Cambridge A Level Mathematics 9709 — 2023 Oct/Nov Paper 4 · Variant 2

1 A block of mass 15kg slides down a line of greatest slope of an inclined plane. The top of the plane is at a vertical height of 1.6m above the level of the bottom of the plane. The speed of the block at the top of the plane is 2ms−1 and the speed of the block at the bottom of the plane is 4ms−1. Find the work done against the resistance to motion of the block. [4] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1 1 2 1 2 B1 For KE at top or bottom.   15  2  = 30    15  4  = 120  Need not be evaluated. 2 2 1 2  15  ( 4 − 2 ) is B0. 2  15 g  1.6  = 240  B1 For PE change. Need not be evaluated. 240 + 30 = 120 + W M1 Attempt at work energy equation; 4 relevant terms; dimensionally correct; allow sign errors. 1 2  15  ( 4 − 2 ) is M0. 2 If W = F times a numerical distance seen, then M0. Work done = 150 J A1 1 Alternative method for Q1 2 2 1.6 *M1 Attempt to use v 2 = u 2 + 2 as with 4 = 2 + 2 a  sin 1.6 1.6 s = or but not sin cos s = 1.6sin or 1.6cos or 1.6. If  is given a value, then M0. Must be using speeds 2 and 4 here. 15 g sin− R = 15a DM1 3 terms; allow sign errors; allow sin/cos mix but weight must be resolved; dimensionally correct. R = 93.75sin A1 R = 93.75cos Must be consistent with their s .  1.6  A1 Work done = 93.75sin = 150 J    sin 4

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2 A 30Å R 2 N m kg 40Å B The diagram shows a smooth ring R, of mass mkg, threaded on a light inextensible string. A horizontal force of magnitude 2N acts on R. The ends of the string are attached to fixed points A and B on a vertical wall. The part AR of the string makes an angle of 30Å with the vertical, the part BR makes an angle of 40Å with the vertical and the string is taut. The ring is in equilibrium. Find the tension in the string and find the value of m. [5] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 Attempt to resolve in at least one direction to form an equation. *M1 Correct number of terms; allow sign errors; allow sin/cos mix; allow with different T ’s. T sin30 + T sin40 −=2 0 A1 If different T ’s then allow M1A1A0 max. T cos30 − T cos40 − mg = 0 A1 Allow with their T. Attempt to solve for T or m DM1 From equation(s) with correct number of relevant terms. Tension T = 1.75, m = 0.0175 A1 T = 1.7501 m= 0.017497 awrt 1.75 for T www, and awrt 0.0175 for m www. 5

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3 120 N 20Å 10 kg 30Å A block of mass 10kg is at rest on a rough plane inclined at an angle of 30Å to the horizontal. A force of 120N is applied to the block at an angle of 20Å above a line of greatest slope (see diagram). There is a force resisting the motion of the block and 200J of work is done against this force when the block has moved a distance of 5m up the plane from rest. Find the speed of the block when it has moved a distance of 5m up the plane from rest. [5] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 3 Work done by 120 N force = 120  5cos20  = 563.81557 . B1 ( PE change = )10 g  5sin30  = 250  B1 For PE change. Attempt at work energy equation M1 4 relevant terms; dimensionally correct; allow sign errors; allow sin/cos mix in relevant resolved terms. 1 2 A1 120  5cos20 − 10 g  5sin30 − 200 =  10  v 2 563.815− 250 − 200 = 5v 2    Speed = 4.77 m s–1 A1 awrt 4.77. 3 Alternative method for Question 3 200 *B1 oe e.g. 5  RF = 200 . Resistive force =  = 40  5 120 cos20 − RF − 10 g sin30 = 10a *M1 4 relevant terms; dimensionally correct; allow sign errors; allow sin/cos mix; allow with their resistive force or just RF. a = 2.276… A1 Allow arwt 2.3 to 2sf from correct work. v 2 = 0 + 2  ( 2.276) 5 DM1 Use of v 2 = u 2 + 2 as using u = 0 , s = 5 and their positive a which has come from a resistive force using work done. Speed = 4.77 m s–1 A1 awrt 4.77. 5

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4 A particle P of mass 0.2kg lies at rest on a rough horizontal plane. A horizontal force of 1.2N is applied to P. (a) Given that P is in limiting equilibrium, find the coefficient of friction between P and the plane. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Given instead that the coefficient of friction between P and the plane is 0.3, find the distance travelled by P in the third second of its motion. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(a) R = 0.2g B1 1.2 =  0.2g M1 Resolve horizontally and using F = µR to get an equation in µ; 2 relevant terms. µ = 0.6 A1 oe 3 4(b) 1.2 − 0.3  0.2 g = 0.2a *M1 Resolve horizontally using Newton’s Second Law; 3 relevant terms; allow sign errors; R = 0.2 g only. a = 3 A1 0.6 = 0.2a only seen, allow with BOD, but if 0.6 as friction being used as resultant force, this is M0A0. 1 2 1 2 DM1 1 2 s3 = 0 + 3 3  = 13.5 s2 = 0 + 3 2  = 6  For use of s = ut + at (or a 2 2 2 complete method) to find a distance at least once with u = 0 and their positive a and t = 2 or t = 3 . Distance = 13.5 −=6 7.5 m A1 www 4

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5 A particle A of mass 0.5kg is projected vertically upwards from horizontal ground with speed 25ms−1. (a) Find the speed of A when it reaches a height of 20m above the ground. 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When A reaches a height of 20m, it collides with a particle B of mass 0.3kg which is moving downwards in the same vertical line as A with speed 32.5ms−1. In the collision between the two particles, B is brought to instantaneous rest. (b) Show that the velocity of A immediately after the collision is 4.5ms−1 downwards. 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(c) Find the time interval between A and B reaching the ground. You should assume that A does not bounce when it reaches the ground. 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Mark scheme: 5(a) v 2 = 252 + 2 ( − g )  20 M1 Use of v 2 = u 2 + 2 as with u = 25 , s = 20 and a =  g . 1 2 1 2 OR using change in KE =  change OR  0.5  v =  0.5  25 − 0.5  g  20 2 2 in PE. Speed = 15 m s-1 A1 2 5(b) Taking up as positive direction: 0.5  15 + 0.3 −( 32.5 ) = 0.5v + 0 or M1 For use of conservation of momentum, 3 non-zero terms, allow sign errors, using their speed Taking down as positive direction: 0.5 −( 15 ) + 0.3  32.5 = 0.5v + 0 15 m s-1. Must show how 2.25 is obtained. [Taking up as positive direction: velocity of A = −4.5 m s−1] A1 Any error seen in calculating v is [Taking down as positive direction: velocity of A = 4.5 m s−1] A0. Speed = 4.5 m s–1 direction downwards Must explicitly say 4.5 m s–1 and downwards. 2 5(c) 1 2 M1 Using constant acceleration Downwards to be positive, for A 20 = 4.5t A + gt A and solve for At formula(e) to get a correct equation 2 in At and solve for At . 1 2 Upwards to be positive, for A −20 = −4.5t A − gt A and solve for At If using quadratic formula, must be 2 the correct formula. If factorising, when brackets expanded, 2 terms correct. 1 2 M1 Using constant acceleration For B 20 = 0 + gt B  t = 2  and solve for Bt formula(e) to get a correct equation 2 in Bt and solve for Bt . At = 1.6 or Bt = 2 A1 Difference = 0.4 s only A1 4

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6 A railway engine of mass 120000kg is towing a coach of mass 60000kg up a straight track inclined at an angle of ! to the horizontal where sin ! = 0.02. There is a light rigid coupling, parallel to the track, connecting the engine and coach. The driving force produced by the engine is 125000N and there are constant resistances to motion of 22000N on the engine and 13000N on the coach. (a) Find the acceleration of the engine and find the tension in the coupling. 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At an instant when the engine is travelling at 30ms−1, it comes to a section of track inclined upwards at an angle " to the horizontal. The power produced by the engine is now 4500000W and, as a result, the engine maintains a constant speed. (b) Assuming that the resistance forces remain unchanged, find the value of ". 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Mark scheme: 6(a) Engine: 125000 − 120000 g  0.02 − 22000 − T = 120000a *M1 Attempt at Newton’s second law at least once; correct number of 125000 − 120000 g sin (1.145..) − 22000 − T = 120000a relevant terms; allow sign errors; 125000 − 24000 − 22000 − T = 120000 a  79000 − T = 120000 a  allow sin/cos mix; allow g missing; a value for or sin must Coach: T − 60000 g  0.02 − 13000 = 60000a be substituted. T − 60000 g sin (1.145 .. ) − 13000 = 60000 a Allow with = 1.1 or better = 1.145991998  . T − 12000 − 13000 = 60000 a  T − 25000 = 60000 a  A1 Any equations correct. System: 125000 − 120000 g  0.02 − 60000 g  0.02 − 22000 − 13000 = (120000 + 60000 ) a A1 Two equations correct. 125000 − 120000 g sin (1.145..) − 60000 g sin (1.145..) − 22000 − 13000 = (120000 + 60000 ) a If using separate equations for  125000 − 24000 − 12000 − 22000 − 13000 = ( 120000 + 60000 ) a  54000 = 180000a  engine and coach and different T ’s, then allow M1A1A0 max. Solve for T or a DM1 Using equations with the correct number of relevant terms. If no working seen, must be solutions to their equation(s) to be awarded M1. Acceleration = 0.3 m s–2 A1 Allow 0.299 from use of = 1.15. and Awrt 43000 to 3sf from correct work. Tension = 43 000 N 5 6(b) 4500000 B1 P Driving force, DF =  = 150000  Use of F = , 30 v oe e.g. DF  30 = 4500000 . Attempt to resolve parallel to the track once if using system equation, twice if using equations for M1 Correct number of relevant terms; engine and coach separately allow sign errors; allow sin/cos mix; allow g missing. Must be correct number of equations depending on method. System: 150000 − 120000 gsin− 60000 gsin− 22000 − 13000 = 0 A1 Allow DF or their DF. or for Engine: 150000 − 120000 gsin− 22000 − T ' = 0 Must be using same T ' . and Coach: T '− 60000 gsin− 13000 = 0 Solve to get = 3.7 A1 3.663058552 awrt 3.7° www. 4

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7 A particle X travels in a straight line. The velocity of X at time t s after leaving a fixed point O is denoted by vms−1, where v = −0.1t3 + 1.8t2 −6t + 5.6. The acceleration of X is zero at t = p and t = q, where p < q. (a) Find the value of p and the value of q. 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It is given that the velocity of X is zero at t = 14. (b) Find the velocities of X at t = p and at t = q, and hence sketch the velocity-time graph for the motion of X for 0 ≤t ≤15. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Find the total distance travelled by X between t = 0 and t = 15. 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Mark scheme: 7(a) 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); allow unsimplified; allow p or q for t . v a = is M0. t  dv  3 −1 2 −1 1−1 2 A1 May be unsimplified.  a = =  3 −0.1t + 2  1.8t − 6t = −0.3t + 3.6t − 6 Allow p or q for .t  dt  dv DM1 Allow p or q for .t Setting a = = 0 and attempt to solve a 3 term quadratic for .t dt Must get 2 values or numerical  dv 2 2  expressions for t from their three a = = 0  3t − 36t + 60 = 0  t − 12t + 20 = 0   term quadratic.  dt  If using quadratic formula, must be the correct formula. If factorising, when brackets expanded, 2 terms correct. p = 2, q = 10 A1 4 7(b) Velocities are 0 m s-1 and 25.6 m s-1 B1 SOI Curve with single minimum turning point followed by single maximum turning point *B1 Ignore placement of graph on axes Not a cusp for the minimum point or maximum point. DB1 All correct in 1st and 4th quadrant. Must go from convex to concave. Need to label 2 and 14 on the t-axis where the curve meets the t-axis. Do not need to show exact velocities at t = 0 or t = 10 or t = 15. Ignore graph outside 0 t 15 . 3 7(c) 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. 0.1 3+1 1.8 2 +1 6 1+1 A1 May be unsimplified. ( s = ) − t + t − t + 5.6t ( + c ) 4 3 2 = −0.025t 4 + 0.6t 3 − 3t 2 + 5.6t [+c] Attempt distance from t = 0 to t = 14 [= 176.4] DM1 Correct use of limits 0 and 14 for their ,s i.e. F (14 ) − F ( 0 ) May see limits 0 to 2 and 2 to 14 used but must be ( F (14 ) − F ( 2 ) ) + ( F ( 2 ) − F ( 0 ) ) . Attempt distance from t = 14 to t = 15 [= (-) 8.025] DM1 Correct use of limits 14 and 15 for their ,s i.e.  ( F (15 ) − F (14 ) ) . 18 For reference F ( 2 ) = = 3.6 , 5 882 F (14 ) = = 176.4 and 5 1347 F (15 ) = = 168.375. 8 7377 A1 www Total distance =176.4 + 8.025 = 184.425 m = Condone 184 or better. 40 7(c) SC for those who show no integration. Max 3 marks. 14 B1  ( −0.1t 3 + 1.8t 2 − 6t + 5.6 ) dt = 176.4 0 15 B1  ( −0.1t 3 + 1.8t 2 − 6t + 5.6 ) dt = −8.025 14 15 OR  ( −0.1t 3 + 1.8t 2 − 6t + 5.6 ) d t = 8.025 14 7377 B1 Condone 184 or better. Total distance =176.4 + 8.025 = 184.425 m = 40 SC for those who show no integration and don’t consider the 2 areas. Max 1 mark. 15 B1 Condone 184 or better. 3 2 7377 0.1t + 1.8t − 6t + 5.6 dt = 184.425 m =  − 40 0 5

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