Motion in a circle

1(a) constant speed or constant magnitude of velocity B1 acceleration (always) perpendicular to velocity B1 1(b)(i) F = mv2 / r C1 or v = rω and F = mrω2 F = 790 × 942 / 318 A1 = 22000 N 1(b)(ii) centripetal acceleration: same B1 maximum speed: greater B1 time taken for one lap of the track: greater B1 © UCLES 2021 Page 7 of 15

1(a) acceleration perpendicular to velocity B1 1(b)(i) decreases B1 1(b)(ii) (acceleration of) 9.8ms–2 is caused by weight of car B1 or centripetal force must be greater than weight of car (acceleration > 9.8ms–2) requires contact force from track B1 or (centripetal force > weight) requires contact force from track 1(c) ½mv 2 = ½mv 2 – mgh C1 Y X a = v2 / r C1 v 2 = 3.82 – 2 × 9.81 × 0.62 so v = 1.5ms–1 A1 Y Y a = 1.52 / 0.31 = 7.3ms–2 (which is less than 9.8ms–2) so no or v = √(9.81 × 0.31) = 1.74 m s–1 so v 2 = 1.742 + 2 × 9.81 × 0.62 (A1) Y X v = 3.9 m s–1 (which is greater than 3.8 m s–1) so no X 1(d) acceleration is independent of mass so makes no difference B1 or mass cancels in the equation so makes no difference © UCLES 2021 Page 7 of 19

1(a) constant speed or constant magnitude of velocity B1 acceleration (always) perpendicular to velocity B1 1(b)(i) F = mv2 / r C1 or v = rω and F = mrω2 F = 790 × 942 / 318 A1 = 22000 N 1(b)(ii) centripetal acceleration: same B1 maximum speed: greater B1 time taken for one lap of the track: greater B1 © UCLES 2021 Page 7 of 15

1(a) angle (subtended at centre of circle) when arc length = radius B1 1(b)  = 2 / T C1 = 2 / (1.0  60  60) A1 = 1.7  10–3 rad s–1 1(c)(i) angle = 1.7  10–3  1400 A1 = 2.4 rad 1(c)(ii) L = arc length / angle C1 = 0.44 / 2.4 or L = 0.44  (3600 / 1400) / 2 L = 0.18 m A1 1(c)(iii) a = r2 C1 = 0.18  (1.745  10–3)2 A1 = 5.5  10–7 m s–2 1(d) centripetal acceleration is negligible compared with acceleration of free fall B1 or numerical comparison establishing answer to (c)(iii) ≪ 9.81 resultant force is negligible compared with weight (of modelling clay) (so variation is negligible) B1 or force exerted by minute hand (approximately) equal (and opposite) to weight of modelling clay © UCLES 2023 Page 6 of 15

1(a) angle (subtended at the centre of a circle) when arc (length) = radius B1 1(b)(i) arrow, labelled V, pointing in NE direction B1 1(b)(ii) arrow, labelled A, pointing in NW direction B1 1(c)(i) v = r C1  = 0.68 / (0.093 – 0.012) A1 = 8.4 rad s–1 1(c)(ii) a = v2 / r or a = r2 C1 a = 0.682 / (0.093 – 0.012) or (0.093 – 0.012)  8.42 A1 = 5.7 m s–2 1(d) angular speed: same for both pieces B1 linear speed: less for second piece than first piece B1 acceleration: less for second piece than first piece B1 © Cambridge University Press & Assessment 2024 Page 6 of 16

1(a) arrow vertically downwards labelled ‘weight’ and arrow perpendicular to cone, inwards and upwards, labelled ‘normal B1 contact force’ 1(b) vertical component of contact force = weight (so no resultant force vertically) B1 horizontal component of contact force is resultant force towards centre (of circle) B1 1(c) a = v2 / r C1 a = g tan 52° C1 9.81  tan 52° = v2 / 0.15 leading to v = 1.4 m s–1 A1 1(d) v = r or a = r2 C1  = 1.4 / 0.15 or √(9.81  tan 52° / 0.15) A1 = 9.3 rads–1 1(e) same resultant force / same acceleration so v2 is proportional to r (so if speed increases radius must also increase) A1 Question Answer Marks

1(a) angle (subtended at centre of a circle) when arc (length) = radius B1 1(b)(i) v = r C1  = 17 / 0.46 A1 = 37 rad s–1 1(b)(ii) T = 2r / v or T = 2 /  C1 = 2 × 0.46 / 17 or 2 / 37 A1 = 0.17 s 1(b)(iii) distance = 2 × 0.038 = 0.24 m A1 1(b)(iv) angle = arc length / radius C1 = 0.24 / 0.15 A1 = 1.6 rad 1(c) point X moves through a smaller distance in the same time B1 or (linear) speed of movement of point X / chain decreases or (linear) speed of (circumference of) both cogs decreases angular speed (of pedals) decreases B1 © Cambridge University Press & Assessment 2025 Page 8 of 18

1(a)(i) radius = 6.37  106  cos 52.2° = 3.90  106 m A1 1(a)(ii) period = 24 hours C1 v = 2r / T C1 or v = r and  = 2 / T v = (2  3.90  106) / (24  60  60) A1 = 280 m s–1 1(b)(i) F = mv2 / r C1 = (58.6  2802) / (3.90  106) A1 = 1.2 N 1(b)(ii) arrow pointing horizontally to the left B1 1(b)(iii) arrow from student pointing along the dotted line, labelled ‘weight’ B1 upwards arrow from student pointing in a direction to the left of normal and above the tangent to the Earth, labelled ‘contact B1 force’ © Cambridge University Press & Assessment 2025 Page 8 of 17