Cambridge A Level Mathematics 9709 — 2024 May/June Paper 4 · Variant 2

1 A cyclist and bicycle have a total mass of 72 kg. The cyclist rides along a horizontal road against a total resistance force of 28 N. Find the total work done by the cyclist to increase his speed from 8 ms -1 to 16 ms -1 while travelling a distance of 100 metres. [3] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 1 Initial KE   2 1 72 8 2304 2     OR Final KE   2 1 72 16 9216 2     OR Work done against resistance   28 100 2800    B1 Correct expression for either KE or correct expression for work done against resistance. For reference,   2 2 1 72 16 8 6912. 2     Attempt at work-energy equation 2 2 1 1 72 8 WD 72 16 28 100 2 2               M1 4 terms; allow sign errors; dimensionally correct. WD 9712J  A1 OE. Condone 9710 J. Do not ISW. Alternative method for Question 1: 2 2 2 2 16 8 16 8 2 100 0.96 2 100              a a (B1) OE, e.g. 192 200 a  . Use of suvat in a complete method to find an expression for a. Must be of the form ' ' a . Attempt at Newton’s second law   DF 28 72 0.96 their        (M1) Three terms; dimensionally correct; allow sign errors; must be using their value of a.   WD 97.12 100 9712    J (A1) OE. Condone 9710 J. Do not ISW. 3

More questions on Energy, work and power

2 A particle P moves in a straight line. At time t s after leaving a point O on the line, P has velocity v ms -1 , where v = 44t - 6t 2 - 36 . (a) Find the set of values of t for which the acceleration of the particle is positive. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find the two values of t at which P returns to O. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 2(a) Attempt to differentiate given v M1 Decrease power by 1 and a change in coefficient in at least one term (which must be the same term); allow unsimplified. Use of v a t  scores M0.   11 44 12 0 3     t t A1 OE, e.g. 44 2 , 3 12 3 , 3.67 or better. Do not allow 11. 3 t  May solve 44 12 0 t   , but final answer must be 11. 3  t If a lower limit included it must be 0. Allow 0 t  or 0 t  . Allow 11 0, 3      or 11 0, 3       . Alternative Method for Question 2(a): Use completing the square to get 2 11 6 3 t                  (M1) OE 11 3 t  (A1) CWO If a lower limit included it must be 0. Allow 0 t  or 0 t  . Allow 11 0, 3      or 11 0, 3       . Question Answer Marks Guidance 2(a) Alternative Method 2 for Question 2(a): Solving 2 44 6 36 0 t t    and find 1 2 2 t t  , or equivalent. (M1) Complete method for finding the value of t at maximum, or use 2 b a  with correct a and b. For reference, 11 67. 3   t 11 3 t  (A1) If a lower limit included it must be 0. Allow 0 t  or 0. t  Allow 11 0, 3      or 11 0, 3       . 2 2(b) Attempt to integrate given v M1 Increase power by 1 and a change in coefficient in at least one term (which must be the same term). Use of s vt  is M0.       1 1 2 1 2 3 44 6 36 22 2 36 1 1 2 1 s t t t c t t t c             A1 Allow unsimplified. 2 3 22 2 36 0 t t t           2, 9 and 0  t ONLY A1 CWO Ignore 0 t  if not rejected. 3

More questions on Kinematics of motion in a straight line

3 P N 40° 25° 2 N i° 10 N 16 N Four coplanar forces of magnitude P N, 10 N, 16 N and 2 N act at a point in the directions shown in the diagram. It is given that the forces are in equilibrium. Find the values of θ and P. [6] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 3 Resolving either direction to get an equation *M1 Correct number of relevant terms; allow sign errors; allow sin/cos mix. 10cos25 2cos40 16sin   9.06307787 1.532088886 16sin 7.530988984 16sin sin 0.4706868115                  A1 10sin25 16cos 2sin40 P     4.226182617 16cos 1.285575219 5.511757837 16cos P P              A1 This may be with their .  Attempt to solve for 1 10cos25 2cos40 sin 16           DM1 From equation(s) with correct number of relevant terms. Must be a numerical expression for .  Attempt to solve for   10sin25 16cos 2sin40 P their    DM1 From equation(s) with correct number of relevant terms. Using their .  Must be a numerical expression for P. 28.1  AND 19.6 P  A1 28.07888819 and 19.6285636. AWRT 28.1 and AWRT 19.6 from correct work. 6

More questions on Forces and equilibrium

4 A car has mass 1400 kg. When the speed of the car is v ms -1 the magnitude of the resistance to motion is kv2 N where k is a constant. (a) The car moves at a constant speed of 24 ms -1 up a hill inclined at an angle of a to the horizontal where sin a = 0.12 . At this speed the magnitude of the resistance to motion is 480 N. (i) Find the value of k. [1] .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... (ii) Find the power of the car’s engine. [3] .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... (b) The car now moves at a constant speed on a straight level road. Given that its engine is working at 54 kW , find this speed. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 4(a)(i) 2 5 24 480 6         k k 576 , 0.833 or better. 1 4(a)(ii) Attempt at Newton’s second law   480 1400 0.12 2160 DF g DF      *M1 3 terms; allow sign errors; allow sin/cos mix. Allow 480 1400 sin6.9 DF g    or better. May see 2 5 24 1400 0.12. 6          DF their g Power   2160 24 their   DB1 For using P = DF x v, where DF is numerical. 51840 W A1 Allow W missing, but if given in kW units must be present. Allow 51.84 kW. Allow 51800, 51.8 kW. 3 4(b) 54000 DF v  and 2 5 6 DF their v       *B1FT FT 5 0. 6  their Get an equation of the form 3 av b  and attempt to solve for v to get a positive value DM1 a and b must both be positive or both negative. Must get to a value for v; if cubic not seen, the cubic may be implied by the correct answer for their equation. Speed = 40.2 m s–1 A1 40.165977. AWRT 40.2 from correct work. 3

More questions on Forces and equilibrium

5 T N 35° 0.8 kg 28° A particle of mass 0.8 kg lies on a rough plane which is inclined at an angle of 28° to the horizontal. The particle is kept in equilibrium by a force of magnitude T N. This force acts at an angle of 35° above a line of greatest slope of the plane (see diagram). The coefficient of friction between the particle and the plane is 0.2 . Find the least and greatest possible values of T. 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Mark scheme: 5 Attempt at resolving perpendicular to plane to get an equation *M1 3 relevant terms; allow sign errors; allow sin/cos mix; allow g missing; 0.8 m  must be used; correct angles must be used. sin35 0.8 cos28   R T g   0.57357 7.06358        R T A1 Attempt to resolve parallel to plane for one of the possible cases to get an equation *M1 3 relevant terms; allow sign errors; allow sin/cos mix; allow g missing; 0.8 m  must be used; correct angles must be used. cos35 0.8 sin 28 T g F     0.81915 3.75577 T F        A1 May use their F. cos35 0.8 sin 28 T g F     0.81915 3.75577 T F        A1 May use their F. Use of 0.2 F R  to get an equation in T only DM1 Dependent on previous 2 M marks. May be implied by correct T value. Allow g missing. If resolved equations incorrect and no working seen, then this mark is implied by the correct T value for their equations. Solve to get 5.53 T  A1 5.534499898 AWRT 5.53 from correct work. Allow 5.54 from correct work. Solve to get 3.33 T  A1 3.326141531 AWRT 3.33 from correct work. Allow 3.32 from correct work. 8

More questions on Forces and equilibrium

6 Three particles A, B and C of masses 5 kg, 1 kg and 2 kg respectively lie at rest in that order on a straight smooth horizontal track XYZ. Initially A is at X, B is at Y and C is at Z. Particle A is projected towards B with a speed of 6 ms -1 and at the same instant C is projected towards B with a speed of v ms -1 . In the subsequent motion, A collides and coalesces with B to form particle D. Particle D then collides and coalesces with C to form particle E and E moves towards Z. 15 - v -1 (a) Show that after the second collision the speed of E is ms . 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(b) The total loss of kinetic energy of the system due to the two collisions is 63 J. Use the result from (a) to show that v = 3 . 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(c) It is given that the distance XY is 36 m and the distance YZ is 98 m. (i) Find the time between the two collisions. 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(ii) Find the time between the instant that A is projected from X and the instant that E reaches Z. [1] .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... ....................................................................................................................................................

Mark scheme: 6(a) Attempt at conservation of momentum for the 1st collision   5 6 5 1 D v       For reference 5.  D v If mgv used, allow M1 M1 A0 max. Attempt at conservation of momentum for the 2nd collision      5 1 2 5 1 2 D E their v v v         DM1 6 non-zero terms; allow sign errors; using correct masses; allow their numerical . D v Allow E v v  for this mark. Note:   5 6 2 5 1 2 E v v    is M2. If mgv used, allow M1 M1 A0 max.  15 4   E v v A1 AG Must in terms of v, as v is given in the question or explicitly defined their letter used as v. Do not allow E v v  for this mark. Any error seen is A0 but condone saying ‘divide by 2’ or equivalent. If mgv used, allow M1 M1 A0 max. 3 Question Answer Marks Guidance 6(b) 2 2 2 1 1 KE 5 6 2 90 2 2 initial v v             2 1 15 KE 5 1 2 2 4 final v          B1 For either KEinitial or KE final correct. Attempt difference in KE is 63 to get an equation   2 2 2 1 1 1 15 5 6 2 5 1 2 63 2 2 2 4 v v                      M1 Using   sum of two initial KE 63.   final KEs Correct number of relevant terms – correct masses, must be adding 2 KE terms for . KEinitial   sum of two initial KEs and KE final coming from use of correct formula and of the correct form. Solve algebraically 2 3 30 117 0 v v    OE to get 3 v  ONLY A1 AG Any error seen is A0. Allow solving correct quadratic expression, rather than correct quadratic equation, for full marks. If 13 v  seen it must be discarded. Must see solving for this mark. A quadratic equation followed by the answer is insufficient. Question Answer Marks Guidance 6(b) Alternative Method for Question 6(b): Using loss of KE in second collision 2 2 2 1 1 1 KE 6 5 2 75 2 2 st after collision v v             2 1 15 KE 5 1 2 2 4 final v          (B1) For either 1 KE st after collision or KE final correct. Attempt difference in KE is   2 2 1 1 63 5 6 6 5 63 15 48 2 2              to get an equation   2 2 2 1 1 1 15 6 5 2 5 1 2 63 15 2 2 2 4 v v                       (M1) Using   1 KE KE 63 1 5 .    st final after collision their Correct number of relevant terms. 1 , KE st after collision KE final and their 15 coming from use of correct formula and of the correct form. Solve algebraically 2 3 30 117 0 v v    OE to get 3 v  ONLY (A1) AG Any error seen is A0. If 13 v  seen it must be discarded. Must see solving for this mark. A quadratic equation followed by the answer is insufficient. Alternative Method 2 for Question 6(b): Verifying that 3  v 2 2 1 1 KE 5 6 2 3 99 2 2 initial      (B1)   2 1 15 3 KE 5 1 2 36 2 4 final           (B1) KE KE 63 initial final   , hence loss in KE is 63 J (B1) Must have a conclusion for this mark. 3 Question Answer Marks Guidance 6(c)(i) Time A to B = 6 s B1 Distance BC    98 3 6 80 their    *B1FT FT their 6 which MUST come from 6 36.  t Use sum of distance moved by D and distance moved by C is 80 m   5 3 80 their t t their       OR use distance moved by C divided by relative velocity   80 5 3 their          DM1 Using D theirv from part (a). 6  D v or 3 and 80 98.  their Time = 10 s A1 Do not ISW. 4 6(c)(ii) 3 10 3 6 6 10 3            32 s B1 1

More questions on Kinematics of motion in a straight line

7 P 2.5 kg 2 m Q 0.5 kg 30° Two particles P and Q of masses 2.5 kg and 0.5 kg respectively are connected by a light inextensible string that passes over a small smooth pulley fixed at the top of a plane inclined at an angle of 30° to the horizontal. Particle P is on the plane and Q hangs below the pulley such that the level of Q is 2 m below the level of P (see diagram). Particle P is released from rest with the string taut and slides down the plane. The plane is rough with coefficient of friction 0.2 between the plane and P. (a) Find the acceleration of P. 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(b) Use an energy method to find the speed of the particles at the instant when they are at the same vertical height. 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Mark scheme: 7(a) 25 3 2.5 cos30 21.65063509 2 R g          B1 Note: 5 3 0.5 cos30 4.330127019. 2    F g Attempt at Newton’s second law *M1 Correct number of dimensionally correct/relevant terms; allow sign errors; allow sin/cos mix. Using this twice to get equations for P and Q; allow different T’s (equations with 0.5 and 2.5). Using once to get a system equation (equation with 0.5 2.5).  EITHER: 0.5 0.5 T g a   AND 2.5 sin30 2.5 g F T a    OR:   2.5 sin30 0.5 2.5 0.5 g F g a     A1 EITHER: Both correct; allow their F; must be the same T. OR: correct system equation; allow their F. Use 0.2 F R  to get an equation in a only   2.5 sin30 0.2 2.5 cos30 0.5 2.5 0.5 g g g a          DM1 Where R is a component of weight of P only; from equation(s) with the correct number of dimensionally correct/relevant terms. 1.06 a  m s–2 A1 Allow 15 5 3 . 6  1.05662433. AWRT 1.06 from correct work. 5 Question Answer Marks Guidance 7(b)   4 sin30 2 3 x x x     OR   2 2 sin30 3 y y y        OR 2 2 1 1 1.056 1.056 sin30 2 1.5886 2 2 t t t                 2 1 2 1.056 1.5886.. sin30 2 3 x             2 1 4 OR 1.056 1.5886.. 2 3 y            *B1 Where x is the distance P moves down the plane or Q moves vertically upwards, or where y is the vertical distance of Q below P’s starting point. Allow 1.3 x  or better; allow 0.67 y  or better. Change in PE       2.5 sin30 0.5 g their x g their x     3 4 g their x        OR       2.5 2 sin30 0.5 2 g their y g their y         3 1 their y g       B1 Using their   2 or 1 or 0 ,  x 0 2;   x or their   2 or 1 or 0 ,  y 0 2.   y WD against friction     0.2 2.5 cos30 4.33 g their x their x        OR WD against friction     0.2 2.5 cos30 2 g their y     B1 Using their   2 or 1 or 0 ,  x 0 2;   x or their   2 or 1 or 0 ,  y 0 2.   y 2 2 1 1 2.5 0.5 2 2 v v       4 4 4 2.5 sin30 0.5 0.2 2.5 cos30 3 3 3 g g g                       OR 2 2 1 1 2.5 0.5 2 2 v v       2 2 2 2.5 2 sin30 0.5 2 0.2 2.5 cos30 2 3 3 3 g g g                          DM1 Attempt at work energy equation; dimensionally correct; 5 relevant terms; allow sign errors; allow sin/cos mix. Must be using correct values of x or y. Question Answer Marks Guidance 1.68 v  A1 1.67859014 AWRT 1.68 from correct work. Question Answer Marks Guidance 7(b) Alternative Method for Question 7(b): Considering energy on Q only Must be using tension and mass 0.5 kg only to be awarded the last 4 marks   4 sin30 2 3 x x x     OR   2 2 sin30 3 y y y        OR 2 2 1 1 1.056 1.056 sin30 2 1.5886 . 2 2 t t t                 2 1 2 1.056 1.5886.. sin30 2 3 x             2 1 4 OR 1.056 1.5886.. 2 3 y            (*B1) Where x is the distance P moves down the plane or Q moves vertically upwards, or where y is the vertical distance of Q below P’s starting point. Allow 1.3 x  or better; allow 0.67 y  or better. Change in PE   0.5 g their x   OR Change in PE     0.5 2 g their y    (B1) Using their   2 or 1 or 0 ,  x 0 2;   x or their   2 or 1 or 0 ,  y 0 2.   y WD by tension    5.528312164 their their x   OR WD by tension       5.528312164 2 their their y    (B1) Using their tension from 7(a) from equation(s) with the correct number of dimensionally correct/relevant terms. Using their   2 or 1 or 0 ,  x 0 2;   x or their   2 or 1 or 0 ,  y 0 2.   y   2 4 1 4 0.5 0.5 5.528312164 3 2 3 g v their                  OR   2 2 1 2 0.5 2 0.5 5.528312164 2 3 2 3 g v their                    (DM1) Using their tension from 7(a) from equation(s) with the correct number of dimensionally correct/relevant terms. Attempt at work energy equation; dimensionally correct; 3 relevant terms; allow sign errors. Must be using correct values of x or y. 1.68 v  (A1) 1.67859014 AWRT 1.68 from correct work. Question Answer Marks Guidance 7(b) Alternative Method 2 for Question 7(b): Considering energy on P only Note: must be using tension and mass 2.5 kg only to be awarded the last 4 marks   4 sin30 2 3 x x x     OR   2 2 sin30 3 y y y        OR 2 2 1 1 1.056 1.056 sin30 2 1.5886 . 2 2 t t t                 2 1 2 1.056 1.5886.. sin30 2 3 x             2 1 4 OR 1.056 1.5886.. 2 3 y            (*B1) Where x is the distance P moves down the plane or Q moves vertically upwards, or where y is the vertical distance of Q below P’s starting point. Allow 1.3 x  or better; allow 0.67 y  or better. Change in PE   2.5 sin30 g their x  OR   2.5 g their y  (B1) Using their   2 or 1 or 0 ,  x 0 2;   x or their   2 or 1 or 0 ,  y 0 2.   y WD by tension    5.528312164 their their x   OR WD against friction   0.2 2.5 cos30 g their x    (B1) Using their tension from 7(a) from equation(s) with the correct number of dimensionally correct/relevant terms. Using their   2 or 1 or 0 ,  x 0 2;   x or their   2 or 1 or 0 ,  y 0 2.   y 2 4 1 2.5 sin30 2.5 3 2 g v             4 4 0.2 2.5 cos30 5.528312164 3 3 g their                 (DM1) Using their tension from 7(a) from equation(s) with the correct number of dimensionally correct/relevant terms. Attempt at work energy equation; dimensionally correct; 4 relevant terms; allow sign errors; allow sin/cos mix. Must be using correct values of x or y. 1.68 v  (A1) 1.67859014 AWRT 1.68 from correct work. Question Answer Marks Guidance 7(b) Special Case for using constant acceleration: Maximum 2 marks   4 sin30 2 3 x x x     OR   2 2 sin30 3 y y y        OR 2 2 1 1 1.056 1.056 sin30 2 1.5886 . 2 2 t t t                 2 1 2 1.056 1.5886.. sin30 2 3 x             2 1 4 OR 1.056 1.5886.. 2 3 y            (B1) Where x is the distance P moves down the plane or Q moves vertically upwards, or where y is the vertical distance of Q below P’s starting point. Allow 1.3 x  or better; allow 0.67 y  or better. 2 4 2 1.06 1.68 3 v v            (B1) 1.67859014 AWRT 1.68 from correct work. 5

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