SI units — Paper 2 · A Level Physics 9702

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) 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

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

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 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 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