SI units — Paper 2 · A Level Physics 9702

1 (a) The current in a wire is I. Charge Q passes one point in the wire in time t. State (i) the relation between I, Q and t, … [1] (ii) which of the quantities I, Q and t are base quantities. … … [2] (b) The current in the wire is due to electrons, each with charge q, that move with speed v along the wire. There are n of these electrons per unit volume. For a wire having a cross-sectional area S, the current I is given by the equation I = nSqv k, where k is a constant. (i) State the units of I, n, S, q and v in terms of the base units. I … n … S … q … v … [3] (ii) By considering the homogeneity of the equation, determine the value of k. k = ………………………….. [2]

8 marks

Mark scheme: 1 (a) (i) Q = It (allow any subject for the equation) B1 [1] (ii) I B1 t B1 [2] (allow 1 mark only if all three quoted) (b) (i) base unit of I is A base unit of n is m–3 (not /m–3) base unit of S is m2 base unit of q is A s (not C) base unit of v is m s–1 (–1 for each error or omission) B3 [3] (ii) A = m–3 m2 A s (m s–1)k M1 e.g. for m: 0 = –3 + 2 + k k = 1 A1 [2] 2

This question in 9702/21 Oct/Nov 2008

1 (a) State the most appropriate instrument, or instruments, for the measurement of the following. (i) the diameter of a wire of diameter about 1 mm … [1] (ii) the resistance of a filament lamp … [1] (iii) the peak value of an alternating voltage … [1] (b) The mass of a cube of aluminium is found to be 580 g with an uncertainty in the measurement of 10 g. Each side of the cube has a length of (6.0 ± 0.1) cm. Calculate the density of aluminium with its uncertainty. Express your answer to an appropriate number of significant figures. density = … ± … g cm–3 [5]

8 marks

Mark scheme: 1 (a) (i) micrometer (screw gauge) / travelling microscope … B1 [1] (ii) either ohm-meter or voltmeter and ammeter or multimeter/avo on ohm setting … B1 [1] (iii) either (calibrated) c.r.o. or a.c. voltmeter and × √2 … B1 [1] (b) density = mass / volume … C1 = 580 / 63 = 2.685 g cm-3 …(allow 2.68, 2.69, 2.7) … A1 % uncertainty in mass = (10 / 580) × 100 = 1.7% … C1 % uncertainty in volume = 3 × (0.1 / 6) × 100 = 5.0% … C1 uncertainty in density = 0.18 g cm-3 density = 2.7 ± 0.2 g cm-3 … A1 [5] (answer 2.69 ± 0.09 g cm-3 scores 4 marks)

This question in 9702/21 May/June 2009

1 A unit is often expressed with a prefix. For example, the gram may be written with the prefix ‘kilo’ as the kilogram. The prefix represents a power-of-ten. In this case, the power-of-ten is 103. Complete Fig. 1.1 to show each prefix with its symbol and power-of-ten. prefix symbol power-of-ten kilo k 103 nano n … centi … 10–2 … M 106 … T 1012 Fig. 1.1 [4]

4 marks

Mark scheme: 1 10–9 …………………………………………….…………..………….………………… … B1 c …………………………………………….…………..………….………………………….. B1 mega ….……………………………………….…………..………….………………………. B1 tera …….……………………………………….…………..………….…………………… … B1 [4]

This question in 9702/21 May/June 2010

1 (a) Two of the SI base quantities are mass and time. State three other SI base quantities. 1. … 2. … 3. … [3] (b) A sphere of radius r is moving at speed v through air of density ρ. The resistive force F acting on the sphere is given by the expression F = Br 2ρvk where B and k are constants without units. (i) State the SI base units of F, ρ and v. F … ρ … v … [3] (ii) Use base units to determine the value of k. k = … [2]

8 marks

Mark scheme: 1 (a) length, current, temperature, amount of substance, (luminous intensity) any three, 1 each B3 [3] (b) (i) F: kg m s–2 B1 ρ: kg m–3 B1 v: m s–1 B1 [3] (ii) some working e.g. kg m s–2 = m2 kg m–3 (m s–1)k M1 hence k = 2 A1 [2] 1

This question in 9702/21 Oct/Nov 2010

2 (a) A sphere of radius R is moving through a fluid with constant speed v. There is a frictional For force F acting on the sphere, which is given by the expression Examiner’s Use F = 6πDRv where D depends on the fluid. (i) Show that the SI base units of the quantity D are kg m–1 s–1. [3] (ii) A raindrop of radius 1.5 mm falls vertically in air at a velocity of 3.7 m s–1. The value of D for air is 6.6 × 10–4 kg m–1 s–1. The density of water is 1000 kg m–3. Calculate 1. the magnitude of the frictional force F, F = … N [1] 2. the acceleration of the raindrop. acceleration = … m s–2 [3] (b) The variation with time t of the speed v of the raindrop in (a) is shown in Fig. 2.1. For Examiner’s Use v 0 0 t Fig. 2.1 (i) State the variation with time of the acceleration of the raindrop. … … … … … [3] (ii) A second raindrop has a radius that is smaller than that given in (a). On Fig. 2.1, sketch the variation of speed with time for this second raindrop. [2]

12 marks

Mark scheme: 2 (a) (i) base units of D: force: kg m s–2 B1 radius: m velocity: m s–1 B1 base units of D: [F / (R × v)] kg m s–2 / (m × m s–1) M1 = kg m–1 s–1 A0 [3] (ii) 1. F = 6π × D × R × v = [6π × 6.6 × 10–4 × 1.5 × 10–3 × 3.7] = 6.9 × 10–5 N A1 [1] 2. mg – F = ma hence a = g – [F / m] m = ρ × V = ρ × 4/3 π R3 = (1.4 × 10–5) C1 a = 9.81 – [6.9 × 10–5] / ρ × 4/3 π × (1.5 × 10–3)3 (9.81 – 4.88) M1 a = 4.9(3) m s–2 A1 [3] (b) (i) a = g at time t = 0 B1 a decreases (as time increases) B1 a goes to zero B1 [3] (ii) Correct shape below original line M1 sketch goes to terminal velocity earlier A1 [2] GCE AS/A LEVEL – May/June 2011 9702 22

This question in 9702/22 May/June 2011

1 (a) (i) State the SI base units of volume. base units of volume … [1] (ii) Show that the SI base units of pressure are kg m–1 s–2. [1] (b) The volume V of liquid that flows through a pipe in time t is given by the equation V π Pr 4 = t 8Cl where P is the pressure difference between the ends of the pipe of radius r and length l. The constant C depends on the frictional effects of the liquid. Determine the base units of C. base units of C … [3]

5 marks

Mark scheme: 1 (a) (i) V units: m3 (allow metres cubed or cubic metres) A1 [1] (ii) Pressure units: kg m s–2 / m2 (allow use of P = ρgh) M1 Units: kg m–1 s–2 A0 [1] (b) V / t units: m3 s–1 B1 Clear substitution of units for P, r4 and l M1 π P r 4 kg m −1 s −2 m 4 C = = 8 V t −1 l m 3 s −1 m Units: kg m–1 s–1 A1 [3] (8 or π in final answer –1. Use of dimensions max 2/3)

This question in 9702/21 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

1 Energy is stored in a metal wire that is extended elastically. (a) Explain what is meant by extended elastically. … … [2] (b) Show that the SІ units of energy per unit volume are kg m–1 s–2. [2] (c) For a wire extended elastically, the elastic energy per unit volume X is given by X = Cε2E where C is a constant, ε is the strain of the wire, and E is the Young modulus of the wire. Show that C has no units. [3]

7 marks

Mark scheme: 1 (a) the wire returns to its original length (not ‘shape’) M1 when the load is removed A1 [2] (b) energy: N m / kg m2 s–2 and volume m3 C1 energy / volume: kg m2 s–2 / m3 M1 energy / volume: kg m–1 s–2 A0 [2] (c) ε has no units B1 E: kg m s–2 m–2 M1 units of RHS: kg m–1 s–2 = LHS units / satisfactory conclusion to show C has no units A1 [3]

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

1 (a) State the SI base units of force. Use … [1] (b) Two wires each of length l are placed parallel to each other a distance x apart, as shown in Fig. 1.1. l I x I Fig. 1.1 Each wire carries a current I. The currents give rise to a force F on each wire given by K I 2l F = x where K is a constant. (i) Determine the SI base units of K. units of K … [2] (ii) On Fig. 1.2, sketch the variation with x of F. The quantities I and l remain constant. F 0 0 x Fig. 1.2 [2] (iii) The current I in both of the wires is varied. For On Fig. 1.3, sketch the variation with I of F. The quantities x and l remain constant. Examiner’sUse F 0 0 I Fig. 1.3 [1]

6 marks

Mark scheme: 1 (a) force: kg m s–2 A1 [1] (b) (i) I2: A2 l: m x: m C1 K: kg m s–2 A–2 A1 [2] (ii) curve of the correct shape (for inverse proportionality) M1 clearly approaching each axis but never touching the axis A1 [2] (iii) curving upwards and through origin A1 [1]

This question in 9702/23 May/June 2013

1 (a) State two SI base units other than the kilogram, metre and second. Use 1. … 2. … [2] (b) A metal wire has original length l0. It is then suspended and hangs vertically as shown in Fig. 1.1. wire Fig. 1.1 The weight of the wire causes it to stretch. The elastic potential energy stored in the wire is E. (i) Show that the SI base units of E are kg m2 s–2. [2] (ii) The elastic potential energy E is given by For Examiner’s E = Cρ2g 2Al03 Use where ρ is the density of the metal, g is the acceleration of free fall, A is the cross-sectional area of the wire and C is a constant. Determine the SI base units of C. SI base units of C … [3]

7 marks

Mark scheme: 1 (a) kelvin / K B1 ampere / amp / A B1 [2] [allow mole / mol and candela / Cd] (b) (i) energy OR work = force × distance [allow any energy expression] C1 units: kg m s–2 × m OR kg (m s–1)2 for ½ mv2 or mc2 M1 (ignore any numerical factor) units: = kg m2 s–2 A0 [2] (ii) units: ρ: kg m–3 g: m s–2 A: m2 l0: m C1 C: kg m2 s–2 / kg2 m–6 m2 s–4 m2 m3 [any subject] C1 = kg–1 m s2 (allow m s2 / kg) A1 [3]

This question in 9702/21 Oct/Nov 2013

1 (a) State two SI base units other than the kilogram, metre and second. Use 1. … 2. … [2] (b) A metal wire has original length l0. It is then suspended and hangs vertically as shown in Fig. 1.1. wire Fig. 1.1 The weight of the wire causes it to stretch. The elastic potential energy stored in the wire is E. (i) Show that the SI base units of E are kg m2 s–2. [2] (ii) The elastic potential energy E is given by For Examiner’s E = Cρ2g 2Al03 Use where ρ is the density of the metal, g is the acceleration of free fall, A is the cross-sectional area of the wire and C is a constant. Determine the SI base units of C. SI base units of C … [3]

7 marks

Mark scheme: 1 (a) kelvin / K B1 ampere / amp / A B1 [2] [allow mole / mol and candela / Cd] (b) (i) energy OR work = force × distance [allow any energy expression] C1 units: kg m s–2 × m OR kg (m s–1)2 for ½ mv2 or mc2 M1 (ignore any numerical factor) units: = kg m2 s–2 A0 [2] (ii) units: ρ: kg m–3 g: m s–2 A: m2 l0: m C1 C: kg m2 s–2 / kg2 m–6 m2 s–4 m2 m3 [any subject] C1 = kg–1 m s2 (allow m s2 / kg) A1 [3]

This question in 9702/22 Oct/Nov 2013

2 The time T for a satellite to orbit the Earth is given by For KR3 Examiner’sUse T = c m M where R is the distance of the satellite from the centre of the Earth, M is the mass of the Earth, and K is a constant. (a) Determine the SI base units of K. SI base units of K … [2] (b) Data for a particular satellite are given in Fig. 2.1. quantity measurement uncertainty T 8.64 × 104 s ± 0.5% R 4.23 × 107 m ± 1% M 6.0 × 1024 kg ± 2% Fig. 2.1 Calculate K and its actual uncertainty in SI units. K = … ± … SI units [4]

6 marks

Mark scheme: 2 (a) SI units for T: s, R: m and M: kg (or seen clearly in formula) C1 s 2kg K = T2 M / R3 units: s2 kg m–3 (allow s2 kg / m3 or 3 ) A1 [2] m (b) % uncertainty in K: 1% (for T) + 3% (for R) + 2% (for M) OR = 6% C1 K = [(86400)2 × 6 × 1024] / (4.23 × 107)3 = 5.918 × 1011 C1 6% of K = 0.355 × 1011 C1 K = (5.9 ± 0.4) × 1011 (SI units) correct power of ten required for both A1 [4] [incorrect % value then max. 1]

This question in 9702/23 Oct/Nov 2013

1 (a) Show that the SІ base units of power are kg m2 s–3. [3] Q (b) The rate of flow of thermal energy in a material is given by t Q CAT = t x where A is the cross-sectional area of the material, T is the temperature difference across the thickness of the material, x is the thickness of the material, C is a constant. Determine the SІ base units of C. base units … [4]

7 marks

Mark scheme: 1 (a) power = energy / time or work done / time B1 force: kg m s–2 (including from mg in mgh or Fv) 1 or kinetic energy ( mv2): kg (m s–1)2 B1 2 (distance: m and (time) –1: s–1) and hence power: kg m s–2 m s–1 = kg m2 s–3 B1 [3] (b) Q / t : kg m2 s–3 C1 A: m2 and x: m and T: K C1 correct substitution into C = (Qx) / tAT or equivalent, or with cancellation C1 units of C : kg m s–3 K–1 A1 [4]

This question in 9702/22 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) The kilogram, metre and second are SI base units. State two other base units. 1. … 2. … [2] (b) Determine the SI base units of (i) stress, SI base units … [2] (ii) the Young modulus. SI base units … [1]

5 marks

Mark scheme: 1 (a) ampere B1 kelvin B1 [2] (allow mole and candela) (b) (i) stress: N m–2 C1 kg m s–2 / m2 = kg m–1 s–2 A1 [2] (ii) Young modulus = stress / strain and strain has no units hence units: kg m–1 s–2 B1 [1]

This question in 9702/23 Oct/Nov 2014

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

4 (a) By reference to the direction of the propagation of energy, state what is meant by a longitudinal wave and by a transverse wave. longitudinal: … … … transverse: … … … [2] (b) The intensity of a sound wave passing through air is given by Ι = Kvρf 2A2 where Ι is the intensity (power per unit area), K is a constant without units, v is the speed of sound, ρ is the density of air, f is the frequency of the wave and A is the amplitude of the wave. Show that both sides of the equation have the same SΙ base units. [3] (c) (i) Describe the Doppler effect. … … [1] (ii) A distant star is moving away from a stationary observer. State the effect of the motion on the light observed from the star. … … … [1] (d) A car travels at a constant speed towards a stationary observer. The horn of the car sounds at a frequency of 510 Hz and the observer hears a frequency of 550 Hz. The speed of sound in air is 340 m s–1. Calculate the speed of the car. speed = … m s–1 [3] [Total: 10]

10 marks

Mark scheme: 4 (a) longitudinal: vibrations/oscillations (of the particles/wave) are parallel to the direction or in the same direction (of the propagation of energy) B1 transverse: vibrations/oscillations (of the particles/wave) are perpendicular to the direction (of the propagation of energy) B1 [2] (b) LHS: intensity = power / area units: kg m s–2 × m × s–1 × m–2 or kg m2 s–3 × m–2 B1 RHS: units: m s–1 × kg m–3 × s–2 × m2 M1 LHS and RHS both kg s–3 A1 [3] (c) (i) change/difference in the observed/apparent frequency when the source is moving (relative to the observer) B1 [1] (ii) wavelength increases/frequency decreases/red shift B1 [1] (d) observed frequency = vfS / (v – vS) C1 550 = (340 × 510) / (340 – vS) C1 vS = 25 (24.7) m s–1 A1 [3]

This question in 9702/22 May/June 2016

1 (a) State two SI base units other than kilogram, metre and second. 1. … 2. … [1] (b) Determine the SI base units of resistivity. base units … [3] (c) (i) A wire of cross-sectional area 1.5 mm2 and length 2.5 m has a resistance of 0.030 Ω. Calculate the resistivity of the material of the wire in nΩ m. resistivity = … nΩ m [3] (ii) 1. State what is meant by precision. … … 2. Explain why the precision in the value of the resistivity is improved by using a micrometer screw gauge rather than a metre rule to measure the diameter of the wire. … … … [2] [Total: 9]

9 marks

Mark scheme: 1(a) kelvin, mole, ampere, candela any two B1 1(b) use of resistivity = RA / l and V = IR (to give ρ = VA/ Il) C1 units of V: (work done / charge) kg m2 s–2 (A s)–1 C1 units of resistivity: (kg m2 s–3 A–1 A–1 m) = kg m3 s–3 A–2 A1 or use of R = ρL / A and P = I2R (gives ρ = PA / I2L) (C1) units of P: kg m2 s–3 (C1) units of resistivity: (kg m2 s–3 × m2) / (A2 × m) = kg m3 s–3 A–2 (A1) 1(c)(i) ρ = (RA/l) C1 = (0.03 × 1.5 × 10–6) / 2.5 (= 1.8 × 10–8) C1 = 18 nΩ m A1 1(c)(ii) 1. precision is determined by the range in the measurements/values/readings/data/results B1 2. metre rule measures to ± 1 mm and micrometer to ± 0.01 mm (so there is less (percentage) uncertainty/random error) B1

This question in 9702/22 May/June 2017

1 (a) The drag force FD acting on a sphere moving through a fluid is given by the expression FD = Kρv 2 where K is a constant, ρ is the density of the fluid and v is the speed of the sphere. Determine the SI base units of K. base units … [3] (b) A ball of weight 1.5 N falls vertically from rest in air. The drag force F D acting on the ball is given by the expression in (a). The ball reaches a constant (terminal) speed of 33 m s–1. Assume that the upthrust acting on the ball is negligible and that the density of the air is uniform. For the instant when the ball is travelling at a speed of 25 m s–1, determine (i) the drag force FD on the ball, FD = … N [2] (ii) the acceleration of the ball. acceleration = … m s–2 [2] (c) Describe the acceleration of the ball in (b) as its speed changes from zero to 33 m s–1. … … … … [3] [Total: 10]

10 marks

Mark scheme: 1(a) C1 units of ρ: kg m–3 and units of v: m s–1 C1 units of K: kg m s–2 / [kg m–3 (m s–1)2] = m2 A1 1(b)(i) Kρ = 1.5 / 332 C1 = 1.38 × 10–3 FD = 1.38 × 10–3 × 252 or FD / 1.5 = 252 / 332 FD = 0.86 N A1 1(b)(ii) a = (1.5 – 0.86) / (1.5 / 9.81) or a = 9.81 – [0.86 / (1.5 / 9.81)] C1 a = 4.2 m s–2 A1 1(c) initial acceleration is g/9.81 (m s–2)/acceleration of free fall B1 acceleration decreases B1 final acceleration is zero B1

This question in 9702/21 Oct/Nov 2017

1 (a) (i) Define power. … … [1] (ii) Show that the SI base units of power are kg m2 s–3. [1] (b) All bodies radiate energy. The power P radiated by a body is given by P = kAT 4 where T is the thermodynamic temperature of the body, A is the surface area of the body and k is a constant. (i) Determine the SI base units of k. base units … [2] (ii) On Fig. 1.1, sketch the variation with T 2 of P. The quantity A remains constant. P 0 0 T 2 Fig. 1.1 [1] [Total: 5]

5 marks

Mark scheme: 1(a)(i) work (done) / time (taken) or energy (transferred) / time (taken) B1 1(a)(ii) Correct substitution of base units of all quantities into any correct equation for power. Examples: (P = E / t or W / t gives) kg m2 s–2 / s = kg m2 s–3 (P = Fs / t or mgh / t gives) kg m s–2 m / s = kg m2 s–3 (P = ½mv2/ t gives) kg (m s–1)2 / s = kg m2 s–3 (P = Fv gives) kg m s–2 m s–1 = kg m2 s–3 (P = VI gives) kg m2 s–2 A–1 s–1 A = kg m2 s–3 A1 1(b)(i) units of A: m2 and units of T: K C1 units of k: kg m2 s–3 / m2 K4 = kg s–3 K–4 A1 1(b)(ii) curve from the origin with increasing gradient B1

This question in 9702/23 Oct/Nov 2017

1 (a) Define force. … [1] (b) State the SI base units of force. … [1] (c) The force F between two point charges is given by Q1Q2 F = 4πr 2ε where Q1 and Q2 are the charges, r is the distance between the charges, ε is a constant that depends on the medium between the charges. Use the above expression to determine the base units of ε. base units … [2] [Total: 4]

4 marks

Mark scheme: 1(a) rate of change of momentum B1 1(b) kg m s–2 A1 1(c) units for Q: A s and for r: m C1 units for ε = (A s × A s) / (kg m s–2 × m2) = A2 kg–1 m–3 s4 A1

This question in 9702/22 May/June 2018

2 (a) The kilogram, metre and second are all SI base units. State two other SI base units. 1. … 2. … [2] (b) A uniform beam AB of length 6.0 m is placed on a horizontal surface and then tilted at an angle of 31° to the horizontal, as shown in Fig. 2.1. 90 N A 6.0 m Y W X 31° B Fig. 2.1 (not to scale) The beam is held in equilibrium by four forces that all act in the same plane. A force of 90 N acts perpendicular to the beam at end A. The weight W of the beam acts at its centre of gravity. A vertical force Y and a horizontal force X both act at end B of the beam. (i) State the name of force X. … [1] (ii) By taking moments about end B, calculate the weight W of the beam. W = … N [2] (iii) Determine the magnitude of force X. magnitude of force X = … N [1] [Total: 6]

6 marks

Mark scheme: 2(a) ampere kelvin (allow mole, candela) any two correct answers, 1 mark each B2 2(b)(i) frictional (force)/friction B1 2(b)(ii) W cos 31° × 3.0 or 90 × 6.0 C1 W cos 31° × 3.0 = 90 × 6.0 W = 210 N A1 2(b)(iii) X = 90 sin 31° = 46 N A1

This question in 9702/22 Oct/Nov 2018

1 (a) Mass, length and time are all SI base quantities. State two other SI base quantities. 1. … 2. … [2] (b) A wire hangs between two fixed points, as shown in Fig. 1.1. fixed fixed horizontal point 17° 17° point 150 N 150 N wire hook rope tyre Fig. 1.1 (not to scale) A child’s swing is made by connecting a car tyre to the wire using a rope and a hook. The system is in equilibrium with the wire hanging at an angle of 17° to the horizontal. The tension in the wire is 150 N. Assume that the rope and hook have negligible weight. (i) Determine the weight of the tyre. weight = … N [2] (ii) The wire has a cross-sectional area of 7.5 mm2 and is made of metal of Young modulus 2.1 × 1011 Pa. The wire obeys Hooke’s law. Calculate, for the wire, 1. the stress, stress = … Pa [2] 2. the strain. strain = … [2] [Total: 8]

8 marks

Mark scheme: 1(a) current temperature (allow amount of substance, luminous intensity) any two correct answers, 1 mark each B2 1(b)(i) W = 2 × (150 × sin 17°) or 2 × (150 × cos 73°) C1 W = 88 N A1 1(b)(ii) 1. σ = F / A C1 = 150 / (7.5 × 10–6) = 2.0 × 107 Pa A1 2. ε = σ / E C1 = 2.0 × 107 / (2.1 × 1011) = 9.5 × 10–5 A1

This question in 9702/23 Oct/Nov 2018

1 (a) The ampere, metre and second are SI base units. State two other SI base units. 1. … 2. … [2] (b) The average drift speed v of electrons moving through a metal conductor is given by the equation: μF v = e where e is the charge on an electron F is a force acting on the electron and μ is a constant. Determine the SI base units of μ. SI base units … [3] [Total: 5]

5 marks

Mark scheme: 1(a) kilogram / kg B1 kelvin / K B1 1(b) units for v: m s–1 and units for F: kg m s–2 C1 units for e: A s C1 units for µ: m s–1 A s / kg m s–2 = A kg–1 s2 A1

This question in 9702/22 Feb/March 2019

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) Determine the SI base units of the moment of a force. SI base units … [1] (b) A uniform square sheet of card ABCD is freely pivoted by a pin at a point P. The card is held in a vertical plane by an external force in the position shown in Fig. 1.1. B 17 cm 45° P A C 4.0 cm G 0.15 N D Fig. 1.1 (not to scale) The card has weight 0.15 N which may be considered to act at the centre of gravity G. Each side of the card has length 17 cm. Point P lies on the horizontal line AC and is 4.0 cm from corner A. Line BD is vertical. The card is released by removing the external force. The card then swings in a vertical plane until it comes to rest. (i) Calculate the magnitude of the resultant moment about point P acting on the card immediately after it is released. moment = … N m [2] (ii) Explain why, when the card has come to rest, its centre of gravity is vertically below point P. … … … … [2] [Total: 5]

5 marks

Mark scheme: 1(a) = kg m2 s–2 A1 1(b)(i) distance of COG from P (= GP) = 17 cos 45° – 4.0 or (144.5)½ – 4.0 (= 8.0 cm) C1 moment = 0.15 × 8.0 × 10–2 = 1.2 × 10–2 N m A1 1(b)(ii) (line of action of) weight acts through pivot/P or distance between (line of action of) weight and pivot/P is zero B1 (so) weight does not have a moment about pivot/P B1

This question in 9702/23 Oct/Nov 2019

1 (a) Length, mass and temperature are all SI base quantities. State two other SI base quantities. 1. … 2. … [2] (b) The acceleration of free fall g may be determined from an oscillating pendulum using the equation 4π2l g = T 2 where l is the length of the pendulum and T is the period of oscillation. In an experiment, the measured values for an oscillating pendulum are l = 1.50 m ± 2% and T = 2.48 s ± 3%. (i) Calculate the acceleration of free fall g. g = … m s–2 [1] (ii) Determine the percentage uncertainty in g. percentage uncertainty = … % [2] (iii) Use your answers in (b)(i) and (b)(ii) to determine the absolute uncertainty of the calculated value of g. absolute uncertainty = … m s–2 [1] [Total: 6]

6 marks

Mark scheme: 1(a) time (electric) current allow amount of substance allow luminous intensity any two of the above quantities, 1 mark each B2 1(b)(i) g = (4π2 × 1.50) / (2.482) = 9.63 m s–2 A1 1(b)(ii) percentage uncertainty = 2 + (3 × 2) or fraction uncertainty = 0.02 + (0.03 × 2) C1 percentage uncertainty = 8% A1 1(b)(iii) absolute uncertainty = 0.08 × 9.6 = 0.8 m s–2 A1

This question in 9702/22 Feb/March 2020

1 (a) Use an expression for work done, in terms of force, to show that the SI base units of energy are kg m2 s–2. [2] (b) (i) The energy E stored in an electrical component is given by Q2 E = 2C where Q is charge and C is a constant. Use this equation and the information in (a) to determine the SI base units of C. SI base units … [2] (ii) Measurements of a constant current in a wire are taken using an analogue ammeter. For these measurements, describe one possible cause of: 1. a random error … … 2. a systematic error. … … [2] [Total: 6]

6 marks

Mark scheme: 1(a) (work =) force × displacement C1 units: kg m s–2 × m = kg m2 s–2 A1 1(b)(i) units of Q: As C1 units of C: kg–1 m–2 A2 s4 A1 1(b)(ii) 1. e.g. reading scale from different angles (wrongly) interpolating between scale readings/divisions B1 2. e.g. zero error wrongly calibrated scale B1

This question in 9702/21 May/June 2020

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

6 The current I in a metal wire is given by the expression I = Anve where v is the average drift speed of the free electrons in the wire and e is the elementary charge. (a) State what is meant by the symbols A and n. A: … n: … [2] (b) Use the above expression to determine the SI base units of e. Show your working. base units … [2] (c) Two lamps P and Q are connected in series to a battery, as shown in Fig. 6.1. P Q Fig. 6.1 The radius of the filament wire of lamp P is twice the radius of the filament wire of lamp Q. The filament wires are made of metals with the same value of n. Calculate the ratio average drift speed of free electrons in filament wire of P . average drift speed of free electrons in filament wire of Q ratio = … [2] [Total: 6]

6 marks

Mark scheme: 6(a) A: cross-sectional area B1 n: number density of free electrons B1 6(b) units of I: A and units of A: m2 and units of v: m s–1 B1 units of e: A / (m2 m–3 m s–1) = A s A1 6(c) ratio = AQ / AP C1 = [πr2] / [π(2r2)] = 0.25 A1

This question in 9702/23 May/June 2020

1 (a) (i) Define the moment of a force about a point. … … [1] (ii) Determine the SI base units of the moment of a force. base units … [1] (b) A uniform rigid rod of length 2.4 m is shown in Fig. 1.1. 2.4 m cross-sectional area A Fig. 1.1 The rod has a weight of 5.2 N and is made of wood of density 790 kg m–3. Calculate the cross-sectional area A, in mm2, of the rod. A = … mm2 [3] (c) A fishing rod AB, made from the rod in (b), is shown in Fig. 1.2. 0.60 m B 0.60 m C T string D 1.20 m 4.6 N 56° stick weight 5.2 N A ground water Fig. 1.2 (not to scale) End A of the rod rests on the ground and a string is attached to the other end B. A support stick exerts a force perpendicular to the rod at point C. The weight of the rod acts at point D. The tension T in the string is in a direction perpendicular to the rod. The rod is in equilibrium and inclined at an angle of 56° to the vertical. The forces and the distances along the rod of points A, B, C and D are shown in Fig. 1.2. (i) Show that the component of the weight that is perpendicular to the rod is 4.3 N. [1] (ii) By taking moments about end A of the rod, calculate the tension T. T = … N [3] [Total: 9]

9 marks

Mark scheme: 1(a)(i) force × perpendicular distance (of line of action of force to the point) B1 1(a)(ii) units: kg m s–2 m = kg m2 s–2 A1 1(b) W = ρVg or W = ρALg C1 A = 5.2 / (790 × 2.4 × 9.81) (= 2.8 × 10–4 (m2)) C1 = 2.8 × 102 mm2 A1 1(c)(i) (component =) 5.2 sin 56° = 4.3 (N) or 5.2 cos 34° = 4.3 (N) A1 1(c)(ii) (T × 2.4) or (4.3 × 1.2) or (4.6 × 1.8) C1 (T × 2.4) + (4.3 × 1.2) = (4.6 × 1.8) C1 T = 1.3 N A1

This question in 9702/21 Oct/Nov 2020

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

2 (a) State what is meant by work done. … … [1] (b) Use your answer in (a) to show that the SI base units of energy are kg m2 s–2. [1] (c) A metal rod is heated at one end so that thermal energy flows to the other end. The thermal energy E that flows through the rod in time t is given by cA(T1 – T2)t E = L where A is the cross-sectional area of the rod, T1 and T2 are the temperatures of the ends of the rod, L is the length of the rod, and c is a constant. Determine the SI base units of c. SI base units … [3] [Total: 5]

5 marks

Mark scheme: 2(a) force × displacement in the direction of the force B1 2(b) units: kg m s–2 × m = kg m2 s–2 A1 2(c) T1: K and T2: K C1 A: m2 and t: s and L: m C1 c = (kg m2 s–2 m) / (m2 K s) = kg m s–3 K–1 A1

This question in 9702/23 Oct/Nov 2021

1 A sphere of radius 2.1 mm falls with terminal (constant) velocity through a liquid, as shown in Fig. 1.1. sphere, radius 2.1 mm weight 7.2 × 10–4 N constant velocity downwards liquid, density ρ Fig. 1.1 Three forces act on the moving sphere. The weight of the sphere is 7.2 × 10–4 N and the upthrust acting on it is 4.8 × 10–4 N. The viscous force FV acting on the sphere is given by FV = krv where r is the radius of the sphere, v is its velocity and k is a constant. The value of k in SI units is 17. (a) Determine the SI base units of k. SI base units … [2] (b) Use the value of the upthrust acting on the sphere to calculate the density ρ of the liquid. ρ = … kg m–3 [3] (c) (i) On the sphere in Fig. 1.1, draw three arrows to show the directions of the weight W, the upthrust U and the viscous force FV. Label these arrows W, U and FV respectively. [1] (ii) Determine the magnitude of the terminal (constant) velocity of the sphere. velocity = … m s–1 [2] [Total: 8]

8 marks

Mark scheme: 1(a) Fv: kg m s–2 k: kg m s–2 / m × m s–1 = kg m–1 s–1 A1 1(b) F = ρgV V = 4 / 3 × π × (2.1 × 10–3)3 (=3.88 × 10–8 m3) C1 ρ = 4.8 × 10–4 / 9.81 × V C1 = 1300 kg m–3 A1 1(c)(i) W downwards, U upwards, Fv upwards B1 1(c)(ii) FV = 7.2 × 10–4 – 4.8 × 10–4 = 2.4 × 10–4 (N) C1 velocity = 2.4 × 10–4 / (17 × 2.1 × 10–3) = 6.7 × 10–3 m s–1 A1

This question in 9702/22 Feb/March 2022

1 (a) In the following list, underline all units that are SI base units. ampere degree Celsius kilogram newton [1] (b) Fig. 1.1 shows a horizontal beam clamped at one end with a block attached to the other end. block direction of oscillations clamp beam Fig. 1.1 The block is made to oscillate vertically. The Young modulus E of the material of the beam is given by kM E = T2 where M is the mass of the block, T is the period of the oscillations and k is a constant. A student determines the values and percentage uncertainties of k, M and T. Table 1.1 lists the percentage uncertainties. Table 1.1 percentage quantity uncertainty k ± 2.1% M ± 0.6% T ± 1.5% The student uses the values of k, M and T to calculate the value of E as 8.245 × 109 Pa. (i) Calculate the percentage uncertainty in the value of E. percentage uncertainty = … % [2] (ii) Use your answer in (b)(i) to determine the value of E, with its absolute uncertainty, to an appropriate number of significant figures. E = ( … ± … ) × 109 Pa [2] [Total: 5]

5 marks

Mark scheme: 1(a) only ampere and kilogram underlined B1 1(b)(i) percentage uncertainty = 2.1 + 0.6 + (1.5  2) C1 = 5.7% A1 1(b)(ii) absolute uncertainty = (5.7 / 100)  8.245  109 ( = 4.7  108 Pa or 0.47  109 Pa) C1 E = (8.2 ± 0.5)  109 Pa A1

This question in 9702/22 May/June 2022

1 (a) State what is meant by work done. … … [1] (b) Use the answer to (a) to determine the SI base units of power. SI base units … [2] (c) The maximum useful output power P of a car travelling on a horizontal road is given by P = v 3 b where v is the maximum speed of the car and b is a constant. For the car, P = 84 kW ± 5% and b = 0.56 ± 7% in SI units. (i) Calculate the value of v. v = … m s–1 [2] (ii) Determine the absolute uncertainty in the value of v. absolute uncertainty = … m s–1 [2] [Total: 7]

7 marks

Mark scheme: Question Answer Marks 1(a) force  displacement in the direction of the force B1 1(b) P = Fs / t C1 = (kg m s–2  m) / s = kg m2 s–3 A1 1(c)(i) 84  103 = v 3  0.56 C1 v = 53 m s–1 A1 1(c)(ii) percentage uncertainty = (5% + 7%) / 3 (= 4%) C1 or fractional uncertainty = (0.05 + 0.07) / 3 (= 0.04) absolute uncertainty = 0.04  53 A1 = () 2 m s–1

This question in 9702/22 Oct/Nov 2022

1 The rate of flow Q of a liquid along a narrow pipe of length L and radius r is given by αr 4 Q = L where α is a constant. An experiment is carried out to determine the value of α. The data from the experiment are shown in Table 1.1. Table 1.1 quantity value percentage uncertainty Q 2.72 × 10–8 m3 s–1 ± 3% r 7.1 × 10–5 m ± 2% L 2.5 × 10–2 m ± 4% (a) Use information in Table 1.1 to show that the SI base unit of α is s–1. [1] (b) Show that the percentage uncertainty in α is 15%. [1] (c) Calculate α with its absolute uncertainty. Give your answer to an appropriate number of significant figures. α = ( … ± … ) × 107 s–1 [3] [Total: 5]

5 marks

Mark scheme: Question Answer Marks 1(a) (SI base unit of =) m3 s–1  m / m4 = s–1 A1 1(b) (percentage uncertainty =) 3 + 4 + 2  4 = 15 (%) A1 1(c)  = QL / r 4 C1 = 2.72  10–8  2.5  10–2 / (7.1  10–5)4 = 2.7  107 absolute uncertainty = 0.15  [2.7  107] C1 = 0.4  107 = (2.7 ± 0.4)  107 s–1 A1

This question in 9702/23 Oct/Nov 2022

1 (a) Underline all the SI base units in the following list. ampere coulomb current kelvin newton [1] (b) A toy car moves in a horizontal straight line. The displacement s of the car is given by the equation v 2 s = 2a where a is the acceleration of the car and v is its final velocity. State two conditions that apply to the motion of the car in order for the above equation to be valid. 1 … 2 … [2] (c) An experiment is performed to determine the acceleration of the car in (b). The following measurements are obtained: s = 3.89 m ± 0.5% v = 2.75 m s–1 ± 0.8%. (i) Calculate the acceleration a of the car. a = … m s–2 [1] (ii) Determine the percentage uncertainty, to two significant figures, in a. percentage uncertainty = … % [2] (iii) Use your answers in (c)(i) and (c)(ii) to determine the absolute uncertainty in the calculated value of a. absolute uncertainty = … m s–2 [1] [Total: 7]

7 marks

Mark scheme: Question Answer Marks 1(a) only ampere and kelvin 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) a = 2.752 / (2  3.89) A1 = 0.97 m s–2 1(c)(ii) percentage uncertainty = (2  0.8) + 0.5 C1 = 2.1% A1 1(c)(iii) absolute uncertainty = (2.1 / 100)  0.97 A1 = 0.02 m s–2

This question in 9702/22 Feb/March 2023

1 (a) (i) Define power. … … [1] (ii) Use the definition of power to show that the SI base units of power are kg m2 s–3. [1] (b) The intensity I of a sound wave moving through a gas is given by I = f 2A2vk where f is the frequency of the wave, A is the amplitude of the wave, v is the speed of the wave and k is a constant that depends on the gas. Determine the SI base units of k. SI base units … [3] [Total: 5]

5 marks

Mark scheme: 1(a)(i) work done per unit time B1 1(a)(ii) (P = W / t gives) units: kg m2 s–2 / s = kg m2 s–3 B1 1(b) (I = P / A so) units of I: kg m2 s–3 / m2 or kg s–3 C1 units of f: s–1 and units of A: m and units of v: m s–1 C1 units of k: kg s–3 / [(s–1)2 m2 m s–1] = kg m–3 A1

This question in 9702/21 May/June 2023

1 (a) (i) Define pressure. … … [1] (ii) Use the answer to (a)(i) to show that the SI base units of pressure are kg m–1 s–2. [1] (b) A horizontal pipe has length L and a circular cross‑section of radius R. A liquid of density ρ flows through the pipe. The mass m of liquid flowing through the pipe in time t is given by π(p2 – p1)R 4ρt m = 8kL where p1 and p2 are the pressures at the ends of the pipe and k is a constant. Determine the SI base units of k. SI base units … [3] (c) An experiment is performed to determine the value of k by measuring the values of the other quantities in the equation in (b). The values of L and R each have a percentage uncertainty of 2%. State and explain, quantitatively, which of these two quantities contributes more to the percentage uncertainty in the calculated value of k. … … … [1] [Total: 6]

6 marks

Mark scheme: 1(a)(i) force / area (normal to the force) B1 1(a)(ii) (p = F / A so units are) kg m s–2 / m2 = kg m–1 s–2 A1 1(b) unit of R: m and unit of t: s and unit of L: m C1 unit of : kg m–3 or  = m / V C1 base units of k: (kg m–1 s–2  m4  kg m–3  s) / (kg  m) = kg m–1 s–1 A1 1(c) R contributes 4  2% or 8% (and L contributes 2%) so R contributes more (to the percentage uncertainty in k) B1

This question in 9702/22 May/June 2023

1 (a) Table 1.1 lists some SI quantities. Complete the table by indicating with a tick (✓) which rows are SI base quantities. Table 1.1 quantity base quantity current energy force mass [1] (b) Use the definition of power to determine its SI base units. SI base units … [2] (c) A light meter is used to measure the intensity of light in a classroom. Daylight is incident normally on the sensor of the meter. The sensor has an area of 2.2 cm2. The reading on the meter is 950 W m–2. Calculate the power of the daylight incident on the sensor. power = … W [3] [Total: 6]

6 marks

Mark scheme: Question Answer Marks 1(a) current and mass only ticked A1 1(b) (power =) work (done) / time C1 units of power = J s–1 A1 = kg m2 s–2 / s = kg m2 s–3 1(c) power = intensity  area C1 = 950  2.2  10–4 C1 = 0.21 W A1

This question in 9702/22 Feb/March 2024

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

1 (a) The list below shows some SI quantities. Underline the quantity that is not an SI base quantity. charge current length time [1] (b) A square solar panel with sides of length 1300 mm is shown in Fig. 1.1. incident light solar panel 1300 mm 1300 mm Fig. 1.1 (not to scale) Light is incident normally on the solar panel. (i) The power of the light incident on the solar panel is 750 W. Calculate the intensity of the light. intensity = … W m–2 [3] (ii) The percentage uncertainty in the incident power is ± 3%. The uncertainty in the length of each side is ± 5 mm. Calculate the percentage uncertainty in the intensity of the light. percentage uncertainty = … % [2] (iii) The useful power output of the solar panel is 160 W. Calculate the percentage efficiency of the solar panel. efficiency = … % [1] (iv) Another square solar panel is placed so that light of the same intensity is incident normally on it. The new panel has shorter sides than the original panel. The new panel has the same power output as the original panel. State and explain whether the efficiency of the new panel is greater than, less than or the same as the efficiency of the original panel. … … … … … [3] [Total: 10]

10 marks

Mark scheme: 1(a) charge underlined (and no others) B1 1(b)(i) I = P / A C1 = 750 / (1300  10–3)2 C1 = 440 W m–2 A1 1(b)(ii) percentage uncertainty = 3 + 2  (5 / 1300)  100 C1 = 3 + 2  0.38 = (±) 4% A1 1(b)(iii) efficiency = useful output power / total input power = (160 / 750)  100 = 21% A1 1(b)(iv) area (of the new panel) is less B1 input power (of the new panel) is less (than the input power of the original panel) (as intensity is constant) B1 (useful power output is unchanged so) efficiency is greater (than the original panel) B1

This question in 9702/22 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

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 Scientists are investigating the variation in air pressure at different locations on a mountain. (a) The scientists take measurements of several physical quantities at each location. Complete Table 1.1 by stating the SI base unit for each quantity and identifying with a tick (3) whether each quantity is a scalar or a vector. Use the space for any working. Table 1.1 quantity measured SI base unit scalar vector air temperature air pressure [2] (b) (i) At one location, the density of the air is 1.1 kg m–3. A spherical weather balloon is filled with a gas and released from rest. The balloon has radius 0.90 m. Calculate the upthrust acting on the balloon when it is released. upthrust = … N [2] (ii) Explain why an upthrust acts on the balloon. … … … … [2] (iii) The balloon has weight 19 N. Calculate the magnitude of the initial acceleration of the balloon. acceleration = … m s–2 [3] (c) A quantity c relating to the motion of the balloon is calculated from three measured quantities k, F and v using the formula 2kF c = . v 2 The percentage uncertainties in the measured quantities are given in Table 1.2. Table 1.2 measured quantity percentage uncertainty k 5% F 3% v 4% The calculated value of c is 1.8. Determine the absolute uncertainty in c. absolute uncertainty = … [2] [Total: 11]

11 marks

Mark scheme: Question Answer Marks 1(a) air temperature: K and air pressure: kg m–1 s–2 B1 scalar only ticked for both air temperature and air pressure B1 1(b)(i) upthrust = 1.1  9.81  (4  0.903 / 3) C1 = 33 N A1 1(b)(ii) (due to difference in height / depth there is a) difference in pressure between top and bottom (of balloon) B1 (due to pressure difference, upwards) B1 force on bottom of balloon is greater (than downwards force on top of balloon, so resultant force is upwards) 1(b)(iii) ()F = 33 – 19 C1 (= 14 N) m = 19 / 9.81 C1 ( = 1.94 kg) a = (33 – 19) / (19 / 9.81) A1 = 7.2 m s–2 1(c) 5 + 3 + (2  4) C1 (= 16%) absolute uncertainty in c = 1.8  0.16 A1 = (±) 0.3

This question in 9702/22 Oct/Nov 2025