Physical quantities — Paper 2 · A Level Physics 9702

1 (a) Derive the SI base unit of force. SI base unit of force = ………………………………… [1] (b) A spherical ball of radius r experiences a resistive force F due to the air as it moves through the air at speed v. The resistive force F is given by the expression F = crv, where c is a constant. Derive the SI base unit of the constant c. SI base unit of c = ………………………………… [1] Use (c) The ball is dropped from rest through a height of 4.5 m. (i) Assuming air resistance to be negligible, calculate the final speed of the ball. speed = …………………………… m s–1 [2] (ii) The ball has mass 15 g and radius 1.2 cm. The numerical value of the constant c in the equation in (b) is equal to 3.2 ×10–4 when measured using the SI system of units. Show quantitatively whether the assumption made in (i) is justified. [3]

7 marks

Mark scheme: 1 (a) kg m s–2 B1 [1] (b) kg m–1 s–1 B1 [1] (c) (i) v2 = 2gs = 2 × 9.8 × 4.5 C1 v = 9.4 m s–1 A1 [2] (ii) either F (= 3.2 × 10–4 × 1.2 × 10–2 × 9.4) = 3.6 × 10–5 N M1 weight of sphere (= mg = 15 × 10–3 × 9.8) = 0.15 N M1 3.6 × 10–5 << 0.15, so justified A1 [3] or mg = crvT (M1) terminal speed = 3.8 × 104 m s–1 (M1) 9.4 << 3.8 × 104, so justified (A1)

This question in 9702/21 May/June 2006

1 The uncalibrated scale and the pointer of a meter are shown in Fig. 1.1. uncalibrated scale Fig. 1.1 The pointer is shown in the zero position. The meter is to be used to indicate the volume of fuel in the tank of a car. A known volume V of fuel is poured into the tank and the deflection θ of the pointer is noted. Fig. 1.2 shows the variation with θ of V. 80 V /103 cm3 60 40 20 0 0 20 40 60 80 100 θ/° Fig. 1.2 Examiner’s Use (a) On Fig. 1.1, (i) calibrate the scale at 20 × 103 cm3 intervals, [2] (ii) mark a possible position for a volume of 1.0 × 105 cm3. [1] (b) Suggest one advantage of this scale, as compared with a uniform scale, for measuring fuel volumes in the tank of the car. … … [1]

4 marks

Mark scheme: 1 (a) (i) all positions (accept 20, 40, 60, 80) marked to within ±5° B2 positions are 40°, 70°, 90° and 102° (-1 for each error or omission) (ii) allow 107° → 113° B1 [3] (b) e.g. more sensitive at low volumes B1 [1] (do not allow reference to ‘accuracy’)

This question in 9702/21 May/June 2007

2 A ball B of mass 1.2 kg travelling at constant velocity collides head-on with a stationary ball S For of mass 3.6 kg, as shown in Fig. 2.1. Examiner’s Use v ball B ball S mass 1.2 kg mass 3.6 kg Fig. 2.1 Frictional forces are negligible. The variation with time t of the velocity v of ball B before, during and after colliding with ball S is shown in Fig. 2.2. +4 +3 v / m s–1 +2 +1 0 0 0.1 0.2 0.3 0.4 0.5 t / s –1 –2 Fig. 2.2 (a) State the significance of positive and negative values for v in Fig. 2.2. … … [1] (b) Use Fig. 2.2 to determine, for ball B during the collision with ball S, For Examiner’s (i) the change in momentum of ball B, Use change in momentum = … N s [3] (ii) the magnitude of the force acting on ball B. force = … N [3] (c) Calculate the speed of ball S after the collision. speed = … m s–1 [2] (d) Using your answer in (c) and information from Fig. 2.2, deduce quantitatively whether For the collision is elastic or inelastic. Examiner’s Use … … [2]

11 marks

Mark scheme: 2 (a) ball moving in opposite direction (after collision) … B1 [1] (b) (i) change in momentum = 1.2 (4.0 + 0.8) … C2 (correct values, 1 mark; correct sign {values added}, 1 mark ) = 5.76 N s …(allow 5.8) … A1 [3] (ii) force = ∆p / ∆t or m∆v / ∆t … C1 = 5.76 / 0.08 or 1.2 × 4.8 / 0.08 … C1 = 72 N … A1 [3] (c) 5.76 = 3.6 × V … C1 V = 1.6 m s-1 … A1 [2] (d) either speed of approach = 4.0 m s-1 and speed of separation = 2.4 m s-1 … M1 not equal and so inelastic … A1 or kinetic energy before = 9.6 J and kinetic energy after collision = 4.99 J … M1 kinetic energy after is less / not conserved so inelastic … A1 [2]

This question in 9702/21 May/June 2009

1 A metal wire has a cross-section of diameter approximately 0.8 mm. (a) State what instrument should be used to measure the diameter of the wire. … [1] (b) State how the instrument in (a) is (i) checked so as to avoid a systematic error in the measurements, … … [1] (ii) used so as to reduce random errors. … … … [2]

4 marks

Mark scheme: 1 (a) micrometer/screw gauge/digital callipers ………………………………………. B1 [1] (b) (i) look/check for zero error ……………………………………………………. B1 [1] (ii) take several readings ……………………………………………………….. M1 around the circumference/along the wire …………………………………. A1 [2]

This question in 9702/22 May/June 2010

1 Make estimates of the following quantities. (a) the thickness of a sheet of paper thickness = … mm [1] (b) the time for sound to travel 100 m in air time = … s [1] (c) the weight of 1000 cm3 of water weight = … N [1]

3 marks

Mark scheme: 1 (a) allow 0.05 mm → 0.15 mm B1 [1] (b) allow 0.25 s → 0.5 s B1 [1] (c) allow 8 N → 12 N B1 [1] ignore number of significant figures

This question in 9702/23 Oct/Nov 2010

1 Measurements made for a sample of metal wire are shown in Fig. 1.1. quantity measurement uncertainty length 1750 mm ± 3 mm diameter 0.38 mm ± 0.01 mm resistance 7.5 Ω ± 0.2 Ω Fig. 1.1 (a) State the appropriate instruments used to make each of these measurements. (i) length … [1] (ii) diameter … [1] (iii) resistance … [1] (b) (i) Show that the resistivity of the metal is calculated to be 4.86 × 10–7 Ω m. [2] (ii) Calculate the uncertainty in the resistivity. uncertainty = ± … Ω m [4] (c) Use the answers in (b) to express the resistivity with its uncertainty to the appropriate For number of significant figures. Examiner’s Use resistivity = … ± … Ω m [1]

10 marks

Mark scheme: 1 (a) (i) metre rule / tape (not ‘rule’) B1 [1] (ii) micrometer (screw gauge) / digital caliper B1 [1] (iii) ammeter and voltmeter / ohmmeter / multimeter on ‘ohm’ setting B1 [1] (b) (i) resistivity = RA / L C1 = [7.5 × π × (0.38 × 10–3)2 / 4] / 1.75 M1 = 4.86 × 10–7 Ω m A0 [2] (ii) (uncertainty in R =) [0.2 / 7.5] × 100 = 2.7% and (uncertainty in L =) [3 / 1750] × 100 = 0.17% C1 (uncertainty in A =) 2 × (0.01 / 0.38) × 100 = 5.3 % C1 total = 8.13% C1 uncertainty = 0.395 × 10–7 (Ω m) A1 [4] (missing 2 factor in uncertainty in A, then allow max 3/4) (c) resistivity = (4.9 × 10–7 ± 0.4 × 10–7) Ω m A1 [1]

This question in 9702/21 May/June 2011

3 A helicopter has a cable hanging from it towards the sea below, as shown in Fig. 3.1. For Examiner’s Use helicopter cable sea Fig. 3.1 A man of mass 80 kg rescues a child of mass 50.5 kg. The two are attached to the cable and are lifted from the sea to the helicopter. The lifting process consists of an initial uniform acceleration followed by a period of constant velocity and then completed by a final uniform deceleration. (a) Calculate the combined weight of the man and child. weight = … N [1] (b) Calculate the tension in the cable during (i) the initial acceleration of 0.570 m s–2, tension = … N [2] (ii) the period of constant velocity of 2.00 m s–1. tension = … N [1] (c) During the final deceleration the tension in the cable is 1240 N. Calculate this For deceleration. Examiner’s Use deceleration = … m s–2 [2] (d) (i) Calculate the time over which the man and child are 1. moving with uniform acceleration, time = … s [1] 2. moving with uniform deceleration. time = … s [1] (ii) The time over which the man and child are moving with constant velocity is 20 s. On Fig. 3.2, sketch a graph to show the variation with time of the velocity of the man and child for the complete lifting process. 2.0 velocity / m s–1 1.0 0 0 5 10 15 20 25 30 35 time / s Fig. 3.2 [2]

10 marks

Mark scheme: 3 (a) weight = m × g = 130.5 × 9.81 = 1280 N A1 [1] (b) (i) F = ma T – 1280 = 130.5 × 0.57 C1 T = 1280 + 74.4 = 1350 N A1 [2] (ii) 1280 N A1 [1] (c) 1240 – 1280 = 130.5 × a C1 a = (–) 0.31 m s–2 A1 [2] (d) (i) 1. 3.5 s A1 [1] 2. 6.5 s A1 [1] GCE AS/A LEVEL – May/June 2011 9702 23 (ii) basic shape M1 correct points A1 [2]

This question in 9702/23 May/June 2011

2 (a) Define For Examiner’s (i) force, Use … … [1] (ii) work done. … … [1] (b) A force F acts on a mass m along a straight line for a distance s. The acceleration of the mass is a and the speed changes from an initial speed u to a final speed v. (i) State the work W done by F. [1] (ii) Use your answer in (i) and an equation of motion to show that kinetic energy of a mass can be given by the expression kinetic energy = ½ × mass × (speed)2. [3] (c) A resultant force of 3800 N causes a car of mass of 1500 kg to accelerate from an initial speed of 15 m s–1 to a final speed of 30 m s–1. (i) Calculate the distance moved by the car during this acceleration. distance = … m [2] (ii) The same force is used to change the speed of the car from 30 m s–1 to 45 m s–1. Explain why the distance moved is not the same as that calculated in (i). … … … [1]

9 marks

Mark scheme: 2 (a) (i) force is rate of change of momentum B1 [1] (ii) work done is the product of the force and the distance moved in the direction of the force B1 [1] (b) (i) W = Fs or W = mas or W = m(v2 – u2) / 2 or W = force × distance s A1 [1] (ii) as = (v2 – u2) / 2 any subject M1 W = mas hence W = m(v2 – u2) / 2 M1 RHS represents terms of energy or with u = 0 KE = ½mv 2 A1 [3] (c) (i) work done = ½ × 1500 × [(30)2 – (15)2] (=506250) C1 distance = WD / F = 506250 / 3800 = 133 m A1 [2] or F = ma a = 2.533 (m s–2) C1 v2 = u2 + 2as s = 133 m A1 (ii) the change in kinetic energy is greater or the work done by the force has to be greater, hence distance is greater (for same force) A1 [1] allow: same acceleration, same time, so greater average speed and greater distance

This question in 9702/22 Oct/Nov 2011

2 The variation with time t of velocity v of a car is shown in Fig. 2.1. For Examiner’s Use stage 1 stage 2 20.0 15.0 v / m s–1 10.0 5.0 0 0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 t / s Fig. 2.1 At time t = 0, the driver sees an obstacle in the road. A short time later, the driver applies the brakes. The car travels in two stages, as shown in Fig. 2.1. (a) Use Fig. 2.1 to describe the velocity of the car in 1. stage 1, … … [1] 2. stage 2. … … [1] (b) (i) Calculate the distance travelled by the car from t = 0 to t = 3.5 s. total distance = … m [2] (ii) The car has a total mass of 1250 kg. Determine the total resistive force acting on For the car in stage 2. Examiner’s Use force = … N [3] (c) For safety reasons drivers are asked to travel at lower speeds. For each stage, describe and explain the effect on the distance travelled for the same car and driver travelling at half the initial speed shown in Fig. 2.1. (i) stage 1: … … … [1] (ii) stage 2: … … … … [2]

10 marks

Mark scheme: 2 (a) 1. constant velocity / speed B1 [1] 2. either constant / uniform decrease (in velocity/speed) or constant rate of decrease (in velocity/speed) B1 [1] (b) (i) distance is area under graph for both stages C1 stage 1: distance (18 × 0.65) = 11.7 (m) stage 2: distance = (9 × [3.5 – 0.65]) = 25.7 (m) total distance = 37.(4) m A1 [2] (–1 for misreading graph) {for stage 2, allow calculation of acceleration (6.32 m s–2) and then s = (18 × 2.85) + ½ × 6.32 (2.85)2 = 25.7 m} (ii) either F = ma or EK = ½mv2 C1 a = (18 – 0)/(3.5 – 0.65) EK = ½ × 1250 × (18)2 C1 F = 1250 × 6.3 =7900 N or F = ½ × 1250 × (18)2 / 25.7 = 7900 N A1 [3] or initial momentum = 1250 × 18 (C1) F = change in momentum / time taken (C1) F = (1250 × 18) / 2.85 = 7900 (A1) (c) (i) stage 1: either half / less distance as speed is half / less or half distance as the time is the same or sensible discussion of reaction time B1 [1] (ii) stage 2: either same acceleration and s = v2 / 2a or v2 is ¼ B1 ¼ of the distance B1 [2] GCE AS/A LEVEL – October/November 2011 9702 23

This question in 9702/23 Oct/Nov 2011

1 The volume V of liquid flowing in time t through a pipe of radius r is given by the equation Use V π Pr 4 = t 8Cl where P is the pressure difference between the ends of the pipe of length l, and C depends on the frictional effects of the liquid. An experiment is performed to determine C. The measurements made are shown in Fig. 1.1. V / 10–6 m3 s–1 P / 103 N m–2 r / mm l / m t 1.20 ± 0.01 2.50 ± 0.05 0.75 ± 0.01 0.250 ± 0.001 Fig. 1.1 (a) Calculate the value of C. C = … N s m–2 [2] (b) Calculate the uncertainty in C. uncertainty = … N s m–2 [3] (c) State the value of C and its uncertainty to the appropriate number of significant figures. C = … ± … N s m–2 [1]

6 marks

Mark scheme: V π P r 1 (a) = t 8C l C = [π × 2.5 × 103 × (0.75 × 10–3)4] / (8 × 1.2 × 10–6 × 0.25) C1 = 1.04 × 10–3 N s m–2 A1 [2] (b) 4 × %r C1 %C = %P + 4 × %r + %V/t + %l = 2% + 5.3% + 0.83% + 0.4% (= 8.6%) A1 ∆C = ± 0.089 × 10–3 N s m–2 A1 [3] (c) C = (1.04 ± 0.09) × 10–3 N s m–2 A1 [1] 2 2

This question in 9702/22 May/June 2012

1 (a) The drag force D on an object of cross-sectional area A, moving with a speed v through Use a fluid of density ρ, is given by 1 2 D = CρAv 2 where C is a constant. Show that C has no unit. [2] (b) A raindrop falls vertically from rest. Assume that air resistance is negligible. (i) On Fig. 1.1, sketch a graph to show the variation with time t of the velocity v of the raindrop for the first 1.0 s of the motion. 10.0 8.0 6.0 v / m s–1 4.0 2.0 0 0 1.0 2.0 3.0 4.0 5.0 t / s Fig. 1.1 [1] (ii) Calculate the velocity of the raindrop after falling 1000 m. velocity = … m s–1 [2] (c) In practice, air resistance on raindrops is not negligible because there is a drag force. For This drag force is given by the expression in (a). Examiner’s Use (i) State an equation relating the forces acting on the raindrop when it is falling at terminal velocity. [1] (ii) The raindrop has mass 1.4 × 10–5 kg and cross-sectional area 7.1 × 10–6 m2. The density of the air is 1.2 kg m–3 and the initial velocity of the raindrop is zero. The value of C is 0.60. 1. Show that the terminal velocity of the raindrop is about 7 m s–1. [2] 2. The raindrop reaches terminal velocity after falling approximately 10 m. On Fig. 1.1, sketch the variation with time t of velocity v for the raindrop. The sketch should include the first 5 s of the motion. [2]

10 marks

Mark scheme: 1 (a) units for D identified as kg m s–2 M1 all other units shown: units for A: m2 units for v2: m2 s–2 units for ρ: kg m–3 kg m s −2 C = with cancelling / simplification to give C no units A1 [2] kg m − 3 m 2 m 2 s − 2 (b) (i) straight line from (0,0) to (1,9.8) ± half a square B1 [1] (ii) ½ mv2 = mgh or using v2 = 2 as C1 v = (2 × 9.81 × 1000)1/2 = 140 m s–1 A1 [2] (c) (i) weight = drag (D) ( + upthrust) B1 [1] Allow mg or W for weight and D or expression for D for drag (ii) 1. mg = 1.4 ×10–5 × 9.81 C1 1.4 × 10–5 × 9.81 = 0.5 × 0.6 × 1.2 × 7.1 × 10–6 × v2 M1 v = 7.33 m s–1 A0 [2] 2. line from (0,0) correct curvature to a horizontal line at velocity of 7 m s–1 M1 line reaches 7 m s–1 between 1.5 s and 3.5 s A1 [2]

This question in 9702/22 Oct/Nov 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) Distinguish between mass and weight. For Examiner’s mass: … Use … weight: … … [2] (b) An object O of mass 4.9 kg is suspended by a rope A that is fixed at point P. The object is pulled to one side and held in equilibrium by a second rope B, as shown in Fig. 2.1. P rope A Ƨ O rope B Fig. 2.1 Rope A is at an angle θ to the horizontal and rope B is horizontal. The tension in rope A is 69 N and the tension in rope B is T. (i) On Fig. 2.1, draw arrows to represent the directions of all the forces acting on object O. [2] (ii) Calculate For Examiner’s 1. the angle θ, Use θ = … ° [3] 2. the tension T. T = … N [2]

9 marks

Mark scheme: 2 (a) mass is the property of a body resisting changes in motion / quantity of matter in a body / measure of inertia to changes in motion B1 weight is the force due to the gravitational field/force due to gravity or gravitational force B1 [2] Allow 1/2 for ‘mass is scalar weight is vector’ (b) (i) arrow vertically down through O B1 tension forces in correct direction on rope B1 [2] (ii) 1. weight = mg = 4.9 × 9.81 (= 48.07) C1 69 sin θ = mg C1 θ = 44.(1)° scale drawing allow ± 2° A1 [3] use of cos or tan 1/3 only 2. T = 69 cos θ C1 = 49.6 / 50 N scale drawing 50 ±2 (2/2) 50 ±4 (1/2) A1 [2] correct answers obtained using scale diagram or triangle of forces will score full marks cos in 1. then sin in 2. (2/2)

This question in 9702/21 May/June 2013

1 (a) Determine the SI base units of power. Use SI base units of power … [3] (b) Fig. 1.1 shows a turbine that is used to generate electrical power from the wind. L wind turbine speed v Fig. 1.1 The power P available from the wind is given by P = CL2ρv 3 where L is the length of each blade of the turbine, ρ is the density of air, v is the wind speed, C is a constant. (i) Show that C has no units. [3] (ii) The length L of each blade of the turbine is 25.0 m and the density ρ of air is 1.30 in For SI units. The constant C is 0.931. Examiner’s The efficiency of the turbine is 55% and the electric power output P is 3.50 × 105 W. Use Calculate the wind speed. wind speed = … m s–1 [3] (iii) Suggest two reasons why the electrical power output of the turbine is less than the power available from the wind. 1. … … 2. … … [2]

11 marks

Mark scheme: 1 (a) power = energy / time C1 = (force × distance / time) = kg m2 s–2 / s C1 = kg m2 s–3 A1 [3] (b) (i) units of L2: m2 and units of ρ : kg m–3 and units of v3: m3 s–3 C1 (C = P / L2 ρ v3) hence units of C: kg m2 s–3 m–2 kg–1 m3 m–3 s3 or any correct statement of component units M1 argument /discussion / cancelling leading to C having no units A1 [3] (ii) power available from wind = 3.5 × 105 × 100 / 55 (= 6.36 × 105) C1 v3 = 3.5 × 105 × 100 / (55 × 0.931 × (25)2 × 1.3) C1 v = 9.4 m s–1 A1 [3] (iii) not all kinetic energy of wind converted to kinetic energy of blades B1 generator / conversion to electrical energy not 100% efficient / heat produced in generator / bearings etc B1 [2] (there must be cause of loss and where located)

This question in 9702/22 May/June 2013

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

3 (a) Define For Examiner’s (i) velocity, Use … … [1] (ii) acceleration. … … [1] (b) A car of mass 1500 kg travels along a straight horizontal road. The variation with time t of the displacement x of the car is shown in Fig. 3.1. 140 120 100 80 x / m 60 40 20 0 0 1.0 2.0 3.0 4.0 5.0 6.0 t / s Fig. 3.1 (i) Use Fig. 3.1 to describe qualitatively the velocity of the car during the first six For seconds of the motion shown. Examiner’s Give reasons for your answers. Use … … … … … [3] (ii) Calculate the average velocity during the time interval t = 0 to t = 1.5 s. average velocity = … m s–1 [1] (iii) Show that the average acceleration between t = 1.5 s and t = 4.0 s is –7.2 m s–2. [2] (iv) Calculate the average force acting on the car between t = 1.5 s and t = 4.0 s. force = … N [2]

10 marks

Mark scheme: 3 (a) (i) velocity = rate of change of displacement OR displacement change / time (taken) A1 [1] (ii) acceleration = rate of change of velocity OR change in velocity / time (taken) A1 [1] (b) (i) initial constant velocity as straight line / gradient constant B1 middle section deceleration/ speed / velocity decreases / slowing down as gradient decreases B1 last section lower velocity (than at start) as gradient (constant and) smaller B1 [3] [special case: all three stages correct descriptions but no reasons 1/3] (ii) velocity = 45 / 1.5 = 30 m s–1 A1 [1] (iii) velocity at 4.0 s is (122 – 98) / 2.0 = 12 (m s–1) (allow 12 to 13) B1 acceleration = (12 – 30) / 2.5 = –7.2 m s–2 (if answer not this value then comment needed to explain why, e.g. difficulty in drawing tangent) A1 [2] (iv) F = ma C1 F = (–)1500 × 7.2 = (–)11000 (10800) N A1 [2]

This question in 9702/23 Oct/Nov 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

1 (a) Underline all the base quantities in the following list. ampere charge current mass second temperature weight [2] (b) The potential energy EP stored in a stretched wire is given by EP = ½Cσ2V where C is a constant, σ is the strain, V is the volume of the wire. Determine the SІ base units of C. base units … [3]

5 marks

Mark scheme: 1 (a) current, mass and temperature two correct 2/2, one omission or error 1/2 A2 [2] (b) σ : no units, V: m3 C1 EP: kg m2 s–2 C1 C: kg m2 s–2 × m–3 = kg m–1 s–2 A1 [3]

This question in 9702/23 May/June 2014

1 (a) Mass, length and time are SІ base quantities. State two other base quantities. 1. … 2. … [2] (b) A mass m is placed on the end of a spring that is hanging vertically, as shown in Fig. 1.1. spring mass m Fig. 1.1 The mass is made to oscillate vertically. The time period of the oscillations of the mass is T. The period T is given by m T = C k where C is a constant and k is the spring constant. Show that C has no units. [3]

5 marks

Mark scheme: 1 (a) temperature B1 current B1 [2] (allow amount of substance and luminous intensity) (b) base units of force constant: kg m s–2 m–1 or kg s–2 B1 base units of time and mass: s and kg C1 base units of C: s (kg s–2 / kg)1/2 cancelling to show no units B1 [3]

This question in 9702/21 Oct/Nov 2014

1 (a) Use the definition of power to show that the SI base units of power are kg m2 s–3. [2] (b) Use an expression for electrical power to determine the SI base units of potential difference. units … [2]

4 marks

Mark scheme: 1 (a) power = work / time or energy / time or (force × distance) / time B1 = kg m s–2 × m s–1 = kg m2 s–3 A1 [2] (b) power = VI [or V2 / R and V = IR or I 2R and V = IR] B1 (units of V:) kg m2 s–3 A–1 B1 [2]

This question in 9702/21 May/June 2015

1 (a) Use the definition of work done to show that the SI base units of energy are kg m2 s−2. [2] (b) Define potential difference. … … [1] (c) Determine the SI base units of resistance. Show your working. units … [3]

6 marks

Mark scheme: 1 (a) (work =) force × distance or force × displacement or (W =) F × d M1 units of work: kg m s–2 × m = kg m2 s–2 A1 [2] work (done) or energy (transform ed) (from electrical to other forms) (b) (p.d. = ) B1 [1] charge (c) R = V / I B1 units of V: kg m2 s–2 / A s and units of I: A C1 or R = P / I2 [or P = VI and V = IR] (B1) units of P: kg m2 s–3 and units of I: A (C1) or R = V2 / P (B1) units of V: kg m2 s–2 / A s and units of P: kg m2 s–3 (C1) units of R: (kg m2 s–2 / A2 s =) kg m2 s–3 A–2 A1 [3]

This question in 9702/22 May/June 2015

3 A rod PQ is attached at P to a vertical wall, as shown in Fig. 3.1. R wire wall F 0.64 m 0.96 m 30° Q P rod W Fig. 3.1 The length of the rod is 1.60 m. The weight W of the rod acts 0.64 m from P. The rod is kept horizontal and in equilibrium by a wire attached to Q and to the wall at R. The wire provides a force F on the rod of 44 N at 30° to the horizontal. (a) Determine (i) the vertical component of F, vertical component = … N [1] (ii) the horizontal component of F. horizontal component = … N [1] (b) By taking moments about P, determine the weight W of the rod. W = … N [2] (c) Explain why the wall must exert a force on the rod at P. … … … [1] (d) On Fig. 3.1, draw an arrow to represent the force acting on the rod at P. Label your arrow with the letter S. [1]

6 marks

Mark scheme: 3 (a) (i) (vertical component = 44 sin 30° =) 22 N A1 [1] (ii) (horizontal component = 44 cos 30° =) 38(.1) N A1 [1] (b) W × 0.64 = 22 × 1.60 C1 (W =) 55 N A1 [2] (c) F has a horizontal component (not balanced by W) or F has 38 N acting horizontally or 38 N acts on wall or vertical component of F does not balance W or F and W do not make a closed triangle of forces B1 [1] (d) line from P in direction towards point on wire vertically above W and direction up B1 [1]

This question in 9702/22 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) State two SΙ base quantities other than mass, length and time. 1. … 2. … [2] (b) A beam is clamped at one end and an object X is attached to the other end of the beam, as shown in Fig. 1.1. l oscillation of X clamp beam object X Fig. 1.1 The object X is made to oscillate vertically. The time period T of the oscillations is given by Ml 3 T = K E where M is the mass of X, l is the length between the clamp and X, E is the Young modulus of the material of the beam and K is a constant. (i) 1. Show that the SΙ base units of the Young modulus are kg m–1 s–2. [1] 2. Determine the SΙ base units of K. SΙ base units of K … [2] (ii) Data in SΙ units for the oscillations of X are shown in Fig. 1.2. quantity value uncertainty T 0.45 ± 2.0% l 0.892 ± 0.2% M 0.2068 ± 0.1% K 1.48 × 105 ± 1.5% Fig. 1.2 Calculate E and its actual uncertainty. E = … ± … kg m–1 s–2 [4]

9 marks

Mark scheme: 1 (a) temperature B1 current B1 [2] (allow amount of substance, luminous intensity) (b) (i) 1. E = (stress / strain =) [force / area] / [extension / original length] units of stress: kg m s–2 / m2 and no units for strain B1 units of E: kg m–1 s–2 A0 [1] 2. units for T: s, l: m and M: kg K 2 = T 2 E / M l 3 hence units: s2 kg m–1 s–2 / kg3 (= m–4) C1 units of K: m–2 A1 [2] (ii) % uncertainty in E = 4% (for T 2) + 0.6% (for l 3) + 0.1% (for M) + 3% (for K 2) = 7.7% B1 E = [(1.48 × 105)2 × 0.2068 × (0.892)3] / (0.45)2 = 1.588 × 1010 C1 7.7% of E = 1.22 × 109 C1 E = (1.6 ± 0.1) × 1010 kg m–1 s–2 A1 [4] 12 12

This question in 9702/21 Oct/Nov 2015

1 (a) The intensity of a progressive wave is defined as the average power transmitted through a surface per unit area. Show that the SI base units of intensity are kg s−3. [2] (b) (i) The intensity I of a sound wave is related to the amplitude x0 of the wave by I = Kρcf 2x02 where ρ is the density of the medium through which the sound is passing, c is the speed of the sound wave, f is the frequency of the sound wave and K is a constant. Show that K has no units. [2] (ii) Calculate the intensity, in pW m−2, of a sound wave where K = 20, ρ = 1.2 in SI base units, c = 330 in SI base units, f = 260 in SI base units and x0 = 0.24 nm. intensity = … pW m−2 [3]

7 marks

Mark scheme: 1 (a) energy or W: kg m2 s–2 or power or P: kg m2 s–3 M1 intensity or I: kg m2 s–2 m–2 s–1 (from use of energy expression) or kg m2 s–3 m–2 (from use of power expression) indication of simplification to kg s–3 A1 [2] (b) (i) ρ: kg m–3, c: m s–1, f: s–1, x0: m M1 substitution of terms in an appropriate equation and simplification to show K has no units A1 [2] (ii) I = 20 × 1.2 × 330 × (260)2 × (0.24 × 10–9)2 C1 = 3.1 × 10–11 (W m–2) C1 = 31 (30.8) pW m–2 A1 [3]

This question in 9702/23 Oct/Nov 2015

1 The speed v of a transverse wave on a uniform string is given by the expression T l v = m where T is the tension in the string, l is its length and m is its mass. An experiment is performed to determine the speed v of the wave. The measurements are shown in Fig. 1.1. quantity measurement uncertainty T 1.8 N ± 5% l 126 cm ± 1% m 5.1 g ± 2% Fig. 1.1 (a) State an appropriate instrument to measure the length l. … [1] (b) (i) Use the data in Fig. 1.1 to calculate the speed v. v = … m s−1 [2] (ii) Use your answer in (b)(i) and the data in Fig. 1.1 to determine the value of v, with its absolute uncertainty, to an appropriate number of significant figures. v = … ± … m s−1 [3] [Total: 6]

6 marks

Mark scheme: 1 (a) metre rule / tape measure B1 (b) (i) v = [(1.8 × 126 × 10–2) / 5.1 × 10–3]1 / 2 C1 = 21.1 (m s–1) A1 (ii) percentage uncertainty = 4% or fractional uncertainty = 0.04 C1 ∆v = 0.04 × 21.1 = 0.84 C1 v = 21.1 ± 0.8 (m s–1) A1

This question in 9702/22 Feb/March 2016

1 (a) Make estimates of (i) the mass, in kg, of a wooden metre rule, mass = … kg [1] (ii) the volume, in cm3, of a cricket ball or a tennis ball. volume = … cm3 [1] (b) A metal wire of length L has a circular cross-section of diameter d, as shown in Fig. 1.1. / G Fig. 1.1 The volume V of the wire is given by the expression πd 2L V = . 4 The diameter, length and mass M are measured to determine the density of the metal of the wire. The measured values are: d = 0.38 ± 0.01 mm, L = 25.0 ± 0.1 cm, M = 0.225 ± 0.001 g. Calculate the density of the metal, with its absolute uncertainty. Give your answer to an appropriate number of significant figures. density = … ± … kg m–3 [5] [Total: 7]

7 marks

Mark scheme: 1 (a) (i) (50 to 200) × 10–3 kg or (0.05 to 0.2) kg B1 [1] (ii) (50 to 300) cm3 B1 [1] (b) density = mass / volume or ρ = M / V C1 V = [π(0.38 × 10–3)2 × 25.0 × 10–2] / 4 (= 2.835 × 10–8 m3) C1 ρ = (0.225 × 10–3) / 2.835 × 10–8 = 7940 (kg m–3) A1 ∆ρ / ρ = 2(0.01/0.38) + (0.1/25.0) + (0.001/0.225) [= 0.061] or %ρ = 5.3% + 0.40% + 0.44% (= 6.1%) C1 ∆ρ = 0.061 × 7940 = 480 (kg m–3) density = (7.9 ± 0.5) × 103 kg m–3 or (7900 ± 500) kg m–3 A1 [5] 1

This question in 9702/21 May/June 2016

7 (a) Electric current is a flow of charge carriers. The charge on the carriers is quantised. Explain what is meant by quantised. … [1] (b) A battery of electromotive force (e.m.f.) 9.0 V and internal resistance 0.25 Ω is connected in series with two identical resistors X and a resistor Y, as shown in Fig. 7.1. battery 9.0 V 0.25 1 X Y X 0.15 1 2.7 1 0.15 1 Fig. 7.1 The resistance of each resistor X is 0.15 Ω and the resistance of resistor Y is 2.7 Ω. (i) Show that the current in the circuit is 2.8 A. [3] (ii) Calculate the potential difference across the battery. potential difference = … V [2] (c) Each resistor X connected in the circuit in (b) is made from a wire with a cross-sectional area of 2.5 mm2. The number of free electrons per unit volume in the wire is 8.5 × 1029 m–3. (i) Calculate the average drift speed of the electrons in X. drift speed = … m s–1 [2] (ii) The two resistors X are replaced by two resistors Z made of the same material and length but with half the diameter. Describe and explain the difference between the average drift speed in Z and that in X. … … … … [2] [Total: 10]

10 marks

Mark scheme: 7 (a) charge exists only in discrete amounts B1 [1] (b) (i) E = I(R + r) or V = IR C1 (total resistance =) 2.7 + 0.30 + 0.25 (= 3.25 Ω) M1 I = 9.0 / (2.7 + 0.30 + 0.25) or 9.0 / 3.25 = 2.8 A A1 [3] (ii) V = IRext C1 = 2.77 × 3.0 or 2.8 × 3.0 or V = E – Ir (C1) = 9.0 – 2.77 × 0.25 or 9.0 – 2.8 × 0.25 V = 8.3 (8.31) V or 8.4 V A1 [2] (c) (i) I = nevA v = 2.77 / (8.5 × 1029 × 1.6 × 10–19 × 2.5 × 10–6) M1 = 8.1 (8.147) × 10–6 m s–1 or 8.2 × 10–6 m s–1 A1 [2] (ii) A reduces by a factor 4 (1/4 less) or resistance of Z goes up by 4× M1 current goes down but by less than a factor of 4 (as total resistance does not go up by a factor of 4) so drift speed goes up A1 [2]

This question in 9702/22 May/June 2016

1 (a) Define density. … … [1] (b) The mass m of a metal sphere is given by the expression πd 3ρ m = 6 where ρ is the density of the metal and d is the diameter of the sphere. Data for the density and the mass are given in Fig. 1.1. quantity value uncertainty ρ 8100 kg m–3 ± 5% m 7.5 kg ± 4% Fig. 1.1 (i) Calculate the diameter d. d = … m [1] (ii) Use your answer in (i) and the data in Fig. 1.1 to determine the value of d, with its absolute uncertainty, to an appropriate number of significant figures. d = … ± … m [3] [Total: 5]

5 marks

Mark scheme: 1 (a) (density =) mass / volume B1 [1] (b) (i) d = [(6 × 7.5) / (π × 8100)]1/3 = 0.12(1) m A1 [1] (ii) percentage uncertainty = (4 + 5) / 3 (= 3%) or fractional uncertainty = (0.04 + 0.05) / 3 (= 0.03) C1 absolute uncertainty (= 0.03 × 0.121) = 0.0036 C1 d = 0.121 ± 0.004 m A1 [3]

This question in 9702/21 Oct/Nov 2016

3 (a) Define velocity. … … [1] (b) A car travels in a straight line up a slope, as shown in Fig. 3.1. ms–1 9.0 car mass 850 kg slope Fig. 3.1 The car has mass 850 kg and travels with a constant speed of 9.0 m s–1. The car’s engine exerts a force on the car of 2.0 kN up the slope. A resistive force FD, due to friction and air resistance, opposes the motion of the car. The variation of FD with the speed v of the car is shown in Fig. 3.2. 0.70 FD / kN 0.60 0.50 0.40 0.30 7 8 9 10 11 12 13 14 15 16 v / m s–1 Fig. 3.2 (i) State and explain whether the car is in equilibrium as it moves up the slope. … … … [2] (ii) Consider the forces that act along the slope. Use data from Fig. 3.2 to determine the component of the weight of the car that acts down the slope. component of weight = … N [2] (iii) Show that the power output of the car is 1.8 × 104 W. [2] (iv) The car now travels along horizontal ground. The output power of the car is maintained at 1.8 × 104 W. The variation of the resistive force FD acting on the car is given in Fig. 3.2. Calculate the acceleration of the car when its speed is 15 m s–1. acceleration = … m s–2 [3] [Total: 10]

10 marks

Mark scheme: 3(a) change of displacement / time (taken) B1 3(b)(i) constant velocity, so resultant force is zero M1 (so car is) in (dynamic) equilibrium A1 3(b)(ii) FD = 0.40 (kN) or 0.40 × 103 (N) C1 component of weight = 2.0 × 103 – 0.40 × 103 = 1.6 × 103 N A1 3(b)(iii) P = Fv C1 = 2.0 ×103 × 9.0 = 1.8 × 104 W A1 3(b)(iv) (driving) force = 1.8 × 104 / 15 (= 1.2 × 103) C1 FD = 0.66 (kN) or 0.66 × 103 (N) C1 acceleration = (1.2 × 103 – 0.66 × 103) / 850 = 0.64 (0.635) m s–2 A1

This question in 9702/22 Feb/March 2017

2 (a) Define velocity. … … [1] (b) A ball of mass 0.45 kg leaves the edge of a table with a horizontal velocity v, as shown in Fig. 2.1. ball v path of ball 1.25 m table 1.50 m floor horizontal Fig. 2.1 The height of the table is 1.25 m. The ball travels a distance of 1.50 m horizontally before hitting the floor. Air resistance is negligible. Calculate, for the ball, (i) the horizontal velocity v as it leaves the table, v = … m s–1 [3] (ii) the velocity just as it hits the floor, magnitude of velocity = … m s–1 angle to the horizontal = … ° [4] (iii) the kinetic energy just as it hits the floor, kinetic energy = … J [2] (iv) the loss in gravitational potential energy as it falls from the table to the floor. loss in potential energy = … J [2] (c) Explain why the kinetic energy of the ball in (b)(iii) does not equal the loss of gravitational potential energy in (b)(iv). … … [1] [Total: 13]

13 marks

Mark scheme: 2(a) rate of change of displacement or change in displacement/time taken B1 2(b)(i) s = ut + ½at2 C1 t = [(2 × 1.25) / 9.81]1/2 (= 0.5048 s) C1 or v2 = u2 + 2as vvert = (2 × 9.81 × 1.25)1/2 (= 4.95) (C1) t = [2s / (u + v)] = 2 × 1.25 / 4.95 (= 0.5048 s) (C1) v = d / t = 1.5 / 0.50(48) = 3.0 (2.97) m s–1 A1 2(b)(ii) vertical velocity = at = 9.81 × 0.5048 (= 4.95) [using t = 0.50 gives 4.9] C1 velocity = [(vh)2 + (vv)2]1/2 C1 = [(2.97)2 + (4.95)2]1/2 = 5.8 (5.79) [using t = 0.50 leads to 5.7] A1 direction (= tan–1 4.95/2.97) = 59° A1 2(b)(iii) kinetic energy = ½mv2 C1 = ½ × 0.45 × (5.8)2 = 7.6 (7.57) J [using t = 0.50 leads to 7.3 J] A1 Question Answer Marks 2(b)(iv) potential energy = mgh C1 = (0.45 × 9.81 × 1.25) = 5.5 (5.52) J A1 2(c) there is KE of the ball at the start/leaving table or the ball has an initial/constant horizontal velocity or the ball has velocity at start/leaving table B1

This question in 9702/22 May/June 2017

1 (a) Define (i) displacement, … … [1] (ii) acceleration. … … [1] (b) A remote-controlled toy car moves up a ramp and travels across a gap to land on another ramp, as illustrated in Fig. 1.1. path of car 5.5 m s–1 car ramp P ramp Q d ground θ Fig. 1.1 The car leaves ramp P with a velocity of 5.5 m s–1 at an angle θ to the horizontal. The horizontal component of the car’s velocity as it leaves the ramp is 4.6 m s–1. The car lands at the top of ramp Q. The tops of both ramps are at the same height and are distance d apart. Air resistance is negligible. (i) Show that the car leaves ramp P with a vertical component of velocity of 3.0 m s–1. [1] (ii) Determine the time taken for the car to travel between the ramps. time taken = … s [2] (iii) Calculate the horizontal distance d between the tops of the ramps. d = … m [1] (iv) Calculate the ratio kinetic energy of the car at its maximum height . kinetic energy of the car as it leaves ramp P ratio = … [3] (c) Ramp Q is removed. The car again leaves ramp P as in (b) and now lands directly on the ground. The car leaves ramp P at time t = 0 and lands on the ground at time t = T. On Fig. 1.2, sketch the variation with time t of the vertical component vy of the car’s velocity from t = 0 to t = T. Numerical values of vy and t are not required. vy 0 0 TT t t Fig. 1.2 [2] [Total: 11]

11 marks

Mark scheme: 1(a)(i) distance in a specified direction (from a point) B1 1(a)(ii) change in velocity / time (taken) B1 1(b)(i) vertical component of velocity = (5.52 – 4.62)1/2 = 3.0 (m s–1) or 5.5 cos θ = 4.6 (so θ = 33.2°) and 5.5 sin 33.2° = 3.0 (m s–1) A1 1(b)(ii) s = ut + ½at 2 0 = (3.0 × t) – (½ × 9.81 × t 2) or v = u + at –3.0 = 3.0 – 9.81t C1 t = 0.61 s A1 1(b)(iii) d = 4.6 × 0.61 = 2.8 m A1 1(b)(iv) E = ½mv2 C1 ratio = (½ × m × 4.62) / (½ × m × 5.52) or ratio = (½ × m × 5.52 – m × 9.81 × 0.459) / (½ × m × 5.52) C1 ratio = 0.70 A1 1(c) straight line from positive value of vy at t = 0 to negative value of vy M1 straight line ends at t = T and final magnitude of vy greater than initial magnitude of vy A1

This question in 9702/21 Oct/Nov 2018

2 (a) Define: (i) displacement … … [1] (ii) acceleration. … … [1] (b) A man wearing a wingsuit glides through the air with a constant velocity of 47 m s–1 at an angle of 24° to the horizontal. The path of the man is shown in Fig. 2.1. 47 m s–1 A man in wingsuit glide path total mass 85 kg h 24° B horizontal Fig. 2.1 (not to scale) The total mass of the man and the wingsuit is 85 kg. The man takes a time of 2.8 minutes to glide from point A to point B. (i) With reference to the motion of the man, state and explain whether he is in equilibrium. … … … … [2] (ii) Show that the difference in height h between points A and B is 3200 m. [1] (iii) For the movement of the man from A to B, determine: 1. the decrease in gravitational potential energy decrease in gravitational potential energy = … J [2] 2. the magnitude of the force on the man due to air resistance. force = … N [2] (iv) The pressure of the still air at A is 63 kPa and at B is 92 kPa. Assume the density of the air is constant between A and B. Determine the density of the air between A and B. density = … kg m–3 [2] [Total: 11]

11 marks

Mark scheme: 2(a)(i) distance in a specified direction (from a point) B1 2(a)(ii) change in velocity / time (taken) B1 2(b)(i) constant velocity so no resultant force B1 no resultant force so in equilibrium B1 2(b)(ii) (difference in height =) 47 × 2.8 × 60 × sin24° = 3200 m A1 Question Answer Marks 2(b)(iii) 1 (∆)E = mg(∆)h = 85 × 9.81 × 3200 C1 = 2.7 × 106 J A1 2 In terms of energy: work done = 2.7 × 106 J force = 2.7 × 106 / (47 × 2.8 × 60) C1 = 340 N A1 In terms of forces: component of weight along path = force due to air resistance force = 85 × 9.81 × sin24° (C1) = 340 N (A1) 2(b)(iv) (∆)p = ρg(∆)h (92 – 63) × 103 = ρ × 9.81 × 3200 C1 ρ = 0.92 kg m–3 A1

This question in 9702/22 Feb/March 2019

1 (a) Define velocity. … … [1] (b) The speed v of a sound wave through a gas of pressure P and density ρ is given by the equation kP v = ρ where k is a constant that has no units. An experiment is performed to determine the value of k. The data from the experiment are shown in Fig. 1.1. quantity value uncertainty v 3.3 × 102 m s−1 ± 3% P 9.9 × 104 Pa ± 2% ρ 1.29 kg m−3 ± 4% Fig. 1.1 (i) Use data from Fig. 1.1 to calculate k. k = … [2] (ii) Use your answer in (b)(i) and data from Fig. 1.1 to determine the value of k, with its absolute uncertainty, to an appropriate number of significant figures. k = … ± … [3] [Total: 6]

6 marks

Mark scheme: 1(a) (velocity =) change in displacement / time (taken) B1 1(b)(i) k = [1.29 × (3.3 × 102)2] / 9.9 × 104 C1 = 1.4 A1 1(b)(ii) percentage uncertainty = (3 × 2) + 4 + 2 (= 12%) or fractional uncertainty = (0.03 × 2) + 0.04 + 0.02 (= 0.12) C1 ∆k = 0.12 × 1.42 = 0.17 (allow to 1 significant figure) C1 k = 1.4 ± 0.2 A1

This question in 9702/21 May/June 2019

4 (a) A spherical oil drop has a radius of 1.2 × 10−6 m. The density of the oil is 940 kg m−3. (i) Show that the mass of the oil drop is 6.8 × 10−15 kg. [2] (ii) The oil drop is charged. Explain why it is impossible for the magnitude of the charge to be 8.0 × 10−20 C. … … [1] (b) The charged oil drop in (a) is in a vacuum between two horizontal metal plates, as illustrated in Fig. 4.1. metal plate + V oil drop, mass 6.8 × 10–15 kg 8.0 mm uniform electric field, metal plate field strength 2.1 × 105 V m–1 Fig. 4.1 The plates are separated by a distance of 8.0 mm. The electric field between the plates is uniform and has a field strength of 2.1 × 105 V m−1. The oil drop moves vertically downwards with a constant speed. (i) Calculate the potential difference V between the plates. V = … V [2] (ii) Explain how the motion of the oil drop shows that it is in equilibrium. … … [1] (iii) Determine the charge on the oil drop. charge = … C sign of charge … [3] (c) The magnitude of the potential difference between the plates in (b) is decreased. (i) Explain why the oil drop accelerates downwards. … … … … [2] (ii) Describe the change to the pattern of the field lines (lines of force) representing the uniform electric field as the potential difference decreases. … … [1] (d) Two types of force, X and Y, can act on an oil drop when it is in air, but cannot act on an oil drop when it is in a vacuum. Force X can act on an oil drop when it is stationary or when it is moving. Force Y can only act on an oil drop when it is moving. State the name of: (i) force X … [1] (ii) force Y. … [1] [Total: 14]

14 marks

Mark scheme: 4(a)(i) C1 = (4 / 3) × π × (1.2 × 10–6)3 × 940 = 6.8 × 10–15 (kg) A1 4(a)(ii) minimum charge (on drop) is 1.6 × 10–19 C B1 4(b)(i) V = Ed C1 V = 2.1 × 105 × 8.0 × 10–3 = 1.7 × 103 V A1 4(b)(ii) constant velocity so no resultant force (so in equilibrium) B1 4(b)(iii) mg = Eq or mg = (V / d)q or F = mg and F = Eq C1 q = (6.8 × 10–15 × 9.81) / 2.1 × 105 = 3.2 × 10–19 C A1 sign of charge is negative A1 4(c)(i) electric force decreases B1 weight > electric force or resultant force acts downwards B1 4(c)(ii) (field line) separation increases B1 4(d)(i) upthrust (force) B1 4(d)(ii) air resistance/drag/viscous (force) B1

This question in 9702/21 May/June 2019

1 (a) Make estimates of: (i) the mass, in g, of a new pencil mass = … g [1] (ii) the wavelength of ultraviolet radiation. wavelength = … m [1] (b) The period T of the oscillations of a mass m suspended from a spring is given by m T = 2π k where k is the spring constant of the spring. The manufacturer of a spring states that it has a spring constant of 25 N m–1 ± 8%. A mass of 200 × 10–3 kg ± 4 × 10–3 kg is suspended from the end of the spring and then made to oscillate. (i) Calculate the period T of the oscillations. T = … s [1] (ii) Determine the value of T, with its absolute uncertainty, to an appropriate number of significant figures. T = … ± … s [3] [Total: 6]

6 marks

Mark scheme: 1(a)(i) A1 1(a)(ii) wavelength in range 1 × 10–8 m to 4 × 10–7 m A1 1(b)(i) T = 2π × (200 × 10–3 / 25)0.5 = 0.56 s A1 1(b)(ii) percentage uncertainty = (2% + 8%) / 2 (= 5%) or fractional uncertainty = (0.02+0.08) / 2 (= 0.05) C1 ∆T = 0.56 × 0.05 = 0.028 (s) C1 T = (0.56 ± 0.03) s A1

This question in 9702/21 Oct/Nov 2019

1 (a) Define velocity. … … [1] (b) The drag force FD acting on a car moving with speed v along a straight horizontal road is given by FD = v 2Ak where k is a constant and A is the cross-sectional area of the car. Determine the SI base units of k. SI base units … [2] (c) The value of k, in SI base units, for the car in (b) is 0.24. The cross-sectional area A of the car is 5.1 m2. The car is travelling with a constant speed along a straight road and the output power of the engine is 4.8 × 104 W. Assume that the output power of the engine is equal to the rate at which the drag force FD is doing work against the car. Determine the speed of the car. speed = … m s–1 [3] [Total: 6]

6 marks

Mark scheme: 1(a) (velocity =) change in displacement / time (taken) B1 1(b) units of F: kg m s–2 C1 units of k: kg m s–2 / [m2 × (m s–1)2] = kg m–3 A1 1(c) P = Fv C1 4.8 × 104 = 0.24 × 5.1 × v 3 C1 v = 34 m s–1 A1

This question in 9702/22 May/June 2020

2 (a) A cylinder is suspended from the end of a string. The cylinder is stationary in water with the axis of the cylinder vertical, as shown in Fig. 2.1. cylinder string weight 0.84 N water h density 1.0 × 103 kg m–3 0.031 m Fig. 2.1 (not to scale) The cylinder has weight 0.84 N, height h and a circular cross-section of diameter 0.031 m. The density of the water is 1.0 × 103 kg m−3. The difference between the pressures on the top and bottom faces of the cylinder is 520 Pa. (i) Calculate the height h of the cylinder. h = … m [2] (ii) Show that the upthrust acting on the cylinder is 0.39 N. [2] (iii) Calculate the tension T in the string. T = … N [1] (b) The string is now used to move the cylinder in (a) vertically upwards through the water. The variation with time t of the velocity v of the cylinder is shown in Fig. 2.2. 12.5 10.0 v / cm s–1 7.5 5.0 2.5 0 0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 t / s Fig. 2.2 (i) Use Fig. 2.2 to determine the acceleration of the cylinder at time t = 2.0 s. acceleration = … m s−2 [2] (ii) The top face of the cylinder is at a depth of 0.32 m below the surface of the water at time t = 0. Use Fig. 2.2 to determine the depth of the top face below the surface of the water at time t = 4.0 s. depth = … m [2] (c) The cylinder in (b) is released from the string at time t = 4.0 s. The cylinder falls, from rest, vertically downwards through the water. Assume that the upthrust acting on the cylinder remains constant as it falls. (i) State the name of the force that acts on the cylinder when it is moving and does not act on the cylinder when it is stationary. … [1] (ii) State and explain the variation, if any, of the acceleration of the cylinder as it falls downwards through the water. … … … [2] [Total: 12]

12 marks

Mark scheme: 2(a)(i) (Δ)p = ρg(Δ)h 520 = 1000 × 9.81 × h C1 h = 0.053 m A1 2(a)(ii) (upthrust =) (Δ)p × A = (Δ)p × π(d / 2)2 or (Δ)p × πr 2 C1 = 520 × π(0.031 / 2)2 = 0.39 (N) A1 2(a)(iii) T = 0.84 – 0.39 = 0.45 N A1 Question Answer Marks 2(b)(i) a = (v – u) / t or (Δ)v / (∆)t or gradient C1 = e.g. 8.0 × 10–2 / 2.0 = 4.0 × 10–2 m s–2 A1 2(b)(ii) distance = (½ × 2.5 × 0.10) + (½ × 1.5 × 0.10) or (½ × 4.0 × 0.10) (= 0.20 (m)) C1 depth = 0.32 – 0.20 = 0.12 m A1 2(c)(i) viscous (force) B1 2(c)(ii) viscous force increases (with speed/time/depth) B1 (so) acceleration decreases B1

This question in 9702/22 Oct/Nov 2020

3 A pendulum consists of a solid sphere suspended by a string from a fixed point P, as shown in Fig. 3.1. P θ string 0.93 m Y sphere h X momentum 0.72 N s Fig. 3.1 (not to scale) The sphere swings from side to side. At one instant the sphere is at its lowest position X, where it has kinetic energy 0.86 J and momentum 0.72 N s in a horizontal direction. A short time later the sphere is at position Y, where it is momentarily stationary at a maximum vertical height h above position X. The string has a fixed length and negligible weight. Air resistance is also negligible. (a) On Fig. 3.1, draw a solid line to represent the displacement of the centre of the sphere at position Y from position X. [1] (b) Show that the mass of the sphere is 0.30 kg. [3] (c) Calculate height h. h = … m [2] (d) The distance between point P and the centre of the sphere is 0.93 m. When the sphere is at position Y, the string is at an angle θ to the vertical. Show that θ is 47°. [1] (e) For the sphere at position Y, calculate the moment of its weight about point P. moment = … N m [2] (f) State and explain whether the sphere is in equilibrium when it is stationary at position Y. … … [1] [Total: 10]

10 marks

Mark scheme: 3(a) solid straight line drawn between centre of sphere at X and at Y B1 3(b) p = mv or 0.72 = mv C1 E = ½mv 2 or 0.86 = ½mv 2 C1 (m =) 0.722 / (2 × 0.86) = 0.30 (kg) or v = 2EK / p v = (0.86 × 2) / 0.72 = 2.4 (to 2 s.f.) m = 0.72 / 2.4 = 0.30 (kg) A1 3(c) (Δ)E = mg(Δ)h C1 h = 0.86 / (0.30 × 9.81) = 0.29 m A1 3(d) cosθ = (0.93 – 0.29) / 0.93 so θ = 47° A1 3(e) moment = (0.30 × 9.81) × 0.93 × (sin 47° or cos 43°) or moment = (0.30 × 9.81) × [0.932 – (0.93 – 0.29)2]0.5 C1 = 2.0 N m A1 Question Answer Mark 3(f) there is a resultant force (acting on sphere) or there is a resultant moment (about P acting on pendulum) (so) not in equilibrium B1

This question in 9702/21 May/June 2021

2 (a) Define acceleration. … … [1] (b) A stone falls vertically from the top of a cliff. Fig. 2.1 shows the variation with time t of the velocity v of the stone. 40 v / m s–1 30 20 10 0 0 5 10 15 20 25 30 t / s Fig. 2.1 (i) Explain, with reference to forces acting on the stone, the shape of the curve in Fig. 2.1. … … … … … [3] (ii) Use Fig. 2.1 to determine the speed of the stone when the resultant force on it is zero. speed = … m s–1 [1] (iii) Use Fig. 2.1 to calculate the approximate height through which the stone falls between t = 0 and t = 30 s. height = … m [3] (iv) On Fig. 2.2, sketch the variation with t of the acceleration a of the stone between t = 0 and t = 30 s. 20 a / m s–2 15 10 5 0 0 5 10 15 20 25 30 t / s Fig. 2.2 [3] [Total: 11]

11 marks

Mark scheme: 2(a) change in velocity / time (taken) B1 2(b)(i) air resistance increases (with speed/with time) B1 resultant force decreases (as speed increases/with time) so acceleration decreases (as speed increases/with time) B1 when air resistance equals the weight the speed/velocity/v becomes constant B1 2(b)(ii) speed = 36 m s–1 A1 2(b)(iii) height given by area under the curve C1 height = 950 m Round to two significant figures and award 2 marks for a value in the range 920–980 m and 1 mark for a value in the range 900–910 m or 990–1000 m. A2 2(b)(iv) line starting at (0, 9.8) B1 curve with negative gradient between t = 0 and t = 20 s B1 line showing zero acceleration between t = 20 s and t = 30 s B1 Question Answer Marks

This question in 9702/23 May/June 2021

1 (a) Define density. … … [1] (b) A smooth pebble, made from uniform rock, has the shape of an elongated sphere as shown in Fig. 1.1. r L Fig. 1.1 The length of the pebble is L. The cross-section of the pebble, in the plane perpendicular to L, is circular with a maximum radius r. A student investigating the density of the rock makes measurements to determine the values of L, r and the mass M of the pebble as follows: L = (0.1242 ± 0.0001) m r = (0.0420 ± 0.0004) m M = (1.072 ± 0.001) kg. (i) State the name of a measuring instrument suitable for making this measurement of L. … [1] (ii) Determine the percentage uncertainty in the measurement of r. percentage uncertainty = … % [1] (c) The density ρ of the rock from which the pebble in (b) is composed is given by Mr n ρ = kL where n is an integer and k is a constant, with no units, that is equal to 2.094. (i) Use SI base units to show that n is equal to –2. [2] (ii) Calculate the percentage uncertainty in ρ. percentage uncertainty = … % [3] (iii) Determine ρ with its absolute uncertainty. Give your values to the appropriate number of significant figures. ρ = ( … ± … ) kg m–3 [3] [Total: 11]

11 marks

Mark scheme: 1(a) mass / volume B1 1(b)(i) (vernier/digital) calipers B1 1(b)(ii) percentage uncertainty = (0.0004 / 0.0420) × 100 = 1% A1 1(c)(i) kg m–3 = kg × mn / m or kg m–3 = kg × mn × m–1 M1 –3 = n – 1 and (so) n = –2 A1 1(c)(ii) (Δρ / ρ) = (ΔM / M) + 2(Δr / r) + (ΔL / L) C1 percentage uncertainty = [(0.001 / 1.072) + 2 × (0.0004 / 0.0420) + (0.0001 / 0.1242)] (× 100) C1 = 0.09% + 2 × 0.95% + 0.08% = 2% A1 1(c)(iii) ρ = (1.072 × 0.0420–2) / (2.094 × 0.1242) = 2337 (kg m–3) C1 ∆ρ = 0.021 × 2337 = 49 (kg m–3) C1 ρ = (2340 ± 50) kg m–3 A1

This question in 9702/21 Oct/Nov 2021

1 (a) A unit may be stated with a prefix that represents a power-of-ten multiple or submultiple. Complete Table 1.1 to show the name and symbol of each prefix and the corresponding power-of-ten multiple or submultiple. Table 1.1 power-of-ten multiple prefix or submultiple kilo (k) 103 tera (T) ( ) 10–12 [2] (b) In the following list, underline all the units that are SI base units. ampere coulomb metre newton [1] (c) The potential difference V between the two ends of a uniform metal wire is given by 4ρLI V = 2 πd where d is the diameter of the wire, I is the current in the wire, L is the length of the wire, and ρ is the resistivity of the metal. For a particular wire, the percentage uncertainties in the values of some of the above quantities are listed in Table 1.2. Table 1.2 quantity percentage uncertainty d ± 3.0% I ± 2.0% L ± 2.5% V ± 3.5% The quantities listed in Table 1.2 have values that are used to calculate ρ as 4.1 × 10–7 Ω m. For this value of ρ, calculate: (i) the percentage uncertainty percentage uncertainty = … % [2] (ii) the absolute uncertainty. absolute uncertainty = … Ω m [1] [Total: 6]

6 marks

Mark scheme: 1(a) 1012 B1 pico (p) B1 1(b) ampere and metre both underlined (and no other units underlined) B1 1(c)(i) percentage uncertainty = 3.5 + (3.0 × 2) + 2.5 + 2.0 C1 = 14% A1 1(c)(ii) absolute uncertainty = 4.1 × 10–7 × 14 / 100 = 6 × 10–8 Ω m A1

This question in 9702/22 Oct/Nov 2021

3 (a) Define velocity. … … [1] (b) A constant driving force of 2400 N acts on a car of mass 1200 kg. The car accelerates from rest in a straight line along a horizontal road. Assume that the resistive forces acting on the car are negligible. (i) Calculate the acceleration of the car. acceleration = … m s–2 [1] (ii) On Fig. 3.1, sketch a graph showing the variation with time t of the velocity v of the car for the first 20 seconds of its motion. Label this line A. 50 40 v / m s–1 30 20 10 0 0 4 8 12 16 20 t / s Fig. 3.1 [2] (c) In reality, a resistive force due to air resistance acts on the car in (b). This resistive force increases with speed until it becomes equal in magnitude to the driving force at time t = 12 s. (i) On Fig. 3.1, sketch a second line to show the variation with time t of the velocity v of the car for the first 20 seconds of its motion. Label this line B. [3] (ii) At time t = 20 s, the driving force is increased to 3000 N and remains constant at this value. Describe how the velocity of the car changes due to this increase in the driving force. … … … … [2] [Total: 9]

9 marks

Mark scheme: 3(a) (velocity =) change in displacement / time B1 3(b)(i) a = 2400 / 1200 = 2.0 m s−2 A1 3(b)(ii) straight line from the origin with positive gradient (labelled A) M1 ending at (20, 40) A1 3(c)(i) line starting at origin (with the same gradient as A) and beneath A at all points B1 gradient decreasing to zero B1 straight horizontal line from t = 12 s and ending at t = 20 s (and labelled B) B1 3(c)(ii) the velocity/speed will increase B1 to a new terminal/constant/maximum velocity/speed B1 or the car has an acceleration (B1) to a new (higher) terminal/constant/maximum velocity/speed (B1)

This question in 9702/23 May/June 2022

1 (a) The boxes in Fig. 1.1 contain terms on the left-hand side and examples of these terms on the right-hand side. Draw a line between each term on the left and the correct example on the right. base quantity coulomb base unit electric current derived quantity force derived unit kilogram Fig. 1.1 [2] (b) A set of experimental measurements is described as precise and not accurate. State what is meant by: (i) precise … … [1] (ii) not accurate. … … [1] (c) An object of mass m travels with speed v in a circle of radius r. The force F acting on the object is given by mv2 F = . r The percentage uncertainties of three of the quantities are given in Table 1.1. Table 1.1 quantity percentage uncertainty F ± 3% m ± 4% r ± 5% The value of v is determined from F, m and r. (i) Calculate the percentage uncertainty in v. percentage uncertainty = … % [2] (ii) The value of v is 15.0 m s–1. Calculate the absolute uncertainty in v. absolute uncertainty = … m s–1 [1] [Total: 7]

7 marks

Mark scheme: Question Answer Marks 1(a) C1 any two joined correctly all four joined correctly A1 1(b)(i) the measurements have a small range B1 1(b)(ii) (average of the) measurements not close to the true value B1 1(c)(i) percentage uncertainty = (3 + 5 + 4) / 2 C1 = 6% A1 1(c)(ii) absolute uncertainty = (6 / 100)  15.0 A1 = 0.9 ms–1

This question in 9702/21 Oct/Nov 2022

1 (a) In the following list, underline all quantities that are SI base quantities. charge electric current force time [1] (b) Under certain conditions, the distance s moved in a straight line by an object in time t is given by s = 12at 2 where a is the acceleration of the object. State two conditions under which the above expression applies to the motion of the object. 1 … 2 … [2] (c) The variation with time t of the velocity v of a car that is moving in a straight line is shown in Fig. 1.1. 25 20 15 v / m s–1 10 5 0 0 2 4 6 8 10 12 14 16 t / s –5 –10 –15 Fig. 1.1 (i) Compare, qualitatively, the acceleration of the car at time t = 8.0 s and at time t = 14.0 s in terms of: ● magnitude … … ● direction. … … [2] (ii) Determine the magnitude of the acceleration of the car at time t = 4.0 s. acceleration = … m s–2 [2] (iii) The car is at point X at time t = 0. Determine the magnitude of the displacement of the car from X at time t = 12.0 s. displacement = … m [2] [Total: 9]

9 marks

Mark scheme: Question Answer Marks 1(a) only electric current and time underlined B1 1(b) initial speed / velocity is zero B1 (non-zero magnitude of) acceleration is constant / uniform (and in a straight line) B1 1(c)(i) • magnitude of acceleration at t = 8.0 s is less than that at t = 14.0 s B1 • direction of acceleration at t = 8.0 s is opposite to that at t = 14.0 s B1 1(c)(ii) a = gradient or a = (v – u) / t or a = ∆v / (∆)t C1 a = e.g. (20 + 10) / 12 or (0 + 10) / 4 or (20 – 0) / (12 – 4) A1 a = 2.5 m s–2 1(c)(iii) displacement = [½  (12 – 4)  20] – [½  4  10] C1 or displacement = (–10  12) + (½  2.5  122) or displacement = (20  12) – (½  2.5  122) or displacement = ½  (20 – 10)  12 displacement = 60 m A1

This question in 9702/22 Oct/Nov 2023

1 The drag force FD acting on an object falling through air is given by 1 FD = CρAv 2 2 where A is the cross-sectional area of the object, v is the velocity of the object in the air, ρ is the density of the air and C is a constant called the drag coefficient. (a) Use SI base units to show that the drag coefficient has no units. [3] (b) Fig. 1.1 shows a sphere falling at terminal velocity in air. sphere terminal velocity Fig. 1.1 Assume that the upthrust on the sphere is negligible. On Fig. 1.1, draw and label arrows to show the directions of the two forces acting on the sphere. [2] (c) The mass of the sphere is 49 g. Calculate the drag force FD acting on the sphere. FD = … N [2] (d) The sphere is falling in air at a terminal velocity of 25 in SI base units. The density of the air is 1.2 in SI base units. The diameter of the sphere is 0.060 in SI base units. Use your answer in (c) to calculate the drag coefficient C for the sphere. C = … [3] [Total: 10]

10 marks

Mark scheme: 1(a) units of FD: kg m s–2 M1 units of  kg m–3 and units of A: m2 and units of v: m s−1 or units of v2: m2 s–2 M1 kg m s–2 = C kg m s–2 and comment ‘(so) C has no units’ / unit terms cancelled or C = kg m s−2 / (kg m–3 m2 m2 s–2) and comment ‘(so) C has no units’ / unit terms cancelled A1 1(b) one arrow vertically downward labelled weight to within 10° of the vertical B1 one arrow vertically upwards labelled drag / drag force / FD / air resistance / viscous force to within 10° of the vertical B1 1(c) (at terminal velocity) FD = mg C1 FD = 0.049  9.81 = 0.48 N A1 1(d) area =   (0.060 / 2)2 C1 0.48 = ½  C  1.2    (0.060 / 2)2  252 C1 C = 0.45 A1

This question in 9702/21 May/June 2024

2 (a) Define velocity. … … [1] (b) A student throws a ball over a vertical wall of height h, as shown in Fig. 2.1. path of ball wall 22 m s–1 ball 40° horizontal h ground 1.2 m 36 m Fig. 2.1 (not to scale) The ball leaves the hand of the student at a height of 1.2 m above the horizontal ground. The ball has an initial velocity of 22 m s–1 at an angle of 40° to the horizontal. The wall is a horizontal distance of 36 m from where the student releases the ball. Air resistance is negligible. (i) Determine the time taken for the ball to reach the wall. time taken = … s [2] (ii) Calculate the vertical component u of the initial velocity of the ball. u = … m s–1 [1] (iii) The ball just goes over the wall. Calculate the height h of the wall. h = … m [3] [Total: 7]

7 marks

Mark scheme: 2(a) change in displacement / time (taken) B1 2(b)(i) horizontal velocity = 22  cos 40° C1 time taken = 36 / (22  cos 40°) = 2.1 s A1 2(b)(ii) u = 22  sin 40° = 14 m s−1 A1 2(b)(iii) s = ut + ½ at2 = (14  2.1) + (½  −9.81  2.12) C1 = 7.8 (m) C1 (therefore) height of wall = 7.8 + 1.2 = 9.0 m A1 Question Answer Marks 2(b)(iii) or other methods possible e.g. time to maximum height = (0 – 14) / –9.81 (= 1.43 s) time from maximum height to wall = 2.1 – 1.43 (= 0.67 s) maximum height above release = (14  1.43) + (½  −9.81  1.432) = 9.99 (m) (C1) height from maximum to wall = 0.5  9.81  0.672 (= 2.2 m) height above release = 9.99 – 2.2 = 7.8 (m) (C1) height of wall = 1.2 + 7.8 = 9.0 m (A1)

This question in 9702/21 May/June 2024

1 The drag force FD acting on a sphere falling through a liquid is given by FD = 6πηr v where r is the radius of the sphere, v is the speed of the sphere in the liquid and η is a property of the liquid called the viscosity. (a) Show that the SI base units of viscosity are kg m–1 s–1. [2] (b) The sphere has a radius of 3.0 cm and is falling vertically downwards at a terminal velocity of 2.0 m s–1 through the liquid. The drag force acting on the sphere is 0.096 N. Calculate the viscosity of the liquid. viscosity = … kg m–1 s–1 [2] (c) The sphere is shown in Fig. 1.1. sphere liquid Fig. 1.1 On Fig. 1.1, draw and label arrows to represent the directions of the three forces acting on the sphere as it falls at terminal velocity through the liquid. [2] (d) (i) The density of the liquid is 920 kg m–3. Show that the upthrust acting on the sphere is 1.0 N. [2] (ii) Calculate the mass of the sphere. mass = … kg [2] [Total: 10]

10 marks

Mark scheme: 1(a) units of F: kg m s–2 C1 units of r: m and units of v: m s–1 units of : kg m s–2 / (m  m s–1) = kg m–1 s–1 A1 1(b) viscosity = 0.096 / (6    0.03  2.0) C1 = 0.085 kg m–1 s–1 A1 1(c) one arrow vertically downwards labelled weight / W B1 arrow(s) vertically upwards labelled U / upthrust and drag / FD/viscous force B1 1(d)(i) V = (4 / 3) r3 C1 upthrust = (4 / 3)    0.033  920  9.81 = 1.0 N A1 1(d)(ii) weight = 1.0 + 0.096 (= 1.096 N) C1 m = 1.096 / 9.81 = 0.11 kg A1

This question in 9702/23 May/June 2024

2 (a) Define displacement from a point. … … [1] (b) An object is projected horizontally at a speed of 6.0 m s–1 from a slope, as shown in Fig. 2.1. 6.0 m s–1 object slope θ horizontal Fig. 2.1 (not to scale) The slope is at an angle θ to the horizontal. Air resistance is negligible. The object lands on the slope a time of 0.71 s later and stops without rolling or bouncing. (i) Determine the horizontal distance travelled by the object. distance = … m [1] (ii) Determine the vertical distance travelled by the object. distance = … m [2] (iii) Use your answers in (b)(i) and (b)(ii) to calculate θ. θ = … ° [2] (iv) Determine the magnitude of the displacement of the object from its original position. displacement = … m [2] (v) By considering energy, calculate the speed of the object just before it lands. speed = … m s–1 [3] [Total: 11]

11 marks

Mark scheme: 2(a) distance (from the point) in a straight line in a given direction B1 2(b)(i) distance = speed  time = 6.0  0.71 = 4.3 m A1 2(b)(ii) s = ut + ½ at2 = ½  9.81  0.712 C1 = 2.5 m A1 2(b)(iii) tan  = 2.5 / 4.3 or hypotenuse = √(4.32 + 2.52) ( = 4.97 m) cos  = 4.3 / 4.97 or sin  = 2.5 / 4.97 C1  = 30° A1 Question Answer Marks 2(b)(iv) displacement = √(4.32 + 2.52) C1 = 4.9 m or 5.0 m A1 or displacement = 2.5 / sin 30° or displacement = 4.3 / cos 30° (C1) = 5.0 m (A1) 2(b)(v) KE = ½mv2 or GPE = mgh C1 initial KE + loss in GPE = final KE (½  m  6.02) + (m  9.81  2.5) = (½  m  v2) C1 v = 9.2 m s–1 A1

This question in 9702/23 May/June 2024

1 (a) Define acceleration. … … [1] (b) A small aircraft is flying horizontally at a speed of 42 m s–1 at a height of 63 m above horizontal ground, as shown in Fig. 1.1. speed 42 m s–1 63 m ground Fig. 1.1 The aircraft drops a small parcel. The parcel is released from the aircraft at the instant shown in Fig. 1.1. Air resistance is negligible. (i) On Fig. 1.1, draw a line to show the path of the parcel as it falls from the aircraft to the ground. [1] (ii) Calculate the time taken from the instant of release to the instant the parcel reaches the ground. time = … s [2] (iii) Calculate the vertical component of the velocity of the parcel immediately before it reaches the ground. vertical component of velocity = … m s–1 [1] (iv) Determine the speed at which the parcel reaches the ground. speed = … m s–1 [2] [Total: 7]

7 marks

Mark scheme: Question Answer Marks 1(a) rate of change of velocity B1 1(b)(i) curved path from aircraft to ground, starting horizontal at aircraft and then with increasing negative gradient as it moves B1 towards the ground 1(b)(ii) s = ut + ½at2 C1 63 = ½  9.81  t2 time = 3.6 s A1 1(b)(iii) v2 = 2  9.81  63 A1 or v = 0 + (9.81  3.6) or 63 = (v  3.6) – (½  9.81  3.62) or 63 = ½  (0 + v)  3.6 v = 35 m s–1 1(b)(iv) speed2 = 352 + 422 C1 speed = 55 m s–1 A1

This question in 9702/23 Oct/Nov 2024

1 (a) Define velocity. … … [1] (b) In an experiment, two objects A and B are released from the side of a building, as shown in Fig. 1.1. building A 10.0 m 3.0 m s–1 B h ground Fig. 1.1 (not to scale) Object A is released from rest at a height of 10.0 m above horizontal ground. Object B is released with an initial upward velocity of 3.0 m s−1 at a height h above the ground. Both objects take the same time to reach the ground and they do not collide with each other. Air resistance is negligible. Calculate h. h = … m [3] (c) In a second experiment, object B is released from the same height as in (b) but with a speed of 6.0 m s−1 at an angle of 60° to the vertical, as shown in Fig. 1.2. 60° building 6.0 m s–1 B Fig. 1.2 (i) State and explain whether the time taken for object B to reach the ground is less than, the same as, or greater than the time taken in the first experiment. … … … [2] (ii) By considering energy, state and explain whether the speed at which object B reaches the ground is less than, the same as, or greater than in the first experiment. … … … [2] [Total: 8]

8 marks

Mark scheme: Question Answer Marks 1(a) Rate of change of displacement B1 1(b) For object A: C1 1 2 s = ut + at 2 2  10 so t = 9.81 =1.4 Then for object B: C1 1 2 s = ut + at 2 h = −(3 1.4) + (0.5  9.81  1.42) = 5.7 m A1 OR (C1) h = −(3 1.4) + 10 = 5.7 m (A1) 1(c)(i) time taken (to reach the ground is) same B1 The initial vertical (component of the) velocity is the same (as in part (1b)) B1 1(c)(ii) The (total) initial energy is greater (than in part (1b)) B1 change in gravitational potential energy is same, so speed is greater B1

This question in 9702/23 May/June 2025