Physics (AS & A Level) Paper 42
Oct / Nov 2016·9702·12 questions
Mock exam: 120 min, official timing
Question paper
Question 1
Question 2
Question 3
Question 4
Question 5
Question 6
Question 7
Question 9
Question 11
Question 12
Question 13
Question 14
Mark scheme
Question 1
1(a)[1]
force per unit mass [1]
1(b)(i)[1]
radius/diameter/size (of Proxima Centauri) ≪ /is much less than 4.0 × 1013 km/separation (of Sun and star) or (because) it is a uniform sphere [1]
1(b)(ii)[4]
1. field strength = GM / x2 = (6.67 × 10–11 × 2.5 × 1029) / (4.0 × 1013 × 103) 2 = 1.0 × 10–14 N kg–1 [2]
2. force = field strength × mass = 1.0 × 10–14 × 2.0 × 1030 or force = GMm / x2 = (6.67 × 10–11 × 2.5 × 1029 × 2.0 × 1030) / (4.0 × 1013 × 103) 2 (C1) = 2.0 × 1016 N [2]
1(c)[2]
force (of 2 × 1016 N) would have little effect on (large) mass of Sun would cause an acceleration of Sun of 1.0 × 10–14 m s–2/very small/negligible acceleration [2]
or many stars all around the Sun (B1) net effect of forces/fields is zero (B1)
Question 2
2(a)(i)[1]
number of moles/amount of substance [1]
2(a)(ii)[1]
kelvin temperature/absolute temperature/thermodynamic temperature [1]
2(b)[3]
pV = nRT 4.9 × 105 × 2.4 × 103 × 10–6 = n × 8.31 × 373 n = 0.38 (mol) number of molecules or N = 0.38 × 6.02 × 1023 = 2.3 × 1023 [3]
or pV = NkT (C1) 4.9 × 105 × 2.4 × 103 × 10–6 = N × 1.38 × 10–23 × 373 (M1) number of molecules or N = 2.3 × 1023 (A1)
2(c)[3]
volume occupied by one molecule = (2.4 × 103) / (2.3 × 1023) = 1.04 × 10–20 cm3 mean spacing = (1.04 × 10–20) 1/3 = 2.2 × 10–7 cm (allow 1 s.f.) [3]
(allow other dimensionally correct methods e.g. V = (4/3)πr3)
Question 3
3(a)[2]
(sum of/total) potential energy and kinetic energy of (all) molecules/particles reference to random (distribution) [2]
3(b)(i)[3]
no heat enters (gas)/leaves (gas)/no heating (of gas) work done by gas (against atmosphere as it expands) internal energy decreases [3]
3(b)(ii)[3]
volume decreases so work done on ice/water (allow work done negligible because ∆V small) heating of ice (to break rigid forces/bonds) internal energy increases [3]
Question 4
4(a)(i)[1]
0.225 s and 0.525 s [1]
4(a)(ii)[2]
period or T = 0.30 s and ω = 2π / T ω = 2π / 0.30 ω = 21 rad s–1 [2]
4(a)(iii)[2]
speed = ωx0 or ω(x0 2 – x2) 1/2 and x = 0 = 20.9 × 2.0 × 10–2 = 0.42 m s–1 [2]
or use of tangent method: correct tangent shown on Fig. 4.2 (C1) working e.g. ∆y / ∆x leading to maximum speed in range (0.38–0.46) m s–1 (A1)
4(b)[3]
sketch: reasonably shaped continuous oval/circle surrounding (0,0) curve passes through (0, 0.42) and (0, –0.42) curve passes through (2.0, 0) and (–2.0, 0) [3]
Question 5
5(a)[2]
transducer/transmitter can be also be used as the receiver or transducer both transmits and receives receives reflected pulses between the emitted pulses (needs to be pulsed) in order to measure/determine depth(s) (needs to be pulsed) to determine nature of boundaries Any three of the above marking points, 1 mark each [2]
5(b)(i)[2]
product of speed of (ultra)sound and density (of medium) reference to speed of sound in medium [2]
5(b)(ii)[2]
if Z1 and Z2 are (nearly) equal, IT / I0 (nearly) equal to 1/unity/(very) little reflection/mostly transmission if Z1 ≫ Z2 or Z1 ≪ Z2 or the difference between Z1 and Z2 is (very) large, then IT / I0 is small/zero/mostly reflection/little transmission [2]
Question 6
6(a)[3]
E = 0 or EA = (–)EB (at x = 11 cm) QA / x2 = QB / (20 – x) 2 = 112 / 92 QA / QB or ratio = 1.5 [3]
or E ∝ Q because r same or E = Q / 4πε0r2 and r same (B1) QA / QB = 48 / 32 (C1) QA / QB or ratio = 1.5 (A1)
6(b)(i)[3]
for max. speed, ∆V = (0.76 – 0.18) V or ∆V = 0.58 V q∆V = ½mv2 2 × (1.60 × 10–19) × 0.58 = ½ × 4 × 1.66 × 10–27 × v2 v2 = 5.59 × 107 v = 7.5 × 103 m s–1 [3]
6(b)(ii)[2]
∆V = 0.22 V 2 × (1.60 × 10–19) × 0.22 = ½ × 4 × 1.66 × 10–27 × v2 v2 = 2.12 × 107 v = 4.6 × 103 m s–1 [2]
Question 7
7(a)(i)[1]
charge / potential (difference) or charge per (unit) potential (difference) [1]
7(a)(ii)[2]
(V = Q / 4πε0r and C = Q / V) for sphere, C (= Q / V) = 4πε0r C = 4π × 8.85 × 10–12 × 12.5 × 10–2 = 1.4 × 10–11 F [2]
7(b)(i)[1]
1 / CT = 1 / 3.0 + 1 / 6.0 CT = 2.0 µF [1]
7(b)(ii)[2]
total charge = charge on 3.0 µF capacitor = 2.0 (µ) × 9.0 = 18 (µC) potential difference = Q / C = 18 (µ)C / 3.0 (µ)F = 6.0 V [2]
or argument based on equal charges: 3.0 × V = 6.0 × (9.0 – V) (C1) V = 6.0 V (A1)
7(b)(iii)[2]
potential difference (= 9.0 – 6.0) = 3.0 V charge (= 3.0 × 2.0 (µ)) = 6.0 µC [2]
Question 9
9(a)[3]
(by Newton’s third law) force on wire is up(wards) by (Fleming’s) left-hand rule/right-hand slap rule to give current in direction left to right shown on diagram [3]
9(b)[4]
force ∝ current or F = BIL or B (= 0.080 / 6.0L) = 1 / 75L maximum current = 2.5 × √2 = 3.54 A maximum force in one direction = (3.54 / 6.0) × 0.080 = 0.047 N difference (= 2 × 0.047) = 0.094 N or force varies from 0.047 N upwards to 0.047 N downwards [4]
10 nuclei emitting r.f. (pulse) Larmor frequency/r.f. frequency emitted/detected depends on magnitude of magnetic field nuclei can be located (within a slice) changing field enables position of detection (slice) to be changed [4]
Question 11
11(a)[2]
(induced) e.m.f. proportional/equal to rate of change of (magnetic) flux (linkage) [2]
11(b)[3]
(for same current) iron core gives large(r) (rates of change of) flux (linkage) e.m.f induced in solenoid is greater (for same current) induced e.m.f. opposes applied e.m.f. so current smaller/acts to reduce current [3]
or same supply so same induced e.m.f. balancing it (B1) (rate of change of) flux linkage is same (M1) smaller current for same flux when core present (A1)
11(c)[2]
e.g. (heating due to) eddy currents in core (heating due to current in) resistance of coils hysteresis losses/losses due to changing magnetic field in core Any two of the above marking points, 1 mark each [2]
Question 12
12(a)(i)[1]
electron diffraction/electron microscope (allow other sensible suggestions) [1]
12(a)(ii)[1]
photoelectric effect/Compton scattering (allow other sensible suggestions) [1]
12(b)(i)[1]
arrow clear from –0.54 eV to –3.40 eV [1]
12(b)(ii)[2]
E = hc / λ or E = hf and c = fλ λ = (6.63 × 10–34 × 3.00 × 108) / [(3.40 – 0.54) × 1.60 × 10–19] = 4.35 × 10–7 m [2]
12(c)(i)[2]
wavelength associated with a particle that is moving/has momentum/has speed/has velocity [2]
12(c)(ii)[2]
λ = h / mv v = (6.63 × 10–34) / (9.11 × 10–31 × 4.35 × 10–7) = 1.67 × 103 m s–1 [2]
13 X-ray image of a (single) slice/cross-section (through the patient) taken from different angles/rotating X-ray (beam) computer is used to form/process/build up/store image 2D image (of the slice) repeated for many/different (neighbouring) slices to build up 3D image [6]
Question 13
X-ray image of a (single) slice or cross-section through the patient. (M1)
Taken from different angles, or rotating X-ray beam. (A1)
Computer is used to form, process, build up or store the image. (B1)
2D image of the slice. (B1)
Repeated for many different neighbouring slices. (M1)
To build up a 3D image. (A1)
Question 14
14(a)(i)[1]
He 4 2 or α 4 2 [1]
14(a)(ii)[1]
n 1 0 [1]
14(b)(i)[2]
∆m = (29.97830 +1.00867) – (26.98153 + 4.00260) = 30.98697 – 30.98413 = 2.84 × 10–3 u [2]
14(b)(ii)[2]
E = c2∆m or mc2 = (3.0 × 108) 2 × 2.84 × 10–3 × 1.66 × 10–27 = 4.2 × 10–13 J [2]
14(c)[2]
mass of products is greater than mass of Al plus α or reaction causes (net) increase in (rest) mass (of the system) α-particle must have at least this amount of kinetic energy [2]