Cambridge A Level Mathematics - Further 9231 — 2025 May/June Paper 3 · Variant 3

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

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

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2 A particle P of mass m is attached to one end of a light elastic string of natural length a and modulus of elasticity mg. The other end of the string is attached to a fixed point O on a rough plane inclined at an angle of 30° to the horizontal. The particle P is held at rest at point O before being released. The frictional force acting on P as it slides down the plane is 11 mg . 30 (a) Find, in terms of a, the distance that P moves down the plane before coming to rest. 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(b) It is given that P remains at rest in this new position. Find, in terms of m and g, the magnitude of the frictional force in this position. 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Mark scheme: 2(a) mg 2 11 M1 Energy equation with all three terms. ( x − a ) = mgx sin30 − 30 mgx Dimensionally correct. 2 a B1 Two terms correct (need not be in equation). A1 Correct energy equation. 15 x 2 − 34ax + 15a 2 = 0 M1 Simplify to three-term homogeneous quadratic equation and attempt to solve. (3 x − 5a )(5 x − 3a ) = 0 x = 53 a A1 CAO, must reject x = 53 a if seen. 5 2(b) T = 23 mg B1 F = T − mg sin30   = 23 mg − 12 mg  M1 Equation for F. Allow sin/cos mix. Allow sign errors. 1 6 mg A1 3

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3 A ball of mass m kg is projected vertically upwards with initial speed U m s -1 and moves under gravity. At time t s after projection, the ball has travelled a distance x m and its speed is v m s -1 . There is a resistive force of magnitude mkv 2 N , where k is a positive constant. 1 g + kU 2 (a) Show that the distance travelled by the ball when it is moving upwards is x = ln f p. 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It is given that k = 0.025 and that U = 20 . (b) Find the time taken for the ball to reach its maximum height. 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Mark scheme: 3(a) 2 dv B1 Correct use of N2L with m seen. − mg − mkv = mv dx v 1 2 M1 Attempt to integrate by separating variables. d x = − ln g + kv d v leading to x = − 2 ( ) + c  . Correct form. Allow missing + c .  2 k  g + kv 1 2 M1 v = U when x = 0 , so c = ln g + kU . ( ) 2 k 1  g + kU 2  A1 AG, shown convincingly. x = ln  2  (metres) 2 k  g + kv  4 3(b) 1 2 dv *M1 Attempt to integrate by separating variables. 10 + 40 v = − dt 40 2  dt = − 400 + v dv 40 −1 1 A1 Correct integration. t = − 20 tan ( 20 v )( +C ) When t = 0 , v = 20 , so C = 12 π DM1 Find C and substitute v = 0 to find value for t . π t = −2tan −1 (0) + 12 t = 1.57 (seconds) A1 CWO, accept final answer of 12 π . 4

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4 B C a E 2a F 2a A D a An object consists of a uniform lamina with a particle attached. The uniform lamina ABCEFD of mass m is formed from a rectangle ABCD and an isosceles triangle CEF, where F is the midpoint of CD. The rectangle has sides AB = 2a and AD = a . The triangle CEF has base a and height 2a. The particle of mass km is attached to the lamina at E. The object rests in a vertical plane with its edge AD on horizontal ground (see diagram). Given that the object is on the point of toppling in its vertical plane about the vertex D, find the value of k. 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Mark scheme: 4 B1 Allow one error in first two columns. Areas 2 a ) 2 a 2a 2 12 a ( 2 + 12 a ( 2 a ) Correct distances about any point. SOI, may be seen in moments equation. Distances from D − 12 a 23 a x 1 1 2 *M1 A moments equation with 3 terms, must be 2 a 2 3a 2 x −2 a 2 ( 2 a ) + 2 ( )( 3 a ) = ( ) 1 dimensionally correct. ( x =  9 a or x = 89 a ) 1 DM1 Moments equation about toppling point D. m ( 9 a ) = km ( 2a ) 2 terms only, OE. Must include k. k = 181 A1 Alternative method for question 4 Ratio of masses 2a 2 12 a ( 2 a ) k ( 2 a 2 + 12 a ( 2 a ) ) B1 CorrectMust simplyratiostoincluding2:1:3k .k. SOI, may be seen in moments equation. 1 2 2 ( 2 a ) = 1( 3 a ) + 3k ( 2a ) *M1 A moments equation about D with 3 terms, must be dimensionally correct. Must include k. DM1 Using correct distances. k = 181 A1 4

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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. 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(b) In the subsequent motion find, in terms of r, the greatest height above O reached by the particle. 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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

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6 A B u θ θ 2u m m Two identical uniform smooth spheres A and B, each with mass m, are moving on a horizontal surface with speeds 2u and u respectively when they collide. Immediately before the collision, the spheres are moving parallel to each other in opposite directions such that their directions of motion each make an angle i with the line of centres (see diagram). As a result of the collision, B moves in a direction which is perpendicular to its initial direction of motion. The coefficient of restitution between the spheres is e. (a) Find an expression for tani in terms of e. 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As a result of the collision, A moves in a direction which is perpendicular to the line of centres. (b) Find the value of i. 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Mark scheme: 6(a) Let v and w denote the horizontal components of the velocities for A and B M1 Allow sign errors, m must be present. respectively after the collision. 2mu cos− mu cos= mw + mv w − v = e (3u cos) M1 Restitution equation: directions must be consistent with those used in momentum equation (signs must be consistent). 1 w = 12 ( 3eu cos + u cos)  = 2 (3e + 1)u cos A1 Combining to find equation with w, e, u, . SOI 1 2 ( 3e + 1) u cos M1 OE, using perpendicular motion of B. = tan u sin tan 2= 12 (3e + 1) M1 Obtaining an expression for tan in terms of e . 1 A1 CWO, must reject negative term if seen. tan= 2 (3e + 1) 6 6(b) w = 3eu cos= u cos M1 Using v = 0 to find a value for e. e = 13 tan= 1 , = 45 A1 Alternative method for question 6(b) w M1 v = 0, w = u cos and tan u sin= u cos= u sintan, tan 2 = 1, = 45 A1 2

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7 A particle P is projected from a point O with speed U at an angle 45° above the horizontal and moves freely under gravity. (a) State the vertical and horizontal components of velocity at time t. 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At time T, particle P is moving at an angle of 60° below the horizontal. U (b) Show that T = ` 2 + 6j. 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At time T, the particle strikes a smooth horizontal plane at a point which is a horizontal distance D from O and a vertical distance H below O. (c) Find the ratio H : D. 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After striking the horizontal plane, P rebounds with speed w. The coefficient of restitution between P and the plane is 2. 3 (d) Find w in terms of U. 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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) yv = U sin45 − gT B1 vx = U cos45 1 7(b) v y B1 SOI = − tan60 v x  U  M1 Substitute values for trigonometric ratios to obtain − gT   a linear equation in U and T only. 2  = − 3  U − 2 gT = − 3 U  U   2  U 2U U A1 AG, shown convincingly. T = 1 + 3 = 1 + 3 = 2 + 6 ( ) ( ) ( ) 2 g 2 g 2 g 3 7(c) U B1 D in terms of U and T . D =  T 2 U 1 2 B1 H in terms of U and T . H =  T − 2 gT With or without modulus sign. 2 U 1 2 M1 OE 1 1  T − 2 gT 2 + 6 Obtain a numerical value, may be un-simplified. 2 U ( 2 2 − 12 ( ) ) 1 − 3 2 = = Division or ratios. 1 U  T 2 2 2 With or without modulus sign. 2 H : D = 1 :1 + 3 A1 OE (for example H : D = 3 − 1:2 ), CWO 7(c) Alternative method for question 7(c) U  U  B1 D in terms of U and g . 2 + 6 D =  ( )  2  2 g  U  U  1  U  2  U 2  B1 H in terms of U and g . 2 + 6 2 + 6 = −  With or without modulus sign. H =  ( )  − 2 g  ( )   2  2 g   2 g   2 g  2 M1 OE U  U  1  U  2 + 6 2 + 6 Obtain a numerical value, may be un-simplified.  ( )  − 2 g  ( )  2  2 g   2 g  1 − 3 Division or ratios. = U  U  2 With or without modulus sign. 2 + 6  ( )  2  2 g  H : D = 1 :1 + 3 A1 OE (for example H : D = 3 − 1:2 ), CWO Alternative method for question 7(c) U U B1 U 2 D = 2 + 6 1 + 3 ( ) D = ( ) 2 2 g 2 g 2 2 B1  6   2    U  =  U  − 2 gH      2   2  U 2 M1 (  ) H = 2 g H : D = 1 :1 + 3 A1 OE (for example H : D = 3 − 1:2 ), CWO 4 7(d) 2 3  2  B1 OE, accept un-simplified. wy =  U  =  U  Allow negative sign. 3 2  3  2 M1 2 2  U  2 2 1 2 w = w y +    = 3 U + 2 U   2  2 1 7 A1 CWO w = U 3 + 2 = U 6 3

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