Cambridge A Level Mathematics - Further 9231 — 2025 Oct/Nov Paper 3 · Variant 3

9231/33/O/N/25 · 7 questions · 50 marks · ≈56 min

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Questions as text

Q1 · Two uniform smooth spheres A and B of equal radii have masses 4m and m respectively

1 Two uniform smooth spheres A and B of equal radii have masses 4m and m respectively. Sphere B is at rest on a smooth horizontal surface. Sphere A is moving on the surface with speed u and collides directly with sphere B. After the collision, the momentum of A is three times the momentum of B. Find the value of the coefficient of restitution e. [4] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1 Let v A and v B be the speeds of A and B respectively after the collision. M1 Conservation of momentum, masses correct. 4 mv A + mv B = 4 mu −v A + vB = eu M1 NEL, signs must be consistent with momentum equation. v A = 34 u , v B = u A1 Use momentum condition after the collision to obtain expressions for vA and vB in terms of u. e = 14 A1 Alternative method for question 1 Let v A and v B be the speeds of A and B respectively after the collision. M1 Conservation of momentum, masses correct. 4 mv A + mv B = 4 mu −v A + vB = eu M1 NEL, signs must be consistent with momentum equation. 4 u (1 + e ) A1 Combine to obtain expressions for v A and v B in terms of u v A = 15 u ( 4 − e ) , v B = 5 and e. 1 5 u ( 4 − e ) 1 A1 Use momentum condition after the collision to solve for e. = 3, e = 4 4 5 u (1 + e ) 4

More questions on Momentum

Q2 · A particle P of mass m is moving in a horizontal circle with angular speed ~ on the…

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 .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

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

More questions on Circular motion

Q3 · A particle P is moving in a straight horizontal line

3 A particle P is moving in a straight horizontal line. At time t s, the displacement of P from a fixed point O on the line is x m and the velocity of P is v ms -1 . The acceleration of P is 1 ( v 2 + 4 ) ms -2 in the 2 direction PO. Initially P is at O and is moving with velocity 2 ms -1 . (a) Find an expression for x in terms of t. 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(b) Find the time when P next goes through O. 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Mark scheme: 3(a) dv 1 2 *M1 Forming differential equation and integrating to a term v + 4 = − 2 ( ) involving an inverse tangent. Condone omission of +A. dt 1 −1 1 1 dv 1 2 to obtain a term v + 4 2 tan ( 2 v ) = − 2 t  + A Allow M1 for integrating = 2 ( ) dt involving an inverse tangent. A1 CWO, condone omission of +A. 1 −1 2 1 DM1 Use initial condition to find constant. t = 0, v = 2 : A = 2 tan ( 2 ) = 8 π dx 1 DM1 Integrate into term involving ln of cosine or sine. v = = 2tan ( 4 π − t ) Accept modulus in ln term. dt 1 OE, for example x = 2ln ( cos ( t − 4 π ) ) + B  1 1 . v = −2tan ( t − 4 π ) , x = 2ln cos ( t − 4 π ) + B x = 0, t = 0: B = ln2 A1 Use initial condition and obtain correct expression. AEF, may see an expression in terms of sec. 1 1  x = 2ln + ln2  = ln cos ( t − 2cos 2 ( t − ( 4 π ) ) ( 4 π ) ) Accept modulus in ln term.   5 3(b) 1 M1 Solve equation to find a value for t. 2ln + ln2 = 0 ( cos ( t − 4 π ) ) t = 12 π A1 CWO 2

More questions on Differential equations

Q4 · One end of a light elastic string of natural length a and modulus of elasticity 5mg is…

4 One end of a light elastic string of natural length a and modulus of elasticity 5mg is attached to a fixed point O. Two particles, P and Q, of masses m and 4m respectively are attached to the other end of the string and they hang vertically in equilibrium. Particle Q is then detached from the string, hence releasing particle P from rest. Find, in terms of a, the length of the string when the speed of particle P is first equal to 7 ag . 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Mark scheme: 4 5mge B1 In equilibrium, = 5mg , e = a . a 7 B1 At least one correct EPE term seen. Let x be the length of the string when speed of P is 5 ag . Energy equation in subsequent motion: M1 Dimensionally correct energy equation with one KE term, 5 mga 2 5 mg ( x − a ) 2 1 7 two EPE terms and at least one GPE term. − = mg ( e + a − x ) + 2 m ( 5 ag ) 2 a 2 a A1 Fully correct energy equation. 2 2 M1 Three-term homogeneous quadratic equation in x and a . 25 x − 60 ax + 27 a = 0 x = 95 a only. A1 Alternative method for question 4 5mge B1 In equilibrium, = 5mg , e = a . a Let x be the extension when P is released. B1 At least one correct EPE term seen. Energy equation in subsequent motion: 2 2 M1 Dimensionally correct energy equation with one KE term, 5mga 5mgx 1 7 − = mg ( e − x ) + 2 m ( 5 ag ) two EPE terms and at least one GPE term. 2 a 2 a A1 Fully correct energy equation. 25 x 2 − 10 ax − 8a 2 = 0 M1 Three-term homogeneous quadratic equation in x and a . 4 a only. A1 ( 5 x + 2 a )( 5 x − 4 a ) = 0, x = 5 Required length of string = 95 a . 6

More questions on Hooke's law

Q5 · Y D C B h 2a 45° O ka A x A uniform lamina OABCD consists of a rectangle OACD and a…

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. 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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. 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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

More questions on Equilibrium of a rigid body

Q6 · 17ag P 60° a O A particle P of mass m is attached to one end of a light inextensible…

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. 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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. 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The mass of particle Q is km and the collision between P and Q is perfectly elastic. (c) Find the value of k. 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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

More questions on Circular motion

Q7 · A particle P is projected from a point O on a horizontal plane and moves freely under…

7 A particle P is projected from a point O on a horizontal plane and moves freely under gravity. The initial velocity of P is 25 ms -1 at an angle i above the horizontal, where tan i = 4 . At point A, the 3 direction of motion of P makes an angle of 45° with the downward vertical through A. (a) By differentiating the equation of the trajectory or otherwise, find the coordinates of A. 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At point A, the particle strikes a fixed smooth barrier, rebounds, and lands on the horizontal plane. The barrier is inclined at an angle of 45° to the horizontal. (b) Find the speed of P immediately before it collides with the barrier. 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(c) Given that the coefficient of restitution between the barrier and the particle is 1, find the horizontal 9 distance travelled by P after it strikes the barrier. 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Mark scheme: 7(a) dy 2 gx M1 Valid attempt at differentiating. = tan− dx 2u 2 cos 2  dy B1 = −1 dx −=1 43 − 452 x A M1 Form equation with –1 and attempt to find a value for x A . x A = 1052 A1 52.5 or 525 . g y A = 354 A1 175 35 dy 8.75 or . SC B1 for 4 OE from setting = +1 . 2g dx Alternative method for question 7(a) Horizontally: vH = 25cos= 15 M1 Both equations, allow sign error. Vertically: Vv = 25sin− gt = 20 − gt vV B1 At A: = − tan 45 = −1 v H 20 − gt 7 M1 With RHS = –1, find value for t and x A or y A = −1, t = 2 15 x A = 1052 A1 52.5 or 525 . g y A = 354 A1 175 35 dy 8.75 or . SC B1 for 4 OE from setting = +1 . 2g dx 5 7(b) Let v be the velocity before the collision at A. B1 u x = u cos= 15 = v x v y = − v x = − 15 B1 2 2 B1 FT Speed: v x + v y = 15 2 (m s–1) 3 7(c) Let w be the velocity after the collision. M1 Correct application of NEL. w = ev = 53 2 A1 2 M1 Substitute values into trajectory equation and solve to find x. 35 gx 9 2 35 , − 4 = x tan45 − 5 x − x − 4 = 0 . OR 2 2 5 2 2 cos45 ) Complete method to find x that firstly determines the time of ( 3 ) ( the flight after the particle strikes the barrier. OR 35 4 = − 53 2  12 2 t + 12 gt 2 , t = 32 5 x = ( w cos45 ) t  = 3  32  x = 52 (metres) A1 4

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