Cambridge A Level Mathematics 9709 — 2024 May/June Paper 4 · Variant 3

1 Two particles P and Q of masses 0.2 kg and 0.5 kg respectively are at rest on a smooth horizontal plane. Particle P is projected with a speed 6 m s -1 directly towards Q. After P and Q collide, P moves with a speed of 1 m s -1 . Find the two possible speeds of Q after the collision. [3] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 1 M1 For attempt at use of conservation of momentum in at least one case. Must have three non-zero terms. Allow sign errors. Must have correct masses with relevant velocities. Their v may be in opposite direction. Speed = 2 m s−1 A1 Do not allow negative. Speed = 2.8[0] m s−1 or 14 5 m s−1 or 4 25 m s−1 A1 OE Do not allow negative. 3

More questions on Momentum

2 30° X N 0.2 kg A particle of mass 0.2 kg is attached to one end of a light inextensible string. The other end of the string is attached to a fixed point on a vertical wall. The particle is held in equilibrium by a force of magnitude X N, perpendicular to the string, with the string taut and making an angle of 30° with the wall (see diagram). Find the tension in the string and the value of X. [3] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 2 M1 Must have correct number of relevant terms (forces must have components as required). Allow sin/cos mix. Allow sign errors. sin30 cos30 0.2 0    X T g M1 Must have correct number of relevant terms (forces must have components as required). Allow sin/cos mix but must be consistent with their other equation. Allow sign errors. 1,  X Tension = 1.73 N [1.7320..] or 3 N A1 For both. Alternative Method for Question 2: Resolving in directions of X and T or triangle of forces 0.2 cos60 0   X g (M1) Must have correct number of relevant terms (forces must have components as required). Allow sin/cos mix. Allow sign errors. 0.2 sin60 0   T g (M1) Must have correct number of relevant terms (forces must have components as required). Allow sin/cos mix but must be consistent with their other equation. Allow sign errors. 1,  X Tension = 1.73 N [1.7320..] or 3 N (A1) For both. Alternative Method for Question 2: Using Lami’s theorem 0.2 sin90 sin150 sin120   g X T (M1M1) First M1 for any two fractions. Second M1 for all three fractions or another pair of fractions. Allow sin120 X and sin150 T for M1 marks. 1,  X Tension = 1.73 N [1.7320..] or 3 N (A1) For both. 3

More questions on Forces and equilibrium

3 A car travels along a straight road with constant acceleration a m s -2 , where a 2 0 . The car passes through points A, B and C in that order. The speed of the car at A is u m s -1 in the direction AB. The distance BC is twice the distance AB. The car takes 8 seconds to travel from A to B and 10 seconds to travel from B to C. (a) Find u in terms of a. [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find the speed of the car at C in terms of a. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 3(a)   2 1 : 8 8 2    AB s u a   8 8 32 or 8 2 u u a u a         2 1 : 2 8 10 10 2      BC s u a a   10 130   u a 8 8 10 or 10 2      u a u a a   2 1 : 3 18 18 2 AC s u a       18 18 162 or 18 2 u u a u a      B1B1 For use of 2 1 2   s ut at or . 2 u v s t B1 for any one correct expression, B2 for two correct expressions. Attempt to solve simultaneously   2 2 1 1 8 10 10 2 8 8 2 2              u a a u a   10 130 2 8 32        u a u a OR 2 2 1 1 18 18 3 8 8 2 2             u a u a   1 8 162 3 8 32        u a u a   2 2 1 3 1 18 18 8 10 10 2 2 2               u a u a a   3 1 8 162 10 130 2           u a u a M1 To obtain an equation in u and a only. Must have come from correct expressions but allow 1 3  instead of 3  or 1 2  instead of 2  or 2 3  instead of 3. 2  Note: M0 for 2 2 1 1 10 10 2 8 8 2 2             u a u a leading to 7 . 3  u a Note: M0 for distance 2  AC AB leading to 49 .  u a 11  u a A1 Question Answer Marks Guidance 3(a) Alternative Method for Question 3(a): Using 2 2 2   v u as    2 2 8 2 u a u s a    or      2 2 18 8 2 2 u a u a s a     or    2 2 18 3 2 u a u s a    (B1B1) B1 for any one correct expression, B2 for two correct expressions.     2 2 2 2 18 8 3 2 2             u a u u a u a a or       2 2 2 2 18 8 8 2 2 2               u a u a u a u a a or       2 2 2 2 18 18 8 3 2 2 2               u a u u a u a a a (M1) To obtain an equation in u and a only. 11  u a (A1) 4 Question Answer Marks Guidance 3(b) 11 18   v a a M1 For use of   v u at or other complete suvat method Using their u in terms of a, e.g.     2 2 2 11 2 18 11 162 841 , v a a a a a              2 2 11 2 18 11 162 841 . v a a a a a            Speed 29a  A1FT FT their expression for v so their u + 18a. Note: If answer to part (a) is 7 , 3 u a  then speed = 47 . 3 a 2

More questions on Kinematics of motion in a straight line

4 A particle travels in a straight line. The velocity of the particle at time t s after leaving a point O is v m s -1 , where v = kt 2 - t4 + 3 . The distance travelled by the particle in the first 2 s of its motion is 6 m. You may assume that v 2 0 in the first 2 s of its motion. (a) Find the value of k. [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find the value of the minimum velocity of the particle. You do not need to show that this velocity is a minimum. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 4(a) For attempt at integration M1* The power of t must increase by 1 with a change of coefficient in the same term. Use of  s vt scores M0.   2 1 1 1 3 2 1 4 1 3 2 3 2 1 2 3                kt t t kt t t c A1 Allow unsimplified.   3 2 1 2 2 2 3 2 0 6 3       k DM1 Use of limits 0 and 2 with 6 to form an equation in k only (without c but allow with  c c ). 3  k A1 4 Question Answer Marks Guidance 4(b) 2 3 4   t Or at min value 4 2 2 3     b t a M1 For attempt at differentiation. Must have expression of the form  at b with 3, a  unless their k = 3 2 . Allow 2 4. kt    2 3 4 0     t 2 3  t A1FT OE FT their k 2 . t their k  Allow without working. 2 2 2 5 3 4 3 3 3 3                   v m s-1 A1 OE Allow 1.67 or better for v. Alternative Method for Question 4(b): Using completing the square Attempt at completing the square (M1) Must have 2 2 3 t       OE, or 2 . 2 t their k        2 2 4 3 3 3 3          t (A1FT) FT their k 2 2 4 3. k t k k          5 3  v m s-1 (A1) OE Allow 1.67 or better. 3

More questions on Integration

5 A van of mass 4500 kg is towing a trailer of mass 750 kg down a straight hill inclined at an angle of i to the horizontal where sin i = 0. 05 . The van and the trailer are connected by a light rigid tow-bar which is parallel to the road. There are constant resistance forces of 2500 N on the van and 300 N on the trailer. (a) It is given that the tension in the tow-bar is 450 N. Find the acceleration of the trailer and the driving force of the van’s engine. [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ On another occasion, the van and trailer ascend a straight hill inclined at an angle of a to the horizontal where sin a = 0.09 . The driving force of the van’s engine is now 9100 N, and the speed of the van at the bottom of the hill is 20 m s -1 . The resistances to motion are unchanged. (b) (i) Find the acceleration of the van and the tension in the tow-bar. [5] .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... (ii) Find the speed of the van when it has travelled a distance of 375 m up the hill. 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Mark scheme: 5(a) Use of Newton’s second law for van or trailer or system Note: Trailer has 4 terms Van has 5 terms System has 7 terms (or 5 if counting van and trailer as one body) M1* Must have correct number of relevant terms. Allow sign errors. Allow sin/cos mix. Allow g missing. Masses must be correct for their equation(s). Forces must have components (or not) as required. Must have either 0.05 or sin2.86 or sin 2.9, not just sin .  Trailer: 450 750 0.05 300 750     g a   525 750  a Van: 4500 0.05 2500 450 4500 D g a        700 4500 D a   System:   4500 0.05 750 0.05 2500 300 4500 750         D g g a   175 5250 D a   A1 For any two correct equations. For attempt to solve for a or D DM1 Must get to ‘a =’ or ‘D =’. Must have correct number of relevant terms in the equation(s) which they are using to find a or D. g must be present. Allow sign errors. Allow sin/cos mix. If no working shown to solve their equations, then their answers should be correct for their equations. 0.7  a m s-2 and 3850  D N A1 4 Question Answer Marks Guidance 5(b)(i) Use of Newton’s second law for van or trailer or system Note: Trailer has 4 terms Van has 5 terms System has 7 terms (or 5 if counting van and trailer as one body) M1* Must have correct number of relevant terms Allow sign errors. Allow sin/cos mix. Allow g missing. Masses must be appropriate for their equation(s). Forces must have components (or not) as required. Must have either 0.09 or sin5.16 or sin 5.2 not just sin .  Trailer: 300 750 0.09 750     T g a   975 750   T a Van: 9100 2500 4500 0.09 4500      g T a   2550 4500   T a System:     9100 2500 300 4500 750 0.09 4500 750        g a   1575 5250  a A1A1 A1 for one correct equation, second A1 for another correct equation. If using Van and Trailer equations, must be using the same T for both to get the second A1. For attempt to solve for a or T DM1 Must get to ‘a =’ or ‘T =’. Must have correct number of relevant terms in the equation(s) which they are using to find a or T. g must be present. Allow sign errors. Allow sin/cos mix. If no working shown to solve their equations, then their answers should be correct for their equations. 1200 T  N and 0.3  a m s-2 A1 5 Question Answer Marks Guidance 5(b)(ii) 2 2 20 2 0.3 375 v their     M1 For use of 2 2 20 2 375    v a or other complete method to find 2 or . v v For info time taken 50 . 3 t  25  v m s-1 A1FT FT their value of a, i.e. 400 750 . v theira    Provided it does not lead to root of negative value. Alternative Method for Question 5(b)(ii): Using energy System:      2 2 1 4500 750 20 4500 750 375 0.09 9100 25 2          v g or Van:     2 2 1 4500 20 4500 375 0.09 9100 2500 1 200 2          v g their OR Trailer:     2 2 1 750 20 750 375 0.09 1 200 300 375 2         v g their (M1) Must include all appropriate terms. Allow sign errors. g must be present. Allow their value of T in place of 1200. 25  v m s-1 (A1FT) FT their value of T if using Van or Trailer. 2

More questions on Kinematics of motion in a straight line

6 A cyclist is travelling along a straight horizontal road. The total mass of the cyclist and her bicycle is 80 kg. There is a constant resistance force of magnitude 32 N to the cyclist’s motion. At an instant when she is travelling at 7 m s -1 , her acceleration is 0 .1 m s -2 . (a) Find the power output of the cyclist. 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(b) Find the steady speed that the cyclist can maintain if her power output and the resistance force are both unchanged. 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The cyclist later descends a straight hill of length 32.2 m, inclined at an angle of sin - 1 a 1 k to the 20 horizontal. Her power output is now 120 W, and the resistance force now has variable magnitude such that the work done against this force in descending the hill is 1128 J. The time taken to descend the hill is 4 s. (c) Given that the speed of the cyclist at the top of the hill is 7.5 m s -1 , find her speed at the bottom of the hill. [6] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 6(a) 7 P DF B1 For OE seen at any point in working. Allow any force term or simply DF, e.g. 32, 80 0.1, 32 80 0.1    , 80 10  etc. 32 80 0.1    D M1 For use of Newton’s second law. Must have correct number of terms. Allow sign errors. [Power ]  280 W A1 3 6(b) [At steady speed driving force =] 280 32  v M1 Attempt at equilibrium equation (a = 0) with their power. Steady speed 8.75  m s-1 or 35 4 m s-1 A1FT OE FT their power from part (a) r 280. 32 thei 2 Question Answer Marks Guidance 6(c) 120 4    480 B1 Work done by cyclist. 2 1 80 2  v 2 40v     or 2 1 80 7.5 2     2250 B1 For at least one KE term. 1 80 32.2 20   g   1288   80 10 1.61   B1 Change in PE. Attempt at work-energy equation M1 Attempt at work energy equation with five relevant terms (four relevant terms plus work done against resistance); dimensionally correct. Allow sign errors. Allow sin/cos mix.   2 2 1 1 120 4 80 32.2 1128 80 7.5 20 2         g v 2 480 1 288 1128 40 2250 v         A1 For correct equation. Speed = 8.5[0] m s-1 or 17 2 A1 OE Use of constant acceleration scores M0 and cannot score B marks if the method leading to their answer only uses constant acceleration. 6

More questions on Energy, work and power

7 A D 30° 30° B C The diagram shows a track ABCD which lies in a vertical plane. The section AB is a straight line inclined at an angle of 30° to the horizontal and is smooth. The section BC is a horizontal straight line and is rough. The section CD is a straight line inclined at an angle of 30° to the horizontal and is rough. The lengths AB, BC and CD are each 2 m. A particle is released from rest at A. The coefficient of friction between the particle and both BC and CD is n. There is no change in the speed of the particle when it passes through either of the points B or C. (a) It is given that n = .01 . Find the distance which the particle has moved up the section CD when its speed is 1 m s -1 . 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(b) It is given instead that with a different value of n the particle travels 1 m up the track from C before it comes instantaneously to rest. Find the value of n and the speed of the particle at the instant that it passes C for the second time. 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Mark scheme: 7(a) For CD cos30 R mg Use of 0.1  F R for either BC or CD   0.1   BC F mg m OR 3 0.1 cos30 2         CD F mg m M1 Note: The first two marks are often gained in the work-energy equation. 0.1 cos30 sin30  mg d mgd 3 5 2                  md A1 For sum of work done by friction and the change in PE. Note: Allow terms on different sides of a work energy equation as long as they have different signs. 2 1 2sin30 0.1 2 0.1 cos30 sin30 1 2        mg mg mg d mgd m 1 10 2 cos30 5 2           m m m d md m M1 Attempt at work energy equation with five relevant terms (dimensionally correct). Allow sign errors. Allow sin/cos errors but must be consistent. Note: Initial PE = mg. 1.28  d m or   15 10 3 97  [1.27854…] A1 ISW if go on to find total distance = 2 + 2 + 1.28 having already found 1.28. Question Answer Marks Guidance 7(a) Alternative Method for Question 7(a): Using Newton’s second law and equations of motion For CD cos30  R mg (B1) May be seen in later working without m. If not seen in working check diagram but must be a reaction force, not a downward component of the weight. Use of 0.1  F R for either BC or CD   0.1   BC F mg m or 3 0.1 cos30 2         CD F mg m (M1) For aCD sin30 0.1 cos30    mg mg ma 3 sin30 0.1 cos30 5.866 5 2                      a g g (A1) For correct equation for a or ma in section CD Note: Allow if acceleration in the opposite sense and both signs positive. For aAB sin30  mg ma ⟹ 5  a ⟹   2 0 2 5 2 20     B v For aBC 0.1   mg ma 1  a so   2 20 2 1 2 16     C v 21 16 2 ( sin30 0.1 cos30)     g g d 3 1 16 2 5 d 2                    (M1) Attempt to find d. Allow sign errors in Newton’s second law. Allow sin/cos errors but must be consistent. Should include a valid attempt at 2 C v to get M1. Must get to final line of working. Note: this mark can be earned even if A0 above. Must have 2 term acceleration though could have sign error. d = 1.28 m or   15 10 3 97  [1.27854…] (A1) ISW if go on to find total distance = 2 + 2 + 1.28 having already found 1.28. Question Answer Marks Guidance 7(a) Alternative Method for the last 2 marks: Using an energy method for the third phase For aAB: sin30  mg ma ⟹ 5  a ⟹   2 0 2 5 2 20     B v For aBC: 0.1   mg ma 1  a so   2 20 2 1 2 16     C v   2 2 1 1 4 sin30 0.1 cos30 2     m mgd mg d (M1) Attempt at work energy equation for the third phase with four relevant terms (dimensionally correct). Allow sign errors. Allow sin/cos errors but must be consistent. Must get to final line. d = 1.28 m or   15 10 3 97  [1.27854…] ignore units (A1) ISW if go on to find total distance = 2 + 2 + 1.28 having already found 1.28. 5 Question Answer Marks Guidance 7(b) 2sin30 2 1 cos30 1sin30        mg mg mg mg 10 20 10 cos30 5 OR 1 0 20 5 3 5               m m m m m m m m M1 Attempt at work energy equation with four relevant terms (dimensionally correct). Allow sign errors. Allow sin/cos errors but must be consistent. 0.174  or   4 3 0.174457 13   A1 1sin30 1 cos30    mg mg 5 5 3       m m M1 For difference between the change in PE and the work done by friction. Note: Allow terms on different sides of a work energy equation as long as both have the same sign. Allow sin/cos errors but must be consistent. Using , their  or the correct value of  to at least 2 sf. Must be as part of an attempt to find speed, not , although this could be the first step. 2 2 1 1 1sin30 1 cos30 5 5 3 2 2              mg mg mv m m mv Speed = 2.64m s-1 [2.64164…] A1 Question Answer Marks Guidance 7(b) Alternative Method for Question 7(b): Newton’s second law and equations of motion For aAB sin30  mg ma ⟹ 5  a ⟹   2 0 2 5 2 20     B v For aBC    mg ma   a g so 2 20 2 2     C v g For aCD 3 sin30 cos30 5.866 5 2                        mg mg ma a   0 20 2 2 2 sin30 cos30 1         g g g 20 40 10 10 3 0           (M1) For attempt at equation for µ. Allow sign errors. Allow sin/cos errors but must be consistent. Must get to fourth line for M1. 4 3 0.174 or [0.174457...] 13    (A1) sin30 cos30 5 5 3 a g g           (M1) For correct equation for a or ma in section CD down plane (weight component – friction). Allow sin/cos errors but must be consistent. Using ,  their  or the correct value of  to at least 2sf. Must be as part of an attempt to find speed, not ,  although this could be the first step. 2 0 2( sin30 cos30) 1      v g g ⇒ Speed = 2.64 m s-1 [2.64164…] (A1) Question Answer Marks Guidance 7(b) Alternative method for last 2 marks of Question 7(b): Using energy at the start – total work done against friction   2 1 2sin30 2 cos30 2 2 mg mg mg mv               (M1) For PEA  total work done against friction [= KEC]. Allow sin/cos errors but must be consistent. Using ,  their  or the correct value of  to at least 2sf. 2 1 10 (20 20 cos30 2            m m m mv 2 1 10 (20 10 3 2            m m m mv Speed = 2.64m s-1 [2.64164…] (A1) 4

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