Cambridge A Level Mathematics 9709 — 2025 Feb/March Paper 4 · Variant 2

1 30 N 40 N i° X N Three coplanar forces of magnitudes 40 N, 30 N and X N act at a point in the directions shown in the diagram. Given that the forces are in equilibrium, find the values of i and X. [4] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1 For resolving and forming an equation in any direction *M1 Allow sin/cos mix; correct number of terms. X cos= 40 X sin= 30 A1 For correct resolving in two directions. 2 2 −1 30 DM1 For attempt to solve for either. X = 40 + 30 or  = tan 40 X = 50 , = 36.9 A1 36.869… AWRT 50.0. Alternative for Q1 using triangle of forces 2 2 M1 For attempt to solve for X using Pythagoras. X = 40 + 30 X = 50 A1 AWRT 50.0. 30 M1 For attempt to solve for θ. tan= 40 = 36.9 A1 AWRT 36.9. Alternative for Q1 using Lami’s Theorem X 30 40 *M1 For any 2 fractions correct. = = sin90 sin (180 − ) sin ( 90 + ) A1 For all 3 fractions correct. −1 30  DM1 Solve for θ.  = tan    40  X = 50 , = 36.9 A1 36.869… AWRT 50.0. 4

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2 A cyclist is travelling along a straight horizontal road at a speed of 4 m s -1 when she passes a point O. She accelerates at a constant rate for a distance of 42 m, reaching a speed of V m s -1. She maintains the speed of V m s -1 for 50 m and then decelerates at 2 m s – 2 before coming to rest. The distance travelled while decelerating is 16 m. (a) Find the value of V. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find the total time for which she is in motion from the instant that she passes O. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ .......................... ............................................................................................................................

Mark scheme: 2(a) 2 B1 For use of constant acceleration to get a correct 0 = V −2 2 16 equation in V only. V = 8 only B1 2 2(b) Acceleration section : M1 For attempt to find an equation in t during acceleration or deceleration or constant speed. ( 4 + ( theirV ) ) 42 = t  t = 7  Using their V , s = 16 , a = −2 , u = 4 . Must lead to a 2 positive t . Deceleration section: 1 2 1 2 0 = ( theirV ) − 2t or 16 = .2t or 16 = ( theirV ) t − .2t  t = 4  M1 For attempt to find an equation in t for the other 2 2 2 sections. 50 For constant speed section t =  t = 6.25  Using their V , s = 16 , a = −2 , u = 4 . Must lead to a theirV positive t . 69 A1 AWRT 17.3 from correct work Total time = s = 17.25 s 4 3

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3 An aeroplane is flying at a constant speed. (a) The aeroplane is flying horizontally. The aeroplane’s engines are producing a constant power of 5500 kW, and the aeroplane experiences a constant horizontal resistance force of 25 kN. Find the speed of the aeroplane. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) The aeroplane then ascends 300 m in 50 s, while maintaining the same speed. The resistance force is no longer constant, and the work done against the resistance force in ascending the 300 m is 270 000 kJ. The mass of the aeroplane is 60 000 kg. Find the average power of the aeroplane’s engines. [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 3(a) 5500 = 25v OR 5 500 000 = 25 000v M1 OE For use of Power = Fv. Allow errors in use of kN and/or kW. Speed = 220 m s−1 A1 2 3(b) Change in PE = 60  g  300 [kJ] OR 60 000  g  300 [J] B1 180 000 kJ or 180 000 000 J. Work done by engines = Power  50 B1 OE Power  50 = 60  g  300 + 270 000  Power  50 = 450 000  M1 For work energy equation with 3 terms; Allow with work done by engines instead of Power  50 ; OR Power  50 = 60 000  g  300 + 270 000 000 Allow sign errors; dimensionally correct.  Power  50 = 450 000 000  Required power = 9000 kW or 9 000 000 W A1 4

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4 0.3 kg 0.1 kg A B x m Two particles A and B have masses 0.3 kg and 0.1 kg respectively. The particles are attached to the ends of a light inextensible string. The string passes over a fixed smooth pulley, and the particles hang vertically below the pulley. Both particles are initially at a height of x m above horizontal ground (see diagram). The system is released from rest. (a) Find the tension in the string and the acceleration of the particles. 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During the subsequent motion, B does not reach the pulley. When A reaches the ground, it comes to rest. (b) Given that the greatest height of B above the ground is 1.2 m, find the value of x. 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Mark scheme: 4(a) Attempt at Newton’s second law for at least one case *M1 Allow g missing. Correct number of terms. Allow sign errors. 0.3 g − T = 0.3a A1 Any 2 consistent equations, e.g. allow −a for a if T − 0.1g = 0.1a consistent. 0.3 g − 0.1g = ( 0.3 + 0.1) a Must be same T if individual particle equations. Attempt to solve for T or a DM1 From equation(s) with correct number of relevant terms. Allow g missing. Must get to ‘T =’ or ‘a =’. If no solving seen, must be correct answers for their equations for this mark. Acceleration = 5 m s−2 Tension = 1.5 N A1 Allow acceleration = –5 m s−2. 4 4(b) 2 *M1 v = 0 + 2  theira  x For use of constant acceleration to find 2v or v in terms of x . Using their a , a  g . 2 DM1 2 2 0 = theirv − 2 g  (1.2 − 2 x ) For use of v = u + 2as to get an equation in x only. Allow a =  g . 2 2 theirv OR 0 = their v − 2 gs and 2 x + s = 1.2 [leading to 2 x + = 1.2 ] 2 g x = 0.48 A1 OE 3

More questions on Kinematics of motion in a straight line

5 P 0.6 kg Q 0.4 kg R 0.8 kg 3 m 3 m Three particles P, Q and R, of masses 0.6 kg, 0.4 kg and 0.8 kg respectively, are at rest in a straight line on a smooth horizontal plane. The distance from P to Q is 3 m, and the distance from Q to R is also 3 m (see diagram). P is projected directly towards Q with speed 3 m s -1. After P and Q collide, P continues to move in the same direction with speed 1.5 m s -1. (a) Find the speed of Q after the collision. 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In the subsequent collision between Q and R, these particles coalesce. (b) Find the speed of the combined particle after this collision. 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(c) Find the time that it takes from when P is initially projected until the instant at which P collides with the combined particle. 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Mark scheme: 5(a) 0.6 =3 0.6 1.5 + 0.4v M1 Attempt at conservation of momentum. 3 non-zero terms. Allow sign errors. Speed = 2.25 m s−1 A1 OE must be positive. Allow max M1A0 if g included with the masses. 2 5(b)  0.4  2.25 = ( 0.4 + 0.8 )  w  speed = 0.75 m s−1 B1FT OE condone including g if already penalised in (a). FT their 2.25. their 2.25 speed = 3 1 5(c) 3  4  *B1FT Q takes = s to reach the point at which R was initially.   their 2.25  3  3 *B1FT  1.5  = 2  their 2.25 3 OR 3 −  1.5  = 1 their 2.25 3 OR  ( their 0.75 ) = 1 their 2.25 Difference in speeds of P and QR = 1.5 − their 0.75  = 0 .75  m s−1 DM1 Dependent on both previous B marks. For attempt to find time.  3  3 −  1.5   their 2.25 so time =   1.5 − their 0.75      3   4  t OR ( their 0.75 ) t   3 −  1.5  = 1.5t  →=   their 2.25   3   3   8  T = OR( their 0.75 )  T   3 = 1.5T  →   their 2.25   3   3 3   3  11  T ' = OR ( their 0.75 )  T '   3 = 1.5  T '  →   their 2.25 3   3  3   3 3   3 3  11  T ' = OR ( their 0.75 )  T '  = 1.5  T   →   their 2.25 3   3 1.5   3  5(c)  3 4 4  A1 Allow 3.67 s. Time = + + = 11s    3 3 3  3 Alternative for Q5(c) 3  4  *B1FT Q takes = s to reach the point at which R was initially.   their 2.25  3  3 *B1FT P takes  = 2  s to reach the point at which R was initially, so combined 1.5 3 3  2  particle has travelled for − = s beyond where R was initially.   1.5 their 2.25  3   3 3  So combined particle is  −   ( their 0.75 ) = 0.5  m beyond  1.5 their 2.25  where R was initially. Difference in speeds of P and QR = 1.5 − their 0.75  = 0 .75  m s−1 M1 Dependent on both previous B marks. For attempt to find time.   3 3   their 0.75 )   −   (  1.5 their 2.25   2  so time =    =    1.5 − their 0.75  3       3 2  11 A1 Allow 3.67 s. Time = + 2 + = s    3 3  3 4

More questions on Kinematics of motion in a straight line

6 X N 12 kg a A block of mass 12 kg is placed on a rough plane inclined at an angle of a to the horizontal, where – 1 a = tan 0 .5. A force of X N is applied to the block, directly up the plane (see diagram). The coefficient of friction between the block and the plane is n. (a) It is given that n = .015 and X = 20 . Find the time that it takes for the block to move 2 m down the plane from rest. 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(b) It is given instead that n ! .015 and that when X = 10 , the block is on the point of moving down the plane. Find the value of n and the value of X for which the block is on the point of moving up the plane. 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Mark scheme: 6(a) −1 2 B1 2 R = 12 g cos tan 0.5 = 12 g  = 12 g  cos26.565  Allow cos27 or better for . = 26.56505118 . ( ) 5 5 For reference R = 48 5 = 107.3312629 , 36 5 F = = 16.09968944 5 1 *M1 For use of N2L with 4 terms. Allow sign errors. Allow 12 g  − 20 − F = 12 a sin/cos mix. Allow g missing 5 1 12 g sin26.565−. 20 − F = 12a Allow sin27 or better for . −1 5 12 g sin tan 0.5 − 20 − F = 12a ( ) Allow their possibly incorrect F . 1 2 DM1 For use of F = 0.15R to get an equation in a only, 12 g  − 20 − 0.15  12 g  = 12 a where R is a component of weight or mass. 5 5 12 g sin26.565−. 20 − 0.15  12 g  cos26.565= 12a 12 g sin tan −1 0.5 − 20 − 0.15  12 g  cos tan −1 0.5 = 12a ( ) ( ) −25 + 21 5 A1 SOI. Allow AWRT 1.5 a = 1.46382 a = 1.46 or a = 15 1 2 DM1 Dependent on both M marks. 2 = 0 + their a t For use of constant acceleration to find t. 2 Allow their a . t = 1.65 s A1 t = 1.65304 Allow 1.66 from using a = 1.46 . 6 6(b) For resolving forces parallel to the slope to form an equation in either case *M1 3 terms; allow sin/cos mix. 1 A1 F = 43.7  43.665. 10 + F − 12 g  = 0 10 + F − 12 g  sin 26.565 =. 0  5 1 Allow sin27 or better for . 10 + F − 12 g  sin tan −1 0.5 = 0  5  ( )  2 AND Allow cos27 or better for . 1 5 X − F − 12 g  = 0 5  X − F − 12 g  sin 26.565=. 0   X − F − 12 g  sin tan −1 0.5 = 0   ( )  Solve for X or  DM1 Solving for  must be using R as a component of weight. From equation(s) with the correct number of relevant terms and no sign errors. X = 97.3 and A1 X = −10 + 48 5 . = 0.407 12 − 5 = . 24 Allow X = 97.4 or 97.5 from correct work. Allow = 0.408 from correct work. 4

More questions on Kinematics of motion in a straight line

7 A particle moves in a straight line. The velocity v m s -1 of the particle t s after leaving a fixed point O is given by v = k ( 20 + pt - 6t 2 ) , where k and p are constants. The acceleration of the particle at t = 1 is 42 m s -2 , and the displacement of the particle from O at t = 1 is 93 m. (a) Show that k = 3 and p = 26 . 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(b) Find the distance moved by the particle between the time at which its acceleration is zero and the time at which its velocity is zero. 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Mark scheme: 7(a) For attempt to differentiate v *M1 Decrease power by 1 and a change in coefficient in at least one term (which must be the same term); v a = is M0 t Substitute a = 42 and t = 1 to get A1 OE; Allow unsimplified. 42 = k p −2 6 11 = ( )  k ( p − 12  1) = kp − 12 k 1 For attempt to integrate v *M1 Increase power by 1 and a change in coefficient in at least one term (which must be the same term) s = vt is M0 Substitute s = 93 and t = 1 to get A1 OE; Allow unsimplified.  20 1 1 1+1 6 2 +1   1 2 3  93 = k   1 + p  1 −  1  = k  20 +1 p  1 −2 1   1 1 + 1 2 + 1   2  solving simultaneously for p or k DM1 Dependent on both previous M marks. Allow sign errors only in solving.   1   20 + p − 2 p − 12 ) and 93 = k   42 = k (   Must be solving the correct equations.   2   Must have c = 0 if evaluated. Must get to ‘p =’ or ‘k =’or attempt to verify for both equations. Working must be seen for this mark. Must see at least one line of working once either p or k have been eliminated. p = 26 k = 3 A1 AG Any error seen is A0 6 7(b) 13 *M1 Using their 2 term linear a that has come from = t  a 0  3 ( 26 − 12t ) = 0 = differentiation to solve for t , which must be positive, 6 using correct p . 26 OE e.g. . 12 2 *M1 Attempt to solve given quadratic expression equated to v = 0  3 20 + 26t − 6t = 0 ( ) 0 using correct p and k . Must get at least 1 t value.  2  A1 If 2 values given, they must be both correct. t = 5 or t = −    3  5 DM1 Dependent on previous 2 M marks. 2 3  Distance =  3 20t + 13t − 2t ( )   13 For using their positive limits correctly in their s 6 which has come from integration. May be implied by correct answer.  4537  A1 4913 Distance = 525 − = 525 − 252.05555 273 m Allow .    18  18 272.944 SCB1 for the last 2 marks for answer without seeing 3 20t + 13t 2 − 2t 3 . ( ) 5

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