Mathematics 9709/42 October/November 2022: question paper and mark scheme

Mark scheme

1 Work done by cyclist = 50 × 100 (= 5000 J) B1 1 M1 Work energy equation. Three terms. Allow sign errors. Their 5000 – 3560 = m × 62 Dimensionally correct. 2 mass = 80 kg A1 3 SC: Acceleration considered as a constant 62 = 0 + 2a × 50 [⇒ a = 0.36 ] M1 From use of v = 6 , u = 0 , s = 50. Must be using 3560 correct suvat formulae. 100 − = m × their 0.36 For equation involving mass using N2L with three 50 terms. Allow sign errors in N2L. mass = 80 kg A1

Mark scheme

2(a) R = 0.4g cos 30 = 2 3 or F or μ R = 0.4 g sin 30 [ = 2] B1 Use of m instead of 0.4 condoned. 0.4g sin 30 – µ 0.4g cos 30 = 0 M1 For using F = µR. Allow sin/cos mix. Both must be different components of their weight only, not a 2 term R. Allow sign errors. Allow g omitted. 4sin 30 1 3 A1 AG (exact answer only) μ = = 3 or . If zero scored then SC B1 for [Angle of friction = 30° 4cos30 3 3 1 so] µ = tan 30 = 3. 3 Allow full marks if using m in place of 0.4 or W in place of mg or 0.4g 1 A0 for μ = 0.577= 3 , but A1(ISW) for 3 1 μ = 3 = 0.577 3 3 2(b) 7.2 – 0.4 g sin 30 – F = 0.4a M1 Newton’s second law. Four terms. Second term must be a component of their weight. F ≠ 0 and F ≠ μ . Allow sin/cos mix. Allow sign errors. F must be a numerical expression May use their F from part (a). a=8 A1 1 M1 For use of constant acceleration formula(e) and solving 1 = 0 + × ( their positive 8) × t 2 for t. a ≠ ±10, a ≠ ± g . 2 Allow if a is negative in part (a) and use |a | here. Time = 0.5 s A1 4

Mark scheme

3 Attempt to resolve either direction M1 Correct number of terms. Allow sin/cos mix. Allow sign errors. Allow g missing. 0.3g + T cosα ° – 4 sin 60°= 0 (T cos α° = 0.464…) A1 OE T sinα ° –4 cos 60°= 0 (T sin α° = 2) A1 OE If the two Ts are different, award maximum A1A0 unless subsequently stated that the two Ts are the same. 4 cos 60° −1 2 M1 Attempt to solve for α. No missing/extra terms. Allow α = tan −1 = tan g missing. Must get to ‘α =’. 4 sin 60° − 0.3g 0.464 4cos 60 M1 OE Attempt to solve for T. No missing/extra terms. T= = ( 4 cos 60° )2 + ( 4 sin 60° − 0.3 g )2 = 22 + ( 0.464)2 sin ( theirα ) Allow g missing. Must get to ‘T =’. Tension = 2.05 N α = 76.9 A1 For both AWRT 2.05, 76.9 (Tension = 2.05314… N α = 76.9356…) Alternative method for Q3 using triangle of forces Attempt at cosine rule from triangle of forces M1 Must use lengths 4 and 0.3g with a suitable angle. Allow g missing. T 2 = 42 + ( 0.3g ) − 2 × 4 × ( 0.3g ) × cos30 A1 2 Tension = 2.05 A1 Tension = 2.05314… AWRT 2.05 Attempt at sin rule M1 Must have angle 30° and another angle in terms of α with correct numerators, but allow g missing. Their T 4 Their T 0.3g A1 Correct. Allow sin α instead of sin (180 − α ) . = or = sin 30 sin (180 − α ) sin 30 sin (α − 30 ) α = 76.9 A1 α = 76.9356… AWRT 76.9 3 Alternative method for Q3 using Lami’s theorem Attempt at Lami’s theorem M1 Must have numerators correct and at least one angle correct. Allow g missing. 4 0.3g T A1 A1 A1 for two parts second A1 for all three. = = sin α sin ( 210 − α ) sin (150 ) 4sin 210 M1 For solving for α using compound angle formula. Must α = tan −1 be correct for their angles. Allow g missing. 0.3g + 4cos 210 4sin (150 ) 0.3g sin (150 ) M1 For solving for T using their α. Allow g missing. T= or T = sin α sin ( 210 − α ) Tension = 2.05 N α = 76.9 A1 For both AWRT 2.05, 76.9 6 SC: Tension and the 4 N force considered in the wrong directions Attempt to resolve either direction M1 Correct number of terms. Allow sin/cos mix. Allow sign errors. Allow g missing. T cos 60° –4 sinα °= 0 A1 For both And: T sin 60° – 4 cos α ° − 0.3g = 0 OE If the two Ts are different, they get SC A0 unless they subsequently state that the two Ts are the same. 2 T cos 60° T sin 60°−0.3g 2 1 2 3 2 M1 OE Attempt to solve for T or α. No missing/extra + = 1 T + T − 3√ 3T + 9 = 16 terms. Allow g missing. Must get to ‘T =’ or ‘α=’. 4 4 4 4 T − 3√ 3T − 7 = 0 T = 6.31( or − 1.11) 2 3 OR: 4 3 sin α − 4cos α = 3 ⇒ 8sin (α − 30 ) = 3 ⇒ α = sin −1 + 30 8 3 T = 6.31N α = 52.0 A1 (T = 6.30617…, α=52.0243…) 6

Mark scheme

4(a) P = D × 15 B1 P For any D. OE including . 15 D – 500 = 1200 × 0.8 (⇒ D = 1460) M1 Attempt at Newton’s second law with three terms. Allow sign errors. Power = 21900 W A1 Allow 21900 without units or 21.9 kW, but not simply 21.9 without units or with wrong units. 3 4(b) 1 1 B1 Sight of both KEs. [Change in KE =] × 1200 × 322 − × 1200 × 152 2 2 [ = 614400 − 135000 = 479400] Work done by engine = 21900 × 53 ( = 1160700) B1ft WD OE e.g. 21900 = 53 FT their 21900. Distance AB = 1362.6 m B1 AG Must come from 1160700 – 500d = 479400 OE e.g. 500d = 681300 . 3

Mark scheme

5(a) T – 40g sin 20 –50 = 40a [T − 136.8 − 50 = 40a ] M1 Attempt at Newton’s second law for at least one case. Allow sign errors. Do not allow g missing. Correct 500 cos15 –80g sin 20 –T = 80a [ 482.96− 273.61− T = 80a ] number of terms. Allow sin/cos mix. 500 cos15 –80 g sin 20 – 40 g sin 20 − 50 = ( 80 + 40 ) a A1 Any 2 equations. [ 482.96− 273.61− 136.8− 50 = 120a ] For attempt to solve for T or a M1 From equation(s) with no missing/extra terms. Allow g missing. Must get to ‘T =’ or ‘α =’. Acceleration = 0.188 ms−2 A1 Allow AWRT 0.19. Tension = 194 N A1 5 5(b) [1.2 = 0 + 0.188t] M1 For use of constant acceleration formula(e) and solving for t with their positive a, leading to a positive value of t a ≠ ±10, a ≠ ± g Allow if a is negative in part (a) and use |a | here. Time = 6.39 s A1 Allow 6.38 Allow 6.32 from a = 0.19 2

Mark scheme

6(a) 0.3 × 2 [ +0] = 0.3 × 0.6 + 0.4 × v M1 For use of conservation of momentum. Must be 3 terms. Allow sign errors. Speed of B = 1.05 ms−1 A1 AG Allow M1 A0 if g included with the masses. 2 6(b) 0.4 × 1.05[ +0] = ( 0.4 + m ) × 0.5 M1 For use of conservation of momentum. Must be 3 terms. Allow sign errors. 11 A1 Allow M1 A0 if g included with the masses. m = 0.44 or 25 2 6(c) 1.2 [m] or 0.9 [m] B1 Must be a distance as some candidates get 1.2 from 0.6 . 0.5 0.5t B1 1 Seen but not + at 2 unless later state that a = 0 . 2 B0 if only 0.5t = 2.1 (may see solving to find t = 3.5) . 0.6t B1 1 Seen but not + at 2 unless later state that a = 0 . 2 Allow B2 in place of second and third B1 marks for ‘difference in speeds is 0.1 [ ms−1]’. Distances equal so 0.6t − 0.9 = 0.5t and solve for t M1 OE Must get to ‘ t = ’. Allow ± their 0.9 but not ±1.2 0.9 or ± 2.1 or ±1.5 . Do not allow 0.6t + 0.5t = ±0.9 . Or t = 0.6 − 0.5 Do not allow M1 if either or both terms include 1 + at 2 unless they state a = 0. 2 Time = 9 s A1 CWO 6(c) Alternative method for question 6(c) using time from start of motion 1 [m] or 1.1[m] B1 0.6T B1 1 Seen but not + aT 2 unless later state that a = 0 . 2 0.5T B1 1 Seen but not + aT 2 unless later state that a = 0 . 2 Allow B2 in place of second and third B1 marks for ‘difference in speeds is 0.1 [ ms−1]’. Distances equal so 0.6T −1.1 = 0.5T and solve for T M1 OE Must get to ‘ T = ’. Allow ± their 1.1 but not ±1.0 1.1 or ± 2.1. Do not allow 0.6T + 0.5T = ±1.1 . Or T = Do not allow M1 if either or both terms include 0.6 − 0.5 1 + aT 2 unless they state a = 0 2 ⇒ Time from BC collision = 11 − 2 = 9 s A1 CWO 6(c) Alternative method for question 6(c) using distance travelled from time when B and C collide 1.2 [m] or 0.9 [m] B1 d + 0.9 B1 FT Allow ± their 0.9 but not ±1.2 or ± 2.1 or ±1.5 . Time taken for A is 0.6 Or d + 0.9 = 0.6t d B1 Time taken for BC is 0.5 Or d = 0.5t d + 0.9 d 4.5 M1 Must get to ‘ t = ’. = d = 4.5 Time = Allow ± their 0.9 but not ±1.2 or ± 2.1. 0.6 0.5 0.5 Time = 9 s A1 CWO 5

Mark scheme

7(a) 0.3 2 3 3 M1 For integration (do not penalise missing c) v= t ( +c ) = 0.2t 2 ( +c ) The power of t must increase by 1 with a change of 1.5 coefficient. Use of v = at scores M0. 8 A1 ISW any extra work using the second equation for a. Velocity = 1.6 = ms−1 5 2 7(b) −k − 2 1 − 1 *M1 For integration. No need for constant. Allow use of v= t [ + d ] = 2kt 2 [ + d ] given value of k = 2.6 . −0.5 The power of t must increase by 1 with a change of coefficient. Use of v = at scores M0. −k − 1 − 1 A1 FT For both equations in k and d ( Allow unsimplified ) . Their 1.6 = × 4 2 + d = 2k × 4 2 + d [Their 1.6 = k + d ] −0.5 −k − 1 − 1 k 0.3 = × 16 2 + d = 2k × 16 2 + d 0.3 = 2 + d −0.5 Attempt to solve for k or d DM1 Or substitute k = 2.6 into both equations and solve both for d (with d ≠ 0 ). Must get to ‘ k = ’ or ‘ d = ’. − 1 −2.6 − 2 1 A1 AG (AG for k, not for the expression). k = 2.6 [v =] 5.2t 2 − 1 or [ v = ] t −1 Allow unsimplified expression for v and/or in terms of −0.5 k. If k is substituted then both equations must be shown 1 − to give a value of d = −1 and getting v = 5.2t 2 − 1 SC B1 for solving the correct equations simultaneously with no working seen and getting correct expression for v . SC A1 for correct expression for v if only the first M1 is scored. 7(b) Alternative method for question 7(b) from using limits −k − 2 1 − 1 *M1 For integration No need for constant. Allow use of v= t [ + d ] = 2kt 2 [ + d ] given value of k = 2.6 . −0.5 The power of t must increase by 1 with a change of coefficient. Use of v = at scores M0. −k − −k 1 − 1 − 1 − 1 A1 FT OE For correct unsimplified equation in k from using ×4 2 − × 16 2 = 1.6 − 0.3 or 2k × 4 2 − 2k × 16 2 = 1.6 − 0.3 limits for v as their 1.6 and 0.3 and limits for t as 4 and −0.5 −0.5 16. Attempt to solve for k DM1 Must be an equation using 0.3 and their k, and 16 and 4, Must be subtracting the limits to form the equation in k , but may have sign errors in their 1.6 − 0.3 . − 1 −2.6 − 2 1 A1 AG (AG for k, not for the expression). k = 2.6, v = 5.2t 2 − 1 or v = t −1 Allow unsimplified expression for v and/or in terms of −0.5 k. 4 7(c) − 1 M1 For solving for T. Must get to ‘ T = ’. Must come from ±5.2T 2 − their 1 = 0 integration, with their 1 from part (b) or found here not equal to zero. Do not allow made up value of d. 676 A1 OE Must be exact. Allow both marks as long as T= or 27.04 expression for v is correct, however obtained in 25 Q7(b). 2 7(d) 4 3 0.2 2 5 27.04 − 1 5.2 2 1 M1 3 − 1 0 0.2t 2 dt = 2.5 t or 4 5.2t 2 − their 1 dt = 0.5 t − their 1× t For integration of 0.2t 2 oe or 5.2t 2 − their 1 . May be in terms of k. Not from any other expression. Their 1 may be zero (or replaced by zero). Their 1 may came from either part (b) or part (c) The power of t must increase by 1 with a change of coefficient in at least one term. 4 0.2 5 5.2 1 27.04 5 4 1 27.04 A1ft For both integrals (unsimplified) No need for limits = t 2 + t 2 − t = 0.08t 2 + 10.4t 2 − t FT non-zero value of d. May be in terms of k. 2.5 0 0.5 4 0 4 Their 1 may came from either part (b) or part (c). = 0.08 × 32 + (10.4 × 5.2 − 27.04 ) − (10.4 × 2 − 4 ) [ = 2.56 + 10.24] M1 For correct use of limits (0 and 4 then 4 and their 27.04) in both of their integrals, which have come from 3 1 − integration of 0.2t 2 and 5.2t 2 − their 1 . Not from any other expression. Their 1 may be zero (or replaced by zero). Their 1 may came from either part (b) or part (c). Allow M1 for d = 0 (the final answer is 35.8 ). 64 A1 oe Awrt 12.8 Allow if using 27(.0) rather than 27.04 = or 12.8 Allow all 4 marks as long as expression for v is 5 correct, however obtained in Q7(b). 7(d) SC for using a calculator to integrate. 4 3 B1 AWRT 2.56 Either 0.2t 2 dt = 2.56 Allow 10.2 0 Must use 27.04 or 27(.0) if latter integral. 27.04 − 1 Or 4 5.2t 2 − 1 dt = 10.24 Total distance = 12.8 m B1 AWRT 12.8. Allow if using 27(.0) rather than 27.04 Allow both B marks as long as expression for v is correct, however obtained in Q7(b). 4