Scalars and vectors — Paper 2 · A Level Physics 9702

4 (a) A stone of mass 56 g is thrown horizontally from the top of a cliff with a speed of 18 m s–1, as illustrated in Fig. 4.1. 18 m s–1 16 m sea level Fig. 4.1 The initial height of the stone above the level of the sea is 16 m. Air resistance may be neglected. (i) Calculate the change in gravitational potential energy of the stone as a result of falling through 16 m. change = … J [2] (ii) Calculate the total kinetic energy of the stone as it reaches the sea. kinetic energy = … J [3] Examiner’s Use (b) Use your answer in (a)(ii) to show that the speed of the stone as it hits the water is approximately 25 m s–1. [1] (c) State the horizontal velocity of the stone as it hits the water. horizontal velocity = … m s–1 [1] (d) (i) On the grid of Fig. 4.2, draw a vector diagram to represent the horizontal velocity and the resultant velocity of the stone as it hits the water. [1] Fig. 4.2 (ii) Use your vector diagram to determine the angle with the horizontal at which the stone hits the water. angle = … ° [2]

10 marks

Mark scheme: 4 (a) (i) (change in) potential energy = mgh C1 = 0.056 × 9.8 × 16 = 8.78 J (allow 8.8) A1 [2] (ii) (initial) kinetic energy = ½mv2 C1 = ½ × 0.056 × 182 = 9.07 J (allow 9.1) C1 total kinetic energy = 8.78 + 9.07 = 17.9 J A1 [3] (b) kinetic energy = ½mv2 17.9 = ½ × 0.056 × v2 and v = 25(.3) m s-1 B1 [1] (c) horizontal velocity = 18 m s-1 B1 [1] (d) (i) correct shape of diagram (two sides of right-angled triangle with correct orientation) B1 (ii) angle = 41° → 48° (allow trig. solution based on diagram) A2 [3] (for angle 38°→ 41° or 48°→ 51°, allow 1 mark)

This question in 9702/21 May/June 2007

3 A small ball is thrown horizontally with a speed of 4.0 m s–1. It falls through a vertical height of For 1.96 m before bouncing off a horizontal plate, as illustrated in Fig. 3.1. Examiner’s Use 4.0 m s–1 1.96 m 0.98 m plate Fig. 3.1 Air resistance is negligible. (a) For the ball, as it hits the horizontal plate, (i) state the magnitude of the horizontal component of its velocity, horizontal velocity = … m s–1 [1] (ii) show that the vertical component of the velocity is 6.2 m s–1. [1] (b) The components of the velocity in (a) are both vectors. For Examiner’s Complete Fig. 3.2 to draw a vector diagram, to scale, to determine the velocity of the Use ball as it hits the horizontal plate. Fig. 3.2 velocity = … m s–1] at … ° to the vertical [3] (c) After bouncing on the plate, the ball rises to a vertical height of 0.98 m. (i) Calculate the vertical component of the velocity of the ball as it leaves the plate. vertical velocity = … m s–1 [2] (ii) The ball of mass 34 g is in contact with the plate for a time of 0.12 s. For Examiner’s Use your answer in (c)(i) and the data in (a)(ii) to calculate, for the ball as it bounces Use on the plate, 1. the change in momentum, change = … kg m s–1 [3] 2. the magnitude of the average force exerted by the plate on the ball due to this momentum change. force = … N [2]

12 marks

Mark scheme: 3 (a) (i) speed = 4.0 m s-1 …(allow 1 s.f.) … A1 [1] (ii) v2 = 2gh = 2 × 9.8 × 1.96 … M1 v = 6.2 m s-1 … A0 [1] (use of g = 10 m s-2 loses the mark) (b) correct basic shape with correct directions for vectors … M1 speed = (7.4 ± 0.2) m s-1 … A1 at (33 ± 2)° to the vertical … A1 [3] (for credit to be awarded, speed and angle must be correct on the diagram – not calculated) GCE A/AS LEVEL – October/November 2009 9702 22 (c) (i) either v2 = 2 × 9.8 × 0.98 or v = 6.2 / √2 … C1 speed = 4.4 m s-1 … A1 [2] (allow calculation of t = 0.447 s, then v = 4.4 m s-1) (ii) 1 momentum = mv … C1 change in momentum = 0.034 (6.2 + 4.4) … C1 = 0.36 kg m s-1 … A1 [3] (use of 0.034 (6.2 - 4.4) loses last two marks) 2 force = ∆p / ∆t …….(however expressed) … C1 0.36 = 0.12 = 3.0 N ……(allow 1 s.f.) … A1 [2] [Total: 12]

This question in 9702/22 Oct/Nov 2009

2 (a) Complete Fig. 2.1 to show whether each of the quantities listed is a vector or a scalar. For Examiner’s Use vector / scalar distance moved … speed … acceleration … Fig. 2.1 [3] (b) A ball falls vertically in air from rest. The variation with time t of the distance d moved by the ball is shown in Fig. 2.2. 5 4 d /m 3 2 1 0 0 0.2 0.4 0.6 0.8 1.0 1.2 t /s Fig. 2.2 (i) By reference to Fig. 2.2, explain how it can be deduced that For Examiner’s 1. the ball is initially at rest, Use … … … [2] 2. air resistance is not negligible. … … [1] (ii) Use Fig. 2.2 to determine the speed of the ball at a time of 0.40 s after it has been released. speed = … m s–1 [2] (iii) On Fig. 2.2, sketch a graph to show the variation with time t of the distance d moved by the ball for negligible air resistance. You are not expected to carry out any further calculations. [3]

11 marks

Mark scheme: 2 (a) scalar …………………………………………………………..………………………… B1 scalar …………………………………………………………..………………………… B1 vector …………………………………………………………..………………………… B1 [3] (b) (i) 1 gradient (of graph) is the speed/velocity (can be scored here or in 2)………. B1 initial gradient is zero …………………………………………………………… B1 [2] 2 gradient (of line/graph) becomes constant ……….……..…………………… B1 [1] (ii) speed = (2.8 ± 0.1) m s–1 ……… ………………………………………………… A2 [2] (if answer > ±0.1 but ≤ ±0.2, then award 1 mark) (iii) curved line never below given line and starts from zero …..………………….. B1 continuous curve with increasing gradient …………………..…………………. B1 line never vertical or straight ………………………………..……………………. B1 [3]

This question in 9702/21 May/June 2010

2 (a) State the two conditions that must be satisfied for a body to be in equilibrium. For Examiner’s 1. … Use … 2. … … [2] (b) Three co-planar forces act on a body that is in equilibrium. (i) Describe how to draw a vector triangle to represent these forces. … … … … … … … [3] (ii) State how the triangle confirms that the forces are in equilibrium. … … [1] (c) A weight of 7.0 N hangs vertically by two strings AB and AC, as shown in Fig. 2.1. For Examiner’s Use B C T1 35° T2 50° A 7.0 N Fig. 2.1 For the weight to be in equilibrium, the tension in string AB is T1 and in string AC it is T2. On Fig. 2.1, draw a vector triangle to determine the magnitudes of T1 and T2. T1 = … N T2 = … N [3] (d) By reference to Fig. 2.1, suggest why the weight could not be supported with the strings AB and AC both horizontal. … … [2]

11 marks

Mark scheme: 2 (a) no resultant force/sum of forces zero B1 no resultant moment/torque/sum of moments/torques zero B1 [2] (b) (i) each force is represented by the side of a triangle/by an arrow M1 in magnitude and direction A1 arrows joined, head to tail B1 [3] (could be shown on a sketch diagram) (ii) if the triangle is ‘closed’ (then the forces are in equilibrium) B1 [1] (c) triangle drawn with correct shape (incorrect arrows loses this mark) B1 T1 = 5.4 ± 0.2 N B1 T2 = 4.0 ± 0.2 N B1 [3] (d) forces in strings would be horizontal B1 (so) no vertical force to support the weight B1 [2]

This question in 9702/23 May/June 2010

1 (a) (i) Distinguish between vector quantities and scalar quantities. … … … [2] (ii) State whether each of the following is a vector quantity or a scalar quantity. 1. temperature … [1] 2. acceleration of free fall … [1] 3. electrical resistance … [1] (b) A block of wood of weight 25 N is held stationary on a slope by means of a string, as For shown in Fig. 1.1. Examiner’s Use string T R 35° slope 25 N Fig. 1.1 The tension in the string is T and the slope pushes on the block with a force R that is normal to the slope. Either by scale drawing on Fig. 1.1 or by calculation, determine the tension T in the string. T = … N [3]

8 marks

Mark scheme: 1 (a) (i) scalar quantity has magnitude (allow size) B1 vector quantity has magnitude and direction B1 [2] (ii) 1. temperature: scalar B1 [1] 2. acceleration: vector B1 [1] 3. resistance: scalar B1 [1] (b) either triangle / parallelogram with correct shape C1 tension = 14 .3 N (allow ± 0.5 N) A2 [3] (if > ±0.5 N but ≤ ±1 N, allow 1 mark) or R = 25 cos 35° (C1) T = R tan 35° (C1) T = 14.3 N (A1) or T = 25 sin 35° (C2) T = 14.3 N (A1) or R and T resolved vertically and horizontally (C2) leading to T = 14.3 N (A1) 1

This question in 9702/22 Oct/Nov 2010

1 (a) Distinguish between scalar quantities and vector quantities. … … … [2] (b) In the following list, underline all the scalar quantities. acceleration force kinetic energy mass power weight [1] (c) A stone is thrown with a horizontal velocity of 20 m s–1 from the top of a cliff 15 m high. The path of the stone is shown in Fig. 1.1. 20 m s–1 cliff 15 m ground Fig. 1.1 Air resistance is negligible. For this stone, (i) calculate the time to fall 15 m, time = … s [2] (ii) calculate the magnitude of the resultant velocity after falling 15 m, resultant velocity = … m s–1 [3] (iii) describe the difference between the displacement of the stone and the distance For that it travels. Examiner’s Use … … … [2]

10 marks

Mark scheme: 1 (a) scalar has only magnitude B1 vector has magnitude and direction B1 [2] (b) kinetic energy, mass, power all three underlined B1 [1] (c) (i) s = ut + ½ at2 15 = 0.5 × 9.81 × t2 C1 T = 1.7 s A1 [2] if g = 10 is used then –1 but only once on paper (ii) vertical component vv: vv2 = u2 + 2as = 0 + 2 × 9.81 × 15 or vv = u + at = 9.81 × 1.7(5) vv =17.16 C1 resultant velocity: v2 = (17.16)2 + (20)2 C1 v = 26 m s–1 A1 [3] If u = 20 is used instead of u = 0 then 0/3 Allow the solution using: initial (potential energy + kinetic energy) = final kinetic energy (iii) distance is the actual path travelled B1 displacement is the straight line distance between start and finish points (in that direction) / minimum distance B1 [2]

This question in 9702/22 May/June 2011

3 A ball is thrown against a vertical wall. The path of the ball is shown in Fig. 3.1. For Examiner’s Use P 15.0 m s–1 wall 60.0° S F 6.15 m 9.95 m Fig. 3.1 (not to scale) The ball is thrown from S with an initial velocity of 15.0 m s–1 at 60.0° to the horizontal. Assume that air resistance is negligible. (a) For the ball at S, calculate (i) its horizontal component of velocity, horizontal component of velocity = … m s–1 [1] (ii) its vertical component of velocity. vertical component of velocity = … m s–1 [1] (b) The horizontal distance from S to the wall is 9.95 m. The ball hits the wall at P with a velocity that is at right angles to the wall. The ball rebounds to a point F that is 6.15 m from the wall. Using your answers in (a), (i) calculate the vertical height gained by the ball when it travels from S to P, height = … m [1] (ii) show that the time taken for the ball to travel from S to P is 1.33 s, For Examiner’s Use [1] (iii) show that the velocity of the ball immediately after rebounding from the wall is about 4.6 m s–1. [1] (c) The mass of the ball is 60 × 10–3 kg. (i) Calculate the change in momentum of the ball as it rebounds from the wall. change in momentum = … N s [2] (ii) State and explain whether the collision is elastic or inelastic. … … … [1]

8 marks

Mark scheme: 3 (a) (i) horizontal velocity = 15 cos 60° = 7.5 m s–1 A1 [1] (ii) vertical velocity = 15 sin 60° = 13 m s–1 A1 [1] (b) (i) v2 = u2 + 2as s = (13)2 / (2 × 9.81) = 8.6(1) m A1 [1] using g = 10 then max. 1 (ii) t = 13 / 9.81 = 1.326 s or t = 9.95 / 7.5 = 1.327 s A1 [1] (iii) velocity = 6.15 / 1.33 M1 = 4.6 m s–1 A0 [1] (c) (i) change in momentum = 60 × 10–3 [–4.6 – 7.5] C1 = (–)0.73 N s A1 [2] (ii) final velocity / kinetic energy is less after the collision or relative speed of separation < relative speed of approach M1 hence inelastic A0 [1] GCE AS/A LEVEL – October/November 2011 9702 21

This question in 9702/21 Oct/Nov 2011

1 (a) Distinguish between scalars and vectors. … … [1] (b) Underline all the vector quantities in the list below. acceleration kinetic energy momentum power weight [2] (c) A force of 7.5 N acts at 40° to the horizontal, as shown in Fig. 1.1. 7.5 N 40° horizontal Fig. 1.1 Calculate the component of the force that acts (i) horizontally, horizontal component = … N [1] (ii) vertically. vertical component = … N [1] (d) Two strings support a load of weight 7.5 N, as shown in Fig. 1.2. For Examiner’s Use T1 T2 50° 40° horizontal 7.5N Fig. 1.2 One string has a tension T1 and is at an angle 50° to the horizontal. The other string has a tension T2 and is at an angle 40° to the horizontal. The object is in equilibrium. Determine the values of T1 and T2 by using a vector triangle or by resolving forces. T1 = … N T2 = … N [4]

9 marks

Mark scheme: 1 (a) scalar has magnitude/size, vector has magnitude/size and direction B1 [1] (b) acceleration, momentum, weight B2 [2] (–1 for each addition or omission but stop at zero) (c) (i) horizontally: 7.5 cos 40° / 7.5 sin 50° = 5.7(45) / 5.75 not 5.8 N A1 [1] (ii) vertically: 7.5 sin 40° / 7.5 cos 50° = 4.8(2) N A1 [1] (d) either correct shaped triangle M1 correct labelling of two forces, three arrows and two angles A1 or correct resolving: T2 cos 40° = T1 cos 50° (B1) T1 sin 50° + T2 sin 40° = 7.5 (B1) T1 = 5.7(45) (N) A1 T2 = 4.8 (N) A1 [4] (allow ± 0.2 N for scale diagram)

This question in 9702/23 Oct/Nov 2011

1 (a) Explain the differences between the quantities distance and displacement. Use … … … [2] (b) State Newton’s first law. … … … [1] (c) Two tugs pull a tanker at constant velocity in the direction XY, as represented in Fig. 1.1. tug 1 T1 X 25.0° tanker Y 15.0° T2 tug 2 Fig. 1.1 Tug 1 pulls the tanker with a force T1 at 25.0° to XY. Tug 2 pulls the tanker with a force of T2 at 15.0° to XY. The resultant force R due to the two tugs is 25.0 × 103 N in the direction XY. (i) By reference to the forces acting on the tanker, explain how the tanker may be described as being in equilibrium. … … … … [2] (ii) 1. Complete Fig. 1.2 to draw a vector triangle for the forces R, T1 and T2. [2] For Examiner’s Use R 25.0 × 103 N Fig. 1.2 2. Use your vector triangle in Fig. 1.2 to determine the magnitude of T1 and of T2. T1 = … N T2 = … N [2]

9 marks

Mark scheme: 1 (a) displacement is a vector, distance is a scalar B1 displacement is straight line between two points / distance is sum of lengths moved / example showing difference B1 [2] (either one of the definitions for the second mark) (b) a body continues at rest or at constant velocity unless acted on by a resultant (external) force B1 [1] (c) (i) sum of T1 and T2 equals frictional force B1 these two forces are in opposite directions B1 [2] (allow for 1/2 for travelling in straight line hence no rotation / no resultant torque) (ii) 1. scale vector triangle with correct orientation / vector triangle with correct orientation both with arrows B1 scale given or mathematical analysis for tensions B1 [2] 2. T1 = 10.1 × 103 (± 0.5 × 103) N A1 T2 = 16.4 × 103 (± 0.5 × 103) N A1 [2]

This question in 9702/23 May/June 2012

1 (a) The spacing between two atoms in a crystal is 3.8 × 10–10 m. State this distance in pm. spacing = … pm [1] (b) Calculate the time of one day in Ms. time = … Ms [1] (c) The distance from the Earth to the Sun is 0.15 Tm. Calculate the time in minutes for light to travel from the Sun to the Earth. time = … min [2] (d) Underline all the vector quantities in the list below. distance energy momentum weight work [1] (e) The velocity vector diagram for an aircraft heading due north is shown to scale in For Fig. 1.1. There is a wind blowing from the north-west. Examiner’s Use wind 45° aircraft Fig. 1.1 The speed of the wind is 36 m s–1 and the speed of the aircraft is 250 m s–1. (i) Draw an arrow on Fig. 1.1 to show the direction of the resultant velocity of the aircraft. [1] (ii) Determine the magnitude of the resultant velocity of the aircraft. resultant velocity = … m s–1 [2]

8 marks

Mark scheme: 1 (a) spacing = 380 or 3.8 × 102 pm B1 [1] (b) time = 24 × 3600 time = 0.086 (0.0864) Ms B1 [1] 1.5 × 10 11 (c) time = distance / speed = C1 3 × 10 8 = 500 (s) = 8.3 min A1 [2] (d) momentum and weight B1 [1] (e) (i) arrow to the right of plane direction (about 4° to 24°) B1 [1] (ii) scale diagram drawn or use of cosine formula v2 = 2502 + 362 – 2 × 250 × 36 × cos 45° or resolving v = [(36 cos 45°)2 + (250 – 36 sin 45°)2]1/2 C1 resultant velocity = 226 (220 – 240 for scale diagram) m s–1 allow one mark for values 210 to 219 or 241 to 250 m s–1 or use of formula (v2 = 51068) v = 230 (226) m s–1 A1 [2]

This question in 9702/23 Oct/Nov 2012

2 (a) A student walks from A to B along the path shown in Fig. 2.1. For Examiner’s Use A B Fig. 2.1 The student takes time t to walk from A to B. (i) State the quantity, apart from t, that must be measured in order to determine the average value of 1. speed, … … [1] 2. velocity. … … [1] (ii) Define acceleration. … [1] (b) A girl falls vertically onto a trampoline, as shown in Fig. 2.2. For Examiner’s Use springy material Fig. 2.2 The trampoline consists of a central section supported by springy material. At time t = 0 the girl starts to fall. The girl hits the trampoline and rebounds vertically. The variation with time t of velocity v of the girl is illustrated in Fig. 2.3. 10.0 8.0 6.0 v / m s–1 4.0 2.0 0 0 0.5 1.0 1.5 2.0 t / s – 2.0 – 4.0 rebound – 6.0 time – 8.0 Fig. 2.3 For the motion of the girl, calculate (i) the distance fallen between time t = 0 and when she hits the trampoline, distance = … m [2] (ii) the average acceleration during the rebound. For Examiner’s Use acceleration = … m s–2 [2] (c) (i) Use Fig. 2.3 to compare, without calculation, the accelerations of the girl before and after the rebound. Explain your answer. … … … [2] (ii) Use Fig. 2.3 to compare, without calculation, the potential energy of the girl at t = 0 and t = 1.85 s. Explain your answer. … … … [2]

11 marks

Mark scheme: 2 (a) (i) 1. distance of path / along line AB B1 [1] 2. shortest distance between AB / distance in straight line between AB or displacement from A to B B1 [1] (ii) acceleration = rate of change of velocity A1 [1] (b) (i) distance = area under line or (v/2)t or s = (8.8)2 / (2 × 9.81) C1 = 8.8 / 2 × 0.90 = 3.96 m or s = 3.95 m = 4(.0) m A1 [2] (ii) acceleration = (– 4.4 – 8.8) / 0.50 C1 = (–) 26(.4) m s–2 A1 [2] (c) (i) the accelerations are constant as straight lines B1 the accelerations are the same as same gradient or no air resistance as acceleration is constant or change of speed in opposite directions (one speeds up one slows down) B1 [2] (ii) area under the lines represents height or KE at trampoline equals PE at maximum height B1 second area is smaller / velocity after rebound smaller hence KE less B1 hence less height means loss in potential energy A0 [2]

This question in 9702/23 May/June 2013

1 (a) (i) Define velocity. … … [1] (ii) Distinguish between speed and velocity. … … [2] (b) A car of mass 1500 kg moves along a straight, horizontal road. The variation with time t of the velocity v for the car is shown in Fig. 1.1. 40 30 v / m s–1 20 10 0 0 1.0 2.0 3.0 4.0 5.0 6.0 t / s Fig. 1.1 The brakes of the car are applied from t = 1.0 s to t = 3.5 s. For the time when the brakes are applied, (i) calculate the distance moved by the car, distance = … m [3] (ii) calculate the magnitude of the resultant force on the car. resultant force = … N [3] (c) The direction of motion of the car in (b) at time t = 2.0 s is shown in Fig. 1.2. direction of motion Fig. 1.2 On Fig. 1.2, show with arrows the directions of the acceleration (label this arrow A) and the resultant force (label this arrow F). [1]

10 marks

Mark scheme: 1 (a) (i) either rate of change of displacement or (change in) displacement / time (taken) B1 [1] (ii) speed has magnitude only B1 velocity has magnitude and direction B1 [2] (u + v ) (b) (i) idea of area under graph / use of s = × t C1 2 (18 + 32) s = × 2.5 C1 2 = 62.5 m A1 [3] (ii) a = (18 – 32) / 2.5 (= –5.6) C1 F = ma C1 F = 1500 × (–) 5.6 = (–) 8400 N A1 [3] (c) arrow labelled A and arrow labelled F both to the left B1 [1]

This question in 9702/21 May/June 2014

2 (a) Explain what is meant by a scalar quantity and by a vector quantity. scalar: … … vector: … … [2] (b) A ball leaves point P at the top of a cliff with a horizontal velocity of 15 m s–1, as shown in Fig. 2.1. ball P 15 m s–1 path of ball 25 m cliff Q ground Fig. 2.1 The height of the cliff is 25 m. The ball hits the ground at point Q. Air resistance is negligible. (i) Calculate the vertical velocity of the ball just before it makes impact with the ground at Q. vertical velocity = … m s–1 [2] (ii) Show that the time taken for the ball to fall to the ground is 2.3 s. [1] (iii) Calculate the magnitude of the displacement of the ball at point Q from point P. displacement = … m [4] (iv) Explain why the distance travelled by the ball is different from the magnitude of the displacement of the ball. … … … [2]

11 marks

Mark scheme: 2 (a) scalar has magnitude only B1 vector has magnitude and direction B1 [2] 1 (b) (i) v2 = 0 + 2 × 9.81 × 25 (or using m v2 = mgh) C1 2 v = 22(.1) m s–1 A1 [2] 1 (ii) 22.1 = 0 + 9.81 × t (or 25 = × 9.81 × t 2) M1 2 t (=22.1 / 9.81) = 2.26 s or t [=(5.097)1/2] = 2.26 s A0 [1] (iii) horizontal distance = 15 × t = 15 × 2.257 = 33.86 (allow 15 × 2.3 = 34.5) C1 (displacement)2 = (horizontal distance)2 + (vertical distance)2 C1 = (25)2 + (33.86)2 C1 displacement = 42 (42.08) m (allow 43 (42.6) m, allow 2 or more s.f.) A1 [4] (iv) distance is the actual (curved) path followed by ball B1 displacement is the straight line / minimum distance P to Q B1 [2]

This question in 9702/23 May/June 2014

3 (a) Force is a vector quantity. State three other vector quantities. 1. … 2. … 3. … [2] (b) Three coplanar forces X, Y and Z act on an object, as shown in Fig. 3.1. Y object θ X Z Fig. 3.1 The force Z is vertical and X is horizontal. The force Y is at an angle θ to the horizontal. The force Z is kept constant at 70 N. In an experiment, the magnitude of force X is varied. The magnitude and direction of force Y are adjusted so that the object remains in equilibrium. Fig. 3.2 shows the variation of the magnitude of force Y with the magnitude of force X. 130 Y / N 110 90 70 50 0 20 40 60 80 100 120 X / N Fig. 3.2 (i) Use Fig. 3.2 to estimate the magnitude of Y for X = 0. Y = … N [1] (ii) State and explain the value of θ for X = 0. … … … [2] (iii) The magnitude of X is increased to 160 N. Use resolution of forces to calculate the value of 1. angle θ, θ = … ° [2] 2. the magnitude of force Y. Y = … N [2] (c) The angle θ decreases as X increases. Explain why the object cannot be in equilibrium for θ = 0. … … … [1]

10 marks

Mark scheme: 3 (a) displacement / velocity / acceleration / momentum / etc. three correct (none wrong) 2, two correct (none or one wrong) 1 A2 [2] (b) (i) Y = 70 N [allow 71 N as +½ small square on graph] A1 [1] (ii) θ = 90° M1 (for equilibrium) the direction of Y must be opposite to Z or using Y sin θ = Z, hence sin θ = 70 / 70 = 1, θ = 90° A1 [2] (iii) 1. Y cos θ = 160 and Y sin θ = 70 C1 tan θ = 70 / 160 hence θ = 23.6° (24°) A1 [2] 2. Y = 160 / cos 23.6° or 70 / sin 23.6° C1 = 174.6 or 175 or 170 N A1 [2] or: 1602 + 702 = Y2 (C1) Y = 174.6 or 175 or 170 N (A1) (c) (equilibrium not possible as) there is no vertical component from Y to balance Z B1 [1]

This question in 9702/23 Oct/Nov 2014

2 (a) Define speed and velocity and use these definitions to explain why one of these quantities is a scalar and the other is a vector. speed: … velocity: … … … [2] (b) A ball is released from rest and falls vertically. The ball hits the ground and rebounds vertically, as shown in Fig. 2.1. initial position ball rebound ground Fig. 2.1 The variation with time t of the velocity v of the ball is shown in Fig. 2.2. 12.0 10.0 8.0 v / m s–1 6.0 4.0 2.0 0 0 1.0 2.0 3.0 t / s – 2.0 – 4.0 – 6.0 – 8.0 – 10.0 Fig. 2.2 Air resistance is negligible. (i) Without calculation, use Fig. 2.2 to describe the variation with time t of the velocity of the ball from t = 0 to t = 2.1 s. … … … … … … [3] (ii) Calculate the acceleration of the ball after it rebounds from the ground. Show your working. acceleration = … m s–2 [3] (iii) Calculate, for the ball, from t = 0 to t = 2.1 s, 1. the distance moved, distance = … m [3] 2. the displacement from the initial position. displacement = … m [2] (iv) On Fig. 2.3, sketch the variation with t of the speed of the ball. 12.0 10.0 8.0 speed / m s–1 6.0 4.0 2.0 0 0 1.0 2.0 3.0 t / s – 2.0 – 4.0 – 6.0 – 8.0 – 10.0 Fig. 2.3 [2]

15 marks

Mark scheme: 2 (a) speed = distance / time and velocity = displacement / time B1 speed is a scalar as distance has no direction and velocity is a vector as displacement has direction B1 [2] (b) (i) constant acceleration or linear/uniform increase in velocity until 1.1 s B1 rebounds or bounces or changes direction B1 decelerates to zero velocity at the same acceleration as initial value B1 [3] (ii) a = (v – u) / t or use of gradient implied C1 = (8.8 + 8.8) / 1.8 or appropriate values from line or = (8.6 + 8.6) / 1.8 B1 = 9.8 (9.78) m s–2 or = 9.6 m s–2 A1 [3] (iii) 1. distance = first area above graph + second area below graph C1 = (1.1 × 10.8) / 2 + (0.9 × 8.8) / 2 (= 5.94 + 3.96) C1 = 9.9 m A1 [3] 2. displacement = first area above graph – second area below graph C1 = (1.1 × 10.8) / 2 – (0.9 × 8.8) / 2 = 2.0 (1.98) m A1 [2] (iv) correct shape with straight lines and all lines above the time axis or all below M1 correct times for zero speeds (0.0, 1.15 s, 2.1 s) and peak speeds (10.8 m s–1 at 1.1 s and 8.8 m s–1 at 1.2 s and 3.0 s) A1 [2]

This question in 9702/21 May/June 2015

1 (a) The distance between the Sun and the Earth is 1.5 × 1011 m. State this distance in Gm. distance = … Gm [1] (b) The distance from the centre of the Earth to a satellite above the equator is 42.3 Mm. The radius of the Earth is 6380 km. A microwave signal is sent from a point on the Earth directly below the satellite. Calculate the time taken for the microwave signal to travel to the satellite and back. time = … s [2] (c) The speed v of a sound wave through a gas of density ρ and pressure P is given by CP v = ρ where C is a constant. Show that C has no unit. [3] (d) Underline all the scalar quantities in the list below. acceleration energy momentum power weight [1] (e) A boat travels across a river in which the water is moving at a speed of 1.8 m s–1. The velocity vectors for the boat and the river water are shown to scale in Fig. 1.1. water velocity 1.8 m s–1 river boat velocity 3.0 m s–1 60° river bank Fig. 1.1 (shown to scale) In still water the speed of the boat is 3.0 m s–1. The boat is directed at an angle of 60° to the river bank. (i) On Fig. 1.1, draw a vector triangle or a scale diagram to show the resultant velocity of the boat. [2] (ii) Determine the magnitude of the resultant velocity of the boat. resultant velocity = … m s–1 [2]

11 marks

Mark scheme: 1 (a) 150 or 1.5 × 102 Gm A1 [1] (b) distance = 2 × (42.3 – 6.38) × 106 (= 7.184 × 107 m) C1 (time =) 7.184 × 107 / (3.0 × 108) = 0.24 (0.239) s A1 [2] (c) units of pressure P: kg m s–2 / m2 = kg m–1 s–2 M1 units of density ρ: kg m–3 and speed v: m s–1 M1 simplification for units of C: C = v2 ρ / P units: (m2 s–2 kg m–3) / kg m–1 s–2 and cancelling to give no units for C A1 [3] (d) energy and power (both underlined and no others) A1 [1] (e) (i) vector triangle of correct orientation M1 three arrows for the velocities in the correct directions A1 [2] (ii) length measured from scale diagram 5.2 ± 0.2 cm or components of boat speed determined parallel and perpendicular to river flow C1 velocity = 2.6 m s–1 (allow ± 0.1 m s–1) A1 [2]

This question in 9702/23 May/June 2015

1 (a) The frequency of an X-ray wave is 4.6 × 1020 Hz. Calculate the wavelength in pm. wavelength = … pm [3] (b) The distance from Earth to a star is 8.5 × 1016 m. Calculate the time for light to travel from the star to Earth in Gs. time = … Gs [2] (c) The following list contains scalar and vector quantities. Underline all the scalar quantities. acceleration force mass power temperature weight [1] (d) A boat is travelling in a flowing river. Fig. 1.1 shows the velocity vectors for the boat and the river water. water velocity 8.0 m s–1 boat velocity 14.0 m s–1 60° east Fig. 1.1 The velocity of the boat in still water is 14.0 m s–1 to the east. The velocity of the water is 8.0 m s–1 from 60° north of east. (i) On Fig. 1.1, draw an arrow to show the direction of the resultant velocity of the boat. [1] (ii) Determine the magnitude of the resultant velocity of the boat. magnitude of velocity = … m s–1 [2]

9 marks

Mark scheme: 1 (a) v = fλ C1 λ = (3.0 × 108) / (4.6 × 1020) C1 ( = 6.52 × 10–13 =) 0.65(2) pm A1 [3] (b) t = (8.5 × 1016) / (3.0 × 108) C1 ( = 2.83 × 108 =) 0.28(3) Gs A1 [2] (c) mass, power and temperature all underlined and no others B1 [1] (d) (i) arrow in the direction 30° to 40° south of east B1 [1] (ii) triangle of velocities completed (i.e. correct scale diagram) or correct working given C1 e.g. [142 + 8.02 – 2(14)(8.0) cos 60°]1/2 or [(14 – 8.0 cos 60°)2 + (8.0 sin 60°)2]1/2 resultant velocity = 12(.2) (or 12.0 to 12.4 from scale diagram) m s–1 A1 [2]

This question in 9702/22 Oct/Nov 2015

2 A ball is thrown from a point P with an initial velocity u of 12 m s–1 at 50° to the horizontal, as illustrated in Fig. 2.1. path of ball Q X =12 m s–1 50° P horizontal Fig. 2.1 The ball reaches maximum height at Q. Air resistance is negligible. (a) Calculate (i) the horizontal component of u, horizontal component = … m s–1 [1] (ii) the vertical component of u. vertical component = … m s–1 [1] (b) Show that the maximum height reached by the ball is 4.3 m. [2] (c) Determine the magnitude of the displacement PQ. displacement = … m [4] [Total: 8]

8 marks

Mark scheme: 2 (a) (i) horizontal component (= 12 cos 50°) = 7.7 m s–1 A1 [1] (ii) vertical component (= 12 sin 50° or 7.7 tan 50°) = 9.2 m s–1 A1 [1] (b) v2 = u2 + 2as and v = 0 or mgh = ½mv2 or s = v2 sin2θ / 2g C1 9.22 = 2 × 9.81 × h hence h = 4.3 (4.31) m A1 [2] alternative methods using time to maximum height of 0.94 s: s = ut + ½at2 and t = 0.94 (s) (C1) s = 9.2 × 0.94 – ½ × 9.81 × 0.942 hence s = 4.3 m (A1) or s = vt – ½at2 and t = 0.94 (s) (C1) s = ½ × 9.81 × 0.942 hence s = 4.3 m (A1) or s = ½(u + v)t and t = 0.94 (s) (C1) s = ½ × 9.2 × 0.94 hence s = 4.3 m (A1) (c) t (= 9.2 / 9.81) = 0.94 (0.938) s C1 horizontal distance = 0.938 × 7.7 (= 7.23 m) C1 displacement = [4.32 + 7.232]1/2 C1 = 8.4 m A1 [4]

This question in 9702/21 May/June 2016

1 (a) A list of quantities that are either scalars or vectors is shown in Fig. 1.1. quantity scalar vector distance ✓ energy momentum power time weight Fig. 1.1 Complete Fig. 1.1 to indicate whether each quantity is a scalar or a vector. One line has been completed as an example. [2] (b) A girl runs 120 m due north in 15 s. She then runs 80 m due east in 12 s. (i) Sketch a vector diagram to show the path taken by the girl. Draw and label her resultant displacement R. north east [1] (ii) Calculate, for the girl, 1. the average speed, average speed = … m s–1 [1] 2. the magnitude of the average velocity v and its angle with respect to the direction of the initial path. magnitude of v = … m s–1 angle = … ° [3] [Total: 7]

7 marks

Mark scheme: 1 (a) scalars: energy, power and time A1 vectors: momentum and weight A1 [2] (b) (i) triangle with right angles between 120 m and 80 m, arrows in correct direction and result displacement from start to finish arrow in correct direction and labelled R B1 [1] (ii) 1. average speed (= 200 / 27) = 7.4 m s–1 A1 [1] 2. resultant displacement (= [1202 + 802]1/2) = 144 (m) C1 average velocity (= 144 / 27) = 5.3(3) m s–1 A1 direction (= tan–1 80 / 120) = 34° (33.7) A1 [3]

This question in 9702/23 May/June 2016

1 (a) Complete Fig. 1.1 by putting a tick (3) in the appropriate column to indicate whether the listed quantities are scalars or vectors. quantity scalar vector acceleration force kinetic energy momentum power work Fig. 1.1 [2] (b) A floating sphere is attached by a cable to the bottom of a river, as shown in Fig. 1.2. solid sphere water surface direction of flow of water cable river bed 75° Fig. 1.2 The sphere is in equilibrium, with the cable at an angle of 75° to the horizontal. Assume that the force on the sphere due to the water flow is in the horizontal direction. The radius of the sphere is 23 cm. The sphere is solid and is made from a material of density 82 kg m–3. (i) Show that the weight of the sphere is 41 N. [2] (ii) The tension in the cable is 290 N. Determine the upthrust acting on the sphere. upthrust = … N [2] (iii) Explain the origin of the upthrust acting on the sphere. … … … [1] [Total: 7]

7 marks

Mark scheme: 1(a) scalars: kinetic energy, power, work A1 vectors: acceleration, force, momentum A1 1(b)(i) mass = volume × density or m = V × ρ = 4/3 π (23 × 10–2)3 × 82 C1 weight = 4/3 π (23 × 10–2)3 × 82 × 9.8 = 41 N A1 1(b)(ii) vertical component of tension = 290 sin75° or 290 cos15° (= 280) C1 upthrust = 290 sin75° + 41 = 320 (321) N A1 1(b)(iii) the water pressure is greater than the air pressure or the pressure on lower surface (of sphere) is greater than the pressure on upper surface (of sphere) B1

This question in 9702/22 Feb/March 2017

1 (a) Two forces, with magnitudes 5.0 N and 12 N, act from the same point on an object. Calculate the magnitude of the resultant force R for the forces acting (i) in opposite directions, R = … N [1] (ii) at right angles to each other. R = … N [1] (b) An object X rests on a smooth horizontal surface. Two horizontal forces act on X as shown in Fig. 1.1. 18 N 115° X 55 N Fig. 1.1 (not to scale) A force of 55 N is applied to the right. A force of 18 N is applied at an angle of 115° to the direction of the 55 N force. (i) Use the resolution of forces or a scale diagram to show that the magnitude of the resultant force acting on X is 65 N. [2] (ii) Determine the angle between the resultant force and the 55 N force. angle = … ° [2] (c) A third force of 80 N is now applied to X in the opposite direction to the resultant force in (b). The mass of X is 2.7 kg. Calculate the magnitude of the acceleration of X. acceleration = … m s–2 [3] [Total: 9]

9 marks

Mark scheme: 1(a)(i) R = 7(.0) N B1 1(a)(ii) R = 13 N B1 1(b)(i) forces resolved: 18 sin 65° (vertical) and 55 + 18 cos 65° (horizontal) or scale drawing: correct triangle drawn for forces B1 F = [(18 sin 65°)2 + (55 + 18 cos 65°)2]1/2 = 65 (64.7) N or scale drawing: scale given, length of resultant given correctly, ± 1 N A1 1(b)(ii) angle = tan–1 [18 sin 65° / (55 + 18 cos 65°)] = tan–1 (16.3 / 62.6) or scale drawing: correct angle measured/direction correct on diagram below the 55 N force C1 angle = 15 (14.6)° (below the 55 N force) or scale drawing: angle = 15° ± 1° A1 1(c) (resultant) force = mass × acceleration C1 80 − 65 = 2.7a C1 a = 5.6 m s–2 [5.7 if 64.7 N used from (i)] A1

This question in 9702/23 May/June 2017

1 (a) Complete Fig. 1.1 to indicate whether each of the quantities is a vector or a scalar. quantity vector or scalar acceleration speed power Fig. 1.1 [2] (b) A ball is projected with a horizontal velocity of 1.1 m s–1 from point A at the edge of a table, as shown in Fig. 1.2. table ball 1.1 m s–1 A path of ball B horizontal ground 0.43 m Fig. 1.2 The ball lands on horizontal ground at point B which is a distance of 0.43 m from the base of the table. Air resistance is negligible. (i) Calculate the time taken for the ball to fall from A to B. time = … s [1] (ii) Use your answer in (b)(i) to determine the height of the table. height = … m [2] (iii) The ball leaves the table at time t = 0. For the motion of the ball between A and B, sketch graphs on Fig. 1.3 to show the variation with time t of 1. the acceleration a of the ball, 2. the vertical component sv of the displacement of the ball from A. Numerical values are not required. a sv 0 0 0 t 0 t Fig. 1.3 [2] (c) A ball of greater mass is projected from the table with the same velocity as the ball in (b). Air resistance is still negligible. State and explain the effect, if any, of the increased mass on the time taken for the ball to fall to the ground. … … [1] [Total: 8]

8 marks

Mark scheme: 1(a) acceleration: vector speed: scalar power: scalar All three correct scores 2 marks. Only two correct scores 1 mark. B2 1(b)(i) time = 0.43 / 1.1 = 0.39 s A1 1(b)(ii) s = ut + ½at 2 = ½ × 9.81 × 0.392 C1 = 0.75 m A1 1(b)(iii) 1 horizontal line at a non-zero value of a. B1 2 curved line from origin with increasing gradient. B1 1(c) acceleration (of free fall) is unchanged / not dependent on mass and so no effect (on time taken). A1

This question in 9702/22 Feb/March 2018

1 (a) State what is meant by a scalar quantity and by a vector quantity. scalar: … … vector: … … [2] (b) Complete Fig. 1.1 to indicate whether each of the quantities is a vector or a scalar. quantity vector or scalar power temperature momentum Fig. 1.1 [2] (c) An aircraft is travelling in wind. Fig. 1.2 shows the velocities for the aircraft in still air and for the wind. west 65° aircraft velocity in still air 95 m s–1 wind velocity 28 m s–1 Fig. 1.2 The velocity of the aircraft in still air is 95 m s–1 to the west. The velocity of the wind is 28 m s–1 from 65° south of east. (i) On Fig. 1.2, draw an arrow, labelled R, in the direction of the resultant velocity of the aircraft. [1] (ii) Determine the magnitude of the resultant velocity of the aircraft. magnitude of velocity = … m s–1 [2] [Total: 7]

7 marks

Mark scheme: 1(a) a scalar has magnitude (only) B1 a vector has magnitude and direction B1 1(b) power: scalar temperature: scalar momentum: vector (two correct 1 mark, all three correct 2 marks) B2 1(c)(i) arrow labelled R in a direction from 5° to 20° north of west B1 1(c)(ii) v2 = 282 + 952 – (2 × 28 × 95 × cos 115°) or v2 = [(95 + 28 cos 65°)2 + (28 sin 65°)2] C1 v = 110 ms–1 (109.8 ms–1) A1 or (scale diagram method) triangle of velocities drawn (C1) v = 110 m s–1 (allow 108–112 m s–1) (A1)

This question in 9702/21 May/June 2018

1 A golfer strikes a ball so that it leaves horizontal ground with a velocity of 6.0 m s–1 at an angle θ to the horizontal, as illustrated in Fig. 1.1. vY 6.0 m s–1 4.8 m s–1 ball θ ground vX Fig. 1.1 (not to scale) The magnitude of the initial vertical component vY of the velocity is 4.8 m s–1. Assume that air resistance is negligible. (a) Show that the magnitude of the initial horizontal component vX of the velocity is 3.6 m s–1. [1] (b) The ball leaves the ground at time t = 0 and reaches its maximum height at t = 0.49 s. On Fig. 1.2, sketch separate lines to show the variation with time t, until the ball returns to the ground, of (i) the vertical component vY of the velocity (label this line Y), [2] (ii) the horizontal component vX of the velocity (label this line X). [2] 5.0 4.0 velocity / m s–1 3.0 2.0 1.0 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 t / s –1.0 –2.0 –3.0 –4.0 –5.0 Fig. 1.2 (c) Calculate the maximum height reached by the ball. maximum height = … m [2] (d) For the movement of the ball from the ground to its maximum height, determine the ratio kinetic energy at maximum height . change in gravitational potential energy ratio = … [4] (e) In practice, significant air resistance acts on the ball. Explain why the actual time taken for the ball to reach maximum height is less than the time calculated when air resistance is assumed to be negligible. … … … [1] [Total: 12]

12 marks

Mark scheme: 1(a) or 6.0 sinθ = 4.8 (so θ = 53.1°) and vx = 6.0 cos 53.1° = 3.6 (m s–1) A1 1(b)(i) straight line from (0, 4.8) to (0.49, 0) M1 straight line continues with same slope to (0.98, –4.8) (labelled Y) A1 1(b)(ii) a horizontal line M1 from (0, 3.6) to (0.98, 3.6) (labelled X) A1 1(c) s = ut + ½at2 = (4.8 × 0.49) + (½ × –9.81 × 0.492) or s = ½(u + v)t or area under graph = ½ × (4.8 + 0) × 0.49 or s = vt – ½at2 = ½ × 9.81 × 0.492 or v2 = u2 + 2as s = 4.82 / (2 × 9.81) C1 s = 1.2 m A1 Question Answer Marks 1(d) (∆)E = mg(∆)h C1 E = ½mv2 C1 ratio = (½ × m × 3.62) / (m × 9.81 × 1.2) or ratio = [(½ × m × 6.02) – (m × 9.81 × 1.2)] / (m × 9.81 × 1.2) or ratio = (½ × m × 3.62) / (½ × m × 4.82) C1 ratio = 0.56 A1 1(e) (force due to) air resistance acts in opposite direction to the velocity or (with air resistance, average) resultant force is larger (than weight) B1

This question in 9702/22 Oct/Nov 2018

1 (a) Distinguish between vector and scalar quantities. … … … [2] (b) The electric field strength E at a distance x from an isolated point charge Q is given by the equation Q E = x 2b where b is a constant. (i) Use the definition of electric field strength to show that E has SI base units of kg m A–1 s–3. [2] (ii) Use the units for E given in (b)(i) to determine the SI base units of b. SI base units of b … [2] [Total: 6]

6 marks

Mark scheme: 1(a) scalar quantity has (only) magnitude B1 vector quantity has magnitude and direction B1 1(b)(i) E = F / Q C1 = kg m s–2 / A s = kg m A–1 s–3 A1 1(b)(ii) b = Q / x 2E = A s / m2 kg m A–1 s–3 C1 = A2 s4 kg–1 m–3 A1

This question in 9702/22 Oct/Nov 2019

1 (a) State one similarity and one difference between distance and displacement. similarity: … … difference: … … [2] (b) A student takes several measurements of the same quantity. This set of measurements has high precision, but low accuracy. Describe what is meant by: (i) high precision … … [1] (ii) low accuracy. … … [1] [Total: 4]

4 marks

Mark scheme: 1(a) similarity: both have magnitude B1 difference: distance is a scalar/does not have direction or displacement is a vector/has direction B1 1(b)(i) the measurements have a small range B1 1(b)(ii) the (average of the) measurements is not close to the true value B1

This question in 9702/23 May/June 2020

1 (a) Complete Table 1.1 by putting a tick (3) in the appropriate column to indicate whether the listed quantities are scalars or vectors. Table 1.1 quantity scalar vector acceleration density temperature momentum [2] (b) A toy train moves along a straight section of track. Fig. 1.1 shows the variation with time t of the distance d moved by the train. 0.6 0.5 d / m 0.4 0.3 0.2 0.1 0 0 1 2 3 t / s Fig. 1.1 (i) Describe qualitatively the motion of the train between time t = 0 and time t = 1.0 s. … … [1] (ii) Determine the speed of the train at time t = 2.0 s. speed = … m s−1 [2] (c) The straight section of track in (b) is part of the loop of track shown in Fig. 1.2. track Fig. 1.2 The train completes exactly one lap of the loop. State and explain the average velocity of the train over the one complete lap. … … … [1] [Total: 6]

6 marks

Mark scheme: 1(a) density and temperature indicated as scalars B1 acceleration and momentum indicated as vectors B1 1(b)(i) decelerates or speed/velocity decreases B1 1(b)(ii) speed = (Δ)d / (Δ)t or gradient C1 = e.g. (0.56 – 0.20) / 1.5 = 0.24 m s–1 A1 1(c) displacement is zero (so) average velocity is zero B1

This question in 9702/22 Oct/Nov 2020

3 A ball is fired horizontally with a speed of 41.0 m s–1 from a stationary cannon at the top of a hill. The ball lands on horizontal ground that is a vertical distance of 57 m below the cannon, as shown in Fig. 3.1. ball, initial speed cannon 41.0 m s–1 path of ball 57 m horizontal ground Fig. 3.1 (not to scale) Assume air resistance is negligible. (a) Show that the time taken for the ball to reach the ground, after being fired, is 3.4 s. [2] (b) Calculate the horizontal distance of the ball from the cannon at the point where the ball lands on the ground. horizontal distance = … m [1] (c) Determine the magnitude of the displacement of the ball from the cannon at the point where the ball lands on the ground. displacement = … m [2] (d) The ball leaves the cannon at time t = 0. On Fig. 3.2, sketch a graph to show the variation of the magnitude v of the vertical component of the velocity of the ball with time t from t = 0 to t = 3.4 s. Numerical values are not required. v 0 0 3.4 t / s Fig. 3.2 [1] (e) The cannon recoils horizontally with a speed of 0.340 m s–1 when it fires the ball. The total mass of the ball and the cannon is 1480 kg. Assume that no external horizontal forces act on the ball-cannon system. Determine, to three significant figures, the mass of the ball. mass = … kg [2] (f) The cannon now fires a ball of smaller mass. Assume that air resistance is still negligible. State and explain the change, if any, to the graph in Fig. 3.2 due to the decreased mass of the ball. … … … [2] [Total: 10]

10 marks

Mark scheme: 3(a) s = ½at 2 C1 57 = ½ × 9.81 × t 2 and t = 3.4 (s) A1 3(b) horizontal distance = 41 × 3.4 = 140 m A1 3(c) (displacement)2 = 572 + 1402 C1 displacement = ( 572 + 1402)0.5 = 150 m A1 3(d) straight line from the origin with positive gradient B1 3(e) (1480 – m) × 0.340 = m × 41.0 C1 m = 12.2 kg A1 or mc 0.34 = mb 41 and mc + mb = 1480 (C1) mc = (41 / 0.34)mb (41 / 0.34)mb + mb = 1480 mb = 12.2 kg (A1) 3(f) acceleration (of free fall) is unchanged/is not dependent on mass M1 (so) no change (to the graph) A1

This question in 9702/23 Oct/Nov 2020

1 (a) Complete Table 1.1 by stating whether each of the quantities is a vector or a scalar. Table 1.1 quantity vector or scalar acceleration power work [2] (b) The variation with time t of the velocity v of an object is shown in Fig. 1.1. 1.50 1.25 1.00 v / m s–1 0.75 0.50 0.25 0 0 2.0 4.0 6.0 8.0 10.0 12.0 t / s Fig. 1.1 (i) Determine the acceleration of the object from time t = 0 to time t = 4.0 s. acceleration = … m s−2 [2] (ii) Determine the distance moved by the object from time t = 0 to time t = 4.0 s. distance = … m [2] (c) (i) Define force. … … [1] (ii) The motion represented in Fig. 1.1 is caused by a resultant force F acting on the object. On Fig. 1.2, sketch the variation of F with time t from t = 0 to t = 12.0 s. Numerical values of F are not required. F 0 00 2.02.0 4.04.0 6.06.0 8.08.0 10.010.0 12.012.0 tt // ss Fig. 1.2 [3] [Total: 10]

10 marks

Mark scheme: 1(a) acceleration: vector work: scalar power: scalar Three correct scores 2 marks. Two correct scores 1 mark. B2 1(b)(i) a = (v – u) / t or a = gradient or a = Δv / (Δ)t e.g. a = (1.40 – 0.70) / 4.0 C1 = 0.18 m s–2 A1 1(b)(ii) distance = 0.5 × (0.70 + 1.40) × 4.0 or (0.70 × 4.0) + (0.5 × 0.70 × 4.0) C1 = 4.2 m A1 1(c)(i) (force equal to) rate of change of momentum B1 1(c)(ii) horizontal line starting from t = 0 and ending at t = 4.0 s at a positive value of F B1 horizontal line starting from t = 4.0 s and ending at t = 8.0 s at F = 0 B1 horizontal line starting from t = 8.0 s and ending at t = 12.0 s at a negative value of F and the magnitude of F is larger than from t = 0 to 4.0 s B1

This question in 9702/22 Feb/March 2021

1 (a) Complete Table 1.1 by stating whether each of the quantities is a vector or a scalar. Table 1.1 quantity vector or scalar acceleration electrical resistance momentum [2] (b) State the conditions for an object to be in equilibrium. … … … … [2] (c) A floating solid cylinder is attached by a wire to the sea bed, as shown in Fig. 1.1. cylinder, cross-sectional weight 28 N area 0.0230 m2 surface of water 0.190 m water, wire density 1.00 × 103 kg m–3 sea bed Fig. 1.1 (not to scale) The density of the water is 1.00 × 103 kg m–3. The base of the cylinder is at a depth of 0.190 m below the surface of the water. The cylinder has a weight of 28 N and a cross-sectional area of 0.0230 m2. The wire and the central axis of the cylinder are both vertical. The cylinder is in equilibrium. (i) Calculate, to three significant figures, the upthrust acting on the cylinder due to the water. upthrust = … N [2] (ii) Show that the tension T in the wire is 15 N. [1] (iii) The wire has a cross-sectional area of 3.2 mm2. Calculate the stress in the wire. stress = … Pa [2] (iv) The surface of the water gradually rises until it is level with the top face of the cylinder. State and explain, qualitatively, the variation of the strain energy stored in the wire as the water surface rises. … … … … [2] [Total: 11]

11 marks

Mark scheme: 1(a) acceleration: vector electrical resistance: scalar momentum: vector 1 mark for two correct, 2 marks for all three correct B2 1(b) resultant force (in any direction) is zero B1 resultant torque/moment (about any point) is zero B1 1(c)(i) upthrust = ρ g (∆)h × A C1 = (1.00 × 103 × 9.81 × 0.190) × 0.0230 = 42.9 N A1 1(c)(ii) (T =) 43 – 28 = 15 (N) or (T =) 42.9 – 28 = 14.9 or 15 (N) A1 1(c)(iii) σ = F / A or T / A C1 = 15 / (3.2 × 10–6) = 4.7 × 106 Pa A1 1(c)(iv) upthrust (on cylinder) increases (and weight constant) B1 tension/stress increases and (so) strain energy increases B1

This question in 9702/22 May/June 2021

1 (a) A property of a vector quantity, that is not a property of a scalar quantity, is direction. For example, velocity has direction but speed does not. (i) State two other scalar quantities and two other vector quantities. scalar quantities: … and … vector quantities: … and … [2] (ii) State two properties that are possessed by both scalar and vector physical quantities. 1. … 2. … [2] (b) A ship at sea is travelling with a velocity of 13 m s–1 in a direction 35° east of north in still water, as shown in Fig. 1.1. N N velocity 13 m s–1 35° W E S Fig. 1.1 (i) Determine the magnitudes of the components of the velocity of the ship in the north and the east directions. north component of velocity = … m s–1 east component of velocity = … m s–1 [2] (ii) The ship now experiences a tidal current. The water in the sea moves with a velocity of 2.7 m s–1 to the west. Calculate the resultant velocity component of the ship in the east direction. resultant east component of velocity = … m s–1 [1] (iii) Use your answers in (b)(i) and (b)(ii) to determine the magnitude of the resultant velocity of the ship. magnitude of resultant velocity = … m s–1 [2] (iv) Use your answers in (b)(i) and (b)(ii) to determine the angle between north and the resultant velocity of the ship. angle = … ° [2] [Total: 11]

11 marks

Mark scheme: 1(a)(i) two correct scalar quantities e.g. time, mass, distance, temperature B1 two correct vector quantities e.g. force, acceleration, velocity, displacement B1 1(a)(ii) magnitude B1 unit B1 1(b)(i) north component of velocity = 11 m s–1 A1 east component of velocity = 7.5 m s–1 A1 1(b)(ii) velocity = 7.5 – 2.7 = 4.8 m s–1 A1 1(b)(iii) velocity = √(112 + 4.82) C1 = 12 m s–1 A1 1(b)(iv) angle = tan–1 (4.8 / 11) C1 = 24° A1 Question Answer Marks

This question in 9702/23 May/June 2021

3 (a) Define velocity. … … [1] (b) A remote-controlled toy aircraft is flying horizontally in a wind. Fig. 3.1 shows the velocity vectors, to scale, of the wind and of the aircraft in still air. north wind velocity 54° 23 m s–1 aircraft velocity in still air 42 m s–1 Fig. 3.1 The velocity of the aircraft in still air is 42 m s–1 to the north. The velocity of the wind is 23 m s–1 in a direction of 54° east of south. Determine the magnitude of the resultant velocity of the aircraft. magnitude of velocity = … m s–1 [2] (c) The engine of the aircraft in (b) stops. The aircraft then glides towards the ground with a constant velocity at an angle θ to the horizontal, as illustrated in Fig. 3.2. X aircraft, 280 m weight 46 N glide path θ horizontal of aircraft Y Fig. 3.2 (not to scale) The aircraft has a weight of 46 N and travels a distance of 280 m from point X to point Y. The change in gravitational potential energy of the aircraft for its movement from X to Y is 6100 J. Assume that there is now no wind. (i) Calculate angle θ. θ = … ° [3] (ii) Calculate the magnitude of the force acting on the aircraft due to air resistance. force = … N [2] (d) The aircraft in (c) travels from X to Y in a time of 14 s. Fig. 3.3 shows that, as the aircraft travels from X to Y, it moves directly towards an observer who is standing on the ground. 280 m X aircraft Y observer ground Fig. 3.3 (not to scale) The aircraft emits sound as it travels from X to Y. The observer hears sound of frequency 450 Hz. The speed of the sound in the air is 340 m s–1. Calculate the frequency of the sound that is emitted by the aircraft. frequency = … Hz [3] [Total: 11]

11 marks

Mark scheme: 3(a) change in displacement / time (taken) B1 3(b) by calculation: v 2 = 422 + 232 – (2 × 42 × 23 × cos 54°) or v 2 = (42 – 23 cos 54°)2 + (23 sin 54°)2 or v 2 = (42 – 23 sin 36°)2 + (23 cos 36°)2 C1 v = 34 m s–1 A1 or by scale diagram: triangle of vector velocities drawn (C1) v = 34 m s–1 (allow ± 1 m s–1 if scale diagram used) (A1) 3(c)(i) (Δ)E = mg(Δ)h or (Δ)E = W(Δ)h C1 h = 6100 / 46 (= 133 m) C1 θ = sin–1 (133 / 280) = 28° A1 3(c)(ii) force = 6100 / 280 or 46 sin 28° C1 = 22 N A1 3(d) v(s) = 280 / 14 (= 20 m s–1) C1 fo = fs v / (v – vs) fs = 450 × (340 – 20) / 340 C1 = 420 Hz A1

This question in 9702/23 Oct/Nov 2021

2 Water leaves the end of a hose pipe at point P with a horizontal velocity of 6.6 m s–1, as shown in Fig. 2.1. hose pipe P 6.6 m s–1 path of water h Q ground 3.5 m Fig. 2.1 (not to scale) Point P is at height h above the ground. The water hits the ground at point Q. The horizontal distance from P to Q is 3.5 m. Air resistance is negligible. Assume that the water between P and Q consists of non-interacting droplets of water and that the only force acting on each droplet is its weight. (a) Explain, briefly, why the horizontal component of the velocity of a droplet of water remains constant as it moves from P to Q. … … [1] (b) Show that the time taken for a droplet of water to move from P to Q is 0.53 s. [1] (c) Calculate height h. h = … m [2] (d) For the movement of a droplet of water from P to Q, state and explain whether the displacement of the droplet is less than, more than or the same as the distance along its path. … … … [1] (e) Calculate the magnitude of the displacement of a droplet of water that moves from P to Q. displacement = … m [2] [Total: 7]

7 marks

Mark scheme: 2(a) force (on droplet of water) in horizontal direction is zero. B1 2(b) (time taken =) 3.5 / 6.6 = 0.53 (s) A1 2(c) s = ut + ½at 2 s = ½ × 9.81 × 0.532 C1 h = 1.4 m A1 Question Answer Marks 2(d) displacement is straight-line distance (from P to Q) so less (than distance along path) or displacement is the shortest distance (from P to Q). B1 2(e) (displacement)2 = 3.52 + 1.42 C1 displacement = 3.8 m A1

This question in 9702/22 Feb/March 2022

1 (a) Define velocity. … … [1] (b) A rock of mass 7.5 kg is projected vertically upwards from the surface of a planet. The rock leaves the surface of the planet with a speed of 4.0 m s–1 at time t = 0. The variation with time t of the velocity v of the rock is shown in Fig. 1.1. 5 4 v / m s–1 3 2 1 0 0 1 2 3 4 t / s –1 –2 –3 Fig. 1.1 Assume that the planet does not have an atmosphere and that the viscous force acting on the rock is always zero. (i) Determine the height of the rock above the surface of the planet at time t = 4.0 s. height = … m [3] (ii) Determine the change in the momentum of the rock from time t = 0 to time t = 4.0 s. change in momentum = … N s [2] (iii) Determine the weight W of the rock on this planet. W = … N [2] (c) In practice, the planet in (b) does have an atmosphere that causes a viscous force to act on the moving rock. State and explain the variation, if any, in the resultant force acting on the rock as it moves vertically upwards. … … … … [2] [Total: 10]

10 marks

Mark scheme: 1(a) change in displacement / time (taken) B1 1(b)(i) (displacement =) area under graph C1 (at t = 4.0 s) v = (–) 2.4 C1 height = ½  2.5  4.0 – ½  1.5  2.4 = 3.2 m A1 1(b)(ii) change in momentum = 7.5 (–4.0 – 2.4) C1 = (–) 48 N s A1 1(b)(iii) W = ∆p / (∆)t or ∆mv / (∆)t C1 = 48 / 4.0 = 12 N A1 or W = ma or mg or m(v – u) / t (C1) = 7.5  1.6 or 7.5  (4 + 2.4) / 4.0 = 12 N (A1) 1(c) speed/velocity decreases so viscous force decreases B1 viscous force decreases (and weight constant) so resultant force decreases B1

This question in 9702/21 May/June 2022

2 A steel ball is projected horizontally from the top of a table, as shown in Fig. 2.1. ball table 4.9 m s–1 path of ball edge of table ground 180 cm Fig. 2.1 (not to scale) The ball is projected horizontally at a speed of 4.9 m s–1. The ball lands on the ground a horizontal distance of 180 cm from the edge of the table. Assume that air resistance is negligible. (a) (i) Calculate the time taken for the ball to reach the ground. time = … s [1] (ii) Calculate the vertical component of the velocity of the ball as it hits the ground. velocity = … m s–1 [2] (iii) Determine the magnitude and the angle to the horizontal of the velocity of the ball as it hits the ground. magnitude of velocity = … m s–1 angle to the horizontal = … ° [3] (b) The ball is projected by means of a compressed spring which is attached to a fixed block as shown in Fig. 2.2. ball x0 frictionless fixed track block spring Fig. 2.2 The ball is placed on a frictionless track in front of the spring. The ball is then pulled back so that the spring has compression x0. When the spring is released, the ball is projected horizontally as shown in Fig. 2.3. ball spring Fig. 2.3 The variation with compression x of the applied force F for the spring is shown in Fig. 2.4. 8 F / N 6 4 2 0 0 2 4 6 8 10 x / cm Fig. 2.4 The ball is a uniform sphere of steel of diameter 0.016 m and mass 0.017 kg. (i) Calculate the density of the steel. density = … kg m–3 [3] (ii) All of the elastic potential energy in the spring is converted into kinetic energy of the ball. The speed of the ball as it leaves the spring is 4.9 m s–1. Show that the maximum elastic potential energy of the spring is 0.20 J. [2] (iii) Use Fig. 2.4 to determine the spring constant k of the spring. k = … N m–1 [2] (iv) Use your answer in (b)(iii) and the value of energy given in (b)(ii) to determine the compression x0 of the spring. x0 = … m [2] (c) The steel ball is replaced by a polystyrene ball of the same diameter but of much lower mass. The spring is given compression x0 and is then released. Air resistance on this ball is not negligible after it leaves the spring. Explain: (i) why this ball leaves the spring with a greater speed than that of the steel ball … … … [1] (ii) why this ball takes a longer time to reach the ground than the steel ball. … … … [1] [Total: 17]

17 marks

Mark scheme: 2(a)(i) t = 1.8 / 4.9 A1 = 0.37 s 2(a)(ii) v = u + at C1 = 9.81  0.37 = 3.6 m s–1 A1 2(a)(iii) v 2 = 3.62 + 4.92 C1 v = 6.1 m s–1 A1 = tan–1 (3.6 / 4.9) A1 = 36° 2(b)(i) = m / V C1 4 C1 V = r3 3 4 A1 = 0.017 / [   (0.016 / 2)3 ] 3 = 7900 kg m–3 2(b)(ii) (E =) ½mv2 C1 (E =) ½  0.017  4.92 = 0.20 (J) A1 2(b)(iii) k = F / x or k = gradient C1 e.g. k = 6.4 / 10  10–2 A1 = 64 N m–1 (allow 63–65 N m–1) 2(b)(iv) E = ½kx2 C1 or E = ½Fx and F = kx x0 = [(2  0.20) / 64]0.5 A1 = 0.079 m or 0.080 m 2(c)(i) same elastic potential energy / same (initial) kinetic energy and (polystyrene ball has) smaller mass (so greater speed) B1 or same (average) force and (polystyrene ball has) smaller mass, (so greater average acceleration so greater speed) 2(c)(ii) (for the polystyrene ball there is) B1 less (average vertical) acceleration / smaller (average vertical component of) resultant force (so takes longer time to reach ground)

This question in 9702/21 Oct/Nov 2022

1 (a) Compare scalar and vector quantities. … … … [2] (b) The radius of a small sphere is determined from a measurement of the volume of the sphere. The sphere is submerged in water, displacing some of the water into a measuring cylinder as shown in Fig. 1.1. measuring cylinder sphere displaced water Fig. 1.1 (not to scale) The measured volume of displaced water is (28.0 ± 0.5) cm3. Calculate: (i) the radius, in cm, of the sphere radius = … cm [1] (ii) the percentage uncertainty in the radius of the sphere. percentage uncertainty = … % [2] [Total: 5]

5 marks

Mark scheme: Question Answer Mark 1(a) scalar and vector have magnitude B1 vector has direction (and scalar does not have direction) B1 1(b)(i) r = [(3  28) / 4]1/3 A1 = 1.9 cm 1(b)(ii) percentage uncertainty in V = (0.5 / 28)  100 C1 ( = 1.79%) percentage uncertainty in r = 1.79 / 3 A1 = 0.6%

This question in 9702/21 Oct/Nov 2023

1 (a) Table 1.1 lists some physical quantities. Complete the table by placing a tick (✓) next to the scalar quantities. Table 1.1 acceleration charge momentum power upthrust [1] (b) A uniform cylinder has diameter D, length L and mass M. The density ρ of the cylinder is given by 4M ρ = . π D2 L Table 1.2 shows the data obtained from an experiment to determine the density of the cylinder. Table 1.2 quantity measurement percentage uncertainty D (26.2 ± 0.1) mm … % L (162 ± 1) mm … % M (247 ± 1) g 0.4% (i) Calculate the percentage uncertainties in D and L. Write your answers in Table 1.2. [1] (ii) Calculate the density of the cylinder. Give your answer to three significant figures. density = … kg m–3 [2] (iii) Calculate the percentage uncertainty in the density. percentage uncertainty = … % [2] [Total: 6]

6 marks

Mark scheme: Question Answer Marks 1(a) charge and power only ticked B1 1(b)(i) %D = 0.4% and %L = 0.6% A1 1(b)(ii) = (4  0.247) / [  (26.2  10–3)2  0.162] C1 = 2.83  103 kg m–3 A1 1(b)(iii) percentage uncertainty = 0.4 + (2  0.4) + 0.6 C1 = 1.8% A1

This question in 9702/23 Oct/Nov 2023

2 A ball on horizontal ground is kicked towards a vertical wall. Fig. 2.1 shows the path of the ball. path of ball horizontal h u ground wall 38° 9.0 m ball Fig. 2.1 (not to scale) The ball has an initial velocity u at an angle of 38° to the ground. The ball travels a horizontal distance of 9.0 m before striking the wall at a height h above the ground. The horizontal component uH of the initial velocity of the ball is 9.5 m s–1. Air resistance is negligible. (a) (i) Show that the time t for the ball to reach the wall is 0.95 s. [1] (ii) Calculate the vertical component uV of the initial velocity of the ball. uV = … m s–1 [2] (iii) Determine h. h = … m [2] (b) The speed of the ball just after striking the wall is less than its speed just before striking the wall. State what this indicates about the nature of the collision of the ball with the wall. … … [1] [Total: 6]

6 marks

Mark scheme: 2(a)(i) (time / t =) 9(.0) / 9.5 = 0.95 (s) A1 2(a)(ii) (uV) = 9.5 tan 38° or 9.5 / tan 52° C1 or 9.5 = u cos 38° and uV = u sin 38° or 9.5 = u cos 38° and uV = (u2 – 9.52)½ uV = 7.4 m s–1 A1 2(a)(iii) s = ut + ½at 2 C1 (h =) 7.4  0.95 – ½  9.81  0.952 h = 2.6 m A1 2(b) (collision is) inelastic B1

This question in 9702/23 Oct/Nov 2023

1 (a) State what is meant by a vector quantity. … … [1] (b) A sphere falls vertically through a liquid that has density 830 kg m–3. The sphere has radius r and constant velocity v, as shown in Fig. 1.1. liquid sphere, radius r v Fig. 1.1 (i) The drag force D acting on the sphere is given by D = 6πrηv where η is a property of the liquid. Determine the SI base units of η. SI base units … [3] (ii) State an equation showing the relationship between the magnitudes of the weight W, drag force D and upthrust U acting on the sphere. … [1] (iii) The volume of the sphere is 4.6 cm3. The drag force D is 0.32 N. Calculate the weight of the sphere. weight = … N [2] [Total: 7]

7 marks

Mark scheme: Question Answer Marks 1(a) a quantity with magnitude and direction B1 1(b)(i) SI base units of D: kg m s–2 C1 SI base units of r: m and v: m s–1 C1 base units of : kg m s–2 / (m  m s–1) A1 = kg m–1 s–1 1(b)(ii) W = U + D A1 1(b)(iii) U = 830  9.81  4.6  (10–2)3 C1 ( = 0.037 N) W = 0.037 + 0.32 A1 = 0.36 N

This question in 9702/22 Oct/Nov 2024

1 (a) Table 1.1 lists some physical quantities. Identify with ticks (✓) which quantities are vectors and which are scalars. Table 1.1 quantity scalar vector acceleration displacement gravitational potential energy speed temperature [2] (b) A constant resultant force F acts on a car of mass m. The car moves from rest with constant acceleration a along horizontal ground. When the car has displacement s, the speed of the car is v. (i) Using the concept of work done on the car, show that the kinetic energy EK of the car is given by the equation 1 EK = mv2. 2 [3] (ii) The mass of the car is 920 kg. At time t = 0, the car is at rest. At time t = 5.8 s, its velocity is 17 m s–1. Calculate the kinetic energy of the car at time t = 5.8 s. kinetic energy = … J [1] (iii) Between time t = 0 and time t = 5.8 s, the work done against resistive forces is 4.7 × 104 J. Determine the average output power of the car during this time. power = … W [3] (iv) At time t = 5.8 s, the speed of the car becomes constant. State and explain whether the output power of the car is greater than, less than or the same as the output power just before t = 5.8 s. … … [1] [Total: 10]

10 marks

Mark scheme: Question Answer Marks 1(a) acceleration and displacement identified as vectors (and no others) B1 speed, temperature and gravitational potential energy identified as scalars (and no others) B1 1(b)(i) W = Fs or W = mas B1 s = v2 / 2a or a = v2 / 2s or as = v2 / 2 B1 W = ma(v2 / 2a) or W = m(v2 / 2s)s or W = m(v2 / 2) B1 and (so EK )= ½mv2 OR (B1) W = Fs or W = mas F = mv / t and s = ½vt (B1) W = mv / t  ½vt and (so EK )= ½mv 2 (B1) OR (B1) W = Fs or W = mas a = v / t and s = ½vt (B1) W = m(v / t)(½vt) and (so EK )= ½mv 2 (B1) OR (B1) W = Fs or W = mas a = v / t and s = ½at2 (B1) W = m(v / t)(½  (v / t)  t2) and (so EK )= ½mv 2 (B1) 1(b)(ii) kinetic energy = ½ mv2 A1 = ½  920  172 = 1.3  105 J 1(b)(iii) P = W / t C1 = (4.7  104 + 1.3  105) / 5.8 C1 = 3.1  104 W A1 1(b)(iv) (at/after t = 5.8 s) B1 the kinetic energy (of the car) does not change / work is done only against resistive forces / no work is done to accelerate (the car) so (power output is) less

This question in 9702/22 May/June 2025

1 A child kicks a ball so that it leaves horizontal ground with a velocity of 28 m s–1 at an angle of 34° to the horizontal, as shown in Fig. 1.1. 28 m s–1 34° ground Fig. 1.1 Air resistance is negligible. The ball leaves the ground at time t = 0. (a) (i) Calculate the horizontal component vH and the vertical component vV of the velocity of the ball immediately after it has left the ground. vH = … m s–1 vV = … m s–1 [2] (ii) Show that the ball reaches its maximum height at time t = 1.6 s. [1] (iii) On Fig. 1.2, sketch the variation of vH with time t between t = 0 and t = 3.2 s. Label your line H. 30 velocity / m s–1 20 10 0 0 0.8 1.6 2.4 3.2 t / s –10 –20 –30 Fig. 1.2 [1] (iv) On Fig. 1.2, sketch the variation of vV with time t between t = 0 and t = 3.2 s. Assume that velocity in the upward direction is positive. Label your line V. [3] (b) The total change in momentum of the ball between leaving the ground at t = 0 and landing on the ground at t = 3.2 s is 13 kg m s–1. (i) Define momentum. … … [1] (ii) Calculate the force that acts on the ball while it is in the air. force = … N [2] (iii) Determine the mass of the ball. mass = … kg [1] [Total: 11]

11 marks

Mark scheme: Question Answer Marks 1(a)(i) vH = 28 × cos 34° A1 = 23 m s–1 vV = 28 × sin 34° A1 = 16 m s–1 1(a)(ii) time = 16 / 9.81 = 1.6 s A1 1(a)(iii) horizontal straight line at v = 23 m s–1 from t = 0 to t = 3.2 s B1 1(a)(iv) straight diagonal line starting at a positive velocity from t = 0 to t = 3.2 s, crossing the time axis B1 line starting at v = 16 m s–1 and ending at v = –16 m s–1 B1 line passing through v = 0 at t = 1.6 s B1 1(b)(i) product of mass and velocity B1 1(b)(ii) F = p / t C1 = 13 / 3.2 A1 = 4.1 N 1(b)(iii) m = p / v A1 = 13 / (2  16) = 0.41 kg or m = F / g = 4.06 / 9.81 = 0.41 kg

This question in 9702/24 Oct/Nov 2025