is increased to ~ , the angle between the upward vertical and the normal 2 reaction of the surface on P becomes i , where tan i = 4 . 2 2 3 ~ Find the ratio 1 . [4] ~ 2 … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: 2  N1 cos1 = mg B1 N2L horizontally and vertically for the first case. 4 g = r12 → N1 sin1 = m ( r sin1 ) 12 5  N 2 cos2 = mg B1 N2L horizontally and vertically for the second case. 3 g = r22 → N2 sin2 = m ( r sin2 )22 5 5 4 g r12 M1 Combine to find ratio. = 5 3 g r22 1 1 A1 2 3 =2 4

This question in 9231/33 Oct/Nov 2025

6 17ag P 60° a O A particle P of mass m is attached to one end of a light inextensible string of length a. The other end of the string is attached to a fixed point O. Initially P is held with the string taut and making an angle of 60° with the upward vertical through O. The particle P is projected perpendicular to the string in a downwards direction with speed 17ag . It then starts to move along a circular path in a vertical plane with centre O (see diagram). At the lowest point of its path, vertically below O, the particle P collides with a stationary particle Q. (a) Find, in terms of a and g, an expression for the speed of P immediately before the collision with Q. [2] … … … … … … … … … … … … … … … … … As a result of the collision, P rebounds and moves back along a circular path with centre O. The string becomes slack when P reaches the point on the circle vertically above O. (b) Find, in terms of a and g, an expression for the speed of P immediately after the collision with Q. [3] … … … … … … … … … … … The mass of particle Q is km and the collision between P and Q is perfectly elastic. (c) Find the value of k. [3] … … … … … … … … … … … …

8 marks

Mark scheme: 6(a) Let v be the speed of P before the collision. M1 Dimensionally correct energy equation with a GPE term and Energy conservation before the collision: two KE terms. 1 2 m (17ag ) + 2 3 mag = 12 mv 2 A1 v = 20ag 2 6(b) Let wP be speed of P after the collision and V be its speed at the top. B1 Dimensionally correct equation. 2 T = 0 must be seen or implied. mV N2L: = ( T ) + mg V = ag    a 1 2 1 2 M1 Dimensionally correct energy equation with a GPE term and Energy equation: 2 mwP = 2 mV + 2mga two KE terms. A1 wP = 5ag 3 6(c) Let wQ the speed of Q after the collision. B1 Momentum conserved (must see masses). mv = kmwQ − mwP wP + wQ = ev  = v  B1 NEL with consistent signs. OR OR Use conservation of kinetic energy. 1 2 mv 2 = 12 mwP2 + 12 kmwQ2 k = 3 B1 Must have e = 1 seen or implied. 3

This question in 9231/33 Oct/Nov 2025