Cambridge A Level Mathematics 9709 — 2024 May/June Paper 4 · Variant 1

1 A car starts from rest and accelerates at 2 ms -2 for 10 s. It then travels at a constant speed for 30 s. The car then uniformly decelerates to rest over a period of 20 s. (a) Sketch a velocity-time graph for the motion of the car. [2] (b) Find the total distance travelled by the car. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 1(a) Trapezium on the t-axis starting at (0, 0) B1 Fully correct with correct labels B1 With height 20, intersecting the t-axis at 60, horizontal line segment between 10 and 40. Axis need not be labelled with t and v but if they are they must be correct. If not labelled, then must assume t is on the horizontal axis. 2 1(b) Area =   1 30 60 20 2   or Area 1 1 10 20 30 20 20 20 2 2         M1 Must be considering the area of a trapezium. Allow a single slip in one term only. Must be adding all terms together if considering two triangles and a rectangle. Using their 20 from (a). If no value for the height shown on the graph in (a), then it must be correct. = 900 m A1FT FT their 20   45 20 their   but A0 FT if using 2. 2

More questions on Kinematics of motion in a straight line

2 y 20 N 60° P x F N Two forces of magnitudes 20 N and F N act at a point P in the directions shown in the diagram. (a) Given that the resultant force has no component in the y-direction, calculate the value of F. 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(b) Given instead that F = 10 , find the magnitude and direction of the resultant force. 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Mark scheme: 2(a) and must be linked to F (can be implied by the correct answer seen only). = 17.3 N A1 AWRT 17.3 (17.320508…) or 10 3. 2 Question Answer Marks Guidance 2(b) For resolving in any direction *M1 Correct number of terms; allow sin/cos mix; allow sign errors. (Horizontal component   cos ) 20cos60 X R     [ 10  ] (Vertical component   sin ) 20sin60 10 Y R      [= 7.3205 ]   A1 For both correct. Magnitude =     2 2 20sin60 10 20cos60   [=12.393136…] DM1 OE – correct number of terms. Angle = 1 20sin60 10 tan 20cos60         [=36.206023…] DM1 OE (e.g. reciprocal) - correct number of terms. Magnitude = 12.4 N and Direction = 36.2° above (positive) x-axis A1 OE for direction e.g. 36.2 anticlockwise from (positive) x-direction, 36.2above the horizontal. Possibly seen on a diagram. (Radians: 0.63191… to 3sf or better) 5

More questions on Forces and equilibrium

3 A train of mass 180 000 kg ascends a straight hill of length 1.5 km, inclined at an angle of 1.5° to the horizontal. As it ascends the hill, the total work done to overcome the resistance to motion is 12 000 kJ and the speed of the train decreases from 45 ms -1 to 40 ms -1 . Find the work done by the engine of the train as it ascends the hill, giving your answer in kJ. [4] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 3 PE gained = 180000 1500sin1.5 [= 70677760.4 ] KEInitial = 2 1 180000 45 2   [ = 182 250 000] KEFinal = 2 1 180000 40 2   [= 144 000 000] B1 For initial KE or final KE (for reference: difference in KE is 38 250 000). For work energy equation:   2 2 1 1 2 2 180000 1500sin1.5 12000 000 180000 45 180000 40 WD g         M1 Correct number of terms; dimensionally correct; allow sign errors and minor slip(s) in values; allow sin/cos mix on PE term.     Work done 70677760.4 12000000 38250000 J 70677.7 12000 38250 kJ       = 44400kJ [44427.7604…] A1 Must be in kJ. Alternative Method for Question 3: Newton’s second law and equations of motion 0.142 a  [= 0.141666…] (B1) Correct acceleration from 2 2 40 45 2 (1500). a   Allow AWRT 0.14  or exact 17 120.  17 120 12 000 000 1500 1500( 180000 sin1.5) 180000 1500 D g      (M1) M1 for applying Newton’s second law parallel to the hill. Correct number of terms, allow sin/cos mix on weight component, dimensionally correct and multiplying both sides by 1500. Allow their a or a for acceleration (and minor slip(s) in values). (B1) Correct weight component multiplied by 1500. ( 1500 ) 44400 WD D   kJ [44427.7604…] (A1) Must be in kJ. 4

More questions on Energy, work and power

4 A car of mass 1700 kg is pulling a trailer of mass 300 kg along a straight horizontal road. The car and trailer are connected by a light inextensible cable which is parallel to the road. There are constant resistances to motion of 400 N on the car and 150 N on the trailer. The power of the car’s engine is 14 000 W. Find the acceleration of the car and the tension in the cable when the speed is 20 ms -1 . [6] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 4 Driving force =   14000 700 20  Attempt at Newton’s second law on car or trailer or system *M1 Correct number of relevant terms (e.g. correct masses); allow sign errors. 700 400 1700 T a    150 300 T a     700 400 150 1700 300 a     A2 A1 for one correct and A2 for any two correct. May have DF (or their DF) for 700 but A0 if using 14000 as DF without first stating DF. Must have same T if using first two equations for A2. Solving for a or T DM1 From equations with the correct number of relevant terms. Acceleration = 3 40 = 0.075 m s-2 For both correct answers. Tension = 172.5 N A1 Condone 173. 6

More questions on Kinematics of motion in a straight line

5 A straight slope of length 60 m is inclined at an angle of 12° to the horizontal. A bobsled starts at the top of the slope with a speed of 5 ms -1 . The bobsled slides directly down the slope. (a) It is given that there is no resistance to the bobsled’s motion. Find its speed when it reaches the bottom of the slope. 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(b) It is given instead that the coefficient of friction between the bobsled and the slope is 0.03 . Find the time that it takes for the bobsled to reach the bottom of the slope. 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Mark scheme: 5(a) 2.08 2.07911 a   B1 From sin12 . mg ma Allow exact (e.g. sin12). a g  2 2 5 2 60 v a    M1 For use of 2 2 2 v u as   with u = 5 and s = 60. Allow sign errors but a must be either sin12 g or cos12 g only. Speed = 16.6m s−1 [16.567861…] A1 AWRT 16.6 Alternative Method for Q5(a) For attempt at work energy equation (M1) 3 terms, dimensionally correct. Allow sign errors; allow sin/cos mix on PE term – condone m missing from all terms. Must be a weight component. 2 2 1 1 5 60sin12 2 2 mv m mg     (A1) Correct equation. (for reference: 60 sin12 = 12.4747…) Speed = 16.6 ms−1 [16.567861…] (A1) AWRT 16.6 3 Question Answer Marks Guidance 5(b) cos12 R mg  B1 Resolving correctly perpendicular to the plane. sin12 mg F ma   *M1 Use of Newton’s second law, correct number of terms; allow sign errors; allow sin/cos mix (must be a weight component). For use of 0.03 F R  to get equation in a (and m) only DM1 Where R is a component of weight only (dimensionally correct but allow sin/cos mix).   1.79 1.78567 . a        2 1 60 5 2 t at   and solve for t DM1 Dependent on previous two M marks. For use of 2 1 2 s ut at   with s = 60, u = 5 and their a or other complete method to find positive value(s) of t. Time = 5.86 s [5.86260…] A1 AWRT 5.86 (from using a = 1.79 or better). AWRT 5.85 (from using a = 1.8). AWRT 5.87 (from correct working). Question Answer Marks Guidance 5(b) Alternative method for Question 5(b): Using energy cos12 R mg  (B1) Resolving correctly perpendicular to the plane. 2 2 1 1 2 2 5 60 sin12 60 mv m mg F       (*M1) Use of work-energy principle, correctly number of relevant terms; allow sign errors; allow sin/cos mix on PE term (must be a weight component). For use of 0.03 F R  to get equation in v (and m) only (DM1) Where R is a component of weight only (dimensionally correct but allow sin/cos mix)   15.5 15.46870 . v       .   1 60 5 2 v t   and solve for t (DM1) Dependent on previous two M marks. Use of   1 2 s u v t   with s = 60, u = 5 and their v or other complete method to find positive value(s) of t. Time = 5.86s [5.86260…] (A1) 5

More questions on Kinematics of motion in a straight line

6 A particle moves in a straight line, starting from a point O. The velocity of the particle at time t s after leaving O is v ms -1 . It is given that v = kt 21 - t2 - 8 , where k is a positive constant. The maximum velocity of the particle is 4.5 ms -1 . (a) Show that k = 10 . 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(b) (i) Verify that v = 0 when t = 1 and t = 16 . 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(ii) Find the distance travelled by the particle in the first 16 s. 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Mark scheme: 6(a) *M1 Decrease power by 1 and a change in coefficient in at least one term (which must be the same term). Therefore, must see 1 1 2 2 d ( ) d kt ct t   with . c k  1 1 1 1 1 2 2 1 1 2 0 2 2 2 a kt t kt         A1 May be unsimplified. 1 2 2 1 0 2 0 2 16 k a kt t            or 1 2 4 k t  DM1 Equate a to 0 and attempt to solve for t or k. Must be of the correct form, e.g. 2 t k   or 1 2 k t   (possibly implied by forming a correct equation in k).   1 2 2 2 2 8 4.5 16 16 k k v k           M1 Or complete method to find k or an equation involving k, e.g. 1 1 2 2 4 2 8 4.5 6.25 t t t t      therefore 1 2 4(6.25) . k  Dependent on both previous M marks. 2 12.5 10 8 k k     A1 AG - Allow verification. Any errors seen is A0. Alternative method for last two marks of Question 6(a): Completing the square method (Let 1 2 x t  ) 2 2 2 2 2 4 2 8 2 2 16 8 k k k k v x x x x                       (M1) Dependent on both previous M marks. so 2 8 12.5 8 k   so 10 k  (A1) AG 5 Question Answer Marks Guidance 6(b)(i) 1 2 10 1 2 1 8 0    and 1 2 10 16 2 16 8 0      Or equate 0 v  and solve for t to get 1 t  and t = 16. B1 AG Any errors seen is B0. 1 6(b)(ii) For integration (do not penalise missing c) *M1 Increase power by 1 and a change in coefficient in at least one term (which must be the same term).       3 3 2 2 2 2 3 2 10 2 20 8 8 2 3 t s t t c t t t c         or       3 3 2 2 2 2 3 2 2 2 8 8 2 3 k t k s t t c t t t c         A1 Allow unsimplified; allow with k not substituted. 3 2 2 20 7 1 1 8 1 0 and 3 3                   3 3 2 2 2 2 20 20 16 16 8 16 1 1 8 1 3 3                      = 128 7 45 3 3         DM1 Attempt at correct use of limits 0 and 1 or 1 and 16 – allow a single slip. For both set of limits applied correctly DM1 Allow at most one slip in both. 142 3 m A1 Allow 47.3 or better (47.3333…) SC if no integration shown: B1 for 7 3  or 45, B1 for 142 3 (so 2 marks max). 5

More questions on Differentiation

7 A particle P of mass 0.2 kg is projected vertically upwards from horizontal ground with speed 25 ms -1 . (a) Show that the speed of P when it reaches 20 m above the ground is 15 ms -1 . 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When P reaches 20 m above the ground it collides with a second particle Q of mass 0.1 kg which is moving downwards at 20 ms -1 . P is brought to instantaneous rest in the collision. (b) Find the velocity of Q immediately after the collision. 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When P reaches the ground it rebounds back directly upwards with half of the speed that it had immediately before hitting the ground. (c) Find the height above the ground at which P and Q next collide. 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Mark scheme: 7(a) 2 2 25 2 20 v g    M1 For use of 2 2 2 v u as   or equivalent to get an equation in v only with 25, 20 u s   and . a g  ⇒ speed = 15 m s–1 A1 AG Allow verification – at least one intermediate step from equation of motion to given result. Any errors seen is A0. Alternative method for Question 7(a): 2 2 1 1 0.2 20 0.2 25 0.2 2 2 g v        (M1) Attempt at energy with 0.2, 20, 25; m h u    correct number of terms, allow sign errors. ⇒ speed = 15 m s-1 (A1) AG Allow verification - at least one intermediate step from conservation of energy to given result. Any errors seen is A0. 2 7(b) 0.2 15 0.1 20 0 0.1v      or 0.2 15 0.1 20 0 0.1V      M1 OE Attempt at conservation of momentum; 3 non-zero terms; allow sign errors – use of 25 is M0. 10 v  m s-1 upwards A1 Must have direction (possibly seen on diagram). If using mgv for momentum, then M1 A0 max. 2 Question Answer Marks Guidance 7(c) Speed of P at impact is 20 m s–1 B1 From 2 0 2 20 v g     (OE) - possibly implied by speed of P after impact being stated at 10. Time to when P reaches ground = 2 s B1 From 2 1 2 20 0 g t     (OE). 2 1 2 10 P s t g t    M1 Distance travelled by P after impact with the ground. Must be using their 10 (speed of P after impact with the ground) and . a g  2 1 2 10 Q s t g t     M1 Distance travelled by Q after P’s impact with the ground. Must be using their ±10 (the speed/vel. of Q after their 2 s ( '10' ( ) 2 g    where ‘10’ is the value from (b)) and . a g  20 1 P Q s s t     A1 Time after P hits the ground to next collision. Height = 5m A1 CWO Question Answer Marks Guidance 7(c) Alternative Method for last 4 marks: 2 1 2 10 Q s t g t    (M1) Expression for the displacement of Q after first impact of P and Q. Must be using their 10 (from (b)) and . a g  2 1 2 10 ( 2) ( 2) P s t g t       (M1) Expression for the displacement of P (for values of 2 t  ) measured from point of first collision between P and Q. Must be using their 2 (time for P to reach ground), their 10 (speed of P after impact with the ground), . a g  20 3 Q P s s t     (A1) Correct time between collisions of P and Q. Height = 5m (A1) CWO 6

More questions on Kinematics of motion in a straight line