Cambridge A Level Mathematics 9709 — 2023 Oct/Nov Paper 4 · Variant 1

1 A particle of mass 1.6kg is projected with a speed of 20ms−1 up a line of greatest slope of a smooth plane inclined at ! to the horizontal, where tan ! = 34. Use an energy method to find the distance the particle moves up the plane before coming to instantaneous rest. [3] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1 3 1 2 B1 For either the correct potential energy or kinetic energy 1.6 g x [=9.6x] or   1.6  20 [= 320] term. Need not be evaluated. 5 2 1 2 3 M1 Attempt at energy equation; 2 relevant terms.  1.6  20 = 1.6 g  x sin where sin= Dimensionally correct but allow sign errors. 2 5 Allow sin/cos mix and sin(36.869…) but sin (oe) must 3 have been substituted. M0 for 1.6 g x . 4 100 A1 Allow 33.3. x = 3 3

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2 A B 35Å 40Å 2.4 kg A particle of mass 2.4kg is held in equilibrium by two light inextensible strings, one of which is attached to point A and the other attached to point B. The strings make angles of 35Å and 40Å with the horizontal (see diagram). Find the tension in each of the two strings. [5] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 Attempt to resolve horizontally or vertically to form an equation. *M1 Correct number of terms; allow sin/cos mix; allow sign errors – do not award this mark if using T for both (see SC later). T1 cos35 = T2 cos40 A1 Must be different Ts. T1 sin35 + T2 sin40 = 2.4 g A1 If same Ts, then SC B2 only for this equation. Attempt to solve for either tension. DM1 From equations with correct number of relevant terms. Must get a value for at least one tension.  cos40  E.g. T2   sin35 + sin40  = 24  cos35  T1 = 20.4 N and T2 = 19.0 N A1 T1 = 19.033621 T2 = 20.353166 awrt 20.4 for T1 www and 19(.0) for T2. 5

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3 v (m s−1) 12.6 0 t (s) 0 8 48 62 70 The diagram shows the velocity-time graph for the motion of a bus. The bus starts from rest and accelerates uniformly for 8 seconds until it reaches a speed of 12.6ms−1. The bus maintains this speed for 40 seconds. It then decelerates uniformly in two stages. Between 48 and 62 seconds the bus decelerates at ams−2 and between 62 and 70 seconds it decelerates at 2ams−2 until coming to rest. (a) Find the distance covered by the bus in the first 8 seconds. 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(c) Find the average speed of the bus for the whole journey. 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Mark scheme: 3(a) Distance = 50.4m B1 252 Allow . 5 1 3(b) v1 = 12.6 − ( 62 − 48 ) a M1 Use of suvat for first section of deceleration. 12.6  (62 − 48) a only. 0 = v1 − 2 a  ( 70 − 62 ) M1 Use of suvat for second section of deceleration. An expression for the velocity at 62 seconds must be 2 a  (70 − 62). a = 0.42 A1 –0.42 scores A0. 3 3(c) Speed at time t = 62 is 6.72 m s-1 B1 This may be seen in part (b) but must be used in part (c) to get this mark. s2 = ( 48 − 8 )  12.6  = 504  B2FT B2 FT for any 2 correct, B1 FT for any 1 correct – follow through their value of v1 where 0  v1  12.6 but must  3381  135.24 or oe s3 = 0.5  (12.6 + their 6.72 )  ( 62 − 48 )  =  have come from the correct equations seen in part (b).  25  Allow correct value of 1v from a = −0.42 where or their 6.72  ( 62 − 48 ) + 0.5  ( 62 − 48 )  (12.6 − their 6.72 ) v1 = 12.6 + (62 − 48) a and v1 = −2 a  (70 − 62).  672  26.88 or oe s4 = 0.5  their 6.72  ( 70 − 62 )  =   25  Average speed =10.236 m s-1 B1 2559 59 Allow 10.2 or better oe e.g. , 10 . 250 250 4

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4 Two particles P and Q, of masses 6kg and 2kg respectively, lie at rest 12.5m apart on a rough horizontal plane. The coefficient of friction between each particle and the plane is 0.4. Particle P is projected towards Q with speed 20ms−1. (a) Show that the speed of P immediately before the collision with Q is 10 3ms−1. 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In the collision P and Q coalesce to form particle R. (b) Find the loss of kinetic energy due to the collision. 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The coefficient of friction between R and the plane is 0.4. (c) Find the distance travelled by particle R before coming to rest. 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Mark scheme: 4(a) −0.4  6 g = 6 a *B1 Resolve horizontally using Newton’s second law; 2 relevant terms; must be either −0.4  6 g = 6 a or 0.4  6 g = 6a. v 2 = 202 + 2 −( 4 )  12.5 DM1 Use complete suvat method to get an equation in v or v 2 – must be using u = 20, s = 12.5 and their a. v 2 = 300  v = 10 3 A1 AG. Condone correct expression for v or v2 followed by correct answer. Alternative method for Question 4(a) RF = 0.4  6 g *B1 Correct application of F = R for P. 0.5 6 20 2 − 0.5 6 v 2 = 12.5  (0.4  6 g ) DM1 3 relevant terms; dimensionally correct; allow sign errors only. v 2 = 300  v = 10 3 A1 AG. Condone correct expression for v or v2 followed by correct answer. 3 4(b) 6  10 3 = ( 6 + 2 ) v ' M1 For use of conservation of momentum, 3 non-zero terms, allow sign errors. Use of 20 is M0. v ' = 7.5 3 A1 12.99038… 1 2 B1 Either initial kinetic energy or final kinetic energy correct. Initial KE = 6 10 3 ( )  = 900  Allow unsimplified. 2 1 2 Final KE = 8 7.5 3 = 675 ( )  2 Loss of KE = 225 J A1 4 4(c) 2 M1 Use complete suvat method to find distance. This must be 0 = their 7.5 3 + 2  ( their − 4 )  s ( ) using their v from part (b), so it is dependent on scoring the first M mark in part (b) and either their a from part (a), or from 0.4  8 g = 8a. [Distance =] 21.1 m A1 21.1 or better (21.09375). 2

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5 P B A 1.6 kg 1.2 kg 50Å 40Å The diagram shows a particle A, of mass 1.2kg, which lies on a plane inclined at an angle of 40Å to the horizontal and a particle B, of mass 1.6kg, which lies on a plane inclined at an angle of 50Å to the horizontal. The particles are connected by a light inextensible string which passes over a small smooth pulley P fixed at the top of the planes. The parts AP and BP of the string are taut and parallel to lines of greatest slope of the respective planes. The two planes are rough, with the same coefficient of friction, -, between the particles and the planes. Find the value of - for which the system is in limiting equilibrium. 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Mark scheme: 5 Resolving parallel to the slope at A or B to form an equation. *M1 Correct number of terms; allow sign errors; allow sin/cos mix. 1.6 g sin50 − T − FB = 0 A1 If using the same Fs, then M1A1A0B1 max. T − FA − 1.2 g sin40 = 0 A1 System equation (must be four different terms): 1.6 g sin50 − FB − FA − 1.2 g sin40 = 0 only scores M1A1A1. Any sign errors scores M1 only. R A = 1.2 g cos40 or RB = 1.6 g cos50 *B1 Either correct. Must be explicitly linked to the correct contact (so could be seen on a diagram), or as part of a resolving parallel to the slope equation(s) (so must be combined with ). FA = 1.2 gcos40 or FB = 1.6 gcos50 *M1 Use of F = Rat either A or B. Must be explicitly linked to the correct contact (could be seen on a diagram) or as part of a resolving parallel to the slope equation(s). Allow sin/cos mix error only. 1.6 g sin50 − 1.6 gcos50 = 1.2 g sin40 + 1.2 gcos40 DM1 Eliminating T, FA and FB to form an equation in only.  1.6 g sin50 − 1.2 g sin40  A1 0.23326119… =  = 0.233    1.2 g cos40 + 1.6 g cos50  7

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7 A particle moves in a straight line starting from a point O before coming to instantaneous rest at a point X. At time t s after leaving O, the velocity vms−1 of the particle is given by v = 7.2t2 0 ≤t ≤2, v = 30.6 −0.9t 2 ≤t ≤8, 1600 v = + kt 8 ≤t, t2 where k is a constant. It is given that there is no instantaneous change in velocity at t = 8. Find the distance OX. 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Mark scheme: 7 1600 *M1 Use velocity at t = 8 to set up a linear equation in k only. 30.6 − 0.9 =8 + 8k 2 Allow a slip in one value or sign only. 8 k = −0.2 A1 1600 DM1 Attempt to find the value of t when the particle comes to + ( their k ) =t 0  t = 2 rest using the correct expression for v, set equal to zero t with their negative value of k. Must find a positive value for t (for reference, t = 20). Attempt to integrate v for one of the 3 intervals *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. 7.2 3 A1 May be unsimplified (for reference, limits are from 0 to s = t ( + c ) 2). 3 0.9 2 A1 May be unsimplified (for reference, limits are from 2 to s = 30.6t − t ( + c ) 8). 2 1600 −1 k 2 A1FT May be unsimplified (for reference limits are from 8 to s = t + t ( + c ) −1 2 20). Follow through their value of k or just k only. Either 19.2 or 156.6 or 86.4 B1 One correct distance found. Allow unsimplified e.g. ( 216 − 59.4 ) or 1  ( 8 − 2 )  ( 28.8 + 23.4 ) etc. 2 B1 This mark can be awarded if no integration is shown oe. Distance = 19.2 + ( 216 − 59.4 ) + ( −120 −−( 206.4 ) ) = 262.2 m e.g. 1311. Condone 262 www. 5 9

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