Cambridge A Level Physics 9702 — 2019 Feb/March Paper 2 · Variant 2

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]

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

More questions on SI units

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

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

More questions on Physical quantities

3 Two balls, X and Y, move along a horizontal frictionless surface, as illustrated in Fig. 3.1. 60° 3.0 m s–1 X A B 9.6 m s–1 Y 2.5 kg Fig. 3.1 (not to scale) Ball X has an initial velocity of 3.0 m s–1 in a direction along line AB. Ball Y has a mass of 2.5 kg and an initial velocity of 9.6 m s–1 in a direction at an angle of 60° to line AB. The two balls collide at point B. The balls stick together and then travel along the horizontal surface in a direction at right-angles to the line AB, as shown in Fig. 3.2. V X Y A B Fig. 3.2 (a) By considering the components of momentum in the direction from A to B, show that ball X has a mass of 4.0 kg. [2] (b) Calculate the common speed V of the two balls after the collision. V = ................................................. m s–1 [2] (c) Determine the difference between the initial kinetic energy of ball X and the initial kinetic energy of ball Y. difference in kinetic energy = ....................................................... J [2] [Total: 6]

Mark scheme: 3(a) C1 (m × 3.0) – (2.5 × 9.6 × cos 60°) = 0 so m = 4.0 (kg) A1 Question Answer Marks 3(b) 2.5 × 9.6 × sin60° = (4.0 + 2.5) × V C1 V = 3.2 m s–1 A1 or use of momentum vector triangle: (4.0 × 3.0)2 + [(4.0 + 2.5) × V]2 = (2.5 × 9.6)2 (C1) V = 3.2 m s–1 (A1) 3(c) E = ½mv 2 difference in EK = ½ × 2.5 × (9.6)2 – ½ × 4.0 × (3.0)2 C1 = 97 J A1

More questions on Linear momentum and its conservation

4 (a) Define electric field strength. ................................................................................................................................................... ...............................................................................................................................................[1] (b) Two very small metal spheres X and Y are connected by an insulating rod of length 72 mm. A side view of this arrangement is shown in Fig. 4.1. +3e uniform electric field, X field strength 5.0 × 104 V m–1 72 mm in vertically upwards direction θ horizontal Z θ SIDE rod VIEW Y –3e Fig. 4.1 (not to scale) Sphere X has a charge of +3e and sphere Y has a charge of –3e, where e is the elementary charge. The rod is held at its mid point Z at an angle θ to the horizontal. The rod and spheres have negligible mass and are in a uniform electric field. The electric field strength is 5.0 × 104 V m–1. The direction of this field is vertically upwards. (i) The electric field is produced by applying a potential difference of 4.0 kV between two charged parallel metal plates. 1. Calculate the separation between the plates. separation = ...................................................... m [2] 2. Describe the arrangement of the two plates. Include in your answer a statement of the sign of the charge on each plate. You may draw on Fig. 4.1. .................................................................................................................................... .................................................................................................................................... .................................................................................................................................... ................................................................................................................................[2] (ii) Determine the magnitude and direction of the force on sphere Y. magnitude = ........................................................... N direction ............................................................... [2] (iii) The electric forces acting on the two spheres form a couple. This couple acts on the rod with a torque of 6.2 × 10–16 N m. Calculate the angle θ of the rod to the horizontal. θ = ........................................................ ° [2] [Total: 9]

Mark scheme: 4(a) force per unit positive charge B1 4(b)(i) 1 E = V / d or E = ∆V / ∆d d = 4.0 × 103 / 5.0 × 104 C1 = 8.0 × 10–2 m A1 2 plates are (in) horizontal (plane) (above and below the rod) B1 top (plate) negative and bottom (plate) positive B1 4(b)(ii) magnitude = 5.0 × 104 × 3 × 1.6 × 10–19 = 2.4 × 10–14 N A1 direction is (vertically) downwards / down B1 Question Answer Marks 4(b)(iii) 6.2 × 10–16 = 2.4 × 10–14 × 72 × 10–3 × cosθ C1 θ = 69° A1

More questions on Uniform electric fields

5 (a) By reference to two waves, state: (i) the principle of superposition ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[2] (ii) what is meant by coherence. ........................................................................................................................................... .......................................................................................................................................[1] (b) Two coherent waves P and Q meet at a point in phase and superpose. Wave P has an amplitude of 1.5 cm and intensity I. The resultant intensity at the point where the waves meet is 3I. Calculate the amplitude of wave Q. amplitude = .................................................... cm [2] (c) The apparatus shown in Fig. 5.1 is used to produce an interference pattern on a screen. laser light a wavelength 680 nm D double-slit screen Fig. 5.1 (not to scale) Light of wavelength 680 nm is incident on a double-slit. The slit separation is a. The separation between adjacent fringes is x. Fringes are viewed on a screen at distance D from the double-slit. Distance D is varied from 2.0 m to 3.5 m. The variation with D of x is shown in Fig. 5.2. 10.0 8.0 x / mm 6.0 4.0 2.0 0 2.0 2.5 3.0 3.5 D / m Fig. 5.2 (i) Use Fig. 5.2 to determine the slit separation a. a = ...................................................... m [3] (ii) The laser is now replaced by another laser that emits light of a shorter wavelength. On Fig. 5.2, sketch a possible line to show the variation with D of x for the fringes that are now produced. [2] [Total: 10]

Mark scheme: 5(a)(i) (two) waves meet/overlap (at a point) B1 (resultant) displacement is sum of the displacement of each wave B1 5(a)(ii) constant phase difference (between the waves) B1 5(b) I ∝ A 2 3I / I = (A + 1.5)2 / 1.52 C1 A = 1.1 cm A1 5(c)(i) λ = ax / D C1 e.g. a = 680 × 10–9 × 2.0 / 4.0 × 10–3 C1 a = 3.4 × 10–4 m A1 5(c)(ii) straight line from positive value on x-axis and always below ‘old’ line B1 straight line with a smaller positive gradient than ‘old’ line B1

More questions on Interference

6 (a) Using energy transformations, describe the electromotive force (e.m.f.) of a battery and the potential difference (p.d.) across a resistor. e.m.f.: ........................................................................................................................................ ................................................................................................................................................... p.d.: ........................................................................................................................................... ...............................................................................................................................................[2] (b) A battery of e.m.f. 6.0 V and negligible internal resistance is connected to a network of resistors and a voltmeter, as shown in Fig. 6.1. Z V 32 Ω 6.0 V Y X 24 Ω Fig. 6.1 Resistor Y has a resistance of 24 Ω and resistor Z has a resistance of 32 Ω. (i) The resistance RX of the variable resistor X is adjusted until the voltmeter reads 4.8 V. Calculate: 1. the current in resistor Z current = ....................................................... A [1] 2. the total power provided by the battery power = ..................................................... W [2] 3. the number of conduction electrons that move through the battery in a time interval of 25 s number = .......................................................... [2] 4. the total resistance of X and Y connected in parallel total resistance = ...................................................... Ω [2] 5. the resistance RX. Ω [2] RX = ...................................................... (ii) The resistance RX is now decreased. State and explain the change, if any, to the reading on the voltmeter. ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[2] [Total: 13]

Mark scheme: 6(a) e.m.f.: energy transferred from chemical to electrical (per unit charge) B1 p.d.: energy transferred from electrical to thermal (per unit charge) B1 Question Answer Marks 6(b)(i) 1 I = 4.8 / 32 = 0.15 A A1 2 P = EI or P = VI or P = I 2R or P = V 2 / R = 6.0 × 0.15 or 0.152 × 40 or 6.02 / 40 C1 = 0.90 W A1 3 number = It / e = [0.15 × 25] / 1.6 × 10–19 C1 = 2.3 × 1019 A1 or Q = 0.15 × 25 (= 3.75) number = 3.75 / 1.6 × 10–19 (C1) = 2.3 × 1019 (A1) 4 4.8 / 6.0 = 32 / (RXY + 32) or 1.2 / 6.0 = RXY / (RXY + 32) or 4.8 / 1.2 = 32 / RXY C1 RXY = 8.0 Ω A1 Alternative methods: RXY = (6.0 – 4.8) / 0.15 or (C1) = 8.0 Ω (A1) or 6.0 = 0.15 (32 + RXY) (C1) RXY = 40 – 32 = 8.0 Ω (A1) Question Answer Marks 6(b)(i) 5 1 / 8.0 = 1 / RX + 1 / 24 C1 Rx = 12 Ω A1 Alternative method: IZ = 4.8 / 32 = 0.15 and IY = 1.2 / 24 = 0.05 IX = 0.15 – 0.05 (= 0.10) (C1) RX = 1.2 / 0.10 = 12 Ω (A1) 6(b)(ii) total resistance decreases M1 (so voltmeter) reading increases A1

More questions on Potential difference and power

7 (a) The names of four particles are listed below. alpha beta-plus neutron proton State the name(s) of the particle(s) in this list that: (i) are not fundamental .......................................................................................................................................[1] (ii) do not experience an electric force when situated in an electric field .......................................................................................................................................[1] (iii) has the largest ratio of charge to mass. .......................................................................................................................................[1] (b) A hadron has a charge of +e where e is the elementary charge. The hadron is composed of only two quarks. One of these quarks is an antidown ( d ) quark. By considering charge, state and explain the name (flavour) of the other quark. ................................................................................................................................................... ...............................................................................................................................................[3] [Total: 6]

Mark scheme: 7(a)(i) alpha, neutron and proton B1 7(a)(ii) neutron B1 7(a)(iii) beta plus or β+ B1 7(b) d has charge (+)⅓ e C1 (so) other quark has charge = e – ⅓ e = (+)⅔ e M1 other quark is an up / u A1

More questions on Fundamental particles