Cambridge A Level Mathematics 9709 — 2021 May/June Paper 4 · Variant 2

1 A particle of mass 0.6 kg is projected with a speed of 4 m s−1 down a line of greatest slope of a smooth plane inclined at 10Å to the horizontal. Use an energy method to find the speed of the particle after it has moved 15 m down the plane. [3] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 2 1 Initial KE 0.6 4 2 = × × [ ] 4.8 = 2 1 Final KE 0.6 2 v = × × PE loss 0.6 15sin10 g = × × [ ] 15.628 = 2 1 0.6 15sin10 0.6 4 2 g × × + × × = 2 1 0.6 2 v × × M1 Apply energy equation, 3 terms, dimensions correct 8.25 v = ms-1 A1 3

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2 34 N 1 30 N ! 26 N Coplanar forces of magnitudes 34 N, 30 N and 26 N act at a point in the directions shown in the diagram. Given that sin ! = 5 and sin 1 = 17,8 find the magnitude and direction of the resultant of the three 13 forces. [6] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 Resolve either horizontally or vertically with correct number of terms. M1 Allow θ and α as in the question for this mark [ ] [ ] 8 5 30 34 26 4 17 13 X = − × − × = A1 Allow X ± as they may resolve forces left or right Allow [ ] 30 34sin 28 26sin 23 X = − − angle 2s.f. or better [ ] [ ] 15 12 34 26 6 17 13 Y = × − × = A1 Allow Y ± as they may resolve forces up or down Allow [ ] 34cos28 26cos23 Y = − angle 2s.f. or better [ ] 2 2 R X Y = + M1 Attempt to solve for the magnitude of the force [ ] 1 tan Y X β −  =     or [ ] 1 tan X Y β −  =     M1 Attempt to solve for the direction of the resultant force Question Answer Marks Guidance 2 cont’d 52 2 13 7.21 R = = = N and 56.3 β = above 30N force or anticlockwise from 30N force A1 Both correct with correct explanation of the direction. Must be a correct and clear explanation. 6

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3 A ring of mass 0.3 kg is threaded on a horizontal rough rod. The coefficient of friction between the ring and the rod is 0.8. A force of magnitude 8 N acts on the ring. This force acts at an angle of 10Å above the horizontal in the vertical plane containing the rod. Find the time taken for the ring to move, from rest, 0.6 m along the rod. [6] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 3 Resolving along or perpendicular to the rod M1 3 terms in either direction 8sin10 0.3 R g + = A1 8cos10 0.3 F a − = A1 0.8 F R = [ ] 1.61081 , 1.28865 R F = … = … M1 Using F = µR, where R is 2 terms involving weight and a component of 8 N. [ 21.966 a = …] 2 1 0.6 21.966 2 t = × × M1 Complete method leading to an equation in t such as s = ut + 1 2 at2 with s = 0.6, u = 0 and using their value of a found from a Newton’s second law with 3 terms, namely, component of 8 N, any friction and 0.3a. 0.234 t = seconds A1 Allow use of 22 a = for M1 and A1 Alternative method for Question 3 Resolving perpendicular to the rod M1 8sin10 0.3 R g + = A1 0.8 F R = [ ] 1.61081 , 1.28865 R F = … = … M1 Using F = µR, where R must involve 0.3g and a component of 8 N. Question Answer Marks Guidance 3 [ ] 2 1 8cos10 0.6 0.6 0.3 5.134 2 F v v × = × + × = B1 Work energy equation to find v after 0.6 metres. ( ) 1 0.6 0 5.134 2 t = + × M1 Using ( ) 1 2 s u v t = + to find t . 0.234 t = seconds A1 6

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4 A particle of mass 12 kg is stationary on a rough plane inclined at an angle of 25Å to the horizontal. A pulling force of magnitude P N acts at an angle of 8Å above a line of greatest slope of the plane. This force is used to keep the particle in equilibrium. The coefficient of friction between the particle and the plane is 0.3. Find the greatest possible value of P. [6] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 4 For resolving either parallel to or perpendicular to the plane M1 Three relevant terms in either equation. cos8 12 sin 25 P F g = + A1 12 cos25 sin8 g R P = + A1 0.3 F R = M1 Use F = 0.3R, where R must involve components of both 12g and P. ( ) cos8 0.3 12 cos25 sin8 12 sin 25 P g P g = − + M1 For attempting to solve for P, using equations with the correct number of relevant terms in both. 80.8 P = A1 From P = 80.755… Allow P⩽ 80.8 If more than one case is considered for direction of friction then a choice must be made for final answer. Alternative mark scheme for Question 4 For resolving forces either vertically or horizontally M1 Correct number of terms in either equation. cos25 sin33 12 sin 25 R P g F + = + A1 Question Answer Marks Guidance 4 cos33 cos25 sin25 P F R = + A1 0.3 F R = M1 Use F = 0.3R Solve a pair of simultaneous equations in P and R May see 97.5 R = M1 For attempting to solve for P, using equations with the correct number of relevant terms. 80.8 P = A1 From P = 80.755… Allow P⩽ 80.8 If more than one case is considered for direction of friction then a choice must be made for final answer. 6

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5 A car of mass 1250 kg is pulling a caravan of mass 800 kg along a straight road. The resistances to the motion of the car and caravan are 440 N and 280 N respectively. The car and caravan are connected by a light rigid tow-bar. (a) The car and caravan move along a horizontal part of the road at a constant speed of 30 m s−1. (i) Calculate, in kW, the power developed by the engine of the car. 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(ii) Given that this power is suddenly decreased by 8 kW, find the instantaneous deceleration of the car and caravan and the tension in the tow-bar. 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(b) The car and caravan now travel along a part of the road inclined at sin−1 0.06 to the horizontal. The car and caravan travel up the incline at constant speed with the engine of the car working at 28 kW. (i) Find this constant speed. 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(ii) Find the increase in the potential energy of the caravan in one minute. 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Mark scheme: 5(a)(i) ( ) 440 280 30 P = + × M1 Using P = Fv with F as total resistance 720 30 21.6 P = × = kW A1 Answer must be in kW 2 Question Answer Marks Guidance 5(a)(ii) 21600 8000 P = − W 21600 8000 13600 DF 453.333.. 30 30 −   = = =     B1 FT Follow through on their power from 5(a)(i) Allow 8000 Driving Force (DF) 266.7 30 = = as the force due to solely to the change in power provided correct equation(s) used. Car: DF 440 1250 T a − − = Caravan: 280 800 T a − = System: ( ) DF 440 280 2050a − + = M1 Apply Newton’s 2nd law to either the car or to the caravan or to the system. Must be correct number of relevant terms. If 8000 DF 30 = is used then the equations must be either DF 2050a − = or 280 800 T a − = Solve for either a or T M1 Using equation(s) with no missing/extra terms, DF 720 ≠ . Solving for a either from the system equation or from the car AND caravan equation. OR solving for T from the car AND caravan equation. 0.13 a = − ms-2 and 176 T = N A1 4 Question Answer Marks Guidance 5(b)(i) System: [ ] DF 720 2050 0.06 1950 g = + × = Car: DF 440 1250 0.06 0 T g − − − × = Caravan: 280 800 0.06 0 T g − − × = M1 Apply Newton’s 2nd law with a = 0, either to the system OR by eliminating T between the equations for the car and the caravan, no extra or missing relevant terms, dimensionally correct, to find DF 1950 28000 v = B1 DF P v = × . 28000 v SOI. 14.4 v = ms–-1 A1 3 Question Answer Marks Guidance 5(b)(ii) PE 800 0.06 800 14.4 60 0.06 g d g = × × = × × × M1 Using PE = mgh with h being height gained in 60 s, using their v PE 414 000 = (J) or PE 414 = kJ A1 Using v = 560/39 = 14.359 Alternative method for Question 5(b)(ii) 28 000 60 PE of Caravan 1250 0.06 720 g d d × = + × × + × and 60 14.359 861.54 d = × = M1 For use of WD P t = × to find an expression for PE of caravan and the distance travelled up the incline in 1 minute. [ ] PE 28 000 60 1250 861.54 0.06 720 861.54 g = × − × × − × PE 414 000 = (J) or PE 414 = kJ A1 2

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6 A particle A is projected vertically upwards from level ground with an initial speed of 30 m s−1. At the same instant a particle B is released from rest 15 m vertically above A. The mass of one of the particles is twice the mass of the other particle. During the subsequent motion A and B collide and coalesce to form particle C. Find the difference between the two possible times at which C hits the ground. 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Mark scheme: 6 ( ) 2 30 5 A s t t = ± − or 2 5 Bs t = ± B1 Use of constant acceleration equations to find expressions for displacements of A or B. 15 A B s s + = leading to 15 30t = leading to 0.5 t = B1 Use 15 A B s s + = to find time at which particles collide. 0.5 t = leading to 25 A v = ± and 5 B v = ± B1 Find speed of particles at t = 0.5 before collision. 0.5 t = leading to 2 1 30 0.5 0.5 13.75 2 A h g   = ± × − × = ±     B1 Find position of A or B at which collision occurs at t = 0.5 Alternatively allow 1.25 B h = ± as displacement of B ( ) ( ) ( ) 25 2 5 3 m m m v × − = → 1 15 v = ( ) ( ) ( ) 25 5 2 3 m m m v −× = → 2 5 v = M1 Use of conservation of momentum, either case, using their and 0 or 30 A B v v ≠ , with 3 terms. A1 Both values of v correct Question Answer Marks Guidance 6 Particle C1 2 13.75 15 5 t t − = − Particle C2 2 13.75 5 5 t t − = − M1 Use of s = ut + 1 2 at2 OE to find t, using either their numerical 1v or numerical 2 v from a relevant conservation of momentum equation. 1 2 , 3.74, 2.23 C C t t = leading to 1 5 3 1.50 T = + − = A1 Find 1 2 C C T t t = − from 1 3.736 Ct = and 2 2.232 Ct = 8 Subscripts 1 and 2 refer to the two cases. Alternative method for the final two marks 1 0 15 gt = − , 2 0 5 gt = − → 1 1.5 t = , 2 0.5 t = Total heights 1 13.75 11.25 25 h = + = Or 2 13.75 1.25 15 h = + = 2 1 25 5T = and 2 2 15 5T = → 1 5 T = , 2 3 T = M1 Use of v u gt = − to find time to highest point for either case and use of 2 2 2 v u gs = − to find total height reached for either case, using either their numerical 1v or numerical 2 v from a relevant conservation of momentum equation. Use 2 1 0 2 s gT = + to find time to reach ground (either case). ( ) 1.5 5 0.5 3 1 5 3 1.50 T = + − + = + − = A1 Find difference in total times ( ) ( ) 1 1 2 2 T t T t T = + − +

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7 A particle P moving in a straight line starts from rest at a point O and comes to rest 16 s later. At time t s after leaving O, the acceleration a m s−2 of P is given by a = 6 + 4t 0 ≤t < 2, a = 14 2 ≤t < 4, a = 16 −2t 4 ≤t ≤16. There is no sudden change in velocity at any instant. (a) Find the values of t when the velocity of P is 55 m s−1. 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(b) Complete the sketch of the velocity-time diagram. [2] v (m s−1) t (s) 0 2 4 16 (c) Find the distance travelled by P when it is decelerating. 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Mark scheme: 7(a) [ ] 2 6 2 v t t c = + + or [ ] 14 v t c = + M1 Attempt to integrate a in Stage 1 or Stage 2 or in Stage 2 for use of v u at = + 2 6 2 v t t = + and 14 8 v t = − or ( ) 2 20 v t = = ( ) 4 20 14 2 48 v t = = + × = A1 Velocity in Stage 1 and Stage 2 correct including correct constant Find v at 2 t = and use 14 v u t = + to find v at 4 t = [ ] 2 16 v t t c = − + *M1 Attempt to integrate a in Stage 3. 2 55 16t t = − DM1 Attempt to solve a relevant 3-term quadratic equation which comes from their 2 term v from Stage 3 equated to 55 and finding two values of t 5 and 11 t t = = only A1 Allow only if c = 0 has been shown correctly. Alternative method for Question 7(a) State or imply that only possible range is 4 ⩽ t ⩽ 16 B1 Allow this method if candidates only consider Stage 3 2 16 v t t c = − + M1 For attempt at integration. c = 0 shown A1 Using 𝑣= 0 at 𝑡= 16 Solve 2 55 16t t = − M1 Must find 2 values of 𝑡 and must be from equating their 2 term v to 55 5 and 11 t t = = only A1 Allow only if c = 0 has been shown correctly. 5 Question Answer Marks Guidance 7(b) Positive quadratic for 0 ⩽ t < 2 through (0,0) joining to the bottom of the given line or Negative quadratic for 4 ⩽ t ⩽ 16 going through the point (16,0) and joining the top of the given line B1 All correct with correct gradients (approx) B1 Negative quadratic must have a maximum. There must be no point of inflexion particularly near 16 t = . Ignore any curve drawn outside 0 ⩽ t ⩽ 16. 2 7(c) ( ) 2 16 d s t t t =  − ( ) 2 3 1 8 3 t t c   = − +     M1 Attempt to integrate their v. 16 2 3 8 1 8 3 s t t   = −     1 2 2048 1365 512 170 3 3 s     = − − −         A1 Correct integral and the correct limits used correctly to find an unsimplified expression for the distance from 8 t = to 16 t = only. 1 3 341 s = B1 Allow s = 341 to 3s.f. If no integration seen (calculator used) allow B1 (max 1 out of 3 marks) 3

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