Edexcel Maths M1

CChhaapptteerr 22 AAnnsswweerrssExercise 2A 1 20 m s12 1.6 m s23 0.625 m s24 26 m5 20 m s16 6 m s1 in direction 7 a 9 m s1b 72 m8 a 3 m s1b m s29 a 9.2 m s1b33.6 m10a 18 km h1b312.5 m11a 8 s b 128 m12a 0.4 m s2b 320 m13a 0.25 m s2b 16 s c 234 m14a 19 m s1b 2.4 m s2c 430 m15a x = 0.25 b 150 m16b 500 mExercise 2B 1 7 m s12m s23 2 m s24 8.5 m s15 2.5 s6 0.175 m s27 a 2.5 m s2b 4.8 s8 a 3.5 m s1b 15.5 m s19 a 54 m b 6 s10a 90 m b 8.49 m s1 (3 s.f.)11a 3.3 s (1 d.p.) b 16.2 m s1 (1 d.p.)12a 4, 8 b t = 4: 4 m s1 in direction, t = 8: 4 m s1 in direction 13a 0.8, 4 b 15.0 m s1 (3 s.f.)14a 2 s b 4 m15a 0.34 m s1b 25.5 s (3 s.f.)16 a P: (4t + t2) m Q: [3(t 1) + 1.8(t 1)2] mb t = 6 c 60 mExercise 2C1 10 m2 3.2 s (2 s.f.)3 1.8 m (2 s.f.)4 4.1 s (2 s.f.)5 41 m (2 s.f.)6 a 29 m (2 s.f.) b 2.4 s (2 s.f.)7 a 5.5 m s1 (2 s.f.) b 20 m s1 (2 s.f.)8 a 40 m s1 (2 s.f.) b 3.7 s (2 s.f.)9 a 39 m s1b 78 m (2 s.f.)10 4.7 m (2 s.f.)11a 3.4 s (2 s.f.) b 29 m (2 s.f.)12 2.8 s (2 s.f.)13a 29 (2 s.f.) b 6 s14 30 m (2 s.f.)15a 5.6 m (2 s.f.) b 3.1 m (2 s.f.)18 a 1.4 s (2 s.f.) b 7.2 m (2 s.f.)Exercise 2D1 a 2.25 m s2b 90 m2 a b 360 m3 a 0.4 m s2b158m s2c 460 m4 ab 2125 m5 ab 100 s6 a 0.8 m s2b 1960 m7 ab T = 320 c 3840 m8 ab 60 sc9 a b m s21011ab 720 m12ab T = 12cExercise 2E1 2 m s22 1.9 s (2 s.f.)3 u = 84 a 23 m (2 s.f.) b 2.1 S (2 s.f.)5 a 28 m s1b 208 m6 0.165 m s2 (3 d.p.)7 a8 ab 315 m c 30 s8 a 4.1 s (2 s.f.) b 40 m s1 (2 s.f.) c air resistance9 a 8 m s1b 1.25 m s2c 204.8 m10a 33 m s1 (2 s.f.) b 3.4 s (2 s.f.)11a 60 m b 100 m12ab75m s2c13a u = 11 b 22 m14ab 230 mc15 1.2 s (2 s.f.)16a 50 s b 24.2 m s1 (3 s.f.)17 h = 39 (2 s.f.)18a 32 m s1b 90 m c 10 s19ab 180 m20ac x = 0.2 d 3 km e 125 sCChhaapptteerr 33 AAnnsswweerrssExercise 3A 1 39.2 N2 50 kg3 112 N4 4.2 N5 0.3 m s26 25 kg7 a 25.6 N b 41.2 N c P is 34 N, Q is 49 N8 a 2.1 kg (2 s.f.) b 1.7kg (2 s.f.)c 0.22kg (2 s.f.)9 a 5.8 m s2b 2.7 m s2c 2.7 m s210a 31.2 N b 39.2 N c 41.2 NExercise 3B1 2.3 N (2 s.f.)2 0.35 N3 a 0.9 m s2b 7120 N c 8560 N4 2.25 N5 a 0.5 m s2b 45 N6 a 4 m s2b 800 N7 a 708 N b 498 Nc She feels lighter.8 a 32 s b 256 mc Air resistance unlikely to be constant.9 a 1.5 m s2b 60 kg c 40 kg10a 2.9 m (2 s.f.) b 3.6 m s1(2 s.f.)c 2.17 s (3 s.f.)Exercise 3C1 a i 11.3 N (3 s.f.) ii 4.10 N (3 s.f.)b i 0 N ii 5 N c i 5.14 N (3 s.f.) ii 6.13 N (3 s.f.)d i 8.66 N (3 s.f.) ii 5 Ne i 3.86 N (3 s.f.) ii 4.60 N (3 s.f.)fi Fcos N ii Fsin N 2 a i 2 N ii 6.93 N (3 s.f.)b i 8.13 N (3 s.f.)ii 10.3 N (3 s.f.)c i Pcos + Q Rsin ii Psin RcosExercise 3D1 a i 3 N ii F = 3 N and body remains at restb i b7 N ii F = 7 N and body remains at restc i 7 N ii F = 7 N and body acceleratesiii 1 m s2d i 6 N ii F = 6 N and body remains at reste i 9 N ii F = 9 N and body remains at rest in limiting equilibriumfi 9 N ii F = 9 N and body accelerates iii 0.6 m s2g i 3 Nii F = 3 N and body remains at resth i 5 N ii F = 5 N and body remains at rest in limiting equilibriumi i 5 N ii F = 5 N and body acceleratesiii 0.2 m s2ji 6 N ii F = 6 N and body acceleratesiii 1.22 m s2 (3 s.f.)k i 5 N ii F = 5 N and body acceleratesiii 3.85 m s2 (3 s.f.)l i 12.7 N (3 s.f.) ii The body accelerates.iii 5.39 m s2 (3 s.f.)2 a R = 88 N, = 0.083 (3 s.f.)b R = 80.679 N, = 0.062 (2 s.f.)c R = 118 N, = 0.13 (2 s.f.)Exercise 3E1 3.35 m s2 (3 s.f.)2 a 27.7 N (3 s.f.) b 2.12 m s23 a 2.43 m s2 (3 s.f.) b 4.93 m s1 (3 s.f.)4 28 N5 0.20 (2 s.f.)6 0.15 (2 s.f.)7 a 88.8 N (3 s.f.) b 0.24 (2 s.f.)8 a1513gb 23.5 m (3 s.f.) c 2.35 s (3 s.f.)d 12 .4 m s1 (3 s.f.)Exercise 3F1 a 4 N b 0.8 N2 a R = 45 b 100 N3 a 3 m s2b 2500 N4 a 33.6 N (3 s.f.) b 2m5 a 0.613 m s2 (3 s.f.) b 27.6 N (3 s.f.)c 39.0 N (3 s.f.)6 2.8 m s17 a 0.569 m s2 (3 s.f.) b 0.56 mg8 a 1.12 m s2b 4100 N9 a 21.9 N b 0.418 (3 s.f.) c 38 N (2 s.f.)10a 2 m s2b 600 N c 100 mExercise 3G1 30 m s12 2.5 m s13 2.59 N s4 6.5m s15 3 m s1Exercise 3H1 4 m s12 292m s13 4.5 m s14 a 232m s1b 232 N s5 a 1 m s1 and direction unchanged b 15 N s6 107 a32u b 8 mu8 Larger 8 m s1 and smaller 4 m s19 a 3 b29mu10a 3 m s1b 4.511a 4 m s1 in same directionb 3 m s1 in opposite direction12a 3 m s1b 6 kgExercise 3I1 a 0.103 kg b 4.103 kg2 0.14 (2 s.f.)3 a21u = v b 6mu4 a 0.22 m (2 s.f.) b2514gc 1.1 m s1 (2 s.f.)5 0.12 (2 s.f.)6 a 9.8 N b 9.8 N7 a 14 m s1b335m s1c 0.75 m (2 s.f.)8 a31 g b 3.6 m s1 (2 s.f.) c 232md iacceleration both masses equalii same tension in string either side of pulley9 a 540 N b 180 Nc 450 N10 1000 N vertically downwards11a 2000 b 36 m12a 1.75 m s1b 0.45 N s13a 2.5 m s1b 15 000 N s14a 0.7 m s1b unchanged c 8.25 N s155416a 1.25 m s1b 0.77 (2 s.f.)17 0.44 (2 s.f.)18a 1.3 N (2 s.f.) b 19 m (2 s.f.)1928520a 830 N (2 s.f.) b 1500 N (2 s.f.) c 1700 N (2 s.f.)d Air resistance would reduce speed of lift as it falls and so impulse would be reduced.21a 18 N (2 s.f.) b 0.12 m s2 (2 s.f.)22a 243 N (3 s.f.) b 3.08 m s2 (3 s.f.)c 36.7 m (3 s.f.)23a 7.5 m s1b 11 000 (2 s.f.)c R could be modelled as varying with speed.24a712g N b 1.225a 3.2 m s2b 5.3 N (2 s.f.) c 0.75 (2 s.f.)d The information that the string is inextenible has been used when, in part c the acceleration of A has been taken as equal to the acceleration of B obtained in part a.26a 18 N (2 s.f.) b 2 c 4.2 N s d72sRReevviieeww EExxeerrcciissee 11 1ms v6O 3 11 s t 1ms v30O 2 s tO 1ms v75 s t24T1tT 4T 50 1ms v20O s t 1ms v16 1ms v25O 20 140 s tExercise A 1 a 1.12 m s2b 31.25 s2 a 3.6 m s2b AC = 760 m BC = 440 m3 a 14.4 b 36 m s14 a 0.5 m s2b 7.5 m s15= 6 m s1 (3 s.f.)7 a 24 b OA = 96 m c 4 s and 12 s8 a 2.5 m s2b 31.7 m s1 (3 s.f.)c 1.69 s (3 s.f.)9 a 6t t2b 7 m c t = 510 a 34 (2 s.f.) b 60 m (2 s.f.)11 a 28 m s1b 5.7 s (2 s.f.)12 2 or 413 a 14 (2 s.f.) b 23 m s1 (2 s.f.)14 10 m (2 s.f.)15 a 28 b 4s16 ab 33 m17 ab 18 s18 a constant acceleration b constant speedc 30.5 m19 ab 0.48 m s2c 250 d 375 s20 ab 8 c 2.5 m s221 a 162 m b 6.2 c 0.56 m s222 a 185 s b 2480 mc23 ab 200 s c 60 s d 50 m s12t 20 O s t1t 1ms v3024O T 60 s tBA1t 1ms v10O312 27 s t 2ms aO38712 s t12 27 35O140501080 220 s t 1ms v128O6 Tbuscycle 1ms v6030O 12 40 64 T s tTrain A Train B1tO 1ms v15T s t34T2t240 4t 1ms v3010240 s tOt 3tT24 ab 10 s c 3425 ab 78 m c 35 sd26 a 1ms vb 5400 m c112 s27 a bus has not overtaken cyclistb s t28 ab 9829 ab 96 s c m s230 ab c 800 m31 66 m s132 a 13 m s1b 2 m s1 in direction 33 6.3 N34 a 2.25 m s1 direction of motion unchangedb 1.5 N s35 a 2.4 m s1 b due west c 3000 kg36 a A2.2 m s1B3 m s1 b 0.4 N sc 1.6 N s37 a 3 m s1 b i m = 3.6 ii 18 N s38 750 N39 a 0.42 N b 2.540 a 2.45 m s2 b 0.2541 0.30 (2 s.f.)42 0.37 (2 s.f.)43 a 3 m s2 b 14.8 m s1(3 s.f.) c 0.1 kgd 3.06 s (3 s.f.)44 a 8.6 m s1 b 24 m c 79.2 m45 520 (2 s.f.)46 a 0.693 m s2 (3 s.f.)b 7430 N(3 s.f.)c 28 kN (2 s.f.)47 a 3.6 m s2b 0.75(2 s.f.) c14 m (2 s.f.)48 a 0.35 (2 s.f.)b normal reaction unchanged hence friction force unchangedc 1500 N (2 s.f.)49 a 15 m s1 b 991(3 s.f.)50 a 22.4 b 4.64(3 s.f.) c 6380 (3 s.f.)d Consider air resistance due to motion under gravity51 a 4.2 m s2 b 3.4 N(2 s.f.) c 2.9 m s1 (2 s.f.)d 0.69 s (2 s.f.)52 a 1.4 m s2b 3.4 N(2 s.f.), 4.2 N53 a i 1050 N ii 390 Nb 3 m s54 a 2.2 m s2(2 s.f.) b 22 N(2 s.f.)c 4.4 m(2 s.f.)55 a mg b 0.693(3 s.f.)c mg vertically downwards56 a 1.2 m s2 b 16 Nc The information that the string is inextensible has been used in assuming that the accelerations of P and Q, and hence of the whole system, are the same.d 3 se 20 m s1

57 a 1.0 m (2 s.f.) b 17 N (2 s.f.)c 26 N, direction bisecting angle ABCd 0.55 (2 s.f.)58 a 11 500 N b 6.2 m s2 c 3700 Nd 31 m s1(2 s.f.)59 a 0.24 6.2 m s2 b 530 N (2 s.f.) c 54 md normal reaction of the road on the car is increased when the towbar breaks80T NT NR NBCChhaapptteerr 44 AAnnsswweerrssExercise 4A 1 a Q 5cos30 = 0b P 5sin30 = 0c Q = 4.33 N P = 2.5 N2 a Q Pcos60 = 0b Psin60 43 = 0c Q = 4 N P = 8 N3 a 9 Pcos30 = 0b Q + Psin30 8 = 0c Q = 2.80 N P = 10.4 N4 a 9 Pcos30 = 0b Q + Psin30 8 = 0c Q = 2.80 N P = 10.4 N5 a 4cos45 + Pcos 7 = 0b 4sin45 Psin = 0 c = 34.1 P = 5.04 N6 a 6cos45 2cos60 Psin = 0b 6sin45 + 2sin60 Pcos 4 = 0c = 58.7 P = 3.80 N7 a Pcos + 8sin40 7cos35 = 0b Psin + 7sin35 8cos40 = 0c = 74.4 (allow 74.3) P = 2.20 (allow 2.19)8 a 9cos40 + 3 Pcos 8sin20 = 0b Psin + 9sin40 8cos20 = 0c = 13.6 P = 7.36 9 a Pcos30 Qcos 45 8cos45 = 0b Psin 30 + Qsin45 8sin45 4 = 0c P = 11.2 (3 s.f.)Q = 5.73 (3 s.f.) 10a Qcos60 Pcos 60 + 5sin45 6sin45 = 0b Psin 60 + Qsin60 5cos45 6cos45= 0c P = 3.784 NQ = 5.198 N 11a Q 10sin45 = 0b P 10cos45= 0c P = 7.07 NQ = 7.07 N 12a Q + 2cos60 6sin60 = 0b P 2sin60 6cos60= 0c P = 4.73Q = 4.20 13a 8sin30 Qcos30 = 0b P Qsin30 8cos30= 0c P = 9.24 NQ = 4.62 N 14a 8cos45 10sin30 Q = 0b P + 8sin45 10cos30= 0c P = 3.00 NQ = 0.657 N 15a 2 + 8sin30 Pcos = 0b 4 8cos30 + Psin = 0c = 26.0 P = 6.68 N Exercise 4B1 34.7 N2 a 20 N b 1.773 14.44 S = 30.4 or 30.5, T = 43.05 a 5.46 N b 0.762 kg6 a 1.46 b 55g7 a 3 N b 2 N8 a 2.6 b 4.49 a F = 19.6m, R = 9.8mb F = 17m (3 s.f.), R = 0c P = 11.2 (3 s.f.)Q = 5.73 (3 s.f.) 10 13.9 N11 39.212 37.2 N (3 s.f.)13 F = 12.25, R = 46.6 (3 s.f.)14 P= 20.4 (3 s.f), R = 0.400Exercise 4C1 0.4462 0.1233 a 1.5 N b not limiting4 a 40b The assumption is that the crate and books may be modelled as a particle.5 a 11.9 b 6.406 0.601 (accept 0.6)7 a 13.3 b F = 3.33, X = 9.548 a 9.97 N down the plane b 22.7 Nc 0.4399 a and bX = 44.8 (accept 44.7), R = 51.310 F = 22.1, T = 102 (3 s.f.)11a T = 3.87 b T = 2.7512 0.758Exercise 4D1 = 52.6, T = 24.72 a and bThe weight of the particle is 80 N and the tension in the second string is 69.3 N (3 s.f.).3 a 6.93 (3 s.f.) b 3.46 (3 s.f.)4 a 43 (to nearest degree)b 53 N (to nearest Newton)5 a 138.2 (1 d.p.) b 8.95 (2 d.p.)6 T = 17.3, S = 21.37 R= 20.7, = 0.24 (2 s.f.)8 = 0.296 (3 s.f.)9 36310 1111a 0.577b The book was modelled as a particle.12a W = 11.4 b R = 13.913 2.2 (1 d.p.)14 0.75 (2 d.p.)15 0.262 (3 s.f.)16ab 40.46 c 0.279 (3 s.f.)17a R = 88.3 b P = 74.7c resultant force 9 N down plane and box will move18 11.019 sin cossin cos20a 15.7 (3 s.f.) b 0.62521 0.577 (3 s.f.)22 0.399 (3 s.f.)23a 0.684 (3 s.f.) b 2.33c As 2.425 > 1.596 the ring is not in equilibrium.CChhaapptteerr 55 AAnnsswweerrssExercise 5A 1 6 Nm clockwise2 10.5 Nm clockwise3 13 Nm anticlockwise4 0 Nm5 10 Nm anticlockwise6 11.6 Nm clockwise7 30.5 Nm anticlockwise8 0 Nm9 13.3 Nm clockwise10 33.8 Nm anticlockwiseExercise 5B1 a 5 Nm anticlockwiseb 13 Nm clockwisec 19 Nm anticlockwised 11 Nm anticlockwisee 4 Nm clockwisef 7 Nm anticlockwise2 a 16 Nm clockwiseb 1 Nm anticlockwisec 10 Nm clockwised 7 Nm clockwisee 0.5 Nm anticlockwisef 9.59 Nm anticlockwiseExercise 5C1 a10 N, 10 N b 15 N, 5 Nc 8.6 N, 11.4 Nd 12.6 N, 7.4 N2 a 7.5, 17.5 b 30, 35c 245, 2 d 49, 1.53 0.5 m from B4 59 N5 31 cm from the broomhead6 16.25 N, 13.75 N7 1.71 m8 59m10 2.05 m11a 15 N b rod will tilt c 3.17 mExercise 5D1 2.4 N, 3.6 N2 3.5 m from A3m from 4 a 29.4 N, 118 N b 4.25 mExercise 5E1 a 105 N b 140 N c 1.03 m2 b 0 x < 3 a 40g b x = c i the weight acts at the centre of the plankii the plank remains straightiii the mans weight acts at a single point4 b W = 790 300x c x = 2.53, W = 305 a 200 N b 21 cm6 a 36 kg b 2.2 m7 a 19.6 N b 58 a 588 N bm9 a 125 N b 1.8 mCChhaapptteerr 66 AAnnsswweerrssExercise 6A 1 8.60 km from starting point on bearing of 0542 10 km, 7.2 km on bearing of 3263 7.43 km, 0624 9.13 km, 3405 31.8 km, 2616 174, 328.67 3.01 km, 220Exercise 6B1 a 2b b d c b d 2b e d + b f d + bg 2d h b i 2d + b j b + 2dk b + d l b d2 a 2m b 2p c m d m e p + mf p + m g p +2m h p m i m pj 2m + p k 2p + m l m 2p3 a 2p b 2r c 2p+ 2r d p + re p + r f r g p h 2r + p4 a +b5 a +bExercise 6C1 4i2 5i + 2j3 3i + j4 2i + 3j5 2i j6 3jExercise 6D1 a 6i + 2j b 10i + 8j c 7j d 10i + je 2i + j f 2i 10j g 14i 7j h 8i + 9j2 a 5 b 10 c 13 d 4.47 (3 s.f.)e 5.83 (3 s.f.) f 8.06 (3 s.f.) g 5.83 (3 s.f.)h 4.12 (3 s.f.)3 a 53.1 above b 53.1 below c 67.4 aboved 63.4 above4 a 149 to the right b 29.7 to the rightc 31.0 to the left d 104 to the left5 a = 5 b = 6 a =b = 1 c s = 1 d t = 7 a 3.61 (3 s.f.), 023 b 4.12 (3 s.f.), 104c 3.61 (3 s.f.), 304 d 2.24 (3 s.f.), 243Exercise 6E1 a 5 m s1b 25 km h1c 5.39 m s1d 8.06 cm s12 a 50 km1b 51.0 m c 4.74 km d 967 cm3 a 5 m s1, 75 m b 5.39 m s1, 16.2 mc 5.39 km h1, 16.2 km d 13 km h1, 6.5 kmExercise 6F1 a 8i + 3j b 2i 7j c 17i + 16j d 7i 13j2 a 2i + 5j b i + 3j c 2i + 4j d 2i 5je 2i 5j3 a 6i 8j b 12i + 9j c 4.5i + 6j d 5i + 5je 4i + 6j f 32i 52j g 43i 23jh 35i + 65j4 a 6i + 12j b 7i + 4j c 2i + 6j d 10i 13je 2i 3j, f 4.61 m s1g 4 h 2.55 a 5i + 12j, 13 m s1b 6i 5j, 7.81 m s1c 2i 5j, 5.39 m s1d 3i 2j, 3.61 m s1e 7i + 9j, 11.4 m s16 4.8i 6.4j7 10.1 m8 2.03 m s19 a 2ti + (500 + 3t)j b 721 m10 a 7ti + (400 + 7t)j, (500 3t)i + 15tjb 350i + 750j11 a (1 + 2t)i + (3 t)j, (5 t)i + (2 + 4t)j b5.39 km12 a 121 m s1, 6.08 m s1b 18i 3j c 15i 12jExercise 6G1 a i 8j b 5i + j c 2i + 5j d 3i + 2j2 a 8.06, 82.9 below b 5.10, 169 abovec 5.39, 68.2 above d 3.61, 146 above3 a 6i, 3i m s2b 3i 2j, (i j) m s2c 3i 2j, ( i j) m s2d i 6j, ( i 3j) m s24 a 5.83 N, 59 b 6.32 N, 18.4c 6.40 N, 38.75 a 5.83 N, 3.83 N b 4.39 N, 5.38 Nc 4.20 N, 6.53 N d 14.4 N, 12.7 Ne 4.54 N, 31.9Exercise 6H1 a p = 2, q = 6 b 6.32 N c 182 a (3 + t)i + (10 + t)j b 4.24 km c 16303 a p= 6ti, q = (12 3t)i + (6 + 6t)j b 38.4 kmc 14 a 3 b 10.2 m s2c 168.75 a 4 i + 2j m s2b 22.4 N c 26 m6 a 031 b a= 6ti, b = 3ti + (10 + 5t)jc 1400 d 14567 a 108 b (2 + 9t)i + (4 3t)j c 41, 238 a 124 b (3 2t)i + (2 + 3t)j c 11.2 m s1d 19 a 9.85 m s1b (3 + 4t)i + (2 + 9t)j c 6.5 sd 7.46 m s110 a (5i + 3j) km h1b (10 + 5t)i + (15 + 3t)j, (16 + 12t)i + 26jc 0525RReevviieeww EExxeerrcciissee 221 a 48 b 41.62 a 40.8 b 22.7 N (3 s.f.)3 a 42.9 b 52.8 N (3 s.f.)4 a 35.1 N b 33.0 N (3 s.f.)5 a 26.1 b 51.4 (3 s.f.)6 a 7.5 b 127 a25mgb748 a 86.6 b 1009 47.5 (1 d.p.)10 a 19.9 N b 3.4611 a 23.0 b 17.6c The friction is not limiting and so equilibrium is maintained.12 a 18.7 b 0.60 (2 s.f.)c Equilibrium is maintained and so the parcel does not move.13 a 1.68 b 0.54814 a 257 (3 s.f.) b 12.5 s15 a 131 N b 209 Nc i Friction acts down the slope,. magnitude 0.4Rii No acceleration so net force on package is zero16 a 0.270 b 3.76 m s2 down the plane17 a 125 b 1.46 m s218 5.6 m19 a 88.2 N b 0.875 m20 m21 a 2 b 0.6 m22 a 911 N b 1176 N c 2.25 m23 a 784 N b 0.5 m24 1.6d25 a 50 N b 1.9 m26 a 0.75 b 24 N c 144 Nd The weight of the rock acts precisely at B.27 a 1.25b The weight of the beam acts through its mid-point at C.c 0.4 m28 a 70 N b 120 cm c 3029 a 0.8b The weight acts through the mid-point of the rod.30a i 7.5 kg ii 477.75 Nb Assumed that the centre of mass acts at the point C.31 a 90 X b 2X 30 c 15 X 90d 7532 a Model the plank as a uniform rod.b 240g c 210g33 a 30 kg b 3.6 kgc i plank is uniform so weight acts through mid-point ii rock is a particle so mass of rock acts through end-point A34 a p = 2, q = 6 b 210 or 6.32 (3 s.f.)c 18 (to nearest degree)35 a 7.55 b 14.836 a 14.8 b 144.237 a 63.4 b 2 + 1 = 0 c 4.47 (3 s.f.)38 a 17.5 (1 d.p.) b 66c P = 3i + 12jQ = 4i + 4j39 a 2i + j b 26.6 c 12.6 m40 a 5.83 b 9.4341 a (2i + 6j) km h1b (3i 4j)+ (2i + 6j)tc = 2 d 40 km h142 a 3i 1.5j b 6.71 c 21i 7j43 a 6.5 km h1 (2 s.f) b 337 c 8.5i + 23jd 11i + (17 + 5t)j e 1512 f 4.72 km44 a 6.08 m s1b 3517 c 5i + 32jd 21 m45 a 9.43 m s1b 2i + j + t(5i + 8j) c1.6 sd 4.25 m s1e friction on field so velocity of ball not constant or vertical component of balls motion or time for player to accelerate46 a velocities destroyer: 10i km h1, cruiser: 20j km h1b position vectors destroyer:10ti = d cruiser: 50i +20tj = cc d2 = 500t21000t +2500d as 44.72 > 40 cruiser will not be able to detect destroyer 47 a 031 (to nearest degree)b (3t 10)i + 5tj c 15.20d d2 = 25t260t +100 e 14.24EExxaammiinnaattiioonn SSttyyllee PPaappeerr( )1ms v128O6 15( ) s t39013848O6 15 36( ) s t( ) m sExercise A 1 a14 msb The direction of motion of P has been changed by the collision.c 3.2 N s .2 a 12 3 \N.b 243 ab 36 s. c 4 a34 g .b For A 28 m =c The accelerations of the particles have the same magnitude.5 a 70.9 , to 3 significant figures.b27.46 ms, to 3 significant figures.6 a21ms3.b 2.5 N,to 2 significant figures.c 0.54, to 2 significant figures.7 a 146.b ( ) 8 2 3 t = + s i i j ,6t = r ic 8 T =d 24 2 \km.8 a400 6N6Wx+ | | |\ ..b( ) 600 230xWx=+c W0x2