Cambridge A Level Mathematics - Further 9231 — 2024 Oct/Nov Paper 3 · Variant 2

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

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

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2 A particle P of mass m is attached to one end of a light elastic spring of natural length a and modulus of elasticity 5mg. The other end of the spring is attached to a fixed point O. The spring hangs vertically with P below O. The particle P is pulled down vertically and released from rest when the length of the spring is 3 a . 2 Find the distance of P below O when P first comes to instantaneous rest. [4] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................. ............................................................................................................................

Mark scheme: 2 1 5 mg   a  2 2  B1 Both terms seen. Extension when P comes to rest is x, EPE loss =      − x  2 a   2     1 5mg   a  2 2   a  M1 At least one EPE term and a GPE term, Energy:      − x  = mg  − x  dimensionally correct. 2 a   2    2    Solve: 20 x 2 − 8ax − a 2 = 0 M1 Obtain homogeneous quadratic equation in x and a Must come from an energy equation involving two EPE terms. Note that the correct case simplifies to a linear 5mg  a  equation  + x  = mg and this scores M1. 2 a  2  a 9 A1 x = − , so distance of P below O is a 10 10 2 Alternative method for question 2 Distance of P below O when it comes to rest comes to rest is h B1 1 5mg   a  2 2  EPE loss = =      − ( h − a )  2 a   2     1 5mg   a  2 2   3  M1 At least one EPE term and a GPE term, Energy:      − ( h − a )  = mg  a − h  dimensionally correct. 2 a   2    2    20h 2 − 48ah + 27 a 2 = 0 M1 Obtain homogeneous quadratic/linear equation in h and a. Must come from an energy equation involving two EPE terms. 9 A1 (10 h − 9 a )( 2 h − 3a ) = 0, h = a 10 4

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3 3u 2u A B θ m m The diagram shows two identical smooth uniform spheres A and B of equal radii and each of mass m. The two spheres are moving on a smooth horizontal surface when they collide with speeds 2u and 3u respectively. Immediately before the collision, A’s direction of motion makes an angle i with the line of centres and B’s direction of motion is perpendicular to that of A. After the collision, B moves perpendicular to the line of centres. The coefficient of restitution between the spheres is 1. 3 (a) Find the value of tani. 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(b) Find the total loss of kinetic energy as a result of the collision. 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(c) Find, in degrees, the angle through which the direction of motion of A is deflected as a result of the collision. 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Mark scheme: 3(a) PCLM along line of centres: −mv = m2u cos− m3u sin B1 Must include m, must have minus sign on RHS, accept positive or negative v. If velocity of B after collision is included, it must be equated to zero before this mark is awarded. NEL: v = eu ( 3sin+ 2cos) M1 Must have plus sign on RHS, accept positive or negative v (sign of v does not need to be consistent with PCLM equation) If velocity of B after collision is included, it must be equated to zero before this mark is awarded. 4 A1 Correct work only, except possibly missing m. Eliminate v: 6sin= 8cos, tan= 3 3 3(b) Only change in KE is along line of centres M1 1 2 2 1 2 Loss = m ( 2u cos) + ( 3u sin) − mv ( ) 2 2 1 2  36 144  12 6  2  72 2 A1 (Note that tan= 23 leads to final answer 1336 mu 2 ) mu  + −  −   = mu 2  25 25  5 5   25   2 Alternative method for question 3(b) Alternative method, using total KE M1 1 2 1 2 Loss in KE = [ m ( 2u ) + m ( 3u ) ] 2 2 Or equivalent, with all necessary terms present  1 2 1 2 1 2  −  mv + m ( 2u sin) + m ( 3u cos)   2 2 2  13 2 181 2 72 2 A1 mu − mu = mu 2 50 25 2 3(c) 6 8 M1 [Components of velocity of A after collision are  u  u so] angle between 5 5 line of centres and A’s direction is . Angle of deflection = 180 – 2tan-1 4/3 = 73.7 A1FT FT their answer to part (a) 2

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4 B D 3a E C ka i 3a A The end A of a uniform rod AB of length 6a and weight W is in contact with a rough vertical wall. One end of a light inextensible string of length 3a is attached to the midpoint C of the rod. The other end of the string is attached to a point D on the wall, vertically above A. The rod is in equilibrium when the angle between the rod and the wall is i, where tan i = 3 . A particle of weight W is attached to the point 2 E on the rod, where the distance AE is equal to ka (3 1 k 1 6) (see diagram). The rod and the string are in a vertical plane perpendicular to the wall. The coefficient of friction between the rod and the wall is 1. The rod is about to slip down the wall. 3 (a) Find the value of k. 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(b) Find, in terms of W, the magnitude of the frictional force between the rod and the wall. 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Mark scheme: 4(a) In this question, allow equivalent marks for resolutions in different directions and B1 moments about other points. Apply the guidance given in the main scheme.  T cos+ F = 2W → T sin= R B1 Moments about A: M1 All relevant terms included, dimensionally correct, T cos 3a sin+ T sin 3a cos= W  3a sin+ W  ka sin forces must be resolved if appropriate. Allow sin/cos mix, allow sign errors. LHS: any equivalent expression, for example 3Ta sin 2, 3Tasin (180 − 2) . OR All relevant terms included, dimensionally correct, Moments about C: forces must be resolved if appropriate. R  3a cos = F  3a sin+ W  ( ka − 3a ) sin Allow sin/cos mix, allow sign errors. T M1 1 [6 aT cos= ( 3 + k ) aW and (3cos+ sin) = 2W give]  and giveand give Use F = R and eliminate T and W to obtain an 3 3 1 expression in k and , dependent on a 12cos= ( 3 + k ) (cos+ sin) 3 dimensionally correct moments equation. k = 5 A1 5 4(b) A complete method to find F in terms of W M1 Any complete method to find F. For example, substitute into moments equation to 2 obtain T [T = 13 W, R = 2W ] . 3 2 A1 Correct. F = W 3 2

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5 A particle P is projected from a point O on horizontal ground with speed u at an angle i above the horizontal, where tan i = 1 . The particle P moves freely under gravity and passes through the point 3 with coordinates (3a, 4 a ) relative to horizontal and vertical axes through O in the plane of the motion. 5 (a) Use the equation of the trajectory to show that u 2 = 25ag . 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At the instant when P is moving horizontally, a particle Q is projected from O with speed V at an angle a above the horizontal. The particles P and Q reach the ground at the same point and at the same time. (b) Express V 2 in the form kag, where k is a rational number. 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Mark scheme: 5(a) 4 1 g 2  1  M1 No (implied) sight of trajectory equation M0. 3a )   1 +  Use correct equation of trajectory: a = 3a  − 2  ( 5 3 2u  9  4 5 ga 2 5 ga 1 2 A1 At least one step of intermediate working must be a = a − , = , u = 25 ga seen. 2 2 5 u u 5 AG 2 5(b) For P, time of flight T and range R B1 For Q, time of flight ½ T and range R  2u sin  10 a Time of flight for P or Q. T = =    g  g 2 3 B1 Range for P. [From motion of P, R =  25ag  = ] 15a g 10 1 2 R M1 Obtain an expression for v cos. May involve u For Q: → R = v cos T , v cos= and . 2 T 2 M1 Obtain an expression for v sin. May involve u 1 1  1  1  0 = v sin T − g  T  , v sin= gT and . 2 2  2  4 2 2 M1 2  2 R   1   5  Square and add: v = + gT = 90ag + ag        T   4   8  2 725 A1 1 v = ag gT 8 tan = 4 = 1 2 R 12 T 5(b) Alternative method for question 5(b) For P, time of flight T and range R M1 For Q, time of flight ½ T and range R Horizontal motion for P and Q T R = u cosT and R = ( v cos) 2 Both. Vertical motion for P and Q M1 gT gT 1 2 u sin= and v sin= Both, may come from using s = ut + at . 2 4 2 Equate two expressions for R: A1 6 v cos= u v cos= 2u cos 10 Equate two expressions for vertical motion: A1 1 1 v sin= u sin v sin= u 2 2 10 2 2  2 1 2  29 2  M1 Square and add: v = u  4cos + sin  = u   4  8  29 725 A1  25ag = ag 8 8 6

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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. 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(b) Find the ratio of the tensions in the string immediately before and immediately after it strikes the peg. 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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

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7 A particle P of mass m kg is held at rest at a point O and released so that it moves vertically under gravity against a resistive force of magnitude 0.1mv 2 N, where v m s -1 is the velocity of P at time t s. (a) Find an expression for v in terms of t. 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The displacement of P from O at time t s is x m. (b) Find an expression for v2 in terms of x. 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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 2 B1 Use of suvat means 0 marks in this part m = mg − 0.1mv Note that no mg term means 0 marks in this part. dt d v Must see m, may be cancelled before a = used d t dv 2 1 2 M1* = 10 − 0.1v = 100 − v ( ) dt 10 dv 1 = dt 100 − v 2 10 Separate variables and integrate. May see partial fractions, but integral is on Formula sheet,  v + 10 allow missing + A for M1 only ln = 2t + A    10 − v  A1 Must see modulus sign Use t = 0, v = 0, A = 0 DM1 Remove logs to obtain v in terms of t M1 2 t A1 −2 t 10 e − 1 10 1 − e ( ) ( ) v = aef v = e 2 t + 1 1 + e −2 t 6 7(b) dv 1 2 M1* Use of suvat means 0 marks in this part v = 100 − v ( ) dx 10 Separate variables and integrate, allow missing + A for M1 only. 1 2 1 − ln 100 − v = x + B ( ) 2 10 A1 For A1, allow missing modulus sign 1 DM1 Use x = 0, v = 0, B = − ln100 2 Remove logs to obtain v 2 in terms of x M1 x A1 AEF − v 2 = 100(1 − e 5 ) Allow 10g instead of 100. 5

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