Cambridge A Level Mathematics 9709 — 2024 Oct/Nov Paper 4 · Variant 2

1 v (m s–1) V 0 t (s) 4 10 T –3 The velocity of a particle moving in a straight line at time t seconds after leaving a fixed point O is v m s -1 . The diagram shows a velocity-time graph which models the motion of the particle from t = 0 to t = T . The graph consists of four straight line segments. The particle accelerates from rest to a speed of V m s -1 over a period of 4 s, and then decelerates at 5 m s -2 to instantaneous rest over a period of 6 s. 3 The particle then travels back towards O, reaching a maximum speed of 3 m s -1 before coming to rest at time t = T . (a) Find the value of V. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Given that the total distance travelled by the particle from t = 0 to t = T is 68 m, find the value of T. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1(a) Use of v = u + at or use of acceleration is gradient of the line M1 5 Use v = 0 , a =  and t = 10 − 4  = 6  . 3 5 0 − V 5 V − 0 5 0 = V −  (10 − 4 ) OR = − OR = . 3 10 − 4 3 4 − 0 2 V = 10 A1 Make sure that 10 comes from a correct initial equation or 0 − V 5 equivalent, if not then A0, e.g. = leading to 10 − 4 3 V = 10 is M1A0. 2 1(b) 1 B1FT SOI Distance in first 10 seconds =  10 their V 1 1 2 OE, e.g. 4 their V +  (10 − 4 )  their V 2 2 If correct distance is 50 m. 1 1 M1 Use distance (68) is total area under graph.  ( T − 10 ) +3  10  their V = 68 2 2 Their 50 must be from 5 their V . OR Allow with -3. 1 1 1 1 1  ( T − 10 ) +3 4 their V +  (10 − 4 )  their V = 68 Allow  X +3  10  their V = 68 and use of 2 2 2 2 2 T = X + 10 . T = 22 A1 WWW Be aware that V = 10 from wrong work in (a) can be awarded maximum of B1M1A0. Special Case for assumption of isosceles triangle for t = 10 to t = T : 1 B1FT SOI Distance in first 10 seconds =  10 their V 1 1 2 OE, e.g. 4 their V +  (10 − 4 )  their V . 2 2 If correct distance is 50 m.  1  T − 10  1  B1 WWW  T = 22    =3  ( 68 − 50 ) →  2  2  2  Special Case for assumption that particle returns to O:  1 1  98 B1 32.66666. T = or 32.7   ( T − 10 )  3 =  68 →  AWRT 32.7.  2 2  3 3

More questions on Kinematics of motion in a straight line

2 A block of mass 20 kg is held at rest at the top of a plane inclined at 30° to the horizontal. The block is projected with speed 5 m s -1 down a line of greatest slope of the plane. There is a resistance force acting on the block. As the block moves 2 m down the plane from its point of projection, the work done against this resistance force is 50 J. Find the speed of the block when it has moved 2 m down the plane. [4] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 2 Change in PE =  20 g sin30  2  = 200  B1 1 2 2 B1 For either expression.   20v  = 10v    1 2 2 Do not allow  20 ( v − 5 ) . 1 2 2   20  5  = 250  2 1 2 1 2 M1 Attempt at work energy equation; 4 terms; dimensionally  20v −  20  5 = 20 g sin30 −2 50 correct; allow sign errors; PE term must include a 2 2 component (allow sin/cos mix). 10v 2 − 250 = 200 − 50    1 2 Do not allow with  20 ( v − 5 ) for KE. 2 Speed = 6.32 m s-1 OR 40 m s-1 A1 OE, e.g. 2 10 . 6.324555…. AWRT 6.32. Special case for assumption of constant resistance force: 50 B1 20a = 20 g sin30 −  → a = 3.75 2 v 2 = 52 + 2 2 3.75  v = 6.32 m s-1 OR 40 m s-1 B1 OE, e.g. 2 10 . 6.324555… AWRT 6.32. 4

More questions on Energy, work and power

3 A cyclist is riding along a straight horizontal road. The total mass of the cyclist and his bicycle is 90 kg. The power exerted by the cyclist is 250 W. At an instant when the cyclist’s speed is 5 m s -1 , his acceleration is 0.1 m s -2 . (a) Find the value of the constant resistance to motion acting on the cyclist. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ The cyclist comes to the bottom of a hill inclined at 2° to the horizontal. (b) Given that the power and resistance to motion are unchanged, find the steady speed which the cyclist could maintain when riding up the hill. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 3(a) 250 B1 For use of power =Fv e.g. 5  PF = 250 . [Peddling force = PF =] 5 their PF − R = 90  0.1 M1 Using their PF  250 . 3 terms; allow sign errors. Dimensionally correct. Resistance = 41 N A1 3 3(b) 250 M1 For attempt at resolving up the hill; 3 terms; allow sign − their 41 − 90 g sin2 = 0 errors; Must be a component of weight (NOT mass) but v allow sin/cos mix; allow use of their 41. Dimensionally correct. 250 Oe eg v = . their 41 + 90 g sin 2 Steady speed = 3.45 m s–1 A1 3.45258… AWRT 3.45. 2

More questions on Newton’s laws of motion

4 P Q i° 45° A B 0.2 kg 0.1 kg The diagram shows two particles, A and B, of masses 0.2 kg and 0.1 kg respectively. The particles are suspended below a horizontal ceiling by two strings, AP and BQ, attached to fixed points P and Q on the ceiling. The particles are connected by a horizontal string, AB. Angle APQ = 45° and BQ P = i° . Each string is light and inextensible. The particles are in equilibrium. (a) Find the value of the tension in the string AB. 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(b) Find the value of i and the tension in the string BQ. 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Mark scheme: 4(a) TAB sin45 = 0.2 g cos45 *B1 BOD for using sin45 instead of cos45, particularly if TAB using wrong components in (b). OR tan 45 = 0.2 g 0.2 g OR tan 45 = TAB OR TAB = TAP cos45 and TAP sin45 = 0.2 g TAB = 2 N DB1 Condone TAB = 0.2 g (with or without working) for full marks. WWW. DO NOT ISW. ALTERNATIVE for 4(a) using LAMI’s THEOREM: TAB 0.2 g B1 = sin135 sin135 TAB = 2 N B1 Condone TAB = 0.2 g (with or without working) for full marks. WWW. DO NOT ISW. 2 4(b) TBQ cos− theirTAB = 0 *M1 For resolving either horizontally OR vertically; 2 terms; allow sin/cos mix; allow their TAB ; TBQ sin− 0.1g = 0 M0 for any use of TAB = TBQ = TAP . A1FT For both. FT their TAB ONLY; TAB  TBQ . Sight of ( theirTAB ) tan= 0.1g is M1 only without seeing an equation for TBQ . −1 0.1g  2 2 DM1 Solve for θ or solve for TBQ from equations with the = tan   or TBQ = ( 0.1g ) + ( theirTAB )  theirTAB  correct number of relevant terms. Using their TAB . = 26.6 AND TBQ = 5 or 2.24 A1 = 26.56505. TBQ = 2.236067 . AWRT 26.6 and 2.24. FIRST ALTERNATIVE for 4(b): TBQ = ( theirTAB ) cos+ 0.1g sin M1 For resolving either parallel to BQ or perpendicular to BQ; 2 terms; allow sign errors on 3 term equation only; allow ( theirTAB ) sin = 0.1g cos sin/cos mix; allow their TAB . Sight of ( theirTAB ) tan= 0.1g is M1 only without seeing an equation for TBQ . M0 for any use of TAB = TBQ = TAP . A1FT For both. FT their TAB ONLY; TAB  TBQ . −1 0.1g  DM1 Solve for θ (or solve for TBQ ) from equations with the = tan      ( theirTAB )  correct number of relevant terms. Using their TAB . 4(b) = 26.6 AND TBQ = 5 or 2.24 A1 = 26.56505. TBQ = 2.236067 . AWRT 26.6 and 2.24. SECOND ALTERNATIVE for 4(b) using triangle of forces: AB ONLY; TBQ2 = ( 0.1g ) 2 + ( theirTAB ) 2 M1 Using their T M0 for any use of TAB = TBQ = TAP . OR 0.1g = TBQ sin () OR ( theirTAB ) = TBQ cos () A1FT For any 2 equations. FT their TAB ONLY; TAB  TBQ . Sight of ( theirTAB ) tan= 0.1g is M1 only without seeing an equation for TBQ . Solve for TBQ or  DM1 Solve for θ or solve for TBQ from equations with the 2 2 correct number of relevant terms. Using their TAB . E.g. TBQ = ( 0.1g ) + ( theirTAB ) −1 0.1g  = tan    theirTAB  = 26.6 AND TBQ = 5 or 2.24 A1 = 26.56505. TBQ = 2.236067 . AWRT 26.6 and 2.24. THIRD ALTERNATIVE for 4(b) using LAMI’s THEOREM: *M1 For any 2 fractions correct; M0 for any use of TAB = TBQ = TAP . AB ONLY. TBQ 0.1g theirTAB TBQ 0.1g theirTAB A1FT For all 3 fractions correct. FT their T = = OR = = sin90 sin (180 − ) sin ( 90 + ) sin90 sin cos Sight of ( theirTAB ) tan= 0.1g is M1 only without seeing an equation for TBQ . 4(b) −1 0.1g  DM1 Solve for θ. = tan    theirTAB  = 26.6 AND TBQ = 5 or 2.24 A1 = 26.56505. TBQ = 2.236067 . AWRT 26.6 and 2.24. FOURTH ALTERNATIVE for 4(b) for resolving on the whole system: TBQ sin+ ( theirTAP ) sin45 = 0.2 g + 0.1g M1 For resolving either horizontally or vertically on the whole system; using theirTAP . TBQ cos= ( theirTAP ) cos45 M0 for any use of TAB = TBQ = TAP . If TAP not found in part (a), must have either TAP sin45 = 0.2 g or ( theirTAB ) = TAP cos45 . Correct number of terms; allow sign errors on the 4 term equation ONLY; allow sin/cos mix. A1FT Both equations correct. May see TBQ sin= 1 and TBQ cos= 2 . Sight of ( theirTAB ) tan= 0.1g is M1 only without seeing an equation for TBQ . −1 0.1g + 0.2 g − ( theirTAP ) sin45  DM1 Solve for θ or solve for TBQ from equations with the = tan      ( theirTAP ) cos45  correct number of relevant terms. Using their TAP . 2 2 or TBQ = ( 0.1g + 0.2 g − ( theirTAP ) sin45 ) + ( ( theirTAP ) cos45 ) = 26.6 AND TBQ = 5 or 2.24 A1 = 26.56505. TBQ = 2.236067 . AWRT 26.6 and 2.24. 4

More questions on Forces and equilibrium

5 Two particles, P and Q, of masses 2m kg and m kg respectively, are held at rest in the same vertical line. The heights of P and Q above horizontal ground are 1 m and 2 m respectively. P is projected vertically upwards with speed 2 m s -1 . At the same instant, Q is released from rest. (a) Find the speed of each particle immediately before they collide. 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(b) It is given that immediately after the collision the downward speed of Q is 3.5 m s -1 . Find the speed of P at the instant that it reaches the ground. 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Mark scheme: 5(a)  1 2  1 2 M1 1 2 OR  For use of s = ut + at at least once with a =  g and  s P up =    2t − gt   Qs down =   gt  2  2 2 u = 0 or u = 2 . Seen anywhere. 1 1  1  M1 Use sP + sQ =  ( 2 − 1) OR 1 ONLY with Ps and Qs of 2t − gt 2 + gt 2 = 2 − 1 =t   2 2  2  the correct form 1 A1 Must be positive. vP = 2 − g = −3 m s -1 so speed = 3 m s -1 2 1 A1 Must be positive. vQ = − g = −5 m s -1 so speed = 5 m s -1 If A0A0, allow SCB1 if both are negative 2 4 5(b) of conservation of momentum; 4 non- zero terms; their 5  = 2 mv + m  3.5 *M1 Use 2 m  ( their 3  vP  ) + m  (  vQ  ) using allow sign errors their 3 m s -1 and their 5 m s -1; and m missing ONLY. Do not allow with v P = 2 or with Qv = 0. Do not allow made up values for 3 and 5. Allow LHS of our equation to have only one term but only if getting vP = 0 in (a) from s P = sQ . v = 3.75 A1 Allow v = −3.75 from correct work.  1   1 1  1  2   1  *B1 This may be seen in part (a), but do not award the mark At t = 2  − g =       , s p =    until stated/used in part (b).   2 2 2    2   4     1 1  1  2   1  = OR         ps = − 3 2 + 2 g  2   4     ( − 3 ) 2 − 2 2   1  =  OR    4  ps =    2 g      1  1  1 2 5  OR  At t = , sQ =   g    =    2  2  2  4   1   1 1  1  2   5  At t = 5  − g =       OR  , sQ =      2 2 2    2   4      52  −0 2   5    OR   sQ =      =    2 g  4     3  so height above ground =    4  5(b) 2 2  3  DM1 Use of v 2 = u 2 + 2 as using their v and a =  g ; their s is v = ( their 3.75 ) + 2 ( g )    4    1    5  either  OE. 2 − 1 +  −   or    2 2 45 OR 0 = ( their 3.75 ) −2 ( g )  s =s   4    4  64 Dependent on the previous M mark and B mark being 2 2  3  45  awarded. 0 +  AND v =  2 ( g )    +     4  64  v = 5.39 m s -1 A1 465 Or ; AWRT 5.39; 5.390964… 4 5

More questions on Kinematics of motion in a straight line

6 A particle, P, travels in a straight line, starting from a point O with velocity 6 m s -1 . The acceleration of P at time t s after leaving O is a m s -2 , where 1 a = -1.5 t 2 for 0 G t G 1, 2 1 - 21 a = 1. 5t - 3t for t 2 1. (a) Find the velocity of P at t = 1. 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(b) Given that there is no change in the velocity of P when t = 1, find an expression for the velocity of P for t 2 1. 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(c) Given that the velocity of P is positive for t G 4 , find the total distance travelled between t = 0 and t = 4 . 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Mark scheme: 6(a) Attempt to integrate a for 0 t 1 M1 Increase power by 1 and a change in coefficient in at least one term (which must be the same term); 3 2 v = at is M0. Expect ( v = ) − t  + c  . 3 3 A1 3 ( v = ) − 1.5 t 2 + 6 = − t 2 + 6 Allow for − t 2 + c and c = 6 seen from CWO. 1.5 Allow unsimplified. Velocity at t = 1 is 5 ms-1 A1 CWO. 3 6(b) Attempt to integrate a for t  1 *M1 Increase power by 1 and a change in coefficient in at least  3 1 3 1  one term (which must be the same term); 1.5 3 v = at is M0.  ( v = ) t 2 − t 2  + c  = t 2 − 6t 2  + c   3 1/ 2   2  Use v = their 5 when t = 1 in attempt to find c DM1 Must get a numerical expression for c ; if no substitution  5 = 1 − 6 + c  seen, c must be correct for their expression for v . Their 5 must not be a made up value. 3 1 A1 OE, but must be a complete (possibly unsimplified) ( v = ) t 2 − 6t 2 + 10 expression. 3 6(c) For attempt at integration of their v for either section *M1 Increase power by 1 and a change in coefficient in at least one term (which must be the same term); Their v ’s have to come from integration. s = vt is M0. Their v ’s have to come from integration. 5 5 A1 For either correct. ( s1 = ) − 1 t 2 + 6t  + c  = − 2 t 2 + 6t  + c  Allow unsimplified. 5 / 2 5 1 52 6 32 2 52 32 ( s2 = ) t − t + 10t  + c  = t − 4t + 10t  + c  5 / 2 3 / 2 5 For use of limits 0 and 1 for 1s and 1 and 4 for 2s DM1 Using correct limits correctly, in expressions that have come from integration (where all powers must have increased by 1 with a change in coefficient for non-linear terms) of a v that came from integration, and using the limits correctly would lead to 2 positive values when used in their expressions. 1s and 2s must have the correct number of non-constant terms and must include a linear term. Total distance = 20 m A1  2   28  s1 =  − +1 6  1  − 0  = = 5.6  .  5   5   2 52 32  s2 =   4 − 4  4 + 10  4  −  5     2 52 32  72  . = = 14.4   1 − 4  1 + 10  1   5   5  If either displacement expression has a constant of integration and is given incorrectly, then award A0 even if 20 m seen. For reference the constant of integration for s 2 is –0.8. 6(c) SC for integration not seen: 28 72 B1 s1 = or s2 = 5 5 Total distance = 20 m B1 4

More questions on Kinematics of motion in a straight line

7 A 0.2 kg i B 0.3 kg 0.25 m Two particles, A and B, of masses 0.2 kg and 0.3 kg respectively, are attached to the ends of a light inextensible string. The string passes over a small fixed smooth pulley which is attached to the bottom of a rough plane inclined at an angle i to the horizontal where sin i = 0. 6 . Particle A lies on the plane, and particle B hangs vertically below the pulley, 0.25 m above horizontal ground. The string between A and the pulley is parallel to a line of greatest slope of the plane (see diagram). The coefficient of friction between A and the plane is 1.125 . Particle A is released from rest. (a) Find the tension in the string and the magnitude of the acceleration of the particles. 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(b) When B reaches the ground, it comes to rest. Find the total distance that A travels down the plane from when it is released until it comes to rest. You may assume that A does not reach the pulley. 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Mark scheme: cos37 or better for 0.8.7(a) R = 0.2 g  0.8  = 1.6  B1 Allow R is a component of weight. F = 1.125 R  = 1.8  *M1 Where Must have 1.125  0.2 g  0.8 or 1.125  0.2 g  0.6 or with using cos37 or sin37 or better for 0.8 and 0.6 respectively. These 2 marks may be embedded in the N2L equation(s). Use of Newton’s second law for A or B or system *M1 Correct number of terms; allow sign errors; allow sin/cos mix. Dimensionally correct. 0.3 g − T = 0.3a A1 For any 2 correct equations. T + 0.2 g  0.6 − F = 0.2a Allow sin37 or better for 0.6. Allow their possibly incorrect F . 0.3 g + 0.2 g  0.6 − F = 0.5a a = 4.8 m s–2 A1 Must be positive. For attempt to solve for T DM1 From equations with the correct number of relevant terms. If a found first then substituting into an equation with the correct number of relevant terms and solving. If resolved equations incorrect and no working seen, then this mark is implied by the correct T value for their equations. Dependent on previous 2 M marks T = 1.56 N A1 7 7(b) [For B or A] v 2 = 0 2 + 2  4.8  0.25 *M1 Use of v 2 = u 2 + 2 as with u = 0 and using  2 2 15  s = 0.25 , their 4.8 , a  g . Must be a complete method   v = 2.4 or v = = 1.549   5  to get an expression for v or 2v . Attempt at Newton’s 2nd Law on A when string becomes slack *M1 3 terms; allow sign errors; allow sin/cos mix; allow their F from part (a); Dimensionally correct; must be  0.2 g  0.6 − 1.125  0.2 g  0.8 = 0.2 a  non-zero using 0.2 for the mass. Allow = 37 or better. For reference a = −3 (or −2.99 if using = 36.9 ). 20 = 2.4 + 2 −( 3)  s =s 0.4 DM1 Using constant acceleration formula(e) using a negative acceleration to get an expression in s only. Dependent on previous 2 M marks. Total distance = 0.25 + 0.4 = 0.65 m A1 AWRT 0.650. Allow 0.651 from use of = 36.9 . ALTERNATIVE for 7(b) using energy: [For B or A] v 2 = 0 2 + 2  4.8  0.25 *M1 Use of v 2 = u 2 + 2 as with u = 0 .  2 2 15  Using s = 0.25 , their 4.8 , a  g .   v = 2.4 or v = = 1.549   5  Must be a complete method to get an expression for v or 2v . For attempt at work energy equation DM1 3 terms; dimensionally correct; allow sin/cos mix in PE term and work done against Friction term; allow sign errors; allow their non-zero F from part (a). 7(b) 1 A1 For correct equation in d only; must be using 0.2 for the  0.2  2.4 + 0.2 g  0.6  d − 1.125  0.2 g  0.8  d = 0 mass. 2   d = 0.4m  Allow = 37 or better. Total distance = 0.25 + 0.4 = 0.65 m A1 AWRT 0.650. Allow 0.651 from use of = 36.9 . 4

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