Cambridge A Level Mathematics 9709 — 2025 Oct/Nov Paper 4 · Variant 5

1 32 N 21 N 35° 47° P N 65° Q N Coplanar forces of magnitudes P N, Q N, 32 N and 21 N act at a point in the directions shown in the diagram. The forces are in equilibrium. Find the value of P and the value of Q. [5] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1 Resolving in any direction to get an equation *M1 Correct number of relevant terms; allow sign errors; allow consistent or inconsistent sin/cos mix. P + 21cos47 − Qcos65 − 32cos35 = 0 A1 This could be with their Q. P + 14.321− Qcos65 − 26.212= 0 21sin47 + 32sin35 − Qsin65 = 0 A1 15.358+ 18.354− Qsin65 = 0 33.712− Qsin65 = 0 Attempt to solve for P = −21cos47 + ( 37.198... ) cos65 + 32cos35 DM1 From equation(s) with correct number of relevant terms. Using their Q. Must be a numerical expression for P . All forces resolved as appropriate but allow consistent or inconsistent sin/cos mix. If no intermediate working to solve their two original equations, can still earn DM1 if answers correct for their equations but DM0 if answers wrong for their equations. P = 27.6 AND Q = 37.2 A1 27.61147… and 37.19804… DM1 may be earned here. 5

More questions on Forces and equilibrium

2 An athlete runs along a straight horizontal road. The athlete starts from rest and accelerates at 0 .5 ms -2 reaching a speed of V ms -1 . The athlete maintains this speed of V ms -1 for T s before decelerating at 0. 2 ms -2 back to rest. The athlete covers a total distance of 1350 m in 246 seconds. (a) Sketch the velocity-time graph for the motion of the athlete. [1] v (m s–1) t (s) O (b) Find T in terms of V, and hence show that 7V 2 - 492V + 2700 = 0 . [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (c) Find the value of V, giving a justification for your answer. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ .......................... ............................................................................................................................

Mark scheme: 2(a) Correct velocity-time graph B1 Trapezium starting and finishing on the horizontal axis. V and 246 labelled but ignore gradients. Ignore any other labelling. Condone label o6 rather than V. 1 2(b) Method using total time first V V *B1 For one correct time in terms of V. Time accelerating is [= 2V] or time decelerating is  = 5V  0.5 0.2 V V B1 OE Allow correct un-simplified expression. T =  246 − −  = 246 − 7V  0.5 0.2 DM1 Set up an equation in V only with the correct number of relevant terms to find total distance. V V Dependent on and seen, possibly embedded. 0.5 0.2 1  V  1  V  e.g. triangle/rectangle/triangle to get first equation   V  + V ( 246 − 7V ) +   V  = 1350 2  0.5  2  0.2  or trapezium to get second equation 1 or ( 246 + 246 − 7V )(V ) = 1350 1 2 2 or use s = ut + at to get the third equation 2 1 2 1 2 or 0 +  0.5  ( 2V ) + V ( 246 − 7V ) + V  5V −  0.2  ( 5V ) = 1350 2 2 or use v = u + 2 as to get the fourth equation 2 2 2 2 or use v = u + at to get the fifth equation V V or + V ( 246 − 7V ) + = 1350 2 ( 0.5 ) 2 ( 0.2 ) or two expressions for T can be equated to form an equation for V to get the sixth equation OE.  V  or 0 = V − 0.2  246 − 246 + 7V −   0.5  2700 − 246V or e.g. 246 − 7V = V 2(b) 7V 2 − 492V + 2700 = 0 A1 AG. No need to see intermediate steps after a correct equation seen but must check that any intermediate steps seen are correct. If any errors seen anywhere then A0. Alternative using total displacement first (trapezium OE) V 2 2 *B1 For distance accelerating/decelerating. Distance accelerating is  = V    2  0.5 V 2 2 or distance decelerating is  = 2.5V  or for trapezium area or could use 2 triangles and   2  0.2 rectangle. 1 ( 246 + T ) V = 1350 2 246 − T or V + VT = 1350 2 1  V  1  V  or alternatively   V  + VT +   V  = 1350 2  0.5  2  0.2   2700 − 246V  B1 OE. Allow correct un-simplified expression. = T = 2700 − 246   V  V  7 2 1350 − V 2 or alternatively (from above alternative) T =  V V 2700 V DM1 Set up an equation in V only with the correct number t1 + T + t 2 = 246  + − 246 + = 246 of relevant terms but all terms present. 0.5 V 0.2 Some candidates are getting to the correct equation for 2700 or 2V + − 246 + 5V = 246 oe T and then substituting this back into the trapezium V equation which simply gets 1350 = 1350, so does not 7 2 1350 − V show the required equation and gets B1B1M0A0. 2 or alternatively (from above alternative) 7 v + = 246 V 2(b) 7V 2 − 492V + 2700 = 0 A1 AG. No need to see intermediate steps after a correct equation seen but must check that any intermediate steps seen are correct. If any errors seen anywhere then A0. Alternative where T is not found V V *B1 For one correct time or distance in terms of V. Time accelerating is [= 2V] or time decelerating is  = 5V  0.5 0.2 V 2 2 V 2 or Distance accelerating is  = V  or distance decelerating is   2  0.5 2  0.2  = 2.5V 2    V 2 V 2 B1 OE. Total distance = 246V − − 2  0.5 2  0.2 V 2 V 2 DM1 Set up an equation in V only with the correct number 246V − − = 1350 of relevant terms but all terms present. 2  0.5 2  0.2 7V 2 − 492V + 2700 = 0 A1 AG. No need to see intermediate steps after a correct equation seen, but must check that any intermediate steps seen are correct. If any errors seen anywhere then A0. 4 2(c) V = 6 ONLY *B1 450 B0B0 if both solutions stated and do not discard 7 450 DB1 Any acceptable reason for rejecting greater root e.g. V  as e.g. this would lead to the athlete running at a constant speed for a not feasible for an athlete to be running at 64.3 metres 7 per second.   450    2700 − 246    Other root must be seen, not e.g. just state T > 0  7   450  negative time  = or 246 − 7   = −204  without mention of other root but if they state 450 ,   450   7       7   7   then say T  0 , so not suitable then award B1. OR the athlete would take longer than 246 seconds to decelerate Calculations not necessary but if calculations are  450 2250  shown they must be correct.  t decelerate =  0.2 = = 321.4  246   7 7   450  OR  t acc + dec =  7 = 450  246   7  2

More questions on Kinematics of motion in a straight line

3 A car of mass 800 kg is moving on a straight road. When the car is moving at a constant speed of 20 ms -1 on a horizontal section of the road, the engine of the car is working at P W. When the car is moving at a constant speed of 12 ms -1 up a section of the road inclined at sin -1 0.15 to the horizontal, the engine of the car is also working at P W. On both sections of the road there is a constant force of magnitude R N resisting the motion of the car. (a) Find the value of R and the value of P. 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(b) Find the acceleration of the car when it is moving at 10 ms -1 up the inclined section of the road with the engine working at 32 kW. 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Mark scheme: 3(a) P P *B1 OE. P = 12F or P = 20F but B0 if erroneous value Either driving force: or 12 20 of F. P *B1 3 − R = 0 or 20 F1 = 12 F2 OE. P = 20R or F1 = F2 . 20 5 P *M1 Attempt at N2L on inclined section of the road; − R − 800 g  0.15 = 0 correct number of relevant terms, allow sign errors 12 and sin/cos mix with components only where needed. or F2 − R − 800 g  0.15 = 0 where F2 is the larger of the two forces Must have more than simply sinor cos. g must P or − R − 800 g  sin8.6 = 0 be present. Resistance term here must be same as that 12 in second B1. or P = 12 ( R + 800 g  0.15 ) or P = 12 ( R + 800 g  sin8.6 ) P P DM1 Eliminate and solve for either P or R. − − 800 g  0.15 = 0  P = Correct number of relevant terms, allow sign errors 12 20 and sin/cos mix. Must have more than simply or F2 − F1 − 800 g  0.15 = 0  P = sinor cos. P P or − − 800 g  sin8.6 = 0  P = If no solving seen, answer must be correct for their 12 20 equations to be awarded DM1. or 20 R − 12 ( R + 800 g  0.15 ) = 0  R = Dependent on B1B1M1 or 20 R − 12 ( R + 800 g  sin8.6 ) = 0  R = P = 36000 and R = 1800 A1 Using angle of 8.63 gives P = 36012 ( 36000 ) and R = 1800.6 (1800) which scores A1. Using an angle of 8.6 does not give answers correct to 3sf so gets A0. 5 3(b) Attempt at N2L to form an equation on inclined section of the road M1 Correct number of relevant terms, allow sign errors and sin/cos mix with components only where needed. Must have more than simply sinor cos. Allow D for driving force. g must be present. 32000 A1FT Correct equation following through their R. − 800 g  0.15 − 1800 = 800 a 10 32000 or − 800 g  sin8.6 − 1800 = 800 a 10 0.25 [m s-2] A1 Condone answer of 0.249 from using an angle of 8.63. Using an angle of 8.6 does not give answer correct to 3sf so gets A0. 3

More questions on Energy, work and power

5 P 6 kg R 2 kg 30° Q 5 kg The diagram shows a particle P of mass 6 kg on a rough plane inclined at an angle of 30° to the horizontal. Two light inextensible strings are attached to P. The strings pass over small smooth pulleys, which are fixed at the ends of the plane. The non-vertical parts of the string are parallel to a line of greatest slope of the plane. Particles Q and R, of masses 5 kg and 2 kg respectively, hang vertically at the ends of the strings. Both strings are taut, and the system is released from rest. It is given that the tension in the string attached to Q is twice the tension in the string attached to R. (a) Find, in terms of g, the tension in each of the strings and the magnitude of the acceleration of the particles. 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(b) Find the coefficient of friction between P and the plane. 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(c) It is given that when the system is released from rest, P is at the midpoint of the plane. In the subsequent motion, R does not reach the pulley at the top of the plane, and P takes 1.5 s to reach the pulley at the bottom of the plane. Find the total length of the plane. 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Mark scheme: 5(a) Attempt N2L for Q and R to form two equations *M1 Correct number of relevant terms in each, allow sign errors and T for both tensions. Masses must be correct. Condone different accelerations. Ignore anything relating to P. For first 4 marks condone substituted value of g. 5 g − TPQ = 5a or 5 g − 2T PR  = 5a A1 Must be using different T’s – possibly with the result that TPQ = 2TPR . TPR − 2 g = 2a or T PR  − 2 g = 2a Condone use of 2TPQ = TPR for this A1. 5 g − TPQ TPR − 2 g or =  = a  Accelerations must be the same. 5 2 Allow TPQ − 5 g = 5a and 2 g − TPR = 2a . 5 g − 4 g = 5a + 4a or 5 g − 2 ( 2 a + 2 g ) = 5 a oe DM1 Use TPQ = 2TPR to obtain an equation in a only or in T 5 g − 2TPR TPR − 2 g only. Or = oe 5 2 Can be implied by correct value of a or TPQ or TPR if no solving seen. 1 10 9.8 9.81 20 40 A1 A1 for any one of the three answers. Second A1 for all a = g Condone or or or 1.11. TPR = g , TPQ = g three. 9 9 9 9 9 9 A1 1 Allow TPR = 2.22 g and TPQ = 4.44 g You may see a = − g if equations ‘backwards’ but 9 1 must get to a = g OE for A1. 9 Tensions must be in terms of g but ISW. 5 Do not allow e.g. TPQ = 5 g − g as final answer . 9 Allow A1 if tensions not labelled. 5 5(b)  RP = 6 g cos30 B1 60 3 Seen or implied by or 30 3 or 3 g 3 but 2 must be the normal reaction. Attempt N2L for P or whole system to form an equation *M1 Correct number of relevant terms (5) (or 4 terms if tensions combined), allow sign errors, allow F for friction but must be using two different expressions for tensions. Must be component of weight (and weights, not masses, if using whole system). Must substitute their tension and acceleration. Must use correct mass(es). You may see an equation such as 5 g − F = 4a , which comes from combining equations for P and N2L for Q or R, and this may be shown in 5(a) rather than 5(b), so in this case link the two parts. 40 20 1 A1FT Correct equation following through their two different For P: 6 gsin30 + their g − their g − F = 6  their g values of T and their a. Do not need a value for F and 9 9 9 allow with wrong F. 1 For system: 5 g + 6 g sin30 − 2 g − F = 13  their g 40 20 9 Note: their g − their g may be combined into 9 9 20 their g 9 410 Note: F = = 45.5555 9 Check very carefully for FT of their a , TPR , TPQ . 40 20 1 DM1 Use of F = R to obtain an equation in  only 6 gsin30 + their g − their g −  6 gcos30 = 6  their g 9 9 9 where R is a component of 6g . 5(b) = 0.877 A1 41 3 0.876717…; . 81 5 5(c) 1 1 2 M1 OE. d =  g  1.5   d = 1.25  2 9 1 2 Use of s = ut + at with u = 0, t = 1.5 and their a 2 from part (a) or other complete method to get an equation in d only. 1 May use x rather than d. 2 Must not use a = 10 unless this is their a from 5(a). 2.5 [m] A1FT FT 2.25a. Alternative using work-energy For 6kg particle M1 Use of work-energy with all dimensionally correct 2 terms present. 1  g  40 g 20 g 6  their  1.5  − 6 g sin30  d = their  d − their  d − their  Terms should have components as required and allow 2  9  9 9 consistent sin/cos mix 6 gcos30  d 1 May use x rather than d.   d = 1.25  2 for whole system 1  g  2  13  their  1.5 − 6 g sin30  d = 5 gd − 2 gd − their  6 gcos30  d   2  9    d = 1.25  2.5 [m] A1FT 2

More questions on Kinematics of motion in a straight line

6 Two particles A and B of masses km and m respectively, where k and m are constants, are free to move in a straight line on a smooth horizontal plane. Particle A is projected towards B with speed 2u and at the same instant B is projected towards A with speed u. The particles collide. After the collision the speed of A is u and both particles move in the same direction as A’s original motion. It is given that 35% of the total kinetic energy is lost in the collision. Find, in terms of u, the speed of B after the collision. 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Mark scheme: 6 km  2u − m  u = km  u + m  vB *M1 Attempt at CLM; correct number of terms and correct masses with relevant velocities but allow sign errors. v B is the speed of B after impact. If g included with the masses then M1A0. vB + u vB A1 Any correct equation linking v B , k and u but with m v B = ( k − )1 u or ku = vB + u or k = or k = + 1 u u eliminated. 1 2 1 2  2 1 2  *B1 KE before impact. 2 kmu + mu KEbefore =  km  ( 2u ) +  m ( u )  =  2 2  2  1 2 1 2 *B1FT KE after impact. KEafter =  km  u +  mvB Allow if their expression for v B is wrong. 2 2 1 2 1 2 2 Or KEafter =  km  u +  mu ( k − 1) Allow B1B1 for the change in KE = 2 2  3 2 1 2 1 2    kmu + mu − mv B   2 2 2   2 1 2  1 2 2 *DM1 Setting up the equation 0.65 KEbefore = KE after OE k + ( k − 1) 0.65   2 kmu + mu  = mu ( )  2  2 with their expression for v B in terms of k substituted.  2 1 2  1 2 2  2 1 2  Must be correct way around and must use 0.65 OE. 2 kmu + − mu 2 kmu + k + ( k − 1) Or  mu  = 0.35   mu  ( ) 2 2 2     Dependent on M1B1B1. 2 2 1 2 2 Or 0.8ku + 0.325u = ( k − 1) u May see 1.6 ku 2 + 0.65u 2 = v B2 leading to 2 1.6ku 2 + 0.65u 2 = ( k − 1) 2 u 2 . 2.6k + 0.65 = k + k 2 − 2k + 1  k 2 − 3.6k + 0.35 = 0 DM1 Re-arrange to a 3-term quadratic in k only – 2 Dependent on M1B1B1M1. or 20k − 72k + 7 = 0 6 k = 3.5 only A1 If k = 0.1 seen must be rejected. Reason not required but if a reason given it must be valid e.g. as k – 1 > 0 or reject later e.g. if k = 0.1  vB = −0.9u but not possible as vB  0 . vB = 2.5u A1 Not dependent on previous A mark. If g included in momentum equation then max M1A0B1B1M1M1A1A0. Ignore the other solution if shown ( vB = −0.9u ) ALTERNATIVE (eliminating k) km  2u − m  u = km  u + m  vB *M1 Attempt at CLM; correct number of terms and correct masses with relevant velocities but allow sign errors - v B is the speed of B after impact. If g included with the masses then M1A0. vB + u vB A1 Any correct equation linking v B , k and u but with m v B = ( k − )1 u or ku = vB + u or k = or k = + 1 u u eliminated. 1 2 1 2  2 1 2  *B1FT KE before impact. 2 kmu + mu KEbefore =  km  ( 2u ) +  m (  u )  =  Allow if their expression for k is wrong. 2 2  2  1  v B + u  2 1 2 2u ) +  mu Or KEbefore =    m  ( 2  u  2 1 2 1 2 *B1FT KE after impact. KEafter =  km  u +  mvB Allow if their expression for k is wrong. 2 2 1  v B + u  2 1 2 or KEafter =    m  u +  mv B Allow B1B1 for the change in KE 2  u  2  3 2 1 2 1 2    kmu + mu −  mv B  .  2 2 2  6  1  vB + u  2 1 2  1  vB + u  2 1 2 *DM1 Setting up the equation 0.65 KEbefore = KE after . m  u +  mvB or 0.65      m  ( 2u ) +  mu  =     2  u  2  2  u  2 OE with expression for k in terms of v B substituted.  1  v B + u  2 1 2   1  v B + u  2 1 2  Must be correct way around and must use 0.65 OE.     m  ( 2u ) +  mu  −     m  u +  mv B  = Dependent on M1B1B1.  2  u  2   2  u  2    1 1  v B + u  May see 1.6 ku 2 + 0.65u 2 = v B 2 oe leading to 0.35   mu 2 2u ) 2 +     m  (  2  u  2   v B + u  2 2 2 u + 0.65u = v B . 1.6   1 1 1    u  or 0.65  mu 2 mu ( v B + u ) + mv B 2  2 mu ( v B + u ) +  =  2  2 2 1.3uv B + 1.625u 2 = 0.5u 2 + 0.5uv B + 0.5v B 2 DM1 Re-arrange to a 3-term quadratic in terms of v B and u 2 2 only.  v B − 1.6uv B − 2.25u = 0 Dependent on M1B1B1M1.  20v B 2 − 32uv B − 45u 2 = 0 v B −0.9u A1 Explicitly seen and rejected. vB = 2.5u A1 Not dependent on previous A mark. If g included in momentum equation, then max Ignore the other solution if shown ( vB = −0.9u ) M1A0B1B1M1M1A1A0. 8

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