Density and pressure — Paper 2 · A Level Physics 9702

1 (a) (i) Define pressure. Use … … [1] (ii) State the units of pressure in base units. … [1] (b) The pressure p at a depth h in an incompressible fluid of density ρ is given by p = ρgh, where g is the acceleration of free fall. Use base units to check the homogeneity of this equation. … … … … [3]

5 marks

Mark scheme: 1 (a) (i) force per unit area (ratio idea essential) B1 (ii) kg m-1 s-2 B1 [2] (b) ρ has base unit kg m-3 B1 g has base unit m s-2 B1 hρg has base unit m × kg m-3 × m s-2 M1 same as pressure QED A0 [3]

This question in 9702/21 Oct/Nov 2005

4 (a) Define density. … … … [1] (b) A U-tube contains some mercury. Water is poured into one arm of the U-tube and oil is poured into the other arm, as shown in Fig. 4.1. oil water 71 cm 53 cm mercury Fig. 4.1 The amounts of oil and water are adjusted until the surface of the mercury in the two arms is at the same horizontal level. (i) State how it is known that the pressure at the base of the column of water is the same as the pressure at the base of the column of oil. … … [1] (ii) The column of water, density 1.0 × 103kg m–3, is 53 cm high. The column of oil is 71 cm high. Calculate the density of the oil. Explain your working. density = ………………………………. kg m–3 [3]

5 marks

Mark scheme: 4 (a) mass per unit volume (ratio idea must be clear, not units) B1 [1] (b) (i) pressure is same at the surface of mercury because at same horizontal level B1 [1] (ii) hρg is same for both B1 53 × 10–2 × 1.0 × 103 × g = 71 × 10–2 × ρ × g C1 ρ = 7.5 × 102 kg m–3 A1 [3] GCE A Level – May/June 2006 9702 02

This question in 9702/21 May/June 2006

3 (a) Define density. … [1] (b) Liquid of density ρ fills a container to a depth h, as illustrated in Fig. 3.1. h area A Fig. 3.1 The container has vertical sides and a base of area A. (i) State, in terms of A, h and ρ, the mass of liquid in the container. … [1] (ii) Hence derive an expression for the pressure p exerted by the liquid on the base of the container. Explain your working. [2] Examiner’s Use (c) The density of liquid water is 1.0 g cm–3. The density of water vapour at atmospheric 1 pressure is approximately g cm–3. 1600 Determine the ratio volume of water vapour (i) , volume of equal mass of liquid water ratio = … [1] mean separation of molecules in water vapour (ii) . mean separation of molecules in liquid water ratio = … [2] (d) State the evidence for (i) the molecules in solids and liquids having approximately the same separation, … … [1] (ii) strong rigid forces between molecules in solids. strong: … rigid: … [2]

10 marks

Mark scheme: 3 (a) mass / volume (ratio idea essential) B1 [1] (b) (i) mass = Ahρ B1 [1] (ii) pressure = force/area B1 weight (of liquid)/force (on base) = Ahρg B1 pressure = hρg A0 [2] (c) (i) ratio = 1600 or 1600:1 A1 [1] (ii) ratio = 3√1600 C1 = 11.7 (allow 12) A1 [2] GCE A/AS LEVEL – May/June 2007 9702 2 (d) (i) density of solids and liquids are (about) equal B1 [1] (ii) strong forces: fixed volume B1 rigid forces: retains shape / does not flow / little deformation B1 [2] (allow 1 mark for fixed volume, fixed shape)

This question in 9702/21 May/June 2007

1 Make reasonable estimates of the following quantities. (a) the frequency of an audible sound wave frequency = … Hz [1] (b) the wavelength, in nm, of ultraviolet radiation wavelength = … nm [1] (c) the mass of a plastic 30 cm ruler mass = … g [1] (d) the density of air at atmospheric pressure density = … kg m–3 [1]

4 marks

Mark scheme: 1 (a) allow anything in range 20 Hz → 20 kHz B1 [1] (b) allow anything in range 10 nm → 400 nm B1 [1] (c) allow anything in range 10 g → 100 g B1 [1] (d) allow anything in range 0.1 kg m–3 → 10 kg m–3 B1 [1]

This question in 9702/21 May/June 2008

3 (a) Distinguish between the moment of a force and the torque of a couple. For Examiner’s moment of a force … Use … … torque of a couple … … … [4] (b) One type of weighing machine, known as a steelyard, is illustrated in Fig. 3.1. 4.8 cm pivot 12 N sliding weight metal rod hook 2.5 N sliding weight Fig. 3.1 The two sliding weights can be moved independently along the rod. With no load on the hook and the sliding weights at the zero mark on the metal rod, the metal rod is horizontal. The hook is 4.8 cm from the pivot. A sack of flour is suspended from the hook. In order to return the metal rod to the horizontal position, the 12 N sliding weight is moved 84 cm along the rod and the 2.5 N weight is moved 72 cm. (i) Calculate the weight of the sack of flour. For Examiner’s Use weight = …………………………N [2] (ii) Suggest why this steelyard would be imprecise when weighing objects with a weight of about 25 N. … … [1]

7 marks

Mark scheme: 3 (a) moment: force × perpendicular distance M1 of force from pivot / axis / point A1 couple: (magnitude of) one force × perpendicular distance M1 between the two forces A1 [4] (penalise the ‘perpendicular’ omission once only) (b) (i) W × 4.8 = (12 × 84) + (2.5 × 72) C1 W = 250 N (248 N) A1 [2] (ii) either friction at the pivot or small movement of weights B1 [1]

This question in 9702/21 Oct/Nov 2008

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

7 (a) Explain the difference in densities in solids, liquids and gases using ideas of the spacing For between molecules. Examiner’s Use … … … … … … … [3] (b) A hydrogen nucleus (proton) may be assumed to be a sphere of radius 1 × 10–15 m. Calculate the density of a hydrogen nucleus. density = … kg m–3 [3] (c) The density of hydrogen gas in a pressurised cylinder is 4 kg m–3. Suggest a reason why this density is much less than your answer in (b). … … … [1]

7 marks

Mark scheme: 7 (a) density in solids and liquids similar M1 spacing in solids and liquids about the same A1 density in gases much less as spacing in gases much greater B1 [3] (b) density = mass / volume C1 mass = 1.67 × 10–27 kg and volume = 4/3 π r 3 C1 density = (1.67 × 10–27) / 4/3 × π × (1.0 × 10–15)3 = 3.99 × 1017 kg m–3 A1 [3] (c) atoms / molecules composed of large amount of empty space / nucleus has very small volume compared to volume of atom / space between atoms in a gas is very large B1 [1]

This question in 9702/23 May/June 2011

1 (a) Define density. … … [1] (b) Explain how the difference in the densities of solids, liquids and gases may be related to the spacing of their molecules. … … … … [2] (c) A paving slab has a mass of 68 kg and dimensions 50 mm × 600 mm × 900 mm. (i) Calculate the density, in kg m–3, of the material from which the paving slab is made. density = … kg m–3 [2] (ii) Calculate the maximum pressure a slab could exert on the ground when resting on one of its surfaces. pressure = … Pa [3]

8 marks

Mark scheme: 1 (a) density = mass / volume B1 [1] (b) density of liquids and solids same order as spacing similar / to about 2× B1 density of gases much less as spacing much more or density of gases much lower hence spacing much more B1 [2] (c) (i) density = 68 / [50 × 600 × 900 × 10–9] C1 = 2520 (allow 2500) kg m–3 A1 [2] (ii) P = F / A C1 = 68 × 9.81 / [50 × 600 × 10–6] C1 = 2.2 × 104 Pa A1 [3]

This question in 9702/21 Oct/Nov 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

3 (a) Show that the pressure P due to a liquid of density ρ is proportional to the depth h below For the surface of the liquid. Examiner’s Use [4] (b) The pressure of the air at the top of a mountain is less than that at the foot of the mountain. Explain why the difference in air pressure is not proportional to the difference in height as suggested by the relationship in (a). … … … … [2]

6 marks

Mark scheme: 3 (a) V = h × A m = V × ρ B1 W = h × A × ρ × g B1 P = F / A B1 P = hρg P is proportional to h if ρ is constant (and g) B1 [4] (b) density changes with height B1 hence density is not constant with link to formula B1 [2] GCE AS/A LEVEL – May/June 2012 9702 23

This question in 9702/23 May/June 2012

3 (a) Define pressure. For Examiner’s … Use … [1] (b) Explain, in terms of the air molecules, why the pressure at the top of a mountain is less than at sea level. … … … … … [3] (c) Fig. 3.1 shows a liquid in a cylindrical container. container liquid 0.250 m Fig. 3.1 The cross-sectional area of the container is 0.450 m2. The height of the column of liquid is 0.250 m and the density of the liquid is 13 600 kg m–3. (i) Calculate the weight of the column of liquid. weight = … N [3] (ii) Calculate the pressure on the base of the container caused by the weight of the For liquid. Examiner’s Use pressure = … Pa [1] (iii) Explain why the pressure exerted on the base of the container is different from the value calculated in (ii). … … [1]

9 marks

Mark scheme: 3 (a) pressure = force / area B1 [1] (b) molecules collide with object / surface and rebound B1 molecules have change in momentum hence force acts B1 fewer molecules per unit volume on top of mountain / temperature is less hence lower speed of molecules B1 hence less pressure A0 [3] (c) (i) ρ = m / V C1 W = Vρg = 0.25 × 0.45 × 9.81 × 13600 C1 = 15000 (15009) N A1 [3] (ii) p = W / A (or using p = ρgh) = 15009 / 0.45 = 3.3 × 104 Pa A1 [1] (iii) pressure will be greater due to the air pressure (acting on the surface of the liquid) B1 [1]

This question in 9702/21 Oct/Nov 2012

4 (a) Define pressure. For Examiner’s … Use … [1] (b) Use the kinetic model to explain the pressure exerted by a gas. … … … … … … [4] (c) Explain whether the collisions between the molecules of an ideal gas are elastic or inelastic. … … … [2]

7 marks

Mark scheme: 4 (a) pressure = force / area (normal to force) A1 [1] (b) molecules/atoms/particles in (constant) random/haphazard motion B1 molecules have a change in momentum when they collide with the walls M1 (force exerted on molecules) therefore force on the walls A1 reference to average force from many molecules/many collisions A1 [4] (c) elastic collision when kinetic energy conserved B1 temperature constant for gas B1 [2]

This question in 9702/21 May/June 2013

1 A cylindrical disc is shown in Fig. 1.1. Use 28 mm 12 mm Fig. 1.1 The disc has diameter 28 mm and thickness 12 mm. The material of the disc has density 6.8 × 103 kg m–3. Calculate, to two significant figures, the weight of the disc. weight = … N [4]

4 marks

Mark scheme: 1 volume = π (14 × 10–3)2 × 12 × 10–3 (=7.389 × 10–6 m3) C1 density = mass / volume [any subject] C1 mass = 6.8 × 103 × 7.389 × 10–6 = 0.0502 weight = mg C1 weight = 0.0502 × 9.81 = 0.49 N (mark not awarded if not to two s.f.) A1 [4]

This question in 9702/23 Oct/Nov 2013

2 A coin is made in the shape of a thin cylinder, as shown in Fig. 2.1. diameter thickness Fig. 2.1 Fig. 2.2 shows the measurements made in order to determine the density ρ of the material used to make the coin. quantity measurement uncertainty mass 9.6 g ± 0.5 g thickness 2.00 mm ± 0.01 mm diameter 22.1 mm ± 0.1 mm Fig. 2.2 (a) Calculate the density ρ in kg m–3. ρ = … kg m–3 [3] (b) (i) Calculate the percentage uncertainty in ρ. percentage uncertainty = … [3] (ii) State the value of ρ with its actual uncertainty. ρ = … ± … kg m–3 [1]

7 marks

Mark scheme: 2 (a) ρ = m / V C1 V = (π d 2 / 4) × t = 7.67 × 10–7 m3 ρ = (9.6 × 10–3) / [π(22.1/2 × 10–3)2 × 2.00 × 10–3] C1 ρ = 12513 kg m–3 (allow 2 or more s.f.) A1 [3] (b) (i) ∆ρ / ρ = ∆m / m + ∆t / t + 2∆d / d C1 = 5.21% + 0.50% + 0.905% [or correct fractional uncertainties] C1 = 6.6% (6.61%) A1 [3] (ii) ρ = 12 500 ± 800 kg m–3 A1 [1]

This question in 9702/22 May/June 2014

2 (a) Define pressure. … [1] (b) A cylinder is placed on a horizontal surface, as shown in Fig. 2.1. diameter cylinder Fig. 2.1 The following measurements were made on the cylinder: mass = 5.09 ± 0.01 kg diameter = 9.4 ± 0.1 cm. (i) Calculate the pressure produced by the cylinder on the surface. pressure = … Pa [3] (ii) Calculate the actual uncertainty in the pressure. actual uncertainty = … Pa [3] (iii) State the pressure, with its actual uncertainty. pressure = … ± … Pa [1]

8 marks

Mark scheme: 2 (a) pressure = force / area (normal to the force) [clear ratio essential] B1 [1] (b) (i) P = mg / A = (5.09 × 9.81) / A C1 A = (πd2 / 4) = π × (9.4 × 10–2)2 / 4 (= 0.00694 m2) C1 P = 49.93 / 0.00694 = 7200 (7195) Pa (minimum of 2 s.f. required) A1 [3] (ii) ∆P / P = ∆m / m + 2∆d / d C1 = 0.01 / 5.09 + (2 × 0.1) / 9.4 (= 0.0020 + 0.021 or 2.3%) C1 ∆P = 170 (165 to 167) Pa A1 [3] (iii) P = 7200 ± 200 Pa A1 [1]

This question in 9702/21 Oct/Nov 2014

4 Fig. 4.1 shows a metal cylinder of height 4.5 cm and base area 24 cm2. metal cylinder 4.5 cm base area 24 cm2 Fig. 4.1 The density of the metal is 7900 kg m–3. (a) Show that the mass of the cylinder is 0.85 kg. [2] (b) The cylinder is placed on a plank, as shown in Fig. 4.2. plank cylinder 40° horizontal Fig. 4.2 The plank is at an angle of 40° to the horizontal. Calculate the pressure on the plank due to the cylinder. pressure = … Pa [3] (c) The cylinder then slides down the plank with a constant acceleration of 3.8 m s–2. A constant frictional force f acts on the cylinder. Calculate the frictional force f. f = … N [3]

8 marks

Mark scheme: 4 (a) density = mass / volume C1 mass = 7900 × 4.5 × 24 × 10–6 = 0.85 (0.853) kg M1 [2] (b) pressure = force / area C1 force = W cos 40° C1 pressure = (0.85 × 9.81 cos 40°) / 24 × 10–4 = 2.7 (2.66) × 103 Pa A1 [3] (c) F = ma C1 W sin 40° – f = ma C1 0.85 × 9.81 × sin 40° – f = 0.85 × 3.8 f (= 5.36 – 3.23) = 2.1 N [5.38 – 3.242 if 0.8532 kg is used for the mass] A1 [3]

This question in 9702/21 Oct/Nov 2015

7 A steel wire of cross-sectional area 15 mm2 has an ultimate tensile stress of 4.5 × 108 N m–2. (a) Calculate the maximum tension that can be applied to the wire. tension = … N [2] (b) The steel of the wire has density 7800 kg m–3. The wire is hung vertically. Calculate the maximum length of the steel wire that could be hung vertically before the wire breaks under its own weight. length = … m [3] Please turn over for Question 8.

5 marks

Mark scheme: 7 (a) stress or σ = F / A C1 max. tension = UTS × A = 4.5 × 108 × 15 × 10–6 = 6800 (6750) N A1 [2] (b) ρ = m / V C1 weight = mg = ρVg = ρALg 6750 = 7.8 × 103 × 15 × 10–6 × L × 9.81 C1 L = 5.9 (5.88) × 103 m A1 or maximum mass = 6750 / 9.81 = 688 kg (C1) mass per unit length = ρA = 0.117 kg m–1 (C1) L = 688 / 0.117 = 5.9 × 103 m (A1) or maximum mass = 6750 / 9.81 = 688 kg (C1) volume = m / ρ = 0.0882 m3 = LA (C1) L = 0.0882 / 15 × 10–6 = 5.9 × 103 m (A1) [3]

This question in 9702/23 Oct/Nov 2015

3 A ball of mass 150 g is at rest on a horizontal floor, as shown in Fig. 3.1. ball floor Fig. 3.1 (a) (i) Calculate the magnitude of the normal contact force from the floor acting on the ball. force = … N [1] (ii) Explain your working in (i). … … … [1] (b) The ball is now lifted above the floor and dropped so that it falls vertically, as illustrated in Fig. 3.2. ball 6.2 m s–1 2.5 m s–1 just before contact just after losing contact Fig. 3.2 Just before contact with the floor, the ball has velocity 6.2 m s–1 downwards. The ball bounces from the floor and its velocity just after losing contact with the floor is 2.5 m s–1 upwards. The ball is in contact with the floor for 0.12 s. (i) State Newton’s second law of motion. … … [1] (ii) Calculate the average resultant force on the ball when it is in contact with the floor. magnitude of force = … N direction of force … [3] (iii) State and explain whether linear momentum is conserved during the collision of the ball with the floor. … … … … [2] [Total: 8]

8 marks

Mark scheme: 3 (a) (i) force (= mg = 0.15 × 9.81) = 1.5 (1.47) N A1 [1] (ii) resultant force (on ball) is zero so normal contact force = weight or the forces are in opposite directions so normal contact force = weight or normal contact force up = weight down A1 [1] (b) (i) (resultant) force proportional/equal to rate of change of momentum B1 [1] (ii) change in momentum = 0.15 × (6.2 + 2.5) (= 1.305 N s) C1 magnitude of force = 1.305 / 0.12 = 11 (10.9) N A1 or (average) acceleration = (6.2 + 2.5) / 0.12 (= 72.5 m s–2) (C1) magnitude of force = 0.15 × 72.5 = 11 (10.9) N (A1) (direction of force is) upwards/up B1 [3] (iii) there is a change/gain in momentum of the floor M1 this is equal (and opposite) to the change/loss in momentum of the ball so momentum is conserved A1 [2] or change of (total) momentum of ball and floor is zero (M1) so momentum is conserved (A1) or (total) momentum of ball and floor before is equal to the (total) momentum of ball and floor after (M1) so momentum is conserved (A1)

This question in 9702/21 May/June 2016

2 (a) Fig. 2.1 shows a liquid in a cylindrical container. F\OLQGULFDO FRQWDLQHU OLTXLG K DUHD $ Fig. 2.1 The cross-sectional area of the container is A. The height of the column of liquid is h and the density of the liquid is ρ. Show that the pressure p due to the liquid on the base of the cylinder is given by p = ρgh. [3] (b) The variation with height h of the total pressure P on the base of the cylinder in (a) is shown in Fig. 2.2. 3.0 3 / 105 Pa 2.0 1.0 0 0 0.5 1.0 1.5 2.0 K / m Fig. 2.2 (i) Explain why the line of the graph in Fig. 2.2 does not pass through the origin (0,0). … … [1] (ii) Use data from Fig. 2.2 to calculate the density of the liquid in the cylinder. density = … kg m–3 [2] [Total: 6]

6 marks

Mark scheme: 2 (a) p = F / A M1 use of m = ρV and use of V = Ah and use of F = mg M1 correct substitution to obtain p = ρgh A1 [3] (b) (i) (when h is zero the pressure is not zero due to) pressure from the air/atmosphere B1 [1] (ii) gradient = ρg or P – 1.0 × 105 = ρgh C1 e.g. ρg = 1.0 × 105 / 0.75 (= 133333) ρ = 133 333 / 9.81 = 14 000 (13 592) kg m–3 A1 [2]

This question in 9702/22 May/June 2016

4 A spring balance is used to weigh a cylinder that is immersed in oil, as shown in Fig. 4.1. spring balance thin wire cross-sectional area 13 cm2 cylinder 5.0 cm oil Fig. 4.1 The reading on the spring balance is 4.8 N. The length of the cylinder is 5.0 cm and the cross- sectional area of the cylinder is 13 cm2. The weight of the cylinder is 5.3 N. (a) The cylinder is in equilibrium when it is immersed in the oil. Explain this in terms of the forces acting on the cylinder. … … [1] (b) Calculate the density of the oil. density = … kg m–3 [3] [Total: 4]

4 marks

Mark scheme: 4 (a) (resultant force = 0) (equilibrium) therefore: weight – upthrust = force from thin wire (allow tension in wire) or 5.3 (N) – upthrust = 4.8 (N) B1 [1] (b) difference in weight = upthrust or upthrust = 0.5 (N) 0.5 = ρghA or m = 0.5 / 9.81 and V = 5.0 × 13 × 10–6 (m3) C1 ρ = 0.5 / (9.81 × 5.0 × 13 × 10–6) C1 = 780 (784) kg m–3 A1 [3]

This question in 9702/23 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

1 (a) (i) Define pressure. … … [1] (ii) Show that the SI base units of pressure are kg m–1 s–2. [1] (b) Gas flows through the narrow end (nozzle) of a pipe. Under certain conditions, the mass m of gas that flows through the nozzle in a short time t is given by m = kC ρP t where k is a constant with no units, C is a quantity that depends on the nozzle size, ρ is the density of the gas arriving at the nozzle, P is the pressure of the gas arriving at the nozzle. Determine the base units of C. base units … [3] [Total: 5]

5 marks

Mark scheme: 1 (a) (i) force / area (normal to the force) B1 [1] (ii) (p = F / A so) units: kg m s–2 / m2 = kg m–1 s–2 A1 [1] allow use of other correct equations: e.g. (∆p = ρg∆h so) kg m–3 m s–2 m = kg m–1 s–2 e.g. (p = W / ∆V so) kg m s–2 m / m3 = kg m–1 s–2 (b) units for m: kg, t: s and ρ: kg m–3 C1 units of C: kg / s (kg m–3 kg m–1 s–2)1/2 or units of C2: kg2 / s2 kg m–3 kg m–1 s–2 C1 units of C: m2 A1 [3]

This question in 9702/22 Oct/Nov 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 9702/23 Oct/Nov 2016

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

7 marks

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

This question in 9702/22 Feb/March 2017

1 (a) Determine the SI base units of stress. Show your working. base units … [2] (b) A beam PQ is clamped so that the beam is horizontal. A mass M of 500 g is hung from end Q and the beam bends slightly, as illustrated in Fig. 1.1. clamp R l horizontal P Q M Fig. 1.1 The length l of the beam from the edge of the clamp R to end Q is 60.0 cm. The width b of the beam is 30.0 mm and the thickness d of the beam is 5.00 mm. The material of the beam has Young modulus E. The mass M is made to oscillate vertically. The time period T of the oscillations is 0.58 s. The period T is given by the expression 4 Ml 3 T = 2π 3 . Ebd (i) Determine E in GPa. E = … GPa [3] (ii) The quantities used to determine E should be measured with accuracy and with precision. 1. Explain the difference between accuracy and precision. accuracy: … … precision: … … [2] 2. In a particular experiment, the quantities l and T are measured with the same percentage uncertainty. State and explain which of these two quantities contributes more to the uncertainty in the value of E. … … … [1] [Total: 8]

8 marks

Mark scheme: 1(a) B1 = kg m–1 s–2 A1 1(b)(i) 0.58 = 2π × [(4 × 0.500 × 0.6003 ) / (E × 0.0300 × 0.005003)]0.5 C1 E = [4π2 × 4 × 0.500 × (0.600)3] / [(0.58)2 × 0.0300 × (0.00500)3] = 1.35 × 1010 (Pa) C1 = 14 (13.5) GPa A1 1(b)(ii)1. (accuracy determined by) the closeness of the value(s)/measurement(s) to the true value B1 (precision determined by) the range of the values/measurements B1 1(b)(ii)2. l is (cubed so) 3 × (percentage/fractional) uncertainty and T is (squared so) 2 × (percentage / fractional) uncertainty and (so) l contributes more B1

This question in 9702/21 May/June 2017

3 (a) A cylinder is made from a material of density 2.7 g cm–3. The cylinder has diameter 2.4 cm and length 5.0 cm. Show that the cylinder has weight 0.60 N. [3] (b) The cylinder in (a) is hung from the end A of a non-uniform bar AB, as shown in Fig. 3.1. 50 cm 20 cm bar 12 cm A B P X cylinder 0.25 N 0.60 N Fig. 3.1 The bar has length 50 cm and has weight 0.25 N. The centre of gravity of the bar is 20 cm from B. The bar is pivoted at P. The pivot is 12 cm from B. An object X is hung from end B. The weight of X is adjusted until the bar is horizontal and in equilibrium. (i) Explain what is meant by centre of gravity. … … [1] (ii) Calculate the weight of X. weight of X = … N [3] (c) The cylinder is now immersed in water, as illustrated in Fig. 3.2. A B P water X 0.25 N Fig. 3.2 An upthrust acts on the cylinder and the bar is not in equilibrium. (i) Explain the origin of the upthrust. … … … … [2] (ii) Explain why the weight of X must be reduced in order to obtain equilibrium for AB. … … … [1] [Total: 10]

10 marks

Mark scheme: 3(a) C1 V = πd2L / 4 or πr2L C1 weight = 2.7 × 103 × π (1.2 × 10–2)2 × 5.0 × 10–2 × 9.81 = 0.60 N A1 3(b)(i) the point from where (all) the weight (of a body) seems to act B1 3(b)(ii) W × 12 C1 (0.25 × 8) + (0.6 × 38) C1 W = (2 + 22.8) / 12 = 2.1 (2.07) N A1 3(c)(i) pressure changes with depth (in water) or pressure on bottom (of cylinder) different from pressure on top B1 pressure on bottom of cylinder greater than pressure on top or force (up) on bottom of cylinder greater than force (down) on top B1 3(c)(ii) anticlockwise moment reduced and reducing the weight of X reduces clockwise moment or anticlockwise moment reduced so clockwise moment now greater than (total) anticlockwise moment B1

This question in 9702/21 May/June 2017

3 A spring is attached at one end to a fixed point and hangs vertically with a cube attached to the other end. The cube is initially held so that the spring has zero extension, as shown in Fig. 3.1. spring with zero extension cube weight 4.0 N 5.1 cm 5.1 cm water 7.0 cm density 1000 kg m–3 Fig. 3.1 Fig. 3.2 The cube has weight 4.0 N and sides of length 5.1 cm. The cube is released and sinks into water as the spring extends. The cube reaches equilibrium with its base at a depth of 7.0 cm below the water surface, as shown in Fig. 3.2. The density of the water is 1000 kg m–3. (a) Calculate the difference in the pressure exerted by the water on the bottom face and on the top face of the cube. difference in pressure = … Pa [2] (b) Use your answer in (a) to show that the upthrust on the cube is 1.3 N. [2] (c) Calculate the force exerted on the spring by the cube when it is in equilibrium in the water. force = … N [1] (d) The spring obeys Hooke’s law and has a spring constant of 30 N m–1. Determine the initial height above the water surface of the base of the cube before it was released. height above surface = … cm [3] (e) The cube in the water is released from the spring. (i) Determine the initial acceleration of the cube. acceleration = … m s–2 [2] (ii) Describe and explain the variation, if any, of the acceleration of the cube as it sinks in the water. … … … [2] [Total: 12]

12 marks

Mark scheme: 3(a) C1 ∆p = 1000 × 9.81 × (7.0 × 10–2 – 1.9 × 10–2) or 686 – 186 = 500 Pa A1 3(b) F = pA or (∆)F = ∆p × A C1 upthrust = 500 × (5.1 × 10–2)2 = 1.3 N or upthrust = (686 – 186) × (5.1 ×10–2)2 = 1.3 N or upthrust = 1000 × 9.81 × 5.1 ×10–2 × (5.1 × 10–2)2 = 1.3 N A1 3(c) force = 4.0 – 1.3 = 2.7 N A1 Question Answer Marks 3(d) extension/x/e = 2.7 / 30 C1 = 0.09 (m) or 9 (cm) C1 height above surface = 9 – 7 = 2 cm A1 3(e)(i) mass = 4.0 / 9.81 C1 acceleration = 2.7 / (4.0 / 9.81) = 6.6 m s–2 A1 3(e)(ii) viscous force increases (and then becomes constant) M1 (weight and upthrust constant so) acceleration decreases (to zero) A1

This question in 9702/22 Oct/Nov 2017

2 A liquid of density ρ fills a container to a depth h, as shown in Fig. 2.1. container liquid h base area A Fig. 2.1 The base of the container has area A. (a) Derive, from the definitions of pressure and density, the equation p = ρgh where p is the pressure exerted by the liquid on the base of the container and g is the acceleration of free fall. [3] (b) A small solid sphere falls with constant velocity through the liquid. (i) State 1. the names of the three forces acting on the sphere, … … 2. a word equation that relates the magnitudes of these forces. … [2] (ii) State and explain the changes in energy that occur as the sphere falls. … … … … [2] (c) The liquid in the container is liquid L. Liquid M is now added to the container. The two liquids do not mix. The total depth of the liquids is 0.17 m. Fig. 2.2 shows how the pressure p inside the liquids varies with height x above the base of the container. 9.25 p / 104 Pa liquid L 9.20 9.15 liquid M 9.10 0 0.05 0.10 0.15 0.20 x / m Fig. 2.2 Use Fig. 2.2 to (i) state the value of atmospheric pressure, atmospheric pressure = … Pa [1] (ii) determine the density of liquid M. density = … kg m–3 [2] [Total: 10]

10 marks

Mark scheme: 2(a) B1 p = F / A or p = W / A B1 p = [ρAhg] / A or p = [ρVg] / [V / h] (so) p = ρgh A1 Question Answer Marks 2(b)(i) 1. weight/gravitational (force) upthrust (force)/buoyancy (force) drag/viscous/frictional (force)/fluid resistance/resistance B1 2. weight = upthrust + viscous (force) B1 2(b)(ii) • decrease in (gravitational) potential energy (of sphere) due to decrease in height (since Ep = mgh) • increase in thermal energy due to work done against viscous force/drag • loss/change of (total) Ep equal to gain/change in thermal energy Any 2 points. B2 2(c)(i) atmospheric pressure = 9.1(0) × 104 Pa A1 2(c)(ii) (∆)p = ρg(∆)h (9.15 – 9.10) × 104 = ρ × 9.81 × (0.17 – 0.10) C1 ρ = 730 (728) kg m–3 A1

This question in 9702/23 Oct/Nov 2017

5 A solid cylinder is lifted out of oil by a wire attached to a motor. Fig. 5.1 shows two different positions X and Y of the cylinder during the lifting process. beam motor wire cylinder at position Y velocity surface of oil 0.020 m s–1 cylinder at position X oil Fig. 5.1 The motor is fixed to an overhead beam. The cylinder has cross-sectional area 0.018 m2, length 1.2 m and weight 560 N. The density of the oil is 940 kg m–3. Throughout the lifting process, the cylinder moves vertically upwards with a constant velocity of 0.020 m s–1. The viscous force of the oil acting on the cylinder is negligible. (a) Calculate the density of the cylinder. density = … kg m–3 [2] (b) For the cylinder at position X, show that the upthrust due to the oil is 200 N. [2] (c) Calculate, for the moving cylinder at position X, (i) the tension in the wire, tension = … N [1] (ii) the power output of the motor. power = … W [2] (d) The cylinder is raised with constant velocity from position X to position Y. (i) State and explain the variation, if any, of the power output of the motor as the cylinder is raised. Numerical values are not required. … … … … … [3] (ii) The rate of energy output of the motor is less than the rate of increase of gravitational potential energy of the cylinder. Without calculation, explain this difference. … … [1] [Total: 11]

11 marks

Mark scheme: 5(a) C1 = (560 / 9.81) / (1.2 × 0.018) = 2600 kg m–3 A1 5(b) (∆)p = 940 × 9.81 × 1.2 C1 (upthrust =) 940 × 9.81 × 1.2 × 0.018 = 200 N A1 5(c)(i) tension = 560 – 200 = 360 N A1 5(c)(ii) P = Fv C1 = 360 × 0.020 = 7.2 W A1 5(d)(i) upthrust decreases B1 tension (in wire) increases M1 power (output of motor) increases A1 5(d)(ii) there is work done (on the cylinder) by the upthrust or GPE of oil decreases (as it fills the space left by cylinder and so total energy is conserved) B1

This question in 9702/22 May/June 2018

2 (a) State what is meant by work done. … … [1] (b) A diver releases a solid sphere of radius 16 cm from the sea bed. The sphere moves vertically upwards towards the surface of the sea. The weight of the sphere is 20 N. The upthrust acting on the sphere is 170 N. The upthrust remains constant as the sphere moves upwards. (i) Calculate the density of the material of the sphere. density = … kg m–3 [2] (ii) Briefly explain the origin of the upthrust acting on the sphere. … … … [1] (iii) Calculate the acceleration of the sphere as it is released from rest. acceleration = … m s–2 [2] (iv) The viscous (drag) force D acting on the sphere is given by D = kr 2v 2 where r is the radius of the sphere and v is its speed. The constant k is equal to 810 kg m–3. Determine the constant (terminal) speed reached by the sphere. speed = … m s–1 [3] (v) The diver releases a different sphere that moves with a constant speed of 6.30 m s–1 directly towards a stationary ship. The sphere emits sound of frequency 4850 Hz. The ship detects sound of frequency 4870 Hz as the sphere moves towards it. Determine, to three significant figures, the speed of the sound in the water. speed = … m s–1 [2] [Total: 11]

11 marks

Mark scheme: 2(a)(i) B1 2(b)(i) ρ = m / V C1 = (20 / 9.81) / (4/3 × π × 0.163) = 120 kg m–3 A1 2(b)(ii) the pressure on the lower surface (of sphere) is greater than the pressure on the upper surface (of sphere) B1 2(b)(iii) a = (170 – 20) / (20 / 9.81) C1 = 74 m s–2 A1 2(b)(iv) D = 170 – 20 (= 150) C1 810 × (0.162) × v2 = 150 C1 v = 2.7 m s–1 A1 2(b)(v) 4870 = (4850 × v) / (v – 6.30) C1 v = 1530 m s–1 A1

This question in 9702/23 May/June 2018

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) The diameter d of a cylinder is measured as 0.0125 m ± 1.6%. Calculate the absolute uncertainty in this measurement. absolute uncertainty = … m [1] (b) The cylinder in (a) stands on a horizontal surface. The pressure p exerted on the surface by the cylinder is given by 4 W p = 2 . π d The measured weight W of the cylinder is 0.38 N ± 2.8%. (i) Calculate the pressure p. p = … N m−2 [1] (ii) Determine the absolute uncertainty in the value of p. absolute uncertainty = … N m−2 [2] [Total: 4]

4 marks

Mark scheme: 1(a) = 2 × 10–4 m A1 1(b)(i) p = (4 × 0.38) / (π × 0.01252) = 3100 N m–2 A1 1(b)(ii) percentage uncertainty = 2.8 + (2 × 1.6) (= 6%) or fractional uncertainty = 0.028 + (2 × 0.016) (= 0.06) C1 absolute uncertainty = 0.06 × 3100 = 190 N m–2 (allow to 1 significant figure) A1

This question in 9702/22 May/June 2019

3 A cylindrical disc of mass 0.24 kg has a circular cross-sectional area A, as shown in Fig. 3.1. cross-sectional force X area A 8.9 N constant 30° speed 0.60 m s–1 disc, disc mass 0.24 kg ground Fig. 3.1 Fig. 3.2 The disc is on horizontal ground, as shown in Fig. 3.2. A force X of magnitude 8.9 N acts on the disc in a direction of 30° to the horizontal. The disc moves at a constant speed of 0.60 m s−1 along the ground. (a) Determine the rate of doing work on the disc by the force X. rate of doing work = … W [2] (b) The force X and the weight of the disc exert a combined pressure on the ground of 3500 Pa. Calculate the cross-sectional area A of the disc. A = … m2 [3] (c) Newton’s third law describes how forces exist in pairs. One such pair of forces is the weight of the disc and another force Y. State: (i) the direction of force Y … [1] (ii) the name of the body on which force Y acts. … [1] [Total: 7]

7 marks

Mark scheme: 3(a) P = Fv C1 P = 8.9 cos 30° × 0.60 = 4.6 W A1 3(b) p = F / A C1 F = 8.9 sin 30° + (0.24 × 9.81) ( = 6.80 N) C1 A = 6.80 / 3500 = 1.9 × 10–3 m2 A1 3(c)(i) upwards/up B1 3(c)(ii) the Earth/planet B1

This question in 9702/23 May/June 2019

3 A small remote-controlled model aircraft has two propellers, each of diameter 16 cm. Fig. 3.1 is a side view of the aircraft when hovering. body of 16 cm 16 cm aircraft propeller propeller air air speed speed 7.6 m s–1 7.6 m s–1 Fig. 3.1 Air is propelled vertically downwards by each propeller so that the aircraft hovers at a fixed position. The density of the air is 1.2 kg m–3. Assume that the air from each propeller moves with a constant speed of 7.6 m s–1 in a uniform cylinder of diameter 16 cm. Also assume that the air above each propeller is stationary. (a) Show that, in a time interval of 3.0 s, the mass of air propelled downwards by one propeller is 0.55 kg. [3] (b) Calculate: (i) the increase in momentum of the mass of air in (a) increase in momentum = … N s [1] (ii) the downward force exerted on this mass of air by the propeller. force = … N [1] (c) State: (i) the upward force acting on one propeller force = … N [1] (ii) the name of the law that explains the relationship between the force in (b)(ii) and the force in (c)(i). … [1] (d) Determine the mass of the aircraft. mass = … kg [1] (e) In order for the aircraft to hover at a very high altitude (height), the propellers must propel the air downwards with a greater speed than when the aircraft hovers at a low altitude. Suggest the reason for this. … … [1] (f) When the aircraft is hovering at a high altitude, an electric fault causes the propellers to stop rotating. The aircraft falls vertically downwards. When the aircraft reaches a constant speed of 22 m s–1, it emits sound of frequency 3.0 kHz from an alarm. The speed of the sound in the air is 340 m s–1. Determine the frequency of the sound heard by a person standing vertically below the falling aircraft. frequency = … Hz [2] [Total: 11]

11 marks

Mark scheme: 3(a) C1 V = π × (0.16 / 2)2 × 7.6 × 3.0 (= 0.458 m3) C1 m = π × (0.16 / 2)2 × 7.6 × 3.0 × 1.2 = 0.55 kg A1 3(b)(i) ∆p = 0.55 × 7.6 = 4.2 N s A1 3(b)(ii) F = 4.2 / 3.0 or 0.55 × 7.6 / 3.0 = 1.4 N A1 3(c)(i) F = 1.4 N A1 3(c)(ii) Newton’s third law (of motion) B1 3(d) 2 × 1.4 = m × 9.81 m = 0.29 kg A1 3(e) the density of air is less at high altitude B1 3(f) fo = fsv / (v – vs) = 3000 × 340 / (340 – 22) C1 = 3200 Hz A1

This question in 9702/21 Oct/Nov 2019

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

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 spring is extended by a force. The variation with extension x of the force F is shown in Fig. 3.1. 8.0 6.0 F / N 4.0 2.0 0 0 1.0 2.0 3.0 4.0 5.0 x / cm Fig. 3.1 (a) State the name of the law that relates the force and extension of the spring shown in Fig. 3.1. … [1] (b) Determine: (i) the spring constant, in N m−1, of the spring spring constant = … N m−1 [2] (ii) the strain energy (elastic potential energy) in the spring when the extension is 4.0 cm. strain energy = … J [2] (c) One end of the spring is attached to a fixed point. A cylinder that is submerged in a liquid is now suspended from the other end of the spring, as shown in Fig. 3.2. fixed point spring, extension 4.0 cm cylinder, cross-sectional area 1.2 × 10–3 m2 cylinder, cylinder, weight 6.20 N length 5.8 cm liquid Fig. 3.2 The cylinder has length 5.8 cm, cross-sectional area 1.2 × 10−3 m2 and weight 6.20 N. The cylinder is in equilibrium when the extension of the spring is 4.0 cm. (i) Show that the upthrust acting on the cylinder is 0.60 N. [1] (ii) Calculate the difference in pressure between the bottom face and the top face of the cylinder. difference in pressure = … Pa [2] (iii) Calculate the density of the liquid. density = … kg m−3 [2] (d) The liquid in (c) is replaced by another liquid of greater density. State the effect, if any, of this change on: (i) the upthrust acting on the cylinder … [1] (ii) the extension of the spring. … [1] [Total: 12]

12 marks

Mark scheme: 3(a) Hooke’s (law) B1 3(b)(i) k = F / x or k = gradient e.g. k = 7.0 / 5.0 × 10–2 C1 = 140 N m–1 A1 3(b)(ii) E = ½ F x or E = ½ k x 2 or E = area under graph = ½ × 5.6 × 4.0 × 10–2 or ½ × 140 × (4.0 × 10–2)2 C1 = 0.11 J A1 3(c)(i) (upthrust =) 6.20 – 5.60 = 0.60 (N) A1 3(c)(ii) Δp = ΔF / A = 0.60 / 1.2 × 10–3 C1 = 500 Pa A1 3(c)(iii) (Δ)p = ρg(Δ)h ρ = 500 / (9.81 × 5.8 × 10–2) C1 = 880 kg m–3 A1 3(d)(i) (upthrust) increases B1 3(d)(ii) (extension) decreases B1

This question in 9702/22 Feb/March 2021

1 (a) Define density. … … [1] (b) Fig. 1.1 shows a solid pyramid with a square base. pyramid, density ρ mass m h x x Fig. 1.1 The mass m of the pyramid is given by 1 m = 3ρhx2 where ρ is the density of the material of the pyramid, h is the height, and x is the length of each side of the base. Measurements are taken as shown in Table 1.1. Table 1.1 percentage quantity measurement uncertainty m 19.5 g ± 2% x 4.0 cm ± 5% h 4.8 cm ± 4% (i) Calculate the absolute uncertainty in length x. absolute uncertainty = … cm [1] (ii) The density ρ is calculated from the measurements in Table 1.1. Determine the percentage uncertainty in the calculated value of ρ. percentage uncertainty = … % [2] (c) The square base of the pyramid in (b) rests on the horizontal surface of a bench. Use data from Table 1.1 to calculate the average pressure of the pyramid on the surface of the bench. The uncertainty in your answer is not required. pressure = … Pa [3] [Total: 7]

7 marks

Mark scheme: 1(a) mass / volume B1 1(b)(i) absolute uncertainty = 4.0 × (5 / 100) = (±) 0.2 cm B1 1(b)(ii) percentage uncertainty = 2 + 4 + (5 × 2) C1 = (±) 16% A1 1(c) p = F / A or p = W / A C1 p = (19.5 × 10–3 × 9.81) / (4.0 × 10–2)2 C1 = 120 Pa A1

This question in 9702/21 May/June 2021

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

11 marks

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

This question in 9702/22 May/June 2021

1 (a) 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 solid cylinder of weight 24 N is made of material of density 850 kg m–3. The cylinder has a length of 0.18 m, as shown in Fig. 1.1. cylinder, cross-sectional area A weight 24 N density 850 kg m–3 length 0.18 m Fig. 1.1 Show that the cross-sectional area A of the cylinder is 0.016 m2. [3] (b) The cylinder in (a) is attached by a spring to the bottom of a rigid container of liquid, as shown in Fig. 1.2. cylinder liquid 0.17 m spring tap container Fig. 1.2 (not to scale) The cylinder is in equilibrium with its bottom face at a depth of 0.17 m below the surface of the liquid. The tension in the spring is 8.0 N. (i) Show that the upthrust acting on the cylinder due to the liquid is 32 N. [1] (ii) Calculate the density of the liquid. density = … kg m–3 [3] (c) Fig. 1.3 shows the variation of the tension F with the length of the spring in (b). 8 6 F / N 4 2 0 0 10 20 30 40 50 60 70 length / cm Fig. 1.3 (i) The tap at the bottom of the container is opened so that a fixed amount of liquid flows out of the container. The cylinder moves downwards so that the tension in the spring changes from 8.0 N to 4.0 N. Determine the change in the elastic potential energy of the spring. change in elastic potential energy = … J [3] (ii) More liquid is let out of the container until the upthrust on the cylinder becomes 24 N. For the upthrust of 24 N, determine the length of the spring. length = … cm [1] [Total: 11]

11 marks

Mark scheme: 1(a) m = ρV or ρAL C1 W = mg C1 (A =) 24 / (9.81 × 850 × 0.18) = 0.016 (m2) A1 or P = F / A (C1) P = ρgh (C1) (A =) 24 / (9.81 × 850 × 0.18) = 0.016 (m2) (A1) 1(b)(i) (upthrust =) 24 + 8(.0) = 32 (N) A1 1(b)(ii) (∆)p = 32 / 0.016 (= 2000) C1 (Δ)p = ρg(Δ)h ρ = 2000 / (9.81 × 0.17) C1 = 1200 kg m–3 A1 1(c)(i) E = ½Fx or E = ½kx 2 or E = area under graph C1 (Δ)E = (½ × 8.0 × 0.40) – (½ × 4.0 × 0.20) or (½ × 20 × 0.402) – (½ × 20 × 0.202) or ½ × (4.0 + 8.0) × 0.20 C1 = 1.2 J A1 1(c)(ii) length = 30 cm 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

3 A jet of water hits a vertical wall at right angles, as shown in Fig. 3.1. horizontal jet of water, vertical density 1.0 × 103 kg m–3 wall water runs velocity cross-sectional area down the wall 5.0 m s–1 1.5 × 10–4 m2 Fig. 3.1 (not to scale) The water hits the vertical wall with a velocity of 5.0 m s–1 in a horizontal direction. The cross-sectional area of the jet is 1.5 × 10–4 m2. The density of the water is 1.0 × 103 kg m–3. The water runs down the wall after hitting it. (a) Show that, over a time of 1.6 s, the mass of water hitting the wall is 1.2 kg. [2] (b) Calculate: (i) the decrease in the horizontal momentum of the mass of water in (a) due to hitting the wall decrease in momentum = … N s [1] (ii) the magnitude of the horizontal force exerted on the water by the wall. force = … N [1] (c) State and explain the magnitude of the horizontal force exerted on the wall by the water. … … [1] (d) Calculate the pressure exerted on the wall by the water. pressure = … Pa [2] [Total: 7]

7 marks

Mark scheme: 3(a) (m =) ρV C1 = 1.0 × 103 × 1.5 × 10–4 × 5.0 × 1.6 = 1.2 (kg) A1 3(b)(i) (∆)p = 1.2 × 5.0 = 6.0 N s A1 3(b)(ii) F = 6.0 / 1.6 or 1.2 × 5.0/1.6 = 3.8 N A1 3(c) Newton’s third law applies (so) 3.8 N. B1 3(d) p = F / A = 3.8 / 1.5 × 10–4 C1 = 2.5 × 104 Pa A1

This question in 9702/22 Feb/March 2022

2 (a) State what is meant by the centre of gravity of an object. … … [1] (b) A non-uniform rod XY is pivoted at point P, as shown in Fig. 2.1. 4.00 m 1.70 m 1.10 m X Y wire P sphere F 44.0 N 3.0 N Fig. 2.1 (not to scale) The rod has length 4.00 m and weight 44.0 N. The centre of gravity of the rod is 1.70 m from end X of the rod. Point P is 1.10 m from end X. A sphere hangs by a wire from end Y of the rod. The weight of the sphere is 3.0 N. The weight of the wire is negligible. A force F is applied vertically downwards at end X so that the horizontal rod is in equilibrium. (i) By taking moments about P, calculate F. F = … N [3] (ii) Calculate the force exerted on the rod by the pivot. force = … N [1] (c) The sphere in (b) is now immersed in a liquid in a container, as shown in Fig. 2.2. P liquid, density 1100 kg m–3 F 44.0 N container Fig. 2.2 The density of the liquid is 1100 kg m–3. The upthrust acting on the sphere due to the liquid is 2.5 N. The magnitude of F is unchanged so that the horizontal rod is not in equilibrium. (i) Use Archimedes’ principle to determine the radius r of the sphere. r = … m [3] (ii) Calculate the magnitude and direction of the resultant moment of the forces on the rod about P. magnitude of resultant moment = … N m direction of resultant moment … [2] [Total: 10]

10 marks

Mark scheme: 2(a) point where (all) the weight (of the object) is taken to act B1 2(b)(i) moments about P are: (F  1.10), (44.0  0.60) and (3.0  2.90) 1 mark for any one correct moment and 2 marks for two correct moments C2 (F  1.10) = (44.0  0.60) + (3.0  2.90) F = 32 N A1 2(b)(ii) force = 32 + 44 + 3 = 79 N A1 2(c)(i) (F = gV) V = 2.5 / (1100  9.81) (= 2.32  10–4) C1 V = (4 / 3)r3 C1 r = [(3  2.32  10–4) / 4]1/3 r = 0.038 m A1 2(c)(ii) resultant moment = 2.5  2.9 = 7.3 N m or resultant moment = F  1.1  (44  0.6 + 0.5  2.9) = 7.3 N m (allow 7.2 N m or 7.4 N m from rounded values of F) A1 direction: anticlockwise A1

This question in 9702/21 May/June 2022

2 A sphere is attached by a metal wire to the horizontal surface at the bottom of a river, as shown in Fig. 2.1. sphere direction of flow of water water wire horizontal 68° surface Fig. 2.1 (not to scale) The sphere is fully submerged and in equilibrium, with the wire at an angle of 68° to the horizontal surface. The weight of the sphere is 32 N. The upthrust acting on the sphere is 280 N. The density of the water is 1.0 × 103 kg m–3. Assume that the force on the sphere due to the water flow is in a horizontal direction. (a) By considering the components of force in the vertical direction, determine the tension in the wire. tension = … N [2] (b) For the sphere, calculate: (i) the volume volume = … m3 [1] (ii) the density. density = … kg m–3 [2] (c) The centre of the sphere is initially at a height of 6.2 m above the horizontal surface. The speed of the water then increases, causing the sphere to move to a different position. This movement of the sphere causes its gravitational potential energy to decrease by 77 J. Calculate the final height of the centre of the sphere above the horizontal surface. height = … m [3] (d) The extension of the wire increases when the sphere changes position as described in (c). The wire obeys Hooke’s law. (i) State a symbol equation that gives the relationship between the tension T in the wire and its extension x. Identify any other symbol that you use. … … [1] (ii) Before the sphere changed position, the initial elastic potential energy of the wire was 0.65 J. The change in position of the sphere causes the extension of the wire to double. Calculate the final elastic potential energy of the wire after the sphere has changed position. final elastic potential energy = … J [2] [Total: 11]

11 marks

Mark scheme: 2(a) T sin 68° + 32 = 280 C1 T = 270 N A1 2(b)(i) F = gV V = 280 / (1.0  103  9.81) = 0.029 m3 A1 2(b)(ii)  = (32 / 9.81) / 0.029 C1 = 110 kg m–3 A1 Question Answer Marks 2(c) (∆)E = mg(∆)h or (∆)E = W(∆)h C1 (∆)h = (–) 77 / 32 C1 (∆)h = (–) 2.4 final height = 6.2 – 2.4 = 3.8 m A1 2(d)(i) T = kx where k is a constant or T = (EA / L)x where A is (cross-sectional) area, E is Young modulus, L is (original/unstretched) length B1 2(d)(ii)  2 1 2 E kx or 1 2 E Fx and F = kx C1 E = 0.65  22 = 2.6 J A1 or 1 2 E Fx    1 0.65 270 2 x and so x = 4.8  10–3 m k = F / x = 270 / 4.8  10–3 = 5.6  104 (C1) Question Answer Marks 2(d)(ii) final      3 1 540 9.6 10 2 E or E =          3 1 270 2 4.8 10 2 2 or E =        2 4 3 1 5.6 10 9.6 10 2 = 2.6 J (A1)

This question in 9702/22 May/June 2022

1 A solid metal sphere has a diameter of (3.42 ± 0.02) cm and a mass of (67 ± 2) g. (a) Calculate the density, in g cm–3, of the metal. density = … g cm–3 [3] (b) Determine the percentage uncertainty in the density. percentage uncertainty = … % [2] [Total: 5]

5 marks

Mark scheme: 1(a) C1 V = (4 / 3)    r3 = (4 / 3)    (3.42 / 2)3 ( = 20.9 cm3) C1  = 67 / 20.9 = 3.2 g cm−3 A1 1(b) % = %m + 3  %d = [(2 / 67)  100] + [3  (0.02 / 3.42)  100] C1 = 3.0% + 3  0.58% = 4.7% or 5% A1

This question in 9702/23 May/June 2022

2 A spherical balloon is filled with a fixed mass of gas. A small block is connected by a string to the balloon, as shown in Fig. 2.1. balloon string block ground Fig. 2.1 (not to scale) The block is held on the ground by an external force so that the string is vertical. The density of the air surrounding the balloon is 1.2 kg m–3. The upthrust acting on the balloon is 0.071 N. The upthrust acting on the string and block is negligible. (a) By using Archimedes’ principle, calculate the radius r of the balloon. r = … m [2] (b) The total weight of the balloon, string and block is 0.053 N. The external force holding the block on the ground is removed so that the released block is lifted vertically upwards by the balloon. Calculate the acceleration of the block immediately after it is released. acceleration = … m s–2 [3] (c) The balloon continues to lift the block. The string breaks as the block is moving vertically upwards with a speed of 1.4 m s–1. After the string breaks, the detached block briefly continues moving upwards before falling vertically downwards to the ground. The block hits the ground with a speed of 3.6 m s–1. Assume that the air resistance on the block is negligible. (i) By considering the motion of the block after the string breaks, calculate the height of the block above the ground when the string breaks. height = … m [2] (ii) The string breaks at time t = 0 and the block hits the ground at time t = T. On Fig. 2.2, sketch a graph to show the variation of the velocity v of the block with time t from t = 0 to t = T. Numerical values of t are not required. Assume that v is positive in the upward direction. 4 3 v / m s–1 2 1 0 0 T t – 1 – 2 – 3 – 4 Fig. 2.2 [2] [Total: 9]

9 marks

Mark scheme: 2(a) F = gV C1 0.071 = 1.2  9.81  V (V = 6.03  10–3 m3) A1 4/3    r 3 = 6.03  10–3 r = 0.11 m 2(b) m = 0.053 / 9.81 C1 ( = 5.4  10–3 kg) F = 0.071 – 0.053 C1 ( = 0.018 N) a = (0.071 – 0.053) / (0.053 / 9.81) A1 = 3.3 m s–2 2(c)(i) v 2 = u 2 + 2as C1 3.62 = (–1.4)2 + 2  9.81  s or 3.62 = 1.42 + 2  9.81  s s = 0.56 m A1 2(c)(ii) single straight line from any positive non-zero value of v at t = 0 to any negative non-zero value of v at t = T B1 line starting at (0, 1.4) and ending at (T, –3.6) B1

This question in 9702/22 Oct/Nov 2022

4 A beaker in air contains a liquid. The base of the beaker is in contact with the liquid and has area A, as shown in Fig. 4.1. beaker liquid h base area A Fig. 4.1 The liquid has density ρ and fills the beaker to a depth h. (a) By using the definitions of pressure and density, show that p = ρ gh where p is the pressure due to the liquid that is exerted on the base of the beaker and g is the acceleration of free fall. [3] (b) Suggest why the equation in (a) does not give the total pressure on the base of the beaker. … … [1] (c) Fig. 4.2 shows the variation of the total pressure inside the liquid with depth x below the surface. 9.66 9.65 pressure / 104 Pa 9.64 9.63 9.62 9.61 9.60 0 1 2 3 4 5 6 7 8 x / cm Fig. 4.2 Determine the density of the liquid. density = … kg m–3 [2] (d) A solid cylinder is held stationary by a wire so that the base of the cylinder is level with the surface of the liquid, as shown in Fig. 4.3. 0.53 N wire cross-sectional area 3.7 × 10– 4 m2 cylinder 4.0 × 10–2 m surface of liquid Fig. 4.3 (not to scale) The cylinder has length 4.0 × 10–2 m and cross‑sectional area 3.7 × 10–4 m2. The tension in the wire is 0.53 N. The cylinder is now lowered and then held stationary by the wire so that the top of the cylinder is level with the surface of the liquid. Calculate the new tension in the wire. tension = … N [2] [Total: 8]

8 marks

Mark scheme: 4(a) B1 p = F / A or p = W / A B1 appropriate algebra leading to p = gh e.g. p = Ahg / A or Vg / A or Vg / (V/h) and (so) p = gh B1 4(b) there is atmospheric / air pressure B1 4(c) ∆p = g∆h e.g. (9.66 – 9.60)  104 / 8.0  10–2 =   9.81 C1  = 760 – 770 kg m–3 A1 4(d) F = gV = 760  9.81  3.7  10–4  4.0  10–2 (= 0.11 N) C1 tension = 0.53 – 0.11 = 0.42 N A1 or F = (∆)p  A = (9.63 – 9.60)  104  3.7  10–4 (= 0.11 N) (C1) tension = 0.53 – 0.11 = 0.42 N (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

2 A sphere floats in equilibrium on the surface of sea water of density 1050 kg m−3, as shown in Fig. 2.1. sphere sea water, density 1050 kg m–3 Fig. 2.1 (a) 21% of the volume of the sphere is below the surface of the water. Calculate the density of the sphere. density = … kg m−3 [2] (b) The sphere is now held so that its entire volume is below the surface of the water. The sphere is then released. (i) Calculate the initial acceleration of the sphere. acceleration = … m s−2 [3] (ii) The sphere accelerates upwards but remains entirely below the surface of the water. State and explain what happens to the acceleration of the sphere as its velocity begins to increase. … … … … [3] [Total: 8]

8 marks

Mark scheme: 2(a) C1 0.21V  1050 ( 9.81) = V ( 9.81)    = 220 kg m–3 A1 2(b)(i) F = 1050  9.81  V or W = 220  9.81  V C1 (V  1050  9.81) – (V  220  9.81) = (V  220)  a C1 a = 37 m s–2 A1 2(b)(ii) the (downward) drag / viscous force increases (with speed) M1 resultant force decreases (as upthrust and weight remain the same) M1 acceleration decreases (as its velocity increases) A1

This question in 9702/23 May/June 2023

2 A high‑altitude balloon is stationary in still air. A solid sphere is suspended from the balloon by a string, as shown in Fig. 2.1. balloon string sphere Fig. 2.1 (not to scale) The volume of the balloon is 7.5 m3. The total weight of the balloon, string and sphere is 65 N. The upthrust acting on the string and sphere is negligible. (a) Calculate the density of the air surrounding the balloon. density = … kg m–3 [2] (b) The string breaks, releasing the sphere. (i) State the magnitude of the acceleration of the sphere immediately after the string breaks. acceleration = … m s–2 [1] (ii) State and explain the variation, if any, in the magnitude of the acceleration of the sphere when it is moving downwards before it reaches terminal (constant) velocity. … … … … … … … [3] (c) The sphere has a mass of 4.0 kg. Calculate the total resistive force acting on the sphere at the instant when its acceleration is 1.9 m s–2. resistive force = … N [2] [Total: 8]

8 marks

Mark scheme: 2(a) 65 = gV C1 or 65 = mg and m = V  = 65 / (9.81  7.5) A1 = 0.88 kg m–3 2(b)(i) acceleration = 9.8 m s–2 A1 2(b)(ii) air resistance (acting on sphere) increases B1 resultant force (on sphere) decreases B1 (magnitude of) acceleration decreases B1 2(c) F = ma C1 = 4.0  1.9 ( = 7.6 N) resistive force = (4.0  9.81) – (4.0  1.9) A1 = 32 N

This question in 9702/22 Oct/Nov 2023

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

6 marks

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

This question in 9702/23 Oct/Nov 2023

3 (a) State the conditions for a system to be in equilibrium. … … … [2] (b) Fig. 3.1 shows an airship in flight. The airship is propelled by identical fans that can be angled to control the motion of the airship. airship fans Fig. 3.1 The upthrust on the airship is 93 000 N. The density of the surrounding air is 1.2 kg m–3. (i) Calculate the volume of air displaced by the airship. volume = … m3 [1] (ii) When fully loaded, the weight of the airship is greater than the upthrust. To maintain horizontal flight, the fans provide a total vertical force of 3.0 × 103 N upwards on the airship. Calculate the mass of the airship. mass = … kg [2] (c) At a certain time, the airship in (b) is stationary. The thrust force exerted by a fan on the airship is 2800 N. To produce this force, a mass of 64 kg of air is propelled through the blades of the fan in a time of 0.50 s. Assume that this air is initially stationary at the entrance to the fan. Calculate: (i) the change in momentum ∆p of the air propelled through the fan blades in this time ∆p = … kg m s–1 [2] (ii) the speed of the air as it leaves the fan speed = … m s–1 [2] (iii) the total kinetic energy of this air due to its movement through the fan. kinetic energy = … J [2] [Total: 11]

11 marks

Mark scheme: 3(a) resultant force (in any direction) is zero B1 resultant moment / torque (about any point) is zero B1 3(b)(i) F = Vg A1 V = 93 000 / (1.2  9.81) = 7900 m3 3(b)(ii) weight = 93 000 + 3(.0)  103 C1 m = (93 000 + 3.0  103) / 9.81 A1 = 9800 kg 3(c)(i) (p) = F()t or F = ∆p / ()t or F = rate of change of momentum or F = m (v – u) / t C1 ∆p = 2800  0.50 A1 = 1400 kg m s–1 3(c)(ii) (p =) mv C1 v = 1400 / 64 A1 = 22 m s–1 3(c)(iii) EK = ½mv2 C1 = ½  64  222 A1 = 15 000 J or EK = p2 / 2m (C1) = 14002 / (2  64) (A1) = 15 000 J

This question in 9702/23 Oct/Nov 2023

2 (a) Define acceleration. … … [1] (b) An Olympic diver stands on a platform above a pool of water, as shown in Fig. 2.1. 5.9 m s–1 diver 60° horizontal platform 9.0 m surface of water 1.2 m Fig. 2.1 (not to scale) When the diver is on the platform his centre of gravity is a vertical height of 9.0 m above the surface of the water. The diver jumps from the platform with a velocity of 5.9 m s–1 at an angle of 60° to the horizontal. Air resistance is negligible. When the diver hits the surface of the water, his centre of gravity is a vertical height of 1.2 m above the surface of the water. Calculate the speed of the diver at the instant he hits the surface of the water. speed = … m s–1 [3] (c) The diver in (b) enters the water and decelerates. (i) Describe and explain the variation of the viscous drag force acting on the diver in the water as he moves downwards. … … … … [2] (ii) The diver has a volume of 7.5 × 10–2 m3 . The density of the water is 1.0 × 103 kg m–3 . Show that the upthrust acting on the diver when he is entirely underwater is 740 N. [1] (iii) At a particular instant when the diver is entirely underwater his horizontal velocity is zero. The viscous drag force acting on him at this instant is 950 N vertically upwards. The diver has mass 78 kg. Determine the magnitude and direction of the acceleration of the diver. acceleration = … m s–2 direction … [4] [Total: 11]

11 marks

Mark scheme: 2(a) rate of change of velocity B1 2(b) ½ m()v2= mg()h C1 v2 = 5.92 + 2  9.81  7.8 C1 v2 = 188 v = 14 m s–1 A1 or by resolving components (C1) Vertically: v2 = u2 + 2as v2 = (5.9sin60)2 +2  –9.81  (1.2–9.0) vv = 13.4 horizontally: (C1) vh = 5.9cos60 vh = 2.95 resultant velocity = √(13.42 + 2.952) (A1) = 14 m s–1 2(c)(i) (As the diver moves down their) speed decreases B1 (So) viscous force / drag (force) decreases B1 2(c)(ii) (F =) gV A1 = 1000  9.81  7.5  10–2 = 740 (N)

This question in 9702/22 Feb/March 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) Define density. … … [1] (b) Fig. 1.1 shows a cuboidal glass block. y z x Fig. 1.1 (not to scale) A student measures the mass m of the block and the side lengths x, y and z. The measurements are shown in Table 1.1. Table 1.1 quantity measurement m (0.243 ± 0.001) kg x (5.41 ± 0.01) cm y (11.09 ± 0.01) cm z (1.62 ± 0.01) cm (i) Determine the density of the glass. density = … kg m–3 [2] (ii) Calculate the percentage uncertainty in the density. percentage uncertainty = … % [3] (c) The true value of the density of the glass is different from the answer in (b)(i) because of a systematic error in the measurements. Suggest one possible cause of this systematic error. … … … [1] [Total: 7]

7 marks

Mark scheme: Question Answer Marks 1(a) mass per unit volume B1 1(b)(i) density = 0.243 / (0.0541  0.1109  0.0162) C1 = 2500 kg m–3 A1 1(b)(ii) (0.001 / 0.243) or (0.01 / 5.41) or (0.01 / 11.09) or (0.01 / 1.62) C1 ( / ) = (m / m) + (x / x) + (y / y) + (z / z) C1 = (0.001 / 0.243) + (0.01 / 5.41) + (0.01 / 11.09) + (0.01 / 1.62) ( = 0.013) percentage uncertainty in  = 0.013  100 A1 = 1.3% (allow 1 s.f. answer of 1%) 1(c) zero error (on calipers / balance) B1 or incorrect calibration (of calipers / balance)

This question in 9702/21 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

2 (a) (i) Define pressure. … … [1] (ii) Explain how hydrostatic pressure results in an upthrust force acting on a solid object immersed in a liquid. … … … … [2] (b) A small steel ball of radius r and mass m falls vertically at terminal speed v through oil. The viscous drag force D that acts on the ball is given by D = 6πη r v where η is a property of the oil called its viscosity. (i) On Fig. 2.1, draw labelled arrows from the ball to show the directions of the three forces that act on the ball as it falls. Fig. 2.1 [3] (ii) Determine the SI base units of η. base units … [2] (c) The oil in (b) has a density of 920 kg m–3 and a viscosity of 4.7 in SI units. The steel ball has a mass of 2.4 × 10–3 kg and a radius of 4.2 × 10–3 m. (i) Show that the upthrust force acting on the ball is 2.8 × 10–3 N. [1] (ii) Determine the terminal speed v of the ball. v = … m s–1 [3] [Total: 12]

12 marks

Mark scheme: 2(a)(i) (normal) force per (unit cross-sectional) area B1 2(a)(ii) (due to difference in depth there is a) difference in pressure between top and bottom (of ball) B1 (due to pressure difference, upwards) B1 force on bottom of ball is greater (than downwards force on top of ball, so resultant force is upwards) 2(b)(i) arrow vertically downwards labelled ‘weight’ B1 arrow vertically upwards labelled ‘upthrust’ B1 arrow vertically upwards labelled ‘(viscous) drag’ B1 2(b)(ii) SI base units of D: kg m s–2 C1 base units of : kg m s–2 / (m  m s–1) A1 = kg m–1 s–1 2(c)(i) upthrust = 920  (4 / 3)    (4.2  10–3)3  9.81 = 2.8  10–3 (N) A1 2(c)(ii) weight = drag + upthrust C1 (2.4  10–3  9.81) = (2.8  10–3) + (6  4.7  4.2  10–3  v) C1 v = 0.056 m s–1 A1

This question in 9702/23 Oct/Nov 2025

2 Fig. 2.1 shows a square metal sheet of non-uniform density, with a thin wooden rod fixed at its centre. One of the corners of the sheet is labelled X. X metal sheet rod Fig. 2.1 The rod has negligible mass. The mass of the metal sheet is 2.8 kg. The rod is supported so that the rod is horizontal and the metal sheet is vertical. (a) Define the torque of a couple. … … … [2] (b) When the rod is supported in such a way that it can rotate freely within its support, the sheet hangs in equilibrium with point X vertically above the rod, as shown in Fig. 2.2. X rod metal sheet Fig. 2.2 On Fig. 2.2, draw a line to indicate the range of possible positions for the centre of gravity of the metal sheet. [1] (c) When a torque of 3.3 N m is applied to the rod, the sheet is held in equilibrium with two of its edges horizontal, as shown in Fig. 2.3. Point X is at the top-left corner. X metal sheet rod Fig. 2.3 (i) Explain whether the torque applied to the rod to hold the sheet in equilibrium is clockwise or anticlockwise. … … [1] (ii) Show that the centre of gravity of the sheet has a horizontal displacement of 0.12 m from the rod. [1] (d) The square metal sheet has an average density of 3000 kg m–3 and a uniform thickness of 4.0 mm. Show that the side length of the sheet is 0.48 m. [3] (e) Use the answer in (b) and the information in (c) and (d) to determine the position of the centre of gravity of the sheet. Indicate this position on Fig. 2.3 with a point labelled Y. [2] [Total: 10]

10 marks

Mark scheme: 2(a) product of force and distance M1 perpendicular distance between the (line of action of the two) forces A1 2(b) straight vertical line drawn from centre to bottom corner B1 2(c)(i) centre of gravity is to the right of the rod so torque is anticlockwise B1 or moment of weight of sheet about rod is clockwise so torque is anticlockwise 2(c)(ii) (horizontal displacement) = 3.3 / (2.8  9.81) = 0.12 (m) A1 2(d)  = m / V C1 V = 4.0  10–3 × (side length)2 C1 A1 0.48m side length =   2.8 / ( 3000  0.0040 )  = 2(e) point lies on a straight line at 45° to the horizontal from the bottom-right corner to the edge of the rod B1 point lies on a vertical line half-way between rod and right-hand edge B1

This question in 9702/24 Oct/Nov 2025