Physics (AS & A Level) Paper 43

gravitational force provides/is the centripetal force GMm / r 2 = mv 2 / r or GMm / r 2 = mrω2 and v = 2πr / T or ω = 2π / T with algebra to T 2 = 4π2r 3 / GM [3]

or acceleration due to gravity is the centripetal acceleration (B1) GM / r 2 = v 2 / r or GM / r 2 = rω2 and v = 2πr / T or ω = 2π / T (M1) with algebra to T 2 = 4π2r3 / GM (A1)

equatorial orbit/orbits (directly) above the equator from west to east [2]

(24 × 3600) 2 = 4π2r 3 / (6.67 × 10–11 × 6.0 × 1024) r 3 = 7.57 × 1022 r = 4.2 × 107 m [2]

(T / 24) 2 = {(2.64 × 107) / (4.23 × 107)}3 = 0.243 T = 12 hours [2]

or k (= T 2 / r 3) = 242 / (4.23 × 107) 3 (B1) k = 7.61 × 10–21 T 2 (= kr 3) = 7.61 × 10–21 × (2.64 × 107) 3 = 140 T = 12 hours (A1)

p ∝ T or pV / T = constant or pV = nRT T (= 5 × 300 =) 1500 K [2]

pV = nRT 1.0 × 105 × 4.0 × 10–4 = n × 8.31 × 300 or 5.0 × 105 × 4.0 × 10–4 = n × 8.31 × 1500 n = 0.016 mol [2]

1. heating/thermal energy supplied 2. work done on/to system [2]

1. 240 J 2. same value as given in 1. (= 240 J) and zero given for 3. 3. zero [3]

2k/m = ω2 ω = 2πf (2 × 64 / 0.810) = (2π × f) 2 leading to f = 2.0 Hz [3]

v0 = ωx0 or v0 = 2πfx0 or v = ω(x0 2 – x2) 1/2 and x = 0 v0 = 2π × 2.0 × 1.6 × 10–2 = 0.20 m s–1 [2]

frequency: reduced/decreased maximum speed: reduced/decreased [2]

in an electric field, charges (in a conductor) would move no movement of charge so zero field strength or charge moves until F = 0 / E = 0 [2]

or charges in metal do not move (B1) no (resultant) force on charges so no (electric) field (B1)

at P, EA = (3.0 × 10–12) / [4πε0(5.0 × 10–2) 2] (= 10.79 N C–1) at P, EB = (12 × 10–12) / [4πε0(10 × 10–2) 2] (= 10.79 N C–1) or (3.0 × 10–12) / [4πε0(5.0 × 10–2) 2] – (12 × 10–12) / [4πε0(10 × 10–2) 2] = 0 or (3.0 × 10–12) / [4πε0(5.0 × 10–2) 2] = (12 × 10–12) / [4πε0(10 × 10–2) 2] (M2) fields due to charged spheres are (equal and) opposite in direction, so E = 0 [3]

potential = 8.99 × 109 {(3.0 × 10–12) / (5.0 × 10–2) + (12 × 10–12) / (10 × 10–2)} = 1.62 V [2]

½mv2 = qV EK = ½ × 107 × 1.66 × 10–27 × v 2 qV = 47 × 1.60 × 10–19 × 1.62 v 2 = 1.37 × 108 v = 1.2 × 104 m s–1 [3]

region (of space) where a force is experienced by a particle [1]

gravitational

gravitational and electric

gravitational, electric and magnetic [3]

force (always) normal to direction of motion (magnitude of) force constant or speed is constant/kinetic energy is constant magnetic force provides/is the centripetal force [3]

mv2 / r = Bqv momentum or p or mv = Bqr [2]

(induced) e.m.f. proportional to rate of change of (magnetic) flux (linkage) [2]

flux linkage = BAN = π × 10–3 × 2.8 × π × (1.6 × 10–2) 2 × 85 = 6.0 × 10–4 Wb [1]

e.m.f. = ∆NΦ / ∆t e.m.f. = (6.0 × 10–4 × 2) / 0.30 e.m.f. = 4.0 mV [2]

sketch: E = 0 for t = 0 → 0.3 s, 0.6 s → 1.0 s, 1.6 s → 2.0 s E = 4 mV for t = 0.3 s → 0.6 s (either polarity) E = 2 mV for t = 1.0 s → 1.6 s with opposite polarity [4]

electromagnetic radiation/photons incident on a surface causes emission of electrons (from the surface) [2]

E = hc / λ = (6.63 × 10–34 × 3.00 × 108) / (436 × 10–9) = 4.56 × 10–19 J (4.6 × 10–19 J) [2]

Φ = hc / λ0 λ0 = (6.63 × 10–34 × 3.00 × 108) / (1.4 × 1.60 × 10–19) = 890 nm [2]

λ0 = (6.63 × 10–34 × 3.00 × 108) / (4.5 × 1.60 × 10–19) = 280 nm [1]

caesium: wavelength of photon less than threshold wavelength (or v.v.) or λ0 = 890 nm > 436 nm so yes tungsten: wavelength of photon greater than threshold wavelength (or v.v.) or λ0 = 280 nm < 436 nm so no [2]

11 in metal, conduction band overlaps valence band/no forbidden band/no band gap as temperature rises, no increase in number of free electrons/charge carriers as temperature rises, lattice vibrations increase (lattice) vibrations restrict movement of electrons/charge carriers (current decreases) so resistance increases [5]

time for number of atoms/nuclei or activity to be reduced to one half reference to (number of…) original nuclide/single isotope or reference to half of original value/initial activity [2]

A = A0 exp(–λt) and either t = t½, A = ½A0 or ½A0 = A0 exp(–λt½) so ln 2 = λt½ (and ln 2 = 0.693), hence 0.693 = λt½ [2]

A = λN N = 200 / (2.1 × 10–6) = 9.52 × 107 mass = (9.52 × 107 × 222 × 10–3) / (6.02 × 1023) or mass = 9.52 × 107 × 222 × 1.66 × 10–27 = 3.5 × 10–17 kg [4]