Physics (AS & A Level) Paper 42

force proportional to product of the (two) masses and inversely proportional to the square of their separation either reference to point masses or separation << ‘size’ of masses [2]

gravitational force provides / is the centripetal force GMm / r 2 = mv 2 / r or GMm / r 2 = mrω 2 and v = rω and algebra leading to v = (GM / r) 1 / 2 [2]

1. vA / vB = (rB / rA) 1 / 2 = (2.2 × 1010 / 1.3 × 108) 1 / 2 = 13 (13.0) [2]

2. v = 2πr / T or v ∝ r / T or vT / r = constant TA / TB = (rA / rB) × (vB / vA) = (1.3 × 108 / 2.2 × 1010) × (1 / 13) = 4.5 (4.54) × 10–4 or T 2 = 4π2r 3 / GM or T 2 ∝ r 3 or T 2 / r 3 = constant (C1) TA / TB = (rA 3 / rB 3) 1 / 2 = [(1.3 × 108) 3 / (2.2 × 1010) 3]1 / 2 (C1) = 4.5 (4.54) × 10–4 (A1) [3]

T = 2π / 1.7 ×10–4 = 3.70 × 104 s TB = 3.70 × 104 / 4.54 × 10–4 = 8.1 × 107 s [2]

If identifies TA as TB then 0 / 2

sum of kinetic and potential energy of atoms/molecules reference to random (distribution) [2]

no forces (of attraction or repulsion) between molecules [1]

pV = NkT or pV = nRT and R = kNA , n = N / NA 1/3 Nm<c 2> = NkT or 1/3 m<c 2> = kT <EK> = 1/2 m<c 2> so <EK> = 3/2 kT [3]

<EK> = 3/2 × 1.38 × 10–23 × (273 + 12) = 5.9 (5.90) × 10–21 J [2]

(use of T = 12 K not T = 285 K scores 0 / 2)

number = (17 / 32) × 6.02 × 1023 = 3.2 (3.20) × 1023 [2]

internal energy = 5.9 × 10–21 × 3.2 × 1023 = 1900 (1890) J [1]

the (thermal) energy per unit mass to raise the temperature of a substance by one degree [2]

(If ratio not clear for M1 mark, allow 1 / 2 marks for an otherwise correct answer)

to allow for / determine / cancel heat transfer to / from tube / surroundings [1]

(do not allow ‘to stop / prevent’ heat loss)

either P = mc∆θ ± h or 44.9 = 1.58 × 10–3 × c × (25.5 – 19.5) ± h or 33.3 = 1.11 × 10–3 × c × (25.5 – 19.5) ± h (44.9 – 33.3) = (1.58 – 1.11) × 10–3 × c × (25.5 – 19.5) c = 4100 (4110) J kg–1 K–1 [3]

(allow 1 / 3 for use of only 33.3 W, 1.11 g s–1 leading to 5000 J kg –1 K –1) (allow 1 / 3 for use of only 44.9 W, 1.58 g s–1 leading to 4740 J kg –1 K –1)

V0 = 27 or Vrms = 19.1 33.3 = 272 / 2R or 33.3 = 19.12 / R R = 11 Ω [3]

amplitude = 1.8 cm and period = 0.30 s [1]

EK = ½m ω 2 (x0 2 – x2) or EK = ½mv2 and v = ± ω √(x0 2 – x2) = ½ × 0.080 × (2π / 0.30) 2 × [(1.8 × 10–2) 2 – (1.2 × 10–2) 2] = 3.2 × 10–3 J [3]

speed = Z / ρ = 1.4 × 106 / 940 (=1490) time = (1.1 × 10–2 × 2) / 1490 = 1.5 × 10–5 s [3]

(time of 7.4 × 10–6 s is one way only and scores 2 / 3 marks) (use of speed of light is wrong physics and scores 0 / 3 marks)

I = I0 exp (–µx) or I2 = I1 exp(–µx) ratio = exp (– 48 × 1.1 × 10–2) = 0.59 [2]

0.33 / 100 = 0.59 × (I3 / I2) × 0.59 ratio = 9.5 × 10–3 or 0.33 / 100 = exp (– 48 × 2.2 × 10–2) × (I3 / I2) (C1) ratio = 9.5 × 10–3 (A1) [2]

ratio I3 / I2 increases [1]

(accept: “there is an increase in the proportion of the intensity that is reflected”)

(capacitance =) charge / potential (difference) [1]

V = V1 + V2 + V3 either Q / C = Q / C1 + Q / C2 + Q / C3 or V / Q = V1 / Q + V2 / Q + V3 / Q and so 1 / C = 1 / C1 + 1 / C2+ 1 / C3 [2]

1. 1 / CT = (1 / 200) + (1 / 600) CT = 150 µF [1]

2. Q = CV = 150 × 10–6 × 12 or 600 × 10–6 × 3.0 or 200 × 10–6 × 9.0 = 1.8 × 10–3 C [1]

3. V = Q / C = 1.8 × 10–3 / 600 × 10–6 or V = [200 / (200 + 600)] × 12 = 3(.0) V [1]

energy = ½CV 2 or energy = ½QV and C = Q / V 1/2 × C × 32 = 2 × 1/2 × C × V 2 V = 2.1 V [3]

direction of force due to electric field opposite to force due to magnetic field electric field is up the page [2]

force due to electric field = force due to magnetic field or Eq = Bqv E = Bv = 9.7 × 10–2 × 1.6 × 105 = 1.6 (1.55) × 104 V m–1 [3]

q / m = v / Br = 1.6 × 105 / (9.7 ×10–2 × 4.0 × 10–2) = 4.1 (4.12) × 107 C kg–1 [3]

m = (3 × 1.60 × 10–19) / (4.12 × 107) m = 1.16 × 10–26 / 1.66 × 10–27 = 7(.0) u (allow 7.1 u) [2]

3 protons, 4 neutrons [1]

change in flux linkage = 40 × (5.0 – 3.0) × 10–6 = 8(.0) × 10–5 Wb [1]

time taken = 8.0 × 10–5 / 5.0 × 10–4 = 0.16 (s) speed = 3.0 × 10–2 / 0.16 = 0.19 (0.188) m s–1 or E = (∆Φ / ∆x) × speed speed = 5.0 × 10–4 / (8.0 ×10–5 / 3.0 ×10–2) (C1) = 0.19 (0.188) m s–1 (A1) [2]

a constant non-zero value of E from 0 to 3 cm and a different constant non-zero value of E from 3 to 6 cm E from 3–6 cm has the opposite sign to and larger value than E from 0–3 cm [2]

minimum frequency for electron(s) to be emitted (from surface) reference to frequency of electromagnetic radiation / photon or frequency causing emission of electron(s) from surface with zero kinetic energy (M1) reference to frequency of electromagnetic radiation / photon (A1) [2]

positive intercept on (1 / λ)-axis (when extrapolated) straight line with positive gradient [2]

gradient = hc where c is the speed of light [1]

maximum kinetic energy when electron emitted from surface energy is required to bring an electron to the surface [2]

each photon has more energy fewer photons per unit time fewer electrons per unit time / less current [3]

the penetration of the beam [1]

either decrease the accelerating voltage or decrease voltage between cathode and anode [1]

advantage: image gives depth / image is 3D / final image can be viewed from any angle disadvantage: greater exposure / more risk to health / more expensive / person must remain stationary [2]

λ = ln2 / T½ = ln2 / (53.3 × 24 × 60 × 60) =1.5 × 10–7 s–1 [1]

A = λN N = 39 × 10–3 / 1.5 × 10–7 = 2.6 ×105 m = (2.6 × 105 / 6.0 × 1023) × 7 × 10–3 or 2.6 × 105 × 1.66 × 10–27 × 7 = 3.0 × 10–21 kg [3]

2 / 39 = exp (–1.5 × 10–7 × t) or 2 / 39 = (1 / 2)[t / (53.3 × 24 × 3600)] t = 2.0 × 107 s [2]