Physics (AS & A Level) Paper 22
Oct / Nov 2016·9702·6 questions
Mock exam: 75 min, official timing
Question paper
Question 1
Question 2
Question 3
Question 4
Question 5
Question 6
Mark scheme
Question 1
1(a)(i)[1]
force / area (normal to the force) [1]
1(a)(ii)[1]
(p = F / A so) units: kg m s–2 / m2 = kg m–1 s–2 [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
1(b)[3]
units for m: kg, t: s and ρ: kg m–3 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 units of C: m2 [3]
Question 2
2(a)[2]
∆E = mg∆h = 0.030 × 9.81 × (–) 0.31 = (–) 0.091 J [2]
2(b)[3]
E = ½mv 2 (initial) E = ½ × 0.030 × 1.32 (= 0.0254) 0.5 × 0.030 × v 2 = (0.5 × 0.030 × 1.32) + (0.030 × 9.81 × 0.31) so v = 2.8 m s–1 or 0.5 × 0.030 × v 2 = (0.0254) + (0.091) so v = 2.8 m s–1 [3]
2(c)(i)[2]
0.096 = 0.030 (v + 2.8) v = 0.40 m s–1 [2]
2(c)(ii)[2]
F = ∆p / (∆)t or F = ma = 0.096 / 20 × 10–3 or 0.030 (0.40 + 2.8) / 20 × 10–3 = 4.8 N [2]
2(d)[1]
kinetic energy (of ball and wall) decreases/changes/not conserved, so inelastic or (relative) speed of approach (of ball and wall) not equal to/greater than (relative) speed of separation, so inelastic. [1]
2(e)[2]
force = work done / distance moved = (0.091 – 0.076) / 0.60 = 0.025 N [2]
Question 3
3(a)[2]
resultant force (in any direction) is zero resultant moment/torque (about any point) is zero [2]
3(b)(i)[1]
force = 33 sin 52° or 33 cos 38° = 26 N [1]
3(b)(ii)[3]
26 × 0.30 or W × 0.20 or 12 × 0.40 26 × 0.30 = (W × 0.20) + (12 × 0.40) W = 15 N [3]
3(c)(i)[2]
E = ∆σ / ∆ε or E = σ / ε ∆σ = 2.0 × 1011 × 7.5 × 10–4 = 1.5 × 108 Pa [2]
3(c)(ii)[3]
∆σ = ∆F / A or σ = F / A A = 78 / 1.5 × 108 (= 5.2 × 10–7 m2) 5.2 × 10–7 = πd 2 / 4 d = 8.1 × 10–4 m [3]
Question 4
4(a)[2]
wave incident on/passes by or through an aperture/edge wave spreads (into geometrical shadow) [2]
4(b)(i)[2]
waves (from slits) overlap (at point X) path difference (from slits to X) is zero/ phase difference (between the two waves) is zero (so constructive interference gives bright fringe) [2]
4(b)(ii)[1]
difference in distances = λ / 2 = 580 / 2 = 290 nm [1]
4(b)(iii)[3]
λ = ax / D D = [0.41 × 10–3 × (2 × 2.0 × 10–3)] / 580 × 10–9 = 2.8 m [3]
4(b)(iv)[2]
same separation/fringe width/number of fringes bright fringe(s)/central bright fringe/(fringe at) X less bright dark fringe(s)/(fringe at) Y/(fringe at) Z brighter contrast between fringes decreases Any two of the above four points, 1 mark each [2]
Question 5
5(a)[2]
total/sum of electromotive forces or e.m.f.s = total/sum of potential differences or p.d.s around a loop/(closed) circuit [2]
5(b)(i)[2]
(current in battery =) current in A + current in B or IA + IB (I =) 0.14 + 0.26 = 0.40 A [2]
5(b)(ii)[2]
E = V + Ir 6.8 = 6.0 + 0.40r or 6.8 = 0.40 (15 + r) r = 2.0 Ω [2]
5(b)(iii)[2]
R = V / I ratio (= RA / RB) = (6.0 / 0.14) / (6.0 / 0.26) = 42.9 / 23.1 or 0.26 / 0.14 = 1.9 (1.86) [2]
5(b)(iv)[4]
1. P = EI or VI or P = I 2R or P = V 2 / R = 6.8 × 0.40 = 0.402 × 17 = 6.82 / 17 = 2.7 W (2.72 W) [2]
2. output power = VI = 6.0 × 0.40 (= 2.40 W) efficiency = (6.0 × 0.40) / (6.8 × 0.40) = 2.40 / 2.72 = 0.88 or 88% (allow 0.89 or 89%) [2]
Question 6
6(a)[1]
hadron not a fundamental particle/lepton is fundamental particle or hadron made of quarks/lepton not made of quarks or strong force/interaction acts on hadrons/does not act on leptons [1]
6(b)(i)[2]
e 0 1 ) (+ or ) + (β 0 1 ) (e 0 0ν [2]
6(b)(ii)[1]
weak (nuclear force / interaction) [1]
6(b)(iii)[3]
6(c)[3]
(quark structure of proton is) up, up, down or uud up/u (quark charge) is (+)⅔(e), down/d (quark charge) is –⅓(e) ⅔e + ⅔e – ⅓e = (+)e [3]