Mathematics 9709/42 May/June 2021: question paper and mark scheme

Mark scheme

1 1 B1 Any one of the three expressions correct Initial KE = × 0.6 × 4 2 [ = 4.8] 2 1 Final KE = × 0.6 × v 2 2 PE loss = 0.6 × g × 15sin10 [ = 15.628] 1 1 M1 Apply energy equation, 3 terms, dimensions correct 0.6 × g × 15sin10 + × 0.6 × 4 2 = × 0.6 × v 2 2 2 v = 8.25 ms-1 A1 3

Mark scheme

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

Mark scheme

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

Mark scheme

4 For resolving either parallel to or perpendicular to the plane M1 Three relevant terms in either equation. P cos8 = F + 12 g sin 25 A1 12 g cos 25 = R + P sin 8 A1 F = 0.3R M1 Use F = 0.3R, where R must involve components of both 12 g and P. P cos8 = 0.3 (12 g cos 25 − P sin 8 ) + 12 g sin 25 M1 For attempting to solve for P, using equations with the correct number of relevant terms in both. P = 80.8 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. R cos 25 + P sin 33 = 12 g + F sin 25 A1 4 P cos33 = F cos 25 + R sin 25 A1 F = 0.3R M1 Use F = 0.3R Solve a pair of simultaneous equations in P and R M1 For attempting to solve for P, using equations with the correct May see R = 97.5 number of relevant terms. P = 80.8 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

Mark scheme

5(a)(i) P = ( 440 + 280 ) × 30 M1 Using P = Fv with F as total resistance P = 720 × 30 = 21.6 kW A1 Answer must be in kW 2 5(a)(ii) P = 21600 − 8000 W B1 FT Follow through on their power from 5(a)(i) 21600 − 8000 13600 8000 DF = = = 453.333.. Allow Driving Force (DF) = = 266.7 as the force due to 30 30 30 solely to the change in power provided correct equation(s) used. Car: DF − 440 − T = 1250a M1 Apply Newton’s 2nd law to either the car or to the caravan or to Caravan: T − 280 = 800a the system. Must be correct number of relevant terms. System: DF − ( 440 + 280 ) = 2050a If DF = 8000 is used then the equations must be either 30 −DF = 2050a or T − 280 = 800a 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. a = −0.13 ms-2 and T = 176 N A1 4 5(b)(i) System: DF = 720 + 2050 g × 0.06 [ = 1950] M1 Apply Newton’s 2nd law with a = 0, either to the system OR by Car: DF − 440 − T − 1250 g × 0.06 = 0 eliminating T between the equations for the car and the caravan, no extra or missing relevant terms, dimensionally Caravan: T − 280 − 800 g × 0.06 = 0 correct, to find DF 1950v = 28000 B1 28000 P = DF × v . SOI. v v = 14.4 ms–-1 A1 3 5(b)(ii) PE = 800 g × d × 0.06 = 800 g × 14.4 × 60 × 0.06 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 g × d × 0.06 + 720 × d M1 For use of WD = P × t to find an expression for PE of caravan and d = 60 ×14.359 = 861.54 and the distance travelled up the incline in 1 minute. [ PE = 28 000 × 60 − 1250 g × 861.54 × 0.06 − 720 × 861.54] A1 PE = 414 000 (J) or PE = 414 kJ 2

Mark scheme

6 ( s A = ± 30t − 5t 2 ) or s = ±5t B 2 B1 Use of constant acceleration equations to find expressions for displacements of A or B. s A + sB = 15 leading to 15 = 30t leading to t = 0.5 B1 Use s A + sB = 15 to find time at which particles collide. t = 0.5 leading to vA = ±25 and vB = ±5 B1 Find speed of particles at t = 0.5 before collision. 1 B1 Find position of A or B at which collision occurs at t = 0.5 t = 0.5 leading to hA = ± 30 × 0.5 − g × 0.52 = ±13.75 Alternatively allow hB = ±1.25 as displacement of B 2 25 × ( 2m ) − 5 ( m ) = ( 3m ) v → v1 = 15 M1 Use of conservation of momentum, either case, using their vA and vB ≠ 0 or 30 , with 3 terms. 25 ( m ) − 5 × ( 2m ) = ( 3m ) v → v2 = 5 A1 Both values of v correct 6 Particle C1 −13.75 = 15t − 5t 2 M1 1 2 Use of s = ut + at OE to find t, using either their numerical Particle C2 −13.75 = 5t − 5t 2 2 v1 or numerical v2 from a relevant conservation of momentum equation. tC1 , tC2 = 3.74, 2.23 leading to T = 1 + 5 − 3 = 1.50 A1 Find T = tC1 − tC2 from tC1 = 3.736 and tC2 = 2.232 8 Subscripts 1 and 2 refer to the two cases. Alternative method for the final two marks 0 = 15 − gt1 , 0 = 5 − gt2 → t1 = 1.5 , t2 = 0.5 M1 Use of v = u − gt to find time to highest point for either case and Total heights h1 = 13.75 + 11.25 = 25 use of v 2 = u 2 − 2 gs to find total height reached for either case, Or h2 = 13.75 + 1.25 = 15 using either their numerical v1 or numerical v2 from a relevant 25 = 5T12 and 15 = 5T2 2 → T1 = 5 , T2 = 3 conservation of momentum equation. 1 Use s = 0 + gT 2 to find time to reach ground (either case). 2 ( ) T = 1.5 + 5 − 0.5 + 3 = 1 + 5 − 3 = 1.50 A1 Find difference in total times T = ( t1 + T1 ) − ( t2 + T2 )

Mark scheme

7(a) v = 6t + 2t 2 [ + c ] M1 Attempt to integrate a in Stage 1 or Stage 2 or in Stage 2 for use of v = u + at or v = 14t [ + c ] v = 6t + 2t 2 and v = 14t − 8 A1 Velocity in Stage 1 and Stage 2 correct including correct or constant v ( t = 2 ) = 20 Find v at t = 2 and use v = u + 14t to find v at t = 4 v ( t = 4 ) = 20 + 14 × 2 = 48 v = 16t − t 2 [ + c ] *M1 Attempt to integrate a in Stage 3. 55 = 16t − t 2 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 t = 5 and t = 11 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 v = 16t − t 2 + c M1 For attempt at integration. c = 0 shown A1 Using 𝑣 = 0 at 𝑡 = 16 Solve 55 = 16t − t 2 M1 Must find 2 values of 𝑡 and must be from equating their 2 term v to 55 t = 5 and t = 11 only A1 Allow only if c = 0 has been shown correctly. 5 7(b) Positive quadratic for 0 ⩽ t < 2 through (0,0) joining to the bottom B1 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 All correct with correct gradients (approx) B1 Negative quadratic must have a maximum. There must be no point of inflexion particularly near t = 16 . Ignore any curve drawn outside 0 ⩽ t ⩽ 16. 2 7(c) 1 M1 Attempt to integrate their v. ( ) s = 16t − t 2 dt = 8t 2 − t 3 ( + c ) 3 1 16 A1 Correct integral and the correct limits used correctly to find an s = 8t 2 − t 3 unsimplified expression for the distance from t = 8 to t = 16 3 8 only. 1 2 s = 2048 − 1365 − 512 − 170 3 3 s = 341 13 B1 Allow s = 341 to 3s.f. If no integration seen (calculator used) allow B1 (max 1 out of 3 marks) 3