Cambridge A Level Mathematics - Further 9231 — 2025 Oct/Nov Paper 3 · Variant 2

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] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

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

More questions on Circular motion

2 One end of a light elastic string of natural length a and modulus of elasticity 2mg is attached to a fixed point A on a rough horizontal surface. The other end of the string is attached to a particle P of mass m. The particle and string rest on the surface. The coefficient of friction between P and the surface is n. The particle P is initially held in equilibrium at a distance 4 a from A. The particle is then released from rest. 3 (a) Given that the string never becomes slack, find the minimum value of n. 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It is now given that n = 1 . 2 (b) Find the extension of the string when the particle comes to rest. 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Mark scheme: 2 B1 Correct EPE seen OR correct work done by2(a) 2 mg 1 EPE = ( 3 a ) friction seen. 2 a 1 Work done by friction = mg ( 3 a ) 2 2 mg 1 1 ( 3 a ) = mg ( 3 a ) M1 Equationwork doneortermsinequalityonly. involving EPE and 2 a = 13 A1 CWO 3 2(b) Let x be the extension in the string when P comes to rest. M1 Dimensionally correct equation with two EPE 2 mg 1 ( 3 a ) 2 = 2 mg x 2 + 12 mg ( 13 a − x ) terms and1 one work done term. 2 a 2 a Use of 2 mgx as work done term can be awarded M1 only. A1 Correct energy equation. May be unsimplified.  18 x 2 − 9ax + a 2 = 0, ( 6 x − a )( 3 x − a ) = 0  x = 16 a only. A1 CWO   Alternative method for question 2(b) Let x be the distance moved by P before coming to rest. M1 Dimensionally correct equation with two EPE 2 2 mg 1 2 2 mg 1 1 ( 3 a ) = ( 3 a − x ) + 2 mgx terms and1 one1 work done term. as work done term can 2 a 2 a Use of 2 mg ( 3 a − x ) be awarded M1 only. A1 Correct energy equation. May be unsimplified. 2 1 A1 CWO 6 x − ax = 0, x = 6 a Extension = 13 a − x = 16 a 3

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3 A uniform lamina OABCD is in the form of a rectangle, OBCD, joined along the edge OB to a quarter circle OAB. The length of DO is ka and the length of OB is a. The lamina rests in a vertical plane with its edge CB on a horizontal surface (see diagram). ka D O A a C B (a) Find, in terms of k, a and r, an expression for the distance of the centre of mass above the horizontal surface. 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The lamina is on the point of toppling about B. (b) Find the value of k. 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Mark scheme: 3(a) Vertical distance of COM of sector from OA: B1 SOI ( 2 a cos ( 14 π ) ) sin ( 14 π ) 4 a Allow unsimplified. = 1 3 ( 4 π ) 3π 2 1 1 2  4 a  2 1 2 M1 3-term dimensionally correct moments y ka ( 2 a ) + 4 π a  a −  = ( ka + 4 π a ) equation.  3π   6ka + 3π a − 4a = 3 ( 4k + π ) y  A1 FT Correct moments equation, allow unsimplified. FT their vertical distance of COM of sector from OA. 6 k + 3π − 4 A1 AEF, but do not allow fractions within y = a 3 ( 4 k + π ) numerator and/or denominator. 4 3(b) 2 1 1 2  4 a  M1 Dimensionally correct moments equation ka ( 2 ka ) = 4 π a   about OB using their horizontal distance of  3π  COM of sector from OB.   k 2 = 23  k = 23 = 0.816 A1 Alternative method for question 3(b) 2 M1 Use of x = ka with their x , which may be 6 k + 3kπ + 4 a ( ) On the point of toppling, x = ka : = ka seen in part 3(a). 3 ( 4 k + π )  6k 2 + 4 = 12k 2  k = 23 = 0.816 A1   2

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4 A particle Q is initially positioned at a distance d vertically above a particle P. Particle P is projected with speed U at an angle a above the horizontal. At the same time, Q is projected at an angle b below the horizontal. Both particles move freely under gravity. The particles collide at time T after the projections. (a) Show that d = UT ( sin a + cos a tan b ) . 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The particles collide when P is at its maximum height. (b) Given that a = 30° and b = 60° , find d in terms of U and g. 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Mark scheme: 4(a) Horizontal velocities are the same: B1 B0 M0 M0 A0 for using U q = U . U q cos = U cos 1 2 1 2 M1 Apply vertical equation of motion on both T + 2 gT d = (U sin) T − 2 gT + ( U q sin) particles and combine.  Ucos  M1 Substitute U q and simplify. d = (Usin) T + sin T  cos  d = UT ( sin+ costan) A1 AG, shown convincingly. 4 4(b) Usin= gT B1 Correct equation for T. 2 M1 Substitute their expression for T in given U sin d = ( sin+ costan) result from part 4(a) OR complete method for g finding d in terms of U and g only. May include unevaluated trigonometric functions. Note: d = 2UT and U = 2 gT . 2 A1 U d = g 3

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5 Two uniform smooth spheres, A and B, of equal radii are on a horizontal surface. They have masses of m and km respectively. Sphere A is moving with speed u at an angle a with the line of centres when it collides with sphere B which is stationary. Immediately after the collision, sphere A moves with speed v at an angle 2a with the line of centres (see diagram). A v B 2α α u m km It is given that tan a = 3 . 4 (a) Find v in terms of u. 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(b) Find the coefficient of restitution between the spheres in terms of k. 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(c) Find the range of possible values of k. 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Mark scheme: 5(a) u sin= v sin ( 2) B1 May be implied by correct answer.   5 3 u = 2425 v  v = 85 u B1 2 5(b) mu cos= mv cos ( 2) + kmw B1 PCLM equation, must include masses.   54 u = 257 v + kw, kw = 85 u  B1 NEL equation. eu cos= w − v cos ( 2)   54 eu = w − 257 v  5 4 5 7 M1 Combine and substitute for v and w to obtain = e +  equation in e and k only. 8k 5 8 25 25 7 A1 25 −k7 e = − AEF, for example . 32 k 32 32 k Not dependent on first B mark. 4 5(c) 25 7 M1 Form inequality using their expression for e. 0 ≤ − ≤ 1 Allow equality or strict inequality. 32k 32 7 k ≤ 25 ≤ 39k M1 Rearrange at least one side to the form ak ≤ b or ak ≥ b . Allow equality or strict inequality. 39 25 ≤ k ≤ 257 A1 CAO 3

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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. 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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

More questions on Circular motion

7 A particle P of mass m kg moving along a rough horizontal table has displacement x m from a fixed point O on the table and velocity v m s -1 at time t s. The particle P is subject to a resistive force of magnitude mgkv N, where k is a positive constant, and a frictional force of magnitude nmg . The particle P is initially at O with speed U m s -1 . 1 kU + n (a) Show that t = ln e o. 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It is given that U = 10 , k = 0.04 and n = .02 . (b) Find the distance P moves before coming to rest. 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(c) Find the average speed of P over the period it is moving. 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Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................

Mark scheme: 7(a) dv B1 Must have m. − mgkv − mg = m dt 1 1 *M1 Separate variables and integrate to a natural t = − dv logarithm term. g  kv +  Ignore modulus signs in this part. 1 t = − ln ( kv + ) + C  gk 1 DM1 Use initial condition to find constant OR use When v = u, t = 0: C = ln ( kU + ) correct limits. gk Dependent on having integrated to obtain a natural logarithm term. 1 1 A1 AG, shown convincingly. t = − ln ( kv + ) + ln ( kU + ) Must see an intermediate step before reaching gk gk given result. 1  kU +  Not dependent on B1. t = ln   gk  kv +   4 7(b) 2 dv M1 Three terms in equation of motion, separating − 5 v − 2 = v variables and expressing integrand in a form dx that can be directly integrated. 5 v 5   5  x = − 2  dv = − 2   1 −  dv May be in terms of k ,U and , for example  v + 5   v + 5  1   k  x = − 1 − dv .    gk v + k     Use of suvat in this part cannot be awarded any credit. x = − 52 v + 252 ln ( v + 5 ) + C  M1 Correct form. May be in terms of k ,U and , for example 1    x = − v + ln  v +  + C . 2 gk gk  k  When x = 0, v = 10: C = 25 − 252 ln (15 ) M1 Use initial condition to find constant OR use correct limits in a definite integral. 25 When v = 0, x = 25 − 252 ln (15 ) + 2 ln ( 5 ) = 11.3 [m] A1 Alternative method for question 7(b) 2 M1 Rearrange answer to part 7(a) in the form − t  15  5 − 5 From part (a), t = 52 ln   , v = 15e v = ae bt + c .  v + 5  Use of suvat in this part cannot be awarded any credit. 75 − 52 t M1 Integrate to correct form. x = − 2 e − 5t  + B  75 75 − 52 t 75 M1 Use initial condition to find constant OR use When t = 0, x = 0, B = 2 , x = − 2 e − 5t + 2 correct limits in a definite integral. When v = 0, t = 2.5ln3, x = 25 − 252 ln3 = 11.3 [m] A1 7(b) Alternative method for question 7(b) 2 t M1 Rearrange answer to part 7(a) in the form 5  15  − 5 , v = 15e − 5 From part (a), t = 2 ln   v = ae bt + c .  v + 5  Use of suvat in this part cannot be awarded any credit. 2.5ln3 − 52 t M1 When v = 0, t = 2.5ln3 so x =  0 (15e − 5 ) dt 75 x =  − − 52 t − 5t  2.5ln3 M1 Integrate to correct form. 2 e   0 75 1 75 25 A1 [m] x = − 2 −3 5 12.5  ln3 −−( 2 ) = 25 − 2 ln3 = 11.3 4 7(c) "11.3" "11.3" M1 Divide their x from part 7(b) by the value of t Average speed = = 1  0.04  10 + 0.2  2.5ln3 when v = 0 using the given answer from part ln   7(a). Condone if their x from part 7(b) is 10  0.04  0.2  obtained using suvat. 4.10 [m s–1] A1 4.102, accept values in the range 4.10 – 4.11. 2

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