Cambridge A Level Mathematics 9709 — 2022 May/June Paper 4 · Variant 2

1 Small smooth spheres A and B, of equal radii and of masses 5kg and 3kg respectively, lie on a smooth horizontal plane. Initially B is at rest and A is moving towards B with speed 8.5ms−1. The spheres collide and after the collision A continues to move in the same direction but with a quarter of the speed of B. (a) Find the speed of B after the collision. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the loss of kinetic energy of the system due to the collision. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 1(a) Conservation of momentum M1 3 terms; allow M1 if speed of A after collision is 1 8.5 4  . Allow 5 8.5 5 3    X Y where X and Y are different which may be seen by later work. If X and Y are subsequently used as being equal then M0. 5 8.5 5 0.25 3    v v A1 OE e.g. 5 8.5 5 3 4    V V Speed of B 1 10 ms  A1 Do not award if 10 from using mgv, maximum 2/3 –10 is A0 as speed required not velocity 3 1(b) KE before   2 1 5 8.5 180.625 2    KE after   2 2 1 1 5 2.5 3 10 15.625 150 165.625 2 2        1 Attempt at any of the 3 terms for KE, using their 1 10 ms Not   2 1 5 3 8.5 2    , not   2 1 5 3 2.5 2    not   2 1 5 3 10 2    unless  X Y seen KE loss   180.625 165.625 15    J A1 Accept ‒15, AWRT 15.0 2

More questions on Momentum

2 Coplanar forces of magnitudes 60N, 20N, 16N and 14N act at a point in the directions shown in the diagram. Find the magnitude and direction of the resultant force. [6] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 Resolving either direction M1 3 terms; allow sign errors and allow sin/cos mix      20cos60 14 16cos50 14.2846 X       A1      60 20sin 60 16sin50 30.42278 Y      A1     2 2 14.2846 30.42278 R     M1 Attempt to solve for R ; one missing term in total   1 1 30.42278 tan tan 2.1297 14.2846                 OR   1 1 14.2846 tan tan 0.4596 30.42278                 M1 Attempt to solve for 𝜃 or 𝛼; one missing term in total R = 33.6 N Direction is 64.8° above the 14 N force or 25.2° above the negative 𝑥-axis or 25.2° left of the 60 N force or bearing 335° or 115° anticlockwise from the positive x -axis A1 Both correct. OE; allow 64.9, 25.1 Giving an angle only is insufficient. Direction may be seen on a diagram, with minimum of arrow on resultant. Arrows on both components only is A0 as it doesn’t show the direction of the resultant. However the direction is stated, it must be able to be drawn uniquely. 6

More questions on Forces and equilibrium

3 Two particles A and B, of masses 2.4kg and 1.2kg respectively, are connected by a light inextensible string which passes over a fixed smooth pulley. A is held at a distance of 2.1m above a horizontal plane and B is 1.5m above the plane. The particles hang vertically and are released from rest. In the subsequent motion A reaches the plane and does not rebound and B does not reach the pulley. (a) Show that the tension in the string before A reaches the plane is 16N and find the magnitude of the acceleration of the particles before A reaches the plane. 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(b) Find the greatest height of B above the plane. 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Mark scheme: 3(a) 1.2 1.2   T g a   2.4 1.2 2.4 1.2    g g a M1 Attempt at Newton’s second law on either particle or the system with correct number of terms; allow sign errors. A1 Any 2 consistent and correct May have an a in opposite direction to our a Attempt to solve for a or T M1 From equation(s) with correct number of relevant terms. If g missing then M0A0M1A0, maximum1/4. Must get  a or  T Must not assume 16  T . May attempt to verify a value of a using 16  T in 2 equations 16  T N and 10 3  a ms-2 A1 Both correct; allow 3.33  a . AG for 16  T . Assuming 16  T and only one equation is M1A0M0A0 maximum 1/4. Withhold A mark if 15.9... 16   T , but condone 1.2 3.33 12 16     T or 24 2.4 3.33 16     T 4 Question Answer Marks Guidance 3(b)   2 10 2 2.1 14 3 v     14 3.741 v        OR 2 1 2.4 2.4 2.1 16 2.1 2 v g       OR 2 2 1 1 2.4 1.2 2.4 2.1 1.2 2.1 2 2 v v g g          M1 Use of suvat or use energy to find v or 2v , using their  a g (unless 10 comes from their attempt at a ) from (a), 2.1  s 0 14 2      g s s or      2 1 1.2 14 1.2 2      g h h M1 Attempt to use 2 2 2   v u as (or other complete method), using  a g , to find additional height after string slack, using their v or 2v .   1.5 2.1 0.7 4.3     s m A1 AWRT 4.3(0); Allow use of 3.33  a to give  4.2993 4.3 0   s Allow use of 3.74  v to give  4.29938 4.3 0   s Alternative for question 3(b) - using energy on particle B 16 2.1 1.2   gH M1 Apply energy to B, 2 terms 2.8  H A1   1.5 2.8 4.3    s m A1 3

More questions on Kinematics of motion in a straight line

4 A particle A, moving along a straight horizontal track with constant speed 8ms−1, passes a fixed point O. Four seconds later, another particle B passes O, moving along a parallel track in the same direction as A. Particle B has speed 20ms−1 when it passes O and has a constant deceleration of 2ms−2. B comes to rest when it returns to O. (a) Find expressions, in terms of t, for the displacement from O of each particle t seconds after B passes O. 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(b) Find the values of t when the particles are the same distance from O. 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(c) On the given axes, sketch the displacement-time graphs for both particles, for values of t from 0 to 20. [3]

Mark scheme: 4(a) Use suvat to find expressions for As or Bs As must be using 8 u and time of 4 t For Bs , using 2 1 2   s ut at with 20  u and 2  a     8 4 32 8     As t t A1 Any unsimplified expression; ISW   2 1 20 2 2    Bs t t A1 Any unsimplified expression; ISW If 0 marks scored then allow SC: B1 for  8  As t and B1 for       2 1 20 4 2 4 2      Bs t t maximum 2/3 3 Question Answer Marks Guidance 4(b)   2 1 8 4 20 2 2     t t t *M1 Equating their expressions for As and Bs to form an equation in t where As is of the form 8 32   t and Bs is of the form 2 1 20 2 2    t t Attempt to solve a 3-term quadratic to find at least one t value DM1 For reference 2 12 32 0    t t Allow if no working seen and have correct real solution(s) to their 3-term quadratic. If working shown and if using the formula, it must be using the correct formula. If factorising must have 3 of the 4 terms correct of    4 8   t t 4  t and 8 A1 If 0 marks scored then allow SC: M1 for     2 1 8 20 4 2 4 2       t t t and A1 for 8  t and 12 maximum 2/3. 3 Question Answer Marks Guidance 4(c) Straight line B1 FT Positive gradient, intersecting positive s axis. Full domain not required. FT if they get 8  As t using the SC in (a) Inverted quadratic, passing through origin. B1 FT Full domain not required but must clearly go beyond the maximum. FT if they get     2 1 20 4 2 4 2      Bs t t using the SC in (a), with curve though positive t axis before turning point. All correct, line through (0, 32), quadratic through (20, 0), intersections indicated at 4  t and 8  t . B1 Intersections must occur before the turning point. 3

More questions on Kinematics of motion in a straight line

5 A block of mass 12kg is placed on a plane which is inclined at an angle of 24Å to the horizontal. A light string, making an angle of 36Å above a line of greatest slope, is attached to the block. The tension in the string is 65N (see diagram). The coefficient of friction between the block and plane is -. The block is in limiting equilibrium and is on the point of sliding up the plane. Find -. 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Mark scheme: 5 Attempt at resolving parallel to the plane otherwise dimensionally correct. 65cos36 12 sin 24 g F    A1 3.777707   F Attempt at resolving perpendicular to the plane *M1 3 terms. Allow sign errors, sin/cos mix. Allow g missing, otherwise dimensionally correct. 12 cos 24 65sin 36 g R    A1 71.419   R Use   F R 65cos36 12 sin 24 52.586 48.808 3.777 12 cos 24 65sin 36 109.625 38.206 71.419 g g                   DM1 To get an equation in  only. Dependent on two previous M marks. Allow g missing 0.0529  A1 Allow AWRT 0.053 Do not accept fractional equivalent. 6

More questions on Forces and equilibrium

7 A particle P moves in a straight line. The velocity vms−1 at time t seconds is given by v = 0.5t for 0 ≤t ≤10, v = 0.25t2 −8t + 60 for 10 ≤t ≤20. (a) Show that there is an instantaneous change in the acceleration of the particle at t = 10. 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(b) Find the total distance covered by P in the interval 0 ≤t ≤20. 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Mark scheme: 7(a)  d 0.5 d   v a t Differentiate to get     d 2 0.25 8 0.5 8 d       v a t t t B1 Allow unsimplified   0.5 10 8 3     a B1 CWO. Do not award final B mark if more than 2 accelerations seen and not discarded, 2/3 maximum Ignore any comments, correct or incorrect 3 Question Answer Marks Guidance 7(b) Get distance in first 10 seconds as 25 B1 From suvat or from 10 0 0.5 d  t t 0  v when 12  t and 20  t B1 SOI Attempt to integrate v   2 0.25 8 60 d        s t t t *M1 For integration, the power of t must increase by 1 in at least 1 term with a change of coefficient in the same term.  s vt is M0     3 2 3 2 0.25 8 1 60 4 60 3 2 12               s t t t c t t t c A1 Allow unsimplified Attempt to evaluate their 3 2 1 4 60 12         t t t for 10  t to 12  t and 12  t to 20  t DM1 Using the correct limits correctly 850 800 14 64 25 288 288 25 51 3 3 3 3                            s m A1 Question Answer Marks Guidance 7(b) Special Case for those who use a calculator to integrate. Maximum 4/6 Get distance in first 10 seconds as 25 B1 From suvat or 10 0 0.5 d  t t 0  v when 12  t and 20  t B1 SOI Either   12 2 10 14 0.25 8 60 d 4.67 3       s t t t Or   20 2 12 64 0.25 8 60 d 21.3 3       s t t t B1 Allow 20 2 10 0.25 8 60 d 26     t t t 14 64 25 51 3 3           s m B1 Allow if 12  t and 20  t not found for 3 marks 6

More questions on Kinematics of motion in a straight line